Reflection Paper About Precious

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Physical Constants

Quantity Electron charge Electron mass Permittivity of free space Permeability of free space Velocity of light

Value e = (1.602 177 33 ± 0.000 000 46) × 10−19 C m = (9.109 389 7 ± 0.000 005 4) × 10−31 kg �0 = 8.854 187 817 × 10−12 F/m µ0 = 4π10−7 H/m c = 2.997 924 58 × 108 m/s

Dielectric Constant (�r� ) and Loss Tangent (� �� /� � )

Material Air Alcohol, ethyl Aluminum oxide Amber Bakelite Barium titanate Carbon dioxide Ferrite (NiZn) Germanium Glass Ice Mica Neoprene Nylon Paper Plexiglas Polyethylene Polypropylene Polystyrene Porcelain (dry process) Pyranol Pyrex glass Quartz (fused) Rubber Silica or SiO2 (fused) Silicon Snow Sodium chloride Soil (dry) Steatite Styrofoam Teﬂon Titanium dioxide Water (distilled) Water (sea) Water (dehydrated) Wood (dry)

� r

�� / �

1.0005 25 8.8 2.7 4.74 1200 1.001 12.4 16 4–7 4.2 5.4 6.6 3.5 3 3.45 2.26 2.25 2.56 6 4.4 4 3.8 2.5–3 3.8 11.8 3.3 5.9 2.8 5.8 1.03 2.1 100 80 1 1.5–4

0.1 0.000 6 0.002 0.022 0.013 0.000 25 0.002 0.05 0.000 6 0.011 0.02 0.008 0.03 0.000 2 0.000 3 0.000 05 0.014 0.000 5 0.000 6 0.000 75 0.002 0.000 75 0.5 0.000 1 0.05 0.003 0.000 1 0.000 3 0.001 5 0.04 4 0 0.01

Conductivity (� )

Material Silver Copper Gold Aluminum Tungsten Zinc Brass Nickel Iron Phosphor bronze Solder Carbon steel German silver Manganin Constantan Germanium Stainless steel

, S/m 6.17 × 107 4.10 × 107 3.82 × 107 1.82 × 107 1.67 × 107 1.5 × 107 1.45 × 107 1.03 × 107 1 × 107 0.7 × 107 0.6 × 107 0.3 × 107 0.227 × 107 0.226 × 107 0.22 × 107 0.11 × 107 5.80 × 107

Material Nichrome Graphite Silicon Ferrite (typical) Water (sea) Limestone Clay Water (fresh) Water (distilled) Soil (sandy) Granite Marble Bakelite Porcelain (dry process) Diamond Polystyrene Quartz

, S/m 0.1 × 107 7 × 104 2300 100 5 10−2 5 × 10−3 10−3 10−4 10−5 10−6 10−8 10−9 10−10 2 × 10−13 10−16 10−17

Relative Permeability (µr )

Material Bismuth Parafﬁn Wood Silver Aluminum Beryllium Nickel chloride Manganese sulfate Nickel Cast iron Cobalt

µr 0.999 998 6 0.999 999 42 0.999 999 5 0.999 999 81 1.000 000 65 1.000 000 79 1.000 04 1.000 1 50 60 60

Material Powdered iron Machine steel Ferrite (typical) Permalloy 45 Transformer iron Silicon iron Iron (pure) Mumetal Sendust Supermalloy

µr 100 300 1000 2500 3000 3500 4000 20 000 30 000 100 000

Engineering Electromagnetics
EIGHT H E D I T I O N

William H. Hayt, Jr.
Late Emeritus Professor Purdue University

John A. Buck

Georgia Institute of Technology

ENGINEERING ELECTROMAGNETICS, EIGHTH EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright C 2012 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions C 2006, 2001, and 1989. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOC/DOC 1 0 9 8 7 6 5 4 3 2 1 ISBN 978-0-07-338066-7 MHID 0-07-338066-0 Vice President & Editor-in-Chief: Marty Lange Vice President EDP/Central Publishing Services: Kimberly Meriwether David Publisher: Raghothaman Srinivasan Senior Sponsoring Editor: Peter E. Massar Senior Marketing Manager: Curt Reynolds Developmental Editor: Darlene M. Schueller Project Manager: Robin A. Reed Design Coordinator: Brenda A. Rolwes Cover Design and Image: Diana Fouts Buyer: Kara Kudronowicz Media Project Manager: Balaji Sundararaman Compositor: Glyph International Typeface: 10.5/12 Times Roman Printer: R.R. Donnelley All credits appearing on page or at the end of the book are considered to be an extension of the copyright page. Library of Congress Cataloging-in-Publication Data

Hayt, William Hart, 1920– Engineering electromagnetics / William H. Hayt, Jr., John A. Buck. — 8th ed. p. cm. Includes bibliographical references and index. ISBN 978–0–07–338066–7 (alk. paper) 1. Electromagnetic theory. I. Buck, John A. II. Title. QC670.H39 2010 530.14 1—dc22 2010048332

www.mhhe.com

To Amanda and Olivia

ABOUT THE AUTHORS

William H. Hayt. Jr. (deceased) received his B.S. and M.S. degrees at Purdue University and his Ph.D. from the University of Illinois. After spending four years in industry, Professor Hayt joined the faculty of Purdue University, where he served as professor and head of the School of Electrical Engineering, and as professor emeritus after retiring in 1986. Professor Hayt’s professional society memberships included Eta Kappa Nu, Tau Beta Pi, Sigma Xi, Sigma Delta Chi, Fellow of IEEE, ASEE, and NAEB. While at Purdue, he received numerous teaching awards, including the university’s Best Teacher Award. He is also listed in Purdue’s Book of Great Teachers, a permanent wall display in the Purdue Memorial Union, dedicated on April 23, 1999. The book bears the names of the inaugural group of 225 faculty members, past and present, who have devoted their lives to excellence in teaching and scholarship. They were chosen by their students and their peers as Purdue’s ﬁnest educators.

A native of Los Angeles, California, John A. Buck received his M.S. and Ph.D. degrees in Electrical Engineering from the University of California at Berkeley in 1977 and 1982, and his B.S. in Engineering from UCLA in 1975. In 1982, he joined the faculty of the School of Electrical and Computer Engineering at Georgia Tech, where he has remained for the past 28 years. His research areas and publications have centered within the ﬁelds of ultrafast switching, nonlinear optics, and optical ﬁber communications. He is the author of the graduate text Fundamentals of Optical Fibers (Wiley Interscience), which is now in its second edition. Awards include three institute teaching awards and the IEEE Third Millenium Medal. When not glued to his computer or conﬁned to the lab, Dr. Buck enjoys music, hiking, and photography.

BRIEF CONTENTS

Preface xii

1 Vector Analysis 1 2 Coulomb’s Law and Electric Field Intensity 26 3 Electric Flux Density, Gauss’s Law, and Divergence 4 Energy and Potential 75 5 Conductors and Dielectrics 109 6 Capacitance 143 7 The Steady Magnetic Field 180 8 Magnetic Forces, Materials, and Inductance 230 9 Time-Varying Fields and Maxwell’s Equations 277 10 Transmission Lines 301 11 The Uniform Plane Wave 367 12 Plane Wave Reﬂection and Dispersion 406 13 Guided Waves 453 14 Electromagnetic Radiation and Antennas 511
Appendix A Vector Analysis 553 Appendix B Units 557 Appendix C Material Constants 562 Appendix D The Uniqueness Theorem 565 Appendix E Origins of the Complex Permittivity 567 Appendix F Answers to Odd-Numbered Problems 574 Index 580

48

v

CONTENTS

Preface xii

Chapter 1 Vector Analysis 1
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 Scalars and Vectors 1 Vector Algebra 2 The Rectangular Coordinate System 3 Vector Components and Unit Vectors 5 The Vector Field 8 The Dot Product 9 The Cross Product 11 Other Coordinate Systems: Circular Cylindrical Coordinates 13 1.9 The Spherical Coordinate System 18 References 22 Chapter 1 Problems 22

Chapter 3 Electric Flux Density, Gauss’s Law, and Divergence 48
3.1 Electric Flux Density 48 3.2 Gauss’s Law 52 3.3 Application of Gauss’s Law: Some Symmetrical Charge Distributions 56 3.4 Application of Gauss’s Law: Differential Volume Element 61 3.5 Divergence and Maxwell’s First Equation 64 3.6 The Vector Operator ∇ and the Divergence Theorem 67 References 70 Chapter 3 Problems 71

Chapter 4 Energy and Potential 75
4.1 Energy Expended in Moving a Point Charge in an Electric Field 76 4.2 The Line Integral 77 4.3 Deﬁnition of Potential Difference and Potential 82 4.4 The Potential Field of a Point Charge 84 4.5 The Potential Field of a System of Charges: Conservative Property 86 4.6 Potential Gradient 90 4.7 The Electric Dipole 95 4.8 Energy Density in the Electrostatic Field 100 References 104 Chapter 4 Problems 105

Chapter 2 Coulomb’s Law and Electric Field Intensity 26
2.1 The Experimental Law of Coulomb 26 2.2 Electric Field Intensity 29 2.3 Field Arising from a Continuous Volume Charge Distribution 33 2.4 Field of a Line Charge 35 2.5 Field of a Sheet of Charge 39 2.6 Streamlines and Sketches of Fields 41 References 44 Chapter 2 Problems 44

vi

Contents

vii

Chapter 5 Conductors and Dielectrics 109
5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 Current and Current Density 110 Continuity of Current 111 Metallic Conductors 114 Conductor Properties and Boundary Conditions 119 The Method of Images 124 Semiconductors 126 The Nature of Dielectric Materials 127 Boundary Conditions for Perfect Dielectric Materials 133 References 137 Chapter 5 Problems 138

7.4 Stokes’ Theorem 202 7.5 Magnetic Flux and Magnetic Flux Density 207 7.6 The Scalar and Vector Magnetic Potentials 210 7.7 Derivation of the Steady-Magnetic-Field Laws 217 References 223 Chapter 7 Problems 223

Chapter 8 Magnetic Forces, Materials, and Inductance 230
8.1 Force on a Moving Charge 230 8.2 Force on a Differential Current Element 232 8.3 Force between Differential Current Elements 236 8.4 Force and Torque on a Closed Circuit 238 8.5 The Nature of Magnetic Materials 244 8.6 Magnetization and Permeability 247 8.7 Magnetic Boundary Conditions 252 8.8 The Magnetic Circuit 255 8.9 Potential Energy and Forces on Magnetic Materials 261 8.10 Inductance and Mutual Inductance 263 References 270 Chapter 8 Problems 270

Chapter 6 Capacitance 143
6.1 6.2 6.3 6.4 6.5 Capacitance Deﬁned 143 Parallel-Plate Capacitor 145 Several Capacitance Examples 147 Capacitance of a Two-Wire Line 150 Using Field Sketches to Estimate Capacitance in Two-Dimensional Problems 154 6.6 Poisson’s and Laplace’s Equations 160 6.7 Examples of the Solution of Laplace’s Equation 162 6.8 Example of the Solution of Poisson’s Equation: the p-n Junction Capacitance 169 References 172 Chapter 6 Problems 173

Chapter 9 Time-Varying Fields and Maxwell’s Equations 277
9.1 9.2 9.3 9.4 9.5 Faraday’s Law 277 Displacement Current 284 Maxwell’s Equations in Point Form 288 Maxwell’s Equations in Integral Form 290 The Retarded Potentials 292 References 296 Chapter 9 Problems 296

Chapter 7 The Steady Magnetic Field 180
7.1 Biot-Savart Law 180 7.2 Amp` re’s Circuital Law 188 e 7.3 Curl 195

viii

Contents

Chapter 10 Transmission Lines 301
10.1 Physical Description of Transmission Line Propagation 302 10.2 The Transmission Line Equations 304 10.3 Lossless Propagation 306 10.4 Lossless Propagation of Sinusoidal Voltages 309 10.5 Complex Analysis of Sinusoidal Waves 311 10.6 Transmission Line Equations and Their Solutions in Phasor Form 313 10.7 Low-Loss Propagation 315 10.8 Power Transmission and The Use of Decibels in Loss Characterization 317 10.9 Wave Reﬂection at Discontinuities 320 10.10 Voltage Standing Wave Ratio 323 10.11 Transmission Lines of Finite Length 327 10.12 Some Transmission Line Examples 330 10.13 Graphical Methods: The Smith Chart 334 10.14 Transient Analysis 345 References 358 Chapter 10 Problems 358

12.3 Wave Reﬂection from Multiple Interfaces 417 12.4 Plane Wave Propagation in General Directions 425 12.5 Plane Wave Reﬂection at Oblique Incidence Angles 428 12.6 Total Reﬂection and Total Transmission of Obliquely Incident Waves 434 12.7 Wave Propagation in Dispersive Media 437 12.8 Pulse Broadening in Dispersive Media 443 References 447 Chapter 12 Problems 448

Chapter 13 Guided Waves 453
13.1 Transmission Line Fields and Primary Constants 453 13.2 Basic Waveguide Operation 463 13.3 Plane Wave Analysis of the Parallel-Plate Waveguide 467 13.4 Parallel-Plate Guide Analysis Using the Wave Equation 476 13.5 Rectangular Waveguides 479 13.6 Planar Dielectric Waveguides 490 13.7 Optical Fiber 496 References 506 Chapter 13 Problems 506

Chapter 11 The Uniform Plane Wave 367
11.1 11.2 11.3 11.4 Wave Propagation in Free Space 367 Wave Propagation in Dielectrics 375 Poynting’s Theorem and Wave Power 384 Propagation in Good Conductors: Skin Effect 387 11.5 Wave Polarization 394 References 401 Chapter 11 Problems 401

Chapter 14 Electromagnetic Radiation and Antennas 511
14.1 Basic Radiation Principles: The Hertzian Dipole 511 14.2 Antenna Speciﬁcations 518 14.3 Magnetic Dipole 523 14.4 Thin Wire Antennas 525 14.5 Arrays of Two Elements 533 14.6 Uniform Linear Arrays 537 14.7 Antennas as Receivers 541 References 548 Chapter 14 Problems 548

Chapter 12 Plane Wave Reflection and Dispersion 406
12.1 Reﬂection of Uniform Plane Waves at Normal Incidence 406 12.2 Standing Wave Ratio 413

Contents

ix

Appendix A Vector Analysis
A.1 A.2 A.3

553

Appendix D The Uniqueness Theorem 565 Appendix E Origins of the Complex Permittivity 567 Appendix F Answers to Odd-Numbered Problems 574 Index 580

General Curvilinear Coordinates 553 Divergence, Gradient, and Curl in General Curvilinear Coordinates 554 Vector Identities 556

Appendix B Units 557 Appendix C Material Constants 562

PREFACE

It has been 52 years since the ﬁrst edition of this book was published, then under the sole authorship of William H. Hayt, Jr. As I was ﬁve years old at that time, this would have meant little to me. But everything changed 15 years later when I used the second edition in a basic electromagnetics course as a college junior. I remember my sense of foreboding at the start of the course, being aware of friends’ horror stories. On ﬁrst opening the book, however, I was pleasantly surprised by the friendly writing style and by the measured approach to the subject, which — at least for me — made it a very readable book, out of which I was able to learn with little help from my professor. I referred to it often while in graduate school, taught from the fourth and ﬁfth editions as a faculty member, and then became coauthor for the sixth and seventh editions on the retirement (and subsequent untimely death) of Bill Hayt. The memories of my time as a beginner are clear, and I have tried to maintain the accessible style that I found so welcome then. Over the 50-year span, the subject matter has not changed, but emphases have. In the universities, the trend continues toward reducing electrical engineering core course allocations to electromagnetics. I have made efforts to streamline the presentation in this new edition to enable the student to get to Maxwell’s equations sooner, and I have added more advanced material. Many of the earlier chapters are now slightly shorter than their counterparts in the seventh edition. This has been done by economizing on the wording, shortening many sections, or by removing some entirely. In some cases, deleted topics have been converted to stand-alone articles and moved to the website, from which they can be downloaded. Major changes include the following: (1) The material on dielectrics, formerly in Chapter 6, has been moved to the end of Chapter 5. (2) The chapter on Poisson’s and Laplace’s equations has been eliminated, retaining only the one-dimensional treatment, which has been moved to the end of Chapter 6. The two-dimensional Laplace equation discussion and that of numerical methods have been moved to the website for the book. (3) The treatment on rectangular waveguides (Chapter 13) has been expanded, presenting the methodology of two-dimensional boundary value problems in that context. (4) The coverage of radiation and antennas has been greatly expanded and now forms the entire Chapter 14. Some 130 new problems have been added throughout. For some of these, I chose particularly good “classic” problems from the earliest editions. I have also adopted a new system in which the approximate level of difﬁculty is indicated beside each problem on a three-level scale. The lowest level is considered a fairly straightforward problem, requiring little work assuming the material is understood; a level 2 problem is conceptually more difﬁcult, and/or may require more work to solve; a level 3 problem is considered either difﬁcult conceptually, or may require extra effort (including possibly the help of a computer) to solve.

x

Preface

xi

As in the previous edition, the transmission lines chapter (10) is stand-alone, and can be read or covered in any part of a course, including the beginning. In it, transmission lines are treated entirely within the context of circuit theory; wave phenomena are introduced and used exclusively in the form of voltages and currents. Inductance and capacitance concepts are treated as known parameters, and so there is no reliance on any other chapter. Field concepts and parameter computation in transmission lines appear in the early part of the waveguides chapter (13), where they play additional roles of helping to introduce waveguiding concepts. The chapters on electromagnetic waves, 11 and 12, retain their independence of transmission line theory in that one can progress from Chapter 9 directly to Chapter 11. By doing this, wave phenomena are introduced from ﬁrst principles but within the context of the uniform plane wave. Chapter 11 refers to Chapter 10 in places where the latter may give additional perspective, along with a little more detail. Nevertheless, all necessary material to learn plane waves without previously studying transmission line waves is found in Chapter 11, should the student or instructor wish to proceed in that order. The new chapter on antennas covers radiation concepts, building on the retarded potential discussion in Chapter 9. The discussion focuses on the dipole antenna, individually and in simple arrays. The last section covers elementary transmit-receive systems, again using the dipole as a vehicle. The book is designed optimally for a two-semester course. As is evident, statics concepts are emphasized and occur ﬁrst in the presentation, but again Chapter 10 (transmission lines) can be read ﬁrst. In a single course that emphasizes dynamics, the transmission lines chapter can be covered initially as mentioned or at any point in the course. One way to cover the statics material more rapidly is by deemphasizing materials properties (assuming these are covered in other courses) and some of the advanced topics. This involves omitting Chapter 1 (assigned to be read as a review), and omitting Sections 2.5, 2.6, 4.7, 4.8, 5.5–5.7, 6.3, 6.4, 6.7, 7.6, 7.7, 8.5, 8.6, 8.8, 8.9, and 9.5. A supplement to this edition is web-based material consisting of the aforementioned articles on special topics in addition to animated demonstrations and interactive programs developed by Natalya Nikolova of McMaster University and Vikram Jandhyala of the University of Washington. Their excellent contributions are geared to the text, and icons appear in the margins whenever an exercise that pertains to the narrative exists. In addition, quizzes are provided to aid in further study. The theme of the text is the same as it has been since the ﬁrst edition of 1958. An inductive approach is used that is consistent with the historical development. In it, the experimental laws are presented as individual concepts that are later uniﬁed in Maxwell’s equations. After the ﬁrst chapter on vector analysis, additional mathematical tools are introduced in the text on an as-needed basis. Throughout every edition, as well as this one, the primary goal has been to enable students to learn independently. Numerous examples, drill problems (usually having multiple parts), end-of-chapter problems, and material on the web site, are provided to facilitate this.

xii

Preface

Answers to the drill problems are given below each problem. Answers to oddnumbered end-of-chapter problems are found in Appendix F. A solutions manual and a set of PowerPoint slides, containing pertinent ﬁgures and equations, are available to instructors. These, along with all other material mentioned previously, can be accessed on the website: www.mhhe.com/haytbuck I would like to acknowledge the valuable input of several people who helped to make this a better edition. Special thanks go to Glenn S. Smith (Georgia Tech), who reviewed the antennas chapter and provided many valuable comments and suggestions. Detailed suggestions and errata were provided by Clive Woods (Louisiana State University), Natalya Nikolova, and Don Davis (Georgia Tech). Accuracy checks on the new problems were carried out by Todd Kaiser (Montana State University) and Steve Weis (Texas Christian University). Other reviewers provided detailed comments and suggestions at the start of the project; many of the suggestions affected the outcome. They include: Sheel Aditya – Nanyang Technological University, Singapore Yaqub M. Amani – SUNY Maritime College Rusnani Arifﬁn – Universiti Teknologi MARA Ezekiel Bahar – University of Nebraska Lincoln Stephen Blank – New York Institute of Technology Thierry Blu – The Chinese University of Hong Kong Jeff Chamberlain – Illinois College Yinchao Chen – University of South Carolina Vladimir Chigrinov – Hong Kong University of Science and Technology Robert Coleman – University of North Carolina Charlotte Wilbur N. Dale Ibrahim Elshaﬁey – King Saud University Wayne Grassel – Point Park University Essam E. Hassan – King Fahd University of Petroleum and Minerals David R. Jackson – University of Houston Karim Y. Kabalan – American University of Beirut Shahwan Victor Khoury, Professor Emeritus – Notre Dame University, Louaize-Zouk Mosbeh, Lebanon Choon S. Lee – Southern Methodist University Mojdeh J. Mardani – University of North Dakota Mohamed Mostafa Morsy – Southern Illinois University Carbondale Sima Noghanian – University of North Dakota W.D. Rawle – Calvin College G¨ n¨ l Sayan – Middle East Technical University o u Fred H. Terry – Professor Emeritus, Christian Brothers University Denise Thorsen – University of Alaska Fairbanks Chi-Ling Wang – Feng-Chia University

Preface

xiii

I also acknowledge the feedback and many comments from students, too numerous to name, including several who have contacted me from afar. I continue to be open and grateful for this feedback and can be reached at john.buck@ece.gatech.edu. Many suggestions were made that I considered constructive and actionable. I regret that not all could be incorporated because of time restrictions. Creating this book was a team effort, involving several outstanding people at McGraw-Hill. These include my publisher, Raghu Srinivasan, and sponsoring editor, Peter Massar, whose vision and encouragement were invaluable, Robin Reed, who deftly coordinated the production phase with excellent ideas and enthusiasm, and Darlene Schueller, who was my guide and conscience from the beginning, providing valuable insights, and jarring me into action when necessary. Typesetting was supervised by Vipra Fauzdar at Glyph International, who employed the best copy editor I ever had, Laura Bowman. Diana Fouts (Georgia Tech) applied her vast artistic skill to designing the cover, as she has done for the previous two editions. Finally, I am, as usual in these projects, grateful to a patient and supportive family, and particularly to my daughter, Amanda, who assisted in preparing the manuscript. John A. Buck Marietta, Georgia December, 2010

On the cover: Radiated intensity patterns for a dipole antenna, showing the cases for which the wavelength is equal to the overall antenna length (red), two-thirds the antenna length (green), and one-half the antenna length (blue).

xiv

Preface

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C H A P T E R

1

Vector Analysis

V

ector analysis is a mathematical subject that is better taught by mathematicians than by engineers. Most junior and senior engineering students have not had the time (or the inclination) to take a course in vector analysis, although it is likely that vector concepts and operations were introduced in the calculus sequence. These are covered in this chapter, and the time devoted to them now should depend on past exposure. The viewpoint here is that of the engineer or physicist and not that of the mathematician. Proofs are indicated rather than rigorously expounded, and physical interpretation is stressed. It is easier for engineers to take a more rigorous course in the mathematics department after they have been presented with a few physical pictures and applications. Vector analysis is a mathematical shorthand. It has some new symbols and some new rules, and it demands concentration and practice. The drill problems, ﬁrst found at the end of Section 1.4, should be considered part of the text and should all be worked. They should not prove to be difﬁcult if the material in the accompanying section of the text has been thoroughly understood. It takes a little longer to “read” the chapter this way, but the investment in time will produce a surprising interest. ■

1.1 SCALARS AND VECTORS
The term scalar refers to a quantity whose value may be represented by a single (positive or negative) real number. The x, y, and z we use in basic algebra are scalars, and the quantities they represent are scalars. If we speak of a body falling a distance L in a time t, or the temperature T at any point in a bowl of soup whose coordinates are x, y, and z, then L , t, T, x, y, and z are all scalars. Other scalar quantities are mass, density, pressure (but not force), volume, volume resistivity, and voltage. A vector quantity has both a magnitude1 and a direction in space. We are concerned with two- and three-dimensional spaces only, but vectors may be deﬁned in
1

We adopt the convention that magnitude infers absolute value; the magnitude of any quantity is, therefore, always positive. 1

2

ENGINEERING ELECTROMAGNETICS

n-dimensional space in more advanced applications. Force, velocity, acceleration, and a straight line from the positive to the negative terminal of a storage battery are examples of vectors. Each quantity is characterized by both a magnitude and a direction. Our work will mainly concern scalar and vector field . A ﬁeld (scalar or vector) may be deﬁned mathematically as some function that connects an arbitrary origin to a general point in space. We usually associate some physical effect with a ﬁeld, such as the force on a compass needle in the earth’s magnetic ﬁeld, or the movement of smoke particles in the ﬁeld deﬁned by the vector velocity of air in some region of space. Note that the ﬁeld concept invariably is related to a region. Some quantity is deﬁned at every point in a region. Both scalar field and vector field exist. The temperature throughout the bowl of soup and the density at any point in the earth are examples of scalar ﬁelds. The gravitational and magnetic ﬁelds of the earth, the voltage gradient in a cable, and the temperature gradient in a soldering-iron tip are examples of vector ﬁelds. The value of a ﬁeld varies in general with both position and time. In this book, as in most others using vector notation, vectors will be indicated by boldface type, for example, A. Scalars are printed in italic type, for example, A. When writing longhand, it is customary to draw a line or an arrow over a vector quantity to show its vector character. (CAUTION: This is the ﬁrst pitfall. Sloppy notation, such as the omission of the line or arrow symbol for a vector, is the major cause of errors in vector analysis.)

1.2 VECTOR ALGEBRA
With the deﬁnition of vectors and vector ﬁelds now established, we may proceed to deﬁne the rules of vector arithmetic, vector algebra, and (later) vector calculus. Some of the rules will be similar to those of scalar algebra, some will differ slightly, and some will be entirely new. To begin, the addition of vectors follows the parallelogram law. Figure 1.1 shows the sum of two vectors, A and B. It is easily seen that A + B = B + A, or that vector addition obeys the commutative law. Vector addition also obeys the associative law, A + (B + C) = (A + B) + C Note that when a vector is drawn as an arrow of ﬁnite length, its location is deﬁned to be at the tail end of the arrow. Coplanar vectors are vectors lying in a common plane, such as those shown in Figure 1.1. Both lie in the plane of the paper and may be added by expressing each vector in terms of “horizontal” and “vertical” components and then adding the corresponding components. Vectors in three dimensions may likewise be added by expressing the vectors in terms of three components and adding the corresponding components. Examples of this process of addition will be given after vector components are discussed in Section 1.4.

CHAPTER 1

Vector Analysis

3

Figure 1.1 Two vectors may be added graphically either by drawing both vectors from a common origin and completing the parallelogram or by beginning the second vector from the head of the first and completing the triangle; either method is easily extended to three or more vectors.

The rule for the subtraction of vectors follows easily from that for addition, for we may always express A−B as A+(−B); the sign, or direction, of the second vector is reversed, and this vector is then added to the ﬁrst by the rule for vector addition. Vectors may be multiplied by scalars. The magnitude of the vector changes, but its direction does not when the scalar is positive, although it reverses direction when multiplied by a negative scalar. Multiplication of a vector by a scalar also obeys the associative and distributive laws of algebra, leading to (r + s)(A + B) = r (A + B) + s(A + B) = r A + r B + sA + sB Division of a vector by a scalar is merely multiplication by the reciprocal of that scalar. The multiplication of a vector by a vector is discussed in Sections 1.6 and 1.7. Two vectors are said to be equal if their difference is zero, or A = B if A − B = 0. In our use of vector ﬁelds we shall always add and subtract vectors that are deﬁned at the same point. For example, the total magnetic ﬁeld about a small horseshoe magnet will be shown to be the sum of the ﬁelds produced by the earth and the permanent magnet; the total ﬁeld at any point is the sum of the individual ﬁelds at that point. If we are not considering a vector fiel , we may add or subtract vectors that are not deﬁned at the same point. For example, the sum of the gravitational force acting on a 150 lb f (pound-force) man at the North Pole and that acting on a 175 lb f person at the South Pole may be obtained by shifting each force vector to the South Pole before addition. The result is a force of 25 lb f directed toward the center of the earth at the South Pole; if we wanted to be difﬁcult, we could just as well describe the force as 25 lb f directed away from the center of the earth (or “upward”) at the North Pole.2

1.3 THE RECTANGULAR COORDINATE SYSTEM
To describe a vector accurately, some speciﬁc lengths, directions, angles, projections, or components must be given. There are three simple methods of doing this, and about eight or ten other methods that are useful in very special cases. We are going

2

Students have argued that the force might be described at the equator as being in a “northerly” direction. They are right, but enough is enough.

4

ENGINEERING ELECTROMAGNETICS

to use only the three simple methods, and the simplest of these is the rectangular, or rectangular cartesian, coordinate system. In the rectangular coordinate system we set up three coordinate axes mutually at right angles to each other and call them the x, y, and z axes. It is customary to choose a right-handed coordinate system, in which a rotation (through the smaller angle) of the x axis into the y axis would cause a right-handed screw to progress in the direction of the z axis. If the right hand is used, then the thumb, foreﬁnger, and middle ﬁnger may be identiﬁed, respectively, as the x, y, and z axes. Figure 1.2a shows a right-handed rectangular coordinate system. A point is located by giving its x, y, and z coordinates. These are, respectively, the distances from the origin to the intersection of perpendicular lines dropped from the point to the x, y, and z axes. An alternative method of interpreting coordinate

Figure 1.2 (a) A right-handed rectangular coordinate system. If the curved fingers of the right hand indicate the direction through which the x axis is turned into coincidence with the y axis, the thumb shows the direction of the z axis. (b) The location of points P(1, 2, 3) and Q(2, −2, 1). (c) The differential volume element in rectangular coordinates; dx, dy, and dz are, in general, independent differentials.

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5

values, which must be used in all other coordinate systems, is to consider the point as being at the common intersection of three surfaces. These are the planes x = constant, y = constant, and z = constant, where the constants are the coordinate values of the point. Figure 1.2b shows points P and Q whose coordinates are (1, 2, 3) and (2, −2, 1), respectively. Point P is therefore located at the common point of intersection of the planes x = 1, y = 2, and z = 3, whereas point Q is located at the intersection of the planes x = 2, y = −2, and z = 1. As we encounter other coordinate systems in Sections 1.8 and 1.9, we expect points to be located at the common intersection of three surfaces, not necessarily planes, but still mutually perpendicular at the point of intersection. If we visualize three planes intersecting at the general point P, whose coordinates are x, y, and z, we may increase each coordinate value by a differential amount and obtain three slightly displaced planes intersecting at point P , whose coordinates are x + d x, y + dy, and z + dz. The six planes deﬁne a rectangular parallelepiped whose volume is dv = d xd ydz; the surfaces have differential areas d S of d xd y, dydz, and dzd x. Finally, the distance d L from P to P is the diagonal of the parallelepiped and has a length of (d x)2 + (dy)2 + (dz)2 . The volume element is shown in Figure 1.2c; point P is indicated, but point P is located at the only invisible corner. All this is familiar from trigonometry or solid geometry and as yet involves only scalar quantities. We will describe vectors in terms of a coordinate system in the next section.

1.4 VECTOR COMPONENTS AND UNIT VECTORS
To describe a vector in the rectangular coordinate system, let us ﬁrst consider a vector r extending outward from the origin. A logical way to identify this vector is by giving the three component vectors, lying along the three coordinate axes, whose vector sum must be the given vector. If the component vectors of the vector r are x, y, and z, then r = x + y + z. The component vectors are shown in Figure 1.3a. Instead of one vector, we now have three, but this is a step forward because the three vectors are of a very simple nature; each is always directed along one of the coordinate axes. The component vectors have magnitudes that depend on the given vector (such as r), but they each have a known and constant direction. This suggests the use of unit vectors having unit magnitude by deﬁnition; these are parallel to the coordinate axes and they point in the direction of increasing coordinate values. We reserve the symbol a for a unit vector and identify its direction by an appropriate subscript. Thus ax , a y , and az are the unit vectors in the rectangular coordinate system.3 They are directed along the x, y, and z axes, respectively, as shown in Figure 1.3b. If the component vector y happens to be two units in magnitude and directed toward increasing values of y, we should then write y = 2a y . A vector r P pointing
3

The symbols i, j, and k are also commonly used for the unit vectors in rectangular coordinates.

6

ENGINEERING ELECTROMAGNETICS

Figure 1.3 (a) The component vectors x, y, and z of vector r. (b) The unit vectors of the rectangular coordinate system have unit magnitude and are directed toward increasing values of their respective variables. (c) The vector R P Q is equal to the vector difference r Q − r P .

from the origin to point P(1, 2, 3) is written r P = ax + 2a y + 3az . The vector from P to Q may be obtained by applying the rule of vector addition. This rule shows that the vector from the origin to P plus the vector from P to Q is equal to the vector from the origin to Q. The desired vector from P(1, 2, 3) to Q(2, −2, 1) is therefore R P Q = r Q − r P = (2 − 1)ax + (−2 − 2)a y + (1 − 3)az The vectors r P , r Q , and R P Q are shown in Figure 1.3c. The last vector does not extend outward from the origin, as did the vector r we initially considered. However, we have already learned that vectors having the same magnitude and pointing in the same direction are equal, so we see that to help our visualization processes we are at liberty to slide any vector over to the origin before = ax − 4a y − 2az

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7

determining its component vectors. Parallelism must, of course, be maintained during the sliding process. If we are discussing a force vector F, or indeed any vector other than a displacement-type vector such as r, the problem arises of providing suitable letters for the three component vectors. It would not do to call them x, y, and z, for these are displacements, or directed distances, and are measured in meters (abbreviated m) or some other unit of length. The problem is most often avoided by using component scalars, simply called components, Fx , Fy , and Fz . The components are the signed magnitudes of the component vectors. We may then write F = Fx ax + Fy a y + Fz az . The component vectors are Fx ax , Fy a y , and Fz az . Any vector B then may be described by B = Bx ax + B y a y + Bz az . The magnitude of B written |B| or simply B, is given by |B| =
2 2 2 B x + B y + Bz

(1)

Each of the three coordinate systems we discuss will have its three fundamental and mutually perpendicular unit vectors that are used to resolve any vector into its component vectors. Unit vectors are not limited to this application. It is helpful to write a unit vector having a speciﬁed direction. This is easily done, for a unit vector in a given direction is merely a vector in that direction divided by its magnitude. A unit vector in the r direction is r/ x 2 + y 2 + z 2 , and a unit vector in the direction of the vector B is aB = B
2 2 2 B x + B y + Bz

=

B |B|

(2)

E X A M P L E 1.1

Specify the unit vector extending from the origin toward the point G(2, −2, −1).
Solution. We ﬁrst construct the vector extending from the origin to point G,

G = 2ax − 2a y − az We continue by ﬁnding the magnitude of G, |G| = aG = (2)2 + (−2)2 + (−1)2 = 3

and ﬁnally expressing the desired unit vector as the quotient, G = 2 ax − 2 a y − 1 az = 0.667ax − 0.667a y − 0.333az 3 3 3 |G| A special symbol is desirable for a unit vector so that its character is immediately apparent. Symbols that have been used are u B , a B , 1B , or even b. We will consistently use the lowercase a with an appropriate subscript.

8

ENGINEERING ELECTROMAGNETICS

[NOTE: Throughout the text, drill problems appear following sections in which a new principle is introduced in order to allow students to test their understanding of the basic fact itself. The problems are useful in gaining familiarity with new terms and ideas and should all be worked. More general problems appear at the ends of the chapters. The answers to the drill problems are given in the same order as the parts of the problem.] D1.1. Given points M(−1, 2, 1), N (3, −3, 0), and P(−2, −3, −4), ﬁnd: (a) R M N ; (b) R M N + R M P ; (c) |r M |; (d) a M P ; (e) |2r P − 3r N |.
Ans. 4ax − 5a y − az ; 3ax − 10a y − 6az ; 2.45; −0.14ax − 0.7a y − 0.7az ; 15.56

1.5 THE VECTOR FIELD
We have deﬁned a vector ﬁeld as a vector function of a position vector. In general, the magnitude and direction of the function will change as we move throughout the region, and the value of the vector function must be determined using the coordinate values of the point in question. Because we have considered only the rectangular coordinate system, we expect the vector to be a function of the variables x, y, and z. If we again represent the position vector as r, then a vector ﬁeld G can be expressed in functional notation as G(r); a scalar ﬁeld T is written as T (r). If we inspect the velocity of the water in the ocean in some region near the surface where tides and currents are important, we might decide to represent it by a velocity vector that is in any direction, even up or down. If the z axis is taken as upward, the x axis in a northerly direction, the y axis to the west, and the origin at the surface, we have a right-handed coordinate system and may write the velocity vector as v = v x ax + v y a y + v z az , or v(r) = v x (r)ax + v y (r)a y + v z (r)az ; each of the components v x , v y , and v z may be a function of the three variables x, y, and z. If we are in some portion of the Gulf Stream where the water is moving only to the north, then v y and v z are zero. Further simplifying assumptions might be made if the velocity falls off with depth and changes very slowly as we move north, south, east, or west. A suitable expression could be v = 2e z/100 ax . We have a velocity of 2 m/s (meters per second) at the surface and a velocity of 0.368 × 2, or 0.736 m/s, at a depth of 100 m (z = −100). The velocity continues to decrease with depth, while maintaining a constant direction. D1.2. A vector ﬁeld S is expressed in rectangular coordinates as S = {125/ [(x − 1)2 + (y − 2)2 + (z + 1)2 ]}{(x − 1)ax + (y − 2)a y + (z + 1)az }. (a) Evaluate S at P(2, 4, 3). (b) Determine a unit vector that gives the direction of S at P. (c) Specify the surface f (x, y, z) on which |S| = 1.
Ans. 5.95ax + 11.90a y + 23.8az ; 0.218ax + 0.436a y + 0.873az ; (x − 1)2 + (y − 2)2 + (z + 1)2 = 125

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9

1.6 THE DOT PRODUCT
We now consider the ﬁrst of two types of vector multiplication. The second type will be discussed in the following section. Given two vectors A and B, the dot product, or scalar product, is deﬁned as the product of the magnitude of A, the magnitude of B, and the cosine of the smaller angle between them, A · B = |A| |B| cos θ AB (3)

The dot appears between the two vectors and should be made heavy for emphasis. The dot, or scalar, product is a scalar, as one of the names implies, and it obeys the commutative law, A·B = B·A (4)

for the sign of the angle does not affect the cosine term. The expression A · B is read “A dot B.” Perhaps the most common application of the dot product is in mechanics, where a constant force F applied over a straight displacement L does an amount of work F L cos θ , which is more easily written F · L. We might anticipate one of the results of Chapter 4 by pointing out that if the force varies along the path, integration is necessary to ﬁnd the total work, and the result becomes Work = F · dL

Another example might be taken from magnetic ﬁelds. The total ﬂux crossing a surface of area S is given by B S if the magnetic ﬂux density B is perpendicular to the surface and uniform over it. We deﬁne a vector surface S as having area for its magnitude and having a direction normal to the surface (avoiding for the moment the problem of which of the two possible normals to take). The ﬂux crossing the surface is then B · S. This expression is valid for any direction of the uniform magnetic ﬂux density. If the ﬂux density is not constant over the surface, the total ﬂux is = B · dS. Integrals of this general form appear in Chapter 3 when we study electric ﬂux density. Finding the angle between two vectors in three-dimensional space is often a job we would prefer to avoid, and for that reason the deﬁnition of the dot product is usually not used in its basic form. A more helpful result is obtained by considering two vectors whose rectangular components are given, such as A = A x ax + A y a y + A z az and B = Bx ax + B y a y + Bz az . The dot product also obeys the distributive law, and, therefore, A · B yields the sum of nine scalar terms, each involving the dot product of two unit vectors. Because the angle between two different unit vectors of the rectangular coordinate system is 90◦ , we then have a x · a y = a y · a x = a x · az = az · a x = a y · az = az · a y = 0

10

ENGINEERING ELECTROMAGNETICS

Figure 1.4 (a) The scalar component of B in the direction of the unit vector a is B · a. (b) The vector component of B in the direction of the unit vector a is (B · a)a.

The remaining three terms involve the dot product of a unit vector with itself, which is unity, giving ﬁnally A · B = A x B x + A y B y + A z Bz which is an expression involving no angles. A vector dotted with itself yields the magnitude squared, or A · A = A2 = |A|2 and any unit vector dotted with itself is unity, aA · aA = 1 One of the most important applications of the dot product is that of ﬁnding the component of a vector in a given direction. Referring to Figure 1.4a, we can obtain the component (scalar) of B in the direction speciﬁed by the unit vector a as The sign of the component is positive if 0 ≤ θ Ba ≤ 90◦ and negative whenever 90◦ ≤ θ Ba ≤ 180◦ . To obtain the component vector of B in the direction of a, we multiply the component (scalar) by a, as illustrated by Figure 1.4b. For example, the component of B in the direction of ax is B · ax = Bx , and the component vector is Bx ax , or (B · ax )ax . Hence, the problem of ﬁnding the component of a vector in any direction becomes the problem of ﬁnding a unit vector in that direction, and that we can do. The geometrical term projection is also used with the dot product. Thus, B · a is the projection of B in the a direction.
E X A M P L E 1.2

(5)

(6)

B · a = |B| |a| cos θ Ba = |B| cos θ Ba

In order to illustrate these deﬁnitions and operations, consider the vector ﬁeld G = yax − 2.5xa y + 3az and the point Q(4, 5, 2). We wish to ﬁnd: G at Q; the scalar component of G at Q in the direction of a N = 1 (2ax + a y − 2az ); the vector component 3 of G at Q in the direction of a N ; and ﬁnally, the angle θGa between G(r Q ) and a N .

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11

Solution. Substituting the coordinates of point Q into the expression for G, we have

G(r Q ) = 5ax − 10a y + 3az Next we ﬁnd the scalar component. Using the dot product, we have G · a N = (5ax − 10a y + 3az ) · 1 (2ax + a y − 2az ) = 1 (10 − 10 − 6) = −2 3 3 The vector component is obtained by multiplying the scalar component by the unit vector in the direction of a N, The angle between G(r Q ) and a N is found from (G · a N )a N = −(2) 1 (2ax + a y − 2az ) = −1.333ax − 0.667a y + 1.333az 3

G · a N = |G| cos θGa √ −2 = 25 + 100 + 9 cos θGa and −2 θGa = cos−1 √ = 99.9◦ 134 D1.3. The three vertices of a triangle are located at A(6, −1, 2), B(−2, 3, −4), and C(−3, 1, 5). Find: (a) R AB ; (b) R AC ; (c) the angle θ B AC at vertex A; (d) the (vector) projection of R AB on R AC .
Ans. −8ax + 4a y − 6az ; −9ax + 2a y + 3az ; 53.6◦ ; −5.94ax + 1.319a y + 1.979az

1.7 THE CROSS PRODUCT
Given two vectors A and B, we now deﬁne the cross product, or vector product, of A and B, written with a cross between the two vectors as A × B and read “A cross B.” The cross product A × B is a vector; the magnitude of A × B is equal to the product of the magnitudes of A, B, and the sine of the smaller angle between A and B; the direction of A×B is perpendicular to the plane containing A and B and is along one of the two possible perpendiculars which is in the direction of advance of a right-handed screw as A is turned into B. This direction is illustrated in Figure 1.5. Remember that either vector may be moved about at will, maintaining its direction constant, until the two vectors have a “common origin.” This determines the plane containing both. However, in most of our applications we will be concerned with vectors deﬁned at the same point. As an equation we can write A × B = a N |A| |B| sin θ AB (7)

where an additional statement, such as that given above, is required to explain the direction of the unit vector a N . The subscript stands for “normal.”

12

ENGINEERING ELECTROMAGNETICS

Figure 1.5 The direction of A × B is in the direction of advance of a right-handed screw as A is turned into B.

Reversing the order of the vectors A and B results in a unit vector in the opposite direction, and we see that the cross product is not commutative, for B×A = −(A×B). If the deﬁnition of the cross product is applied to the unit vectors ax and a y , we ﬁnd ax × a y = az , for each vector has unit magnitude, the two vectors are perpendicular, and the rotation of ax into a y indicates the positive z direction by the deﬁnition of a right-handed coordinate system. In a similar way, a y × az = ax and az × ax = a y . Note the alphabetic symmetry. As long as the three vectors ax , a y , and az are written in order (and assuming that ax follows az , like three elephants in a circle holding tails, so that we could also write a y , az , ax or az , ax , a y ), then the cross and equal sign may be placed in either of the two vacant spaces. As a matter of fact, it is now simpler to deﬁne a right-handed rectangular coordinate system by saying that ax × a y = az . A simple example of the use of the cross product may be taken from geometry or trigonometry. To ﬁnd the area of a parallelogram, the product of the lengths of two adjacent sides is multiplied by the sine of the angle between them. Using vector notation for the two sides, we then may express the (scalar) area as the magnitude of A × B, or |A × B|. The cross product may be used to replace the right-hand rule familiar to all electrical engineers. Consider the force on a straight conductor of length L, where the direction assigned to L corresponds to the direction of the steady current I , and a uniform magnetic ﬁeld of ﬂux density B is present. Using vector notation, we may write the result neatly as F = I L × B. This relationship will be obtained later in Chapter 9. The evaluation of a cross product by means of its deﬁnition turns out to be more work than the evaluation of the dot product from its deﬁnition, for not only must we ﬁnd the angle between the vectors, but we must also ﬁnd an expression for the

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13

unit vector a N . This work may be avoided by using rectangular components for the two vectors A and B and expanding the cross product as a sum of nine simpler cross products, each involving two unit vectors, A × B = A x B x a x × a x + A x B y a x × a y + A x Bz a x × a z + A y B x a y × a x + A y B y a y × a y + A y Bz a y × a z + A z B x a z × a x + A z B y a z × a y + A z Bz a z × a z

We have already found that ax × a y = az , a y × az = ax , and az × ax = a y . The three remaining terms are zero, for the cross product of any vector with itself is zero, since the included angle is zero. These results may be combined to give A × B = (A y Bz − A z B y )ax + (A z Bx − A x Bz )a y + (A x B y − A y Bx )az or written as a determinant in a more easily remembered form, ax A × B = Ax Bx ax 2 −4 ay −3 −2 az 1 5 ay Ay By az Az Bz (8)

(9)

Thus, if A = 2ax − 3a y + az and B = −4ax − 2a y + 5az , we have A×B =

= [(−3)(5) − (1(−2)]ax − [(2)(5) − (1)(−4)]a y + [(2)(−2) − (−3)(−4)]az = −13ax − 14a y − 16az D1.4. The three vertices of a triangle are located at A(6, −1, 2), B(−2, 3, −4), and C(−3, 1, 5). Find: (a) R AB × R AC ; (b) the area of the triangle; (c) a unit vector perpendicular to the plane in which the triangle is located.
Ans. 24ax + 78a y + 20az ; 42.0; 0.286ax + 0.928a y + 0.238az

1.8 OTHER COORDINATE SYSTEMS: CIRCULAR CYLINDRICAL COORDINATES
The rectangular coordinate system is generally the one in which students prefer to work every problem. This often means a lot more work, because many problems possess a type of symmetry that pleads for a more logical treatment. It is easier to do now, once and for all, the work required to become familiar with cylindrical and spherical coordinates, instead of applying an equal or greater effort to every problem involving cylindrical or spherical symmetry later. With this in mind, we will take a careful and unhurried look at cylindrical and spherical coordinates.

14

ENGINEERING ELECTROMAGNETICS

The circular cylindrical coordinate system is the three-dimensional version of the polar coordinates of analytic geometry. In polar coordinates, a point is located in a plane by giving both its distance ρ from the origin and the angle φ between the line from the point to the origin and an arbitrary radial line, taken as φ = 0.4 In circular cylindrical coordinates, we also specify the distance z of the point from an arbitrary z = 0 reference plane that is perpendicular to the line ρ = 0. For simplicity, we usually refer to circular cylindrical coordinates simply as cylindrical coordinates. This will not cause any confusion in reading this book, but it is only fair to point out that there are such systems as elliptic cylindrical coordinates, hyperbolic cylindrical coordinates, parabolic cylindrical coordinates, and others. We no longer set up three axes as with rectangular coordinates, but we must instead consider any point as the intersection of three mutually perpendicular surfaces. These surfaces are a circular cylinder (ρ = constant), a plane (φ = constant), and another plane (z = constant). This corresponds to the location of a point in a rectangular coordinate system by the intersection of three planes (x = constant, y = constant, and z = constant). The three surfaces of circular cylindrical coordinates are shown in Figure 1.6a. Note that three such surfaces may be passed through any point, unless it lies on the z axis, in which case one plane sufﬁces. Three unit vectors must also be deﬁned, but we may no longer direct them along the “coordinate axes,” for such axes exist only in rectangular coordinates. Instead, we take a broader view of the unit vectors in rectangular coordinates and realize that they are directed toward increasing coordinate values and are perpendicular to the surface on which that coordinate value is constant (i.e., the unit vector ax is normal to the plane x = constant and points toward larger values of x). In a corresponding way we may now deﬁne three unit vectors in cylindrical coordinates, aρ , aφ , and az . The unit vector aρ at a point P(ρ1 , φ1 , z 1 ) is directed radially outward, normal to the cylindrical surface ρ = ρ1 . It lies in the planes φ = φ1 and z = z 1 . The unit vector aφ is normal to the plane φ = φ1 , points in the direction of increasing φ, lies in the plane z = z 1 , and is tangent to the cylindrical surface ρ = ρ1 . The unit vector az is the same as the unit vector az of the rectangular coordinate system. Figure 1.6b shows the three vectors in cylindrical coordinates. In rectangular coordinates, the unit vectors are not functions of the coordinates. Two of the unit vectors in cylindrical coordinates, aρ and aφ , however, do vary with the coordinate φ, as their directions change. In integration or differentiation with respect to φ, then, aρ and aφ must not be treated as constants. The unit vectors are again mutually perpendicular, for each is normal to one of the three mutually perpendicular surfaces, and we may deﬁne a right-handed cylindrical

4

The two variables of polar coordinates are commonly called r and θ . With three coordinates, however, it is more common to use ρ for the radius variable of cylindrical coordinates and r for the (different) radius variable of spherical coordinates. Also, the angle variable of cylindrical coordinates is customarily called φ because everyone uses θ for a different angle in spherical coordinates. The angle φ is common to both cylindrical and spherical coordinates. See?

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15

Figure 1.6 (a) The three mutually perpendicular surfaces of the circular cylindrical coordinate system. (b) The three unit vectors of the circular cylindrical coordinate system. (c) The differential volume unit in the circular cylindrical coordinate system; dρ, ρdφ, and dz are all elements of length.

coordinate system as one in which aρ × aφ = az , or (for those who have ﬂexible ﬁngers) as one in which the thumb, foreﬁnger, and middle ﬁnger point in the direction of increasing ρ, φ, and z, respectively. A differential volume element in cylindrical coordinates may be obtained by increasing ρ, φ, and z by the differential increments dρ, dφ, and dz. The two cylinders of radius ρ and ρ + dρ, the two radial planes at angles φ and φ + dφ, and the two “horizontal” planes at “elevations” z and z + dz now enclose a small volume, as shown in Figure 1.6c, having the shape of a truncated wedge. As the volume element becomes very small, its shape approaches that of a rectangular parallelepiped having sides of length dρ, ρdφ, and dz. Note that dρ and dz are dimensionally lengths, but dφ is not; ρdφ is the length. The surfaces have areas of ρ dρ dφ, dρ dz, and ρ dφ dz, and the volume becomes ρ dρ dφ dz.

16

ENGINEERING ELECTROMAGNETICS

Figure 1.7 The relationship between the rectangular variables x, y, z and the cylindrical coordinate variables ρ, φ, z. There is no change in the variable z between the two systems.

The variables of the rectangular and cylindrical coordinate systems are easily related to each other. Referring to Figure 1.7, we see that x = ρ cos φ y = ρ sin φ z=z

(10)

From the other viewpoint, we may express the cylindrical variables in terms of x, y, and z: x 2 + y2 y φ = tan−1 x z=z ρ= (ρ ≥ 0) (11)

We consider the variable ρ to be positive or zero, thus using only the positive sign for the radical in (11). The proper value of the angle φ is determined by inspecting the signs of x and y. Thus, if x = −3 and y = 4, we ﬁnd that the point lies in the second quadrant so that ρ = 5 and φ = 126.9◦ . For x = 3 and y = −4, we have φ = −53.1◦ or 306.9◦ , whichever is more convenient. Using (10) or (11), scalar functions given in one coordinate system are easily transformed into the other system. A vector function in one coordinate system, however, requires two steps in order to transform it to another coordinate system, because a different set of component

CHAPTER 1

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17

vectors is generally required. That is, we may be given a rectangular vector A = A x a x + A y a y + A z az where each component is given as a function of x, y, and z, and we need a vector in cylindrical coordinates A = Aρ aρ + Aφ aφ + A z az where each component is given as a function of ρ, φ, and z. To ﬁnd any desired component of a vector, we recall from the discussion of the dot product that a component in a desired direction may be obtained by taking the dot product of the vector and a unit vector in the desired direction. Hence, Aρ = A · aρ Expanding these dot products, we have Aφ = (A x ax + A y a y + A z az ) · aφ = A x ax · aφ + A y a y · aφ and A z = (A x ax + A y a y + A z az ) · az = A z az · az = A z (14) Aρ = (A x ax + A y a y + A z az ) · aρ = A x ax · aρ + A y a y · aρ (12) (13) and Aφ = A · aφ

since az · aρ and az · aφ are zero. In order to complete the transformation of the components, it is necessary to know the dot products ax · aρ , a y · aρ , ax · aφ , and a y · aφ . Applying the deﬁnition of the dot product, we see that since we are concerned with unit vectors, the result is merely the cosine of the angle between the two unit vectors in question. Referring to Figure 1.7 and thinking mightily, we identify the angle between ax and aρ as φ, and thus ax · aρ = cos φ, but the angle between a y and aρ is 90◦ − φ, and a y · aρ = cos (90◦ − φ) = sin φ. The remaining dot products of the unit vectors are found in a similar manner, and the results are tabulated as functions of φ in Table 1.1. Transforming vectors from rectangular to cylindrical coordinates or vice versa is therefore accomplished by using (10) or (11) to change variables, and by using the dot products of the unit vectors given in Table 1.1 to change components. The two steps may be taken in either order.

Table 1.1

Dot products of unit vectors in cylindrical and rectangular coordinate systems

aρ ax · ay · az · cos φ sin φ 0

aφ − sin φ cos φ 0

az 0 0 1

18

ENGINEERING ELECTROMAGNETICS

E X A M P L E 1.3

Transform the vector B = yax − xa y + zaz into cylindrical coordinates.
Solution. The new components are

Bρ = B · aρ = y(ax · aρ ) − x(a y · aρ )

Bφ = B · aφ = y(ax · aφ ) − x(a y · aφ ) Thus,

= y cos φ − x sin φ = ρ sin φ cos φ − ρ cos φ sin φ = 0 = −y sin φ − x cos φ = −ρ sin2 φ − ρ cos2 φ = −ρ

B = −ρaφ + zaz D1.5. (a) Give the rectangular coordinates of the point C(ρ = 4.4, φ = −115◦ , z = 2). (b) Give the cylindrical coordinates of the point D(x = −3.1, y = 2.6, z = −3). (c) Specify the distance from C to D.
Ans. C(x = −1.860, y = −3.99, z = 2); D(ρ = 4.05, φ = 140.0◦ , z = −3); 8.36

D1.6. Transform to cylindrical coordinates: (a) F = 10ax −8a y +6az at point P(10, −8, 6); (b) G = (2x + y)ax − (y − 4x)a y at point Q(ρ, φ, z). (c) Give the rectangular components of the vector H = 20aρ − 10aφ + 3az at P(x = 5, y = 2, z = −1).
Ans. 12.81aρ + 6az ; (2ρ cos2 φ − ρ sin2 φ + 5ρ sin φ cos φ)aρ + (4ρ cos2 φ − ρ sin2 φ − 3ρ sin φ cos φ)aφ ; Hx = 22.3, Hy = −1.857, Hz = 3

1.9 THE SPHERICAL COORDINATE SYSTEM
We have no two-dimensional coordinate system to help us understand the threedimensional spherical coordinate system, as we have for the circular cylindrical coordinate system. In certain respects we can draw on our knowledge of the latitudeand-longitude system of locating a place on the surface of the earth, but usually we consider only points on the surface and not those below or above ground. Let us start by building a spherical coordinate system on the three rectangular axes (Figure 1.8a). We ﬁrst deﬁne the distance from the origin to any point as r . The surface r = constant is a sphere. The second coordinate is an angle θ between the z axis and the line drawn from the origin to the point in question. The surface θ = constant is a cone, and the two surfaces, cone and sphere, are everywhere perpendicular along their intersection, which is a circle of radius r sin θ . The coordinate θ corresponds to latitude,

CHAPTER 1

Vector Analysis

19

Figure 1.8 (a) The three spherical coordinates. (b) The three mutually perpendicular surfaces of the spherical coordinate system. (c) The three unit vectors of spherical coordinates: ar × aθ = aφ . (d) The differential volume element in the spherical coordinate system.

except that latitude is measured from the equator and θ is measured from the “North Pole.” The third coordinate φ is also an angle and is exactly the same as the angle φ of cylindrical coordinates. It is the angle between the x axis and the projection in the z = 0 plane of the line drawn from the origin to the point. It corresponds to the angle of longitude, but the angle φ increases to the “east.” The surface φ = constant is a plane passing through the θ = 0 line (or the z axis). We again consider any point as the intersection of three mutually perpendicular surfaces—a sphere, a cone, and a plane—each oriented in the manner just described. The three surfaces are shown in Figure 1.8b. Three unit vectors may again be deﬁned at any point. Each unit vector is perpendicular to one of the three mutually perpendicular surfaces and oriented in that

20

ENGINEERING ELECTROMAGNETICS

direction in which the coordinate increases. The unit vector ar is directed radially outward, normal to the sphere r = constant, and lies in the cone θ = constant and the plane φ = constant. The unit vector aθ is normal to the conical surface, lies in the plane, and is tangent to the sphere. It is directed along a line of “longitude” and points “south.” The third unit vector aφ is the same as in cylindrical coordinates, being normal to the plane and tangent to both the cone and the sphere. It is directed to the “east.” The three unit vectors are shown in Figure 1.8c. They are, of course, mutually perpendicular, and a right-handed coordinate system is deﬁned by causing ar × aθ = aφ . Our system is right-handed, as an inspection of Figure 1.8c will show, on application of the deﬁnition of the cross product. The right-hand rule identiﬁes the thumb, foreﬁnger, and middle ﬁnger with the direction of increasing r , θ, and φ, respectively. (Note that the identiﬁcation in cylindrical coordinates was with ρ, φ, and z, and in rectangular coordinates with x, y, and z.) A differential volume element may be constructed in spherical coordinates by increasing r , θ, and φ by dr , dθ , and dφ, as shown in Figure 1.8d. The distance between the two spherical surfaces of radius r and r + dr is dr ; the distance between the two cones having generating angles of θ and θ + dθ is r dθ; and the distance between the two radial planes at angles φ and φ + dφ is found to be r sin θ dφ, after a few moments of trigonometric thought. The surfaces have areas of r dr dθ, r sin θ dr dφ, and r 2 sin θ dθ dφ, and the volume is r 2 sin θ dr dθ dφ. The transformation of scalars from the rectangular to the spherical coordinate system is easily made by using Figure 1.8a to relate the two sets of variables: x = r sin θ cos φ y = r sin θ sin φ z = r cos θ

(15)

The transformation in the reverse direction is achieved with the help of x 2 + y2 + z2 z θ = cos−1 x 2 + y2 + z2 y φ = tan−1 x r = (r ≥ 0)

(0◦ ≤ θ ≤ 180◦ )

(16)

The radius variable r is nonnegative, and θ is restricted to the range from 0◦ to 180◦ , inclusive. The angles are placed in the proper quadrants by inspecting the signs of x, y, and z. The transformation of vectors requires us to determine the products of the unit vectors in rectangular and spherical coordinates. We work out these products from Figure 1.8c and a pinch of trigonometry. Because the dot product of any spherical unit vector with any rectangular unit vector is the component of the spherical

CHAPTER 1

Vector Analysis

21

Table 1.2

Dot products of unit vectors in spherical and rectangular coordinate systems

ar ax · ay · az · sin θ cos φ sin θ sin φ cos θ

aθ cos θ cos φ cos θ sin φ − sin θ

aφ − sin φ cos φ 0

vector in the direction of the rectangular vector, the dot products with az are found to be az · aθ = −sin θ az · ar = cos θ

The dot products involving ax and a y require ﬁrst the projection of the spherical unit vector on the x y plane and then the projection onto the desired axis. For example, ar · ax is obtained by projecting ar onto the x y plane, giving sin θ , and then projecting sin θ on the x axis, which yields sin θ cos φ. The other dot products are found in a like manner, and all are shown in Table 1.2.
E X A M P L E 1.4

a z · aφ = 0

We illustrate this procedure by transforming the vector ﬁeld G = (x z/y)ax into spherical components and variables.
Solution. We ﬁnd the three spherical components by dotting G with the appropriate

unit vectors, and we change variables during the procedure: xz xz G r = G · ar = ax · ar = sin θ cos φ y y cos2 φ = r sin θ cos θ sin φ xz xz a x · aθ = cos θ cos φ G θ = G · aθ = y y cos2 φ = r cos2 θ sin φ xz xz a x · aφ = (−sin φ) Gφ = G · aφ = y y = −r cos θ cos φ Collecting these results, we have G = r cos θ cos φ (sin θ cot φ ar + cos θ cot φ aθ − aφ ) Appendix A describes the general curvilinear coordinate system of which the rectangular, circular cylindrical, and spherical coordinate systems are special cases. The ﬁrst section of this appendix could well be scanned now.

22

ENGINEERING ELECTROMAGNETICS

D1.7. Given the two points, C(−3, 2, 1) and D(r = 5, θ = 20◦ , φ = − 70◦ ), ﬁnd: (a) the spherical coordinates of C; (b) the rectangular coordinates of D; (c) the distance from C to D.
Ans. C(r = 3.74, θ = 74.5◦ , φ = 146.3◦ ); D(x = 0.585, y = −1.607, z = 4.70); 6.29

D1.8. Transform the following vectors to spherical coordinates at the points given: (a) 10ax at P(x = −3, y = 2, z = 4); (b) 10a y at Q(ρ = 5, φ = 30◦ , z = 4); (c) 10az at M(r = 4, θ = 110◦ , φ = 120◦ ).
Ans. −5.57ar − 6.18aθ − 5.55aφ ; 3.90ar + 3.12aθ + 8.66aφ ; −3.42ar − 9.40aθ

REFERENCES
1. Grossman, S. I. Calculus. 3d ed. Orlando, Fla.: Academic Press and Harcourt Brace Jovanovich, 1984. Vector algebra and cylindrical and spherical coordinates appear in Chapter 17, and vector calculus is introduced in Chapter 20. 2. Spiegel, M. R. Vector Analysis. Schaum Outline Series. New York: McGraw-Hill, 1959. A large number of examples and problems with answers are provided in this concise, inexpensive member of an outline series. 3. Swokowski, E. W. Calculus with Analytic Geometry. 3d ed. Boston: Prindle, Weber, & Schmidt, 1984. Vector algebra and the cylindrical and spherical coordinate systems are discussed in Chapter 14, and vector calculus appears in Chapter 18. 4. Thomas, G. B., Jr., and R. L. Finney: Calculus and Analytic Geometry. 6th ed. Reading, Mass.: Addison-Wesley Publishing Company, 1984. Vector algebra and the three coordinate systems we use are discussed in Chapter 13. Other vector operations are discussed in Chapters 15 and 17.

CHAPTER 1 PROBLEMS
1.1 Given the vectors M = −10ax + 4a y − 8az and N = 8ax + 7a y − 2az , ﬁnd: (a) a unit vector in the direction of −M + 2N; (b) the magnitude of 5ax + N − 3M; (c) |M||2N|(M + N).

1.2

Vector A extends from the origin to (1, 2, 3), and vector B extends from the origin to (2, 3, −2). Find (a) the unit vector in the direction of (A − B); (b) the unit vector in the direction of the line extending from the origin to the midpoint of the line joining the ends of A and B. The vector from the origin to point A is given as (6, −2, −4), and the unit vector directed from the origin toward point B is (2, −2, 1)/3. If points A and B are ten units apart, ﬁnd the coordinates of point B.

1.3

CHAPTER 1

Vector Analysis

23

1.4

A circle, centered at the origin with a radius of 2 units, lies in the x y plane. Determine the unit vector in√ rectangular components that lies in the x y plane, is tangent to the circle at (− 3,1, 0), and is in the general direction of increasing values of y. A vector ﬁeld is speciﬁed as G = 24x yax + 12(x 2 + 2)a y + 18z 2 az . Given two points, P(1, 2, −1) and Q(−2, 1, 3), ﬁnd (a) G at P; (b) a unit vector in the direction of G at Q; (c) a unit vector directed from Q toward P; (d) the equation of the surface on which |G| = 60.

1.5

1.6

Find the acute angle between the two vectors A = 2ax + a y + 3az and B = ax − 3a y + 2az by using the deﬁnition of (a) the dot product; (b) the cross product.

1.7

1.8

Given the vector ﬁeld E = 4zy 2 cos 2xax + 2zy sin 2xa y + y 2 sin 2xaz for the region |x|, |y|, and |z| less than 2, ﬁnd (a) the surfaces on which E y = 0; (b) the region in which E y = E z ; (c) the region in which E = 0. Demonstrate the ambiguity that results when the cross product is used to ﬁnd the angle between two vectors by ﬁnding the angle between A = 3ax − 2a y + 4az and B = 2ax + a y − 2az . Does this ambiguity exist when the dot product is used? A ﬁeld is given as G = [25/(x 2 + y 2 )](xax + ya y ). Find (a) a unit vector in the direction of G at P(3, 4, −2); (b) the angle between G and ax at P; (c) the value of the following double integral on the plane y = 7.
4 0 0 2

1.9

G · a y dzd x

1.10 By expressing diagonals as vectors and using the deﬁnition of the dot product, ﬁnd the smaller angle between any two diagonals of a cube, where each diagonal connects diametrically opposite corners and passes through the center of the cube. 1.11 Given the points M(0.1, −0.2, −0.1), N (−0.2, 0.1, 0.3), and P(0.4, 0, 0.1), ﬁnd (a) the vector R M N ; (b) the dot product R M N · R M P ; (c) the scalar projection of R M N on R M P ; (d) the angle between R M N and R M P . 1.12 Write an expression in rectangular components for the vector that extends from (x1 , y1 , z 1 ) to (x2 , y2 , z 2 ) and determine the magnitude of this vector. 1.13 Find (a) the vector component of F = 10ax − 6a y + 5az that is parallel to G = 0.1ax + 0.2a y + 0.3az ; (b) the vector component of F that is perpendicular to G; (c) the vector component of G that is perpendicular to F. 1.14 Given that A + B + C = 0, where the three vectors represent line segments and extend from a common origin, must the three vectors be coplanar? If A + B + C + D = 0, are the four vectors coplanar?

24

ENGINEERING ELECTROMAGNETICS

1.15 Three vectors extending from the origin are given as r1 = (7, 3, −2), r2 = (−2, 7, −3), and r3 = (0, 2, 3). Find (a) a unit vector perpendicular to both r1 and r2 ; (b) a unit vector perpendicular to the vectors r1 − r2 and r2 − r3 ; (c) the area of the triangle deﬁned by r1 and r2 ; (d) the area of the triangle deﬁned by the heads of r1 , r2 , and r3 . 1.16 If A represents a vector one unit long directed due east, B represents a vector three units long directed due north, and A + B = 2C − D and 2A − B = C + 2D, determine the length and direction of C.

1.17 Point A(−4, 2, 5) and the two vectors, R AM = (20, 18 − 10) and R AN = (−10, 8, 15), deﬁne a triangle. Find (a) a unit vector perpendicular to the triangle; (b) a unit vector in the plane of the triangle and perpendicular to R AN ; (c) a unit vector in the plane of the triangle that bisects the interior angle at A.

1.18 A certain vector ﬁeld is given as G = (y + 1)ax + xa y . (a) Determine G at the point (3, −2, 4); (b) obtain a unit vector deﬁning the direction of G at (3, −2, 4).

1.19 (a) Express the ﬁeld D = (x 2 + y 2 )−1 (xax + ya y ) in cylindrical components and cylindrical variables. (b) Evaluate D at the point where ρ = 2, φ = 0.2π , and z = 5, expressing the result in cylindrical and rectangular components. 1.20 If the three sides of a triangle are represented by vectors A, B, and C, all directed counterclockwise, show that |C|2 = (A + B) · (A + B) and expand the product to obtain the law of cosines.

1.21 Express in cylindrical components: (a) the vector from C(3, 2, −7) to D(−1, −4, 2); (b) a unit vector at D directed toward C; (c) a unit vector at D directed toward the origin. 1.22 A sphere of radius a, centered at the origin, rotates about the z axis at angular velocity rad/s. The rotation direction is clockwise when one is looking in the positive z direction. (a) Using spherical components, write an expression for the velocity ﬁeld, v, that gives the tangential velocity at any point within the sphere; (b) convert to rectangular components. 1.23 The surfaces ρ = 3, ρ = 5, φ = 100◦ , φ = 130◦ , z = 3, and z = 4.5 deﬁne a closed surface. Find (a) the enclosed volume; (b) the total area of the enclosing surface; (c) the total length of the twelve edges of the surfaces; (d) the length of the longest straight line that lies entirely within the volume. 1.24 Two unit vectors, a1 and a2 , lie in the x y plane and pass through the origin. They make angles φ1 and φ2 , respectively, with the x axis (a) Express each vector in rectangular components; (b) take the dot product and verify the trigonometric identity, cos(φ1 − φ2 ) = cos φ1 cos φ2 + sin φ1 sin φ2 ; (c) take the cross product and verify the trigonometric identity sin(φ2 − φ1 ) = sin φ2 cos φ1 − cos φ2 sin φ1 .

CHAPTER 1

Vector Analysis

25

1.25 Given point P(r = 0.8, θ = 30◦ , φ = 45◦ ) and E = 1/r 2 [cos φ ar + (sin φ/ sin θ) aφ ], ﬁnd (a) E at P; (b) |E| at P; (c) a unit vector in the direction of E at P. 1.26 Express the uniform vector ﬁeld F = 5ax in (a) cylindrical components; (b) spherical components. 1.27 The surfaces r = 2 and 4, θ = 30◦ and 50◦ , and φ = 20◦ and 60◦ identify a closed surface. Find (a) the enclosed volume; (b) the total area of the enclosing surface; (c) the total length of the twelve edges of the surface; (d) the length of the longest straight line that lies entirely within the surface. 1.28 State whether or not A = B and, if not, what conditions are imposed on A and B when (a) A · ax = B · ax ; (b) A × ax = B × ax ; (c) A · ax = B · ax and A × ax = B × ax ; (d) A · C = B · C and A × C = B × C where C is any vector except C = 0.

1.29 Express the unit vector ax in spherical components at the point: (a) r = 2, θ = 1 rad, φ = 0.8 rad; (b) x = 3, y = 2, z = −1; (c) ρ = 2.5, φ = 0.7 rad, z = 1.5.

1.30 Consider a problem analogous to the varying wind velocities encountered by transcontinental aircraft. We assume a constant altitude, a plane earth, a ﬂight along the x axis from 0 to 10 units, no vertical velocity component, and no change in wind velocity with time. Assume ax to be directed to the east and a y to the north. The wind velocity at the operating altitude is assumed to be: (0.01x 2 − 0.08x + 0.66)ax − (0.05x − 0.4)a y 1 + 0.5y 2 Determine the location and magnitude of (a) the maximum tailwind encountered; (b) repeat for headwind; (c) repeat for crosswind; (d) Would more favorable tailwinds be available at some other latitude? If so, where? v(x, y) =

2

C H A P T E R

Coulomb’s Law and Electric Field Intensity

H

aving formulated the language of vector analysis in the ﬁrst chapter, we next establish and describe a few basic principles of electricity. In this chapter, we introduce Coulomb’s electrostatic force law and then formulate this in a general way using ﬁeld theory. The tools that will be developed can be used to solve any problem in which forces between static charges are to be evaluated or to determine the electric ﬁeld that is associated with any charge distribution. Initially, we will restrict the study to ﬁelds in vacuum or free space; this would apply to media such as air and other gases. Other materials are introduced in Chapters 5 and 6 and time-varying ﬁelds are introduced in Chapter 9. ■

2.1 THE EXPERIMENTAL LAW OF COULOMB
Records from at least 600 B.C. show evidence of the knowledge of static electricity. The Greeks were responsible for the term electricity, derived from their word for amber, and they spent many leisure hours rubbing a small piece of amber on their sleeves and observing how it would then attract pieces of ﬂuff and stuff. However, their main interest lay in philosophy and logic, not in experimental science, and it was many centuries before the attracting effect was considered to be anything other than magic or a “life force.” Dr. Gilbert, physician to Her Majesty the Queen of England, was the ﬁrst to do any true experimental work with this effect, and in 1600 he stated that glass, sulfur, amber, and other materials, which he named, would “not only draw to themselves straws and chaff, but all metals, wood, leaves, stone, earths, even water and oil.” Shortly thereafter, an ofﬁcer in the French Army Engineers, Colonel Charles Coulomb, performed an elaborate series of experiments using a delicate torsion balance, invented by himself, to determine quantitatively the force exerted between two objects, each having a static charge of electricity. His published result bears a great similarity to Newton’s gravitational law (discovered about a hundred years earlier).
26

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

27

Coulomb stated that the force between two very small objects separated in a vacuum or free space by a distance, which is large compared to their size, is proportional to the charge on each and inversely proportional to the square of the distance between them, or Q1 Q2 F =k R2 where Q 1 and Q 2 are the positive or negative quantities of charge, R is the separation, and k is a proportionality constant. If the International System of Units1 (SI) is used, Q is measured in coulombs (C), R is in meters (m), and the force should be newtons (N). This will be achieved if the constant of proportionality k is written as 1 k= 4π 0 The new constant 0 is called the permittivity of free space and has magnitude, measured in farads per meter (F/m),
0

= 8.854 × 10−12 = ˙

1 10−9 F/m 36π

(1)

The quantity 0 is not dimensionless, for Coulomb’s law shows that it has the label C2 /N · m2 . We will later deﬁne the farad and show that it has the dimensions C2 /N · m; we have anticipated this deﬁnition by using the unit F/m in equation (1). Coulomb’s law is now F= Q1 Q2 4π 0 R 2 (2)

The coulomb is an extremely large unit of charge, for the smallest known quantity of charge is that of the electron (negative) or proton (positive), given in SI units as 1.602 × 10−19 C; hence a negative charge of one coulomb represents about 6 × 1018 electrons.2 Coulomb’s law shows that the force between two charges of one coulomb each, separated by one meter, is 9 × 109 N, or about one million tons. The electron has a rest mass of 9.109 × 10−31 kg and has a radius of the order of magnitude of 3.8 × 10−15 m. This does not mean that the electron is spherical in shape, but merely describes the size of the region in which a slowly moving electron has the greatest probability of being found. All other known charged particles, including the proton, have larger masses and larger radii, and occupy a probabilistic volume larger than does the electron. In order to write the vector form of (2), we need the additional fact (furnished also by Colonel Coulomb) that the force acts along the line joining the two charges

1

The International System of Units (an mks system) is described in Appendix B. Abbreviations for the units are given in Table B.1. Conversions to other systems of units are given in Table B.2, while the preﬁxes designating powers of ten in SI appear in Table B.3. The charge and mass of an electron and other physical constants are tabulated in Table C.4 of Appendix C.

2

28

ENGINEERING ELECTROMAGNETICS

Figure 2.1 If Q 1 and Q 2 have like signs, the vector force F2 on Q 2 is in the same direction as the vector R12 .

and is repulsive if the charges are alike in sign or attractive if they are of opposite sign. Let the vector r1 locate Q 1 , whereas r2 locates Q 2 . Then the vector R12 = r2 − r1 represents the directed line segment from Q 1 to Q 2 , as shown in Figure 2.1. The vector F2 is the force on Q 2 and is shown for the case where Q 1 and Q 2 have the same sign. The vector form of Coulomb’s law is Q1 Q2 a (3) F2 = 2 12 4π 0 R12 where a12 = a unit vector in the direction of R12 , or a12 =
E X A M P L E 2.1

R12 r2 − r1 R12 = = |R12 | R12 |r2 − r1 |

(4)

We illustrate the use of the vector form of Coulomb’s law by locating a charge of Q 1 = 3 × 10−4 C at M(1, 2, 3) and a charge of Q 2 = −10−4 C at N (2, 0, 5) in a vacuum. We desire the force exerted on Q 2 by Q 1 .
Solution. We use (3) and (4) to obtain the vector force. The vector R12 is

leading to |R12 | = 3, and the unit vector, a12 = 1 (ax − 2a y + 2az ). Thus, 3 F2 = 3 × 10−4 (−10−4 ) 4π (1/36π )10−9 × 32 ax − 2a y + 2az 3 N

R12 = r2 − r1 = (2 − 1)ax + (0 − 2)a y + (5 − 3)az = ax − 2a y + 2az

= −30

ax − 2a y + 2az 3

The magnitude of the force is 30 N, and the direction is speciﬁed by the unit vector, which has been left in parentheses to display the magnitude of the force. The force on Q 2 may also be considered as three component forces, F2 = −10ax + 20a y − 20az

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

29

The force expressed by Coulomb’s law is a mutual force, for each of the two charges experiences a force of the same magnitude, although of opposite direction. We might equally well have written F1 = −F2 = Q1 Q2 Q1 Q2 a =− a 2 21 2 12 4π 0 R12 4π 0 R12 (5)

Coulomb’s law is linear, for if we multiply Q 1 by a factor n, the force on Q 2 is also multiplied by the same factor n. It is also true that the force on a charge in the presence of several other charges is the sum of the forces on that charge due to each of the other charges acting alone. D2.1. A charge Q A = −20 µC is located at A(−6, 4, 7), and a charge Q B = 50 µC is at B(5, 8, −2) in free space. If distances are given in meters, ﬁnd: (a) R AB ; (b) R AB . Determine the vector force exerted on Q A by Q B if 0 = (c) 10−9 /(36π) F/m; (d) 8.854 × 10−12 F/m.
Ans. 11ax + 4a y − 9az m; 14.76 m; 30.76ax + 11.184a y − 25.16az mN; 30.72ax + 11.169a y − 25.13az mN

2.2 ELECTRIC FIELD INTENSITY
If we now consider one charge ﬁxed in position, say Q 1 , and move a second charge slowly around, we note that there exists everywhere a force on this second charge; in other words, this second charge is displaying the existence of a force fiel that is associated with charge, Q 1 . Call this second charge a test charge Q t . The force on it is given by Coulomb’s law, Q1 Qt a Ft = 2 1t 4π 0 R1t Writing this force as a force per unit charge gives the electric fiel intensity, E1 arising from Q 1 : Ft Q1 E1 = = a1t (6) 2 Q1 4π 0 R1t E1 is interpreted as the vector force, arising from charge Q 1 , that acts on a unit positive test charge. More generally, we write the deﬁning expression: E= Ft Qt (7)

in which E, a vector function, is the electric ﬁeld intensity evaluated at the test charge location that arises from all other charges in the vicinity—meaning the electric ﬁeld arising from the test charge itself is not included in E. The units of E would be in force per unit charge (newtons per coulomb). Again anticipating a new dimensional quantity, the volt (V), having the label of joules per

30

ENGINEERING ELECTROMAGNETICS

coulomb (J/C), or newton-meters per coulomb (N · m/C), we measure electric ﬁeld intensity in the practical units of volts per meter (V/m). Now, we dispense with most of the subscripts in (6), reserving the right to use them again any time there is a possibility of misunderstanding. The electric ﬁeld of a single point charge becomes: E= Q 4π
0R 2

aR

(8)

We remember that R is the magnitude of the vector R, the directed line segment from the point at which the point charge Q is located to the point at which E is desired, and a R is a unit vector in the R direction.3 We arbitrarily locate Q 1 at the center of a spherical coordinate system. The unit vector a R then becomes the radial unit vector ar , and R is r . Hence Q1 E= ar (9) 4π 0r 2 The ﬁeld has a single radial component, and its inverse-square-law relationship is quite obvious. If we consider a charge that is not at the origin of our coordinate system, the ﬁeld no longer possesses spherical symmetry, and we might as well use rectangular coordinates. For a charge Q located at the source point r = x ax + y a y + z az , as illustrated in Figure 2.2, we ﬁnd the ﬁeld at a general ﬁeld point r = xax + ya y + zaz by expressing R as r − r , and then E(r) = = Q r−r Q(r − r ) = 4π 0 |r − r |2 |r − r | 4π 0 |r − r |3

Q[(x − x )ax + (y − y )a y + (z − z )az ] 4π 0 [(x − x )2 + (y − y )2 + (z − z )2 ]3/2

(10)

Earlier, we deﬁned a vector ﬁeld as a vector function of a position vector, and this is emphasized by letting E be symbolized in functional notation by E(r). Because the coulomb forces are linear, the electric ﬁeld intensity arising from two point charges, Q 1 at r1 and Q 2 at r2 , is the sum of the forces on Q t caused by Q 1 and Q 2 acting alone, or E(r) = Q1 Q2 a1 + a2 4π 0 |r − r1 |2 4π 0 |r − r2 |2

where a1 and a2 are unit vectors in the direction of (r − r1 ) and (r − r2 ), respectively. The vectors r, r1 , r2 , r − r1 , r − r2 , a1 , and a2 are shown in Figure 2.3.

3

We ﬁrmly intend to avoid confusing r and ar with R and a R . The ﬁrst two refer speciﬁcally to the spherical coordinate system, whereas R and a R do not refer to any coordinate system—the choice is still available to us.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

31

Figure 2.2 The vector r locates the point charge Q, the vector r identifies the general point in space P(x, y, z), and the vector R from Q to P(x, y, z) is then R = r − r .

Figure 2.3 The vector addition of the total electric field intensity at P due to Q 1 and Q 2 is made possible by the linearity of Coulomb’s law.

32

ENGINEERING ELECTROMAGNETICS

If we add more charges at other positions, the ﬁeld due to n point charges is E(r) = Qm am 4π 0 |r − rm |2 m=1 n (11)

E X A M P L E 2.2

In order to illustrate the application of (11), we ﬁnd E at P(1, 1, 1) caused by four identical 3-nC (nanocoulomb) charges located at P1 (1, 1, 0), P2 (−1, 1, 0), P3 (−1, −1, 0), and P4 (1, −1, 0), as shown in Figure 2.4. The magnitudes are: |r − r1 | = 1, |r − r2 | = 5, |r − r3 | = 3, and |r − r4 | = 5. Because Q/4π 0 = 3 × 10−9 /(4π × 8.854 × 10−12 ) = 26.96 V · m, we may now use (11) to obtain E = 26.96 or E = 6.82ax + 6.82a y + 32.8az V/m D2.2. A charge of −0.3 µC is located at A(25, −30, 15) (in cm), and a second charge of 0.5 µC is at B(−10, 8, 12) cm. Find E at: (a) the origin; (b) P(15, 20, 50) cm.
Ans. 92.3ax − 77.6a y − 94.2az kV/m; 11.9ax − 0.519a y + 12.4az kV/m Solution. We ﬁnd that r = ax + a y + az , r√= ax + a y , and thus r − r1 =√ z . a 1

az 1 2ax + az 1 + √ √ 2 1 1 5 5

2

+

2ax + 2a y + az 1 2a y + az 1 + √ √ 2 3 3 5 5

2

Figure 2.4 A symmetrical distribution of four identical 3-nC point charges produces a field at P, E = 6.82ax + 6.82a y + 32.8az V/m.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

33

D2.3. Evaluate the sums: (a)
Ans. 2.52; 0.176

4 1 + (−1)m (0.1)m + 1 ; (b) m2 + 1 (4 + m 2 )1.5 m=0 m=1 5

2.3 FIELD ARISING FROM A CONTINUOUS VOLUME CHARGE DISTRIBUTION
If we now visualize a region of space ﬁlled with a tremendous number of charges separated by minute distances, we see that we can replace this distribution of very small particles with a smooth continuous distribution described by a volume charge density, just as we describe water as having a density of 1 g/cm3 (gram per cubic centimeter) even though it consists of atomic- and molecular-sized particles. We can do this only if we are uninterested in the small irregularities (or ripples) in the ﬁeld as we move from electron to electron or if we care little that the mass of the water actually increases in small but ﬁnite steps as each new molecule is added. This is really no limitation at all, because the end results for electrical engineers are almost always in terms of a current in a receiving antenna, a voltage in an electronic circuit, or a charge on a capacitor, or in general in terms of some large-scale macroscopic phenomenon. It is very seldom that we must know a current electron by electron.4 We denote volume charge density by ρν , having the units of coulombs per cubic meter (C/m3 ). The small amount of charge Q in a small volume ν is Q = ρν ν and we may deﬁne ρν mathematically by using a limiting process on (12), ρν = lim Q ν (13) (12)

ν→0

The total charge within some ﬁnite volume is obtained by integrating throughout that volume, Q= ρν dν vol (14)

Only one integral sign is customarily indicated, but the differential dν signiﬁes integration throughout a volume, and hence a triple integration.

4

A study of the noise generated by electrons in semiconductors and resistors, however, requires just such an examination of the charge through statistical analysis.

34

ENGINEERING ELECTROMAGNETICS

E X A M P L E 2.3

As an example of the evaluation of a volume integral, we ﬁnd the total charge contained in a 2-cm length of the electron beam shown in Figure 2.5.
Solution. From the illustration, we see that the charge density is

ρν = −5 × 10−6 e−10 Q=
0.04 0.02 0 2π 0 0.01

5

ρz

C/m2

The volume differential in cylindrical coordinates is given in Section 1.8; therefore, −5 × 10−6 e−10
5

ρz

ρ dρ dφ dz

We integrate ﬁrst with respect to φ because it is so easy, Q=
0.04 0.02 0 0.01

−10−5 π e−10

5

ρz

ρ dρ dz

and then with respect to z, because this will simplify the last integration with respect to ρ, Q= =
0.01 0 0.01 0

−10−5 π −105 ρz ρ dρ e −105 ρ

z=0.04 z=0.02

−10−5 π (e−2000ρ − e−4000ρ )dρ

Figure 2.5 The total charge contained within the right circular cylinder may be obtained by evaluatingQ = vol ρν dν.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

35

Finally, Q = −10−10 π Q = −10−10 π e−2000ρ e−4000ρ − −2000 −4000 1 1 − 2000 4000 =
0.01 0

−π = 0.0785 pC 40

where pC indicates picocoulombs. The incremental contribution to the electric ﬁeld intensity at r produced by an incremental charge Q at r is ρν ν r−r r−r Q = E(r) = 4π 0 |r − r |2 |r − r | 4π 0 |r − r |2 |r − r | If we sum the contributions of all the volume charge in a given region and let the volume element ν approach zero as the number of these elements becomes inﬁnite, the summation becomes an integral, ρν (r ) dν r−r (15) E(r) = 2 vol 4π 0 |r − r | |r − r |

This is again a triple integral, and (except in Drill Problem 2.4) we shall do our best to avoid actually performing the integration. The signiﬁcance of the various quantities under the integral sign of (15) might stand a little review. The vector r from the origin locates the ﬁeld point where E is being determined, whereas the vector r extends from the origin to the source point where ρv (r )dν is located. The scalar distance between the source point and the ﬁeld point is |r − r |, and the fraction (r − r )/|r − r | is a unit vector directed from source point to ﬁeld point. The variables of integration are x , y , and z in rectangular coordinates. D2.4. Calculate the total charge within each of the indicated volumes: (a) 0.1 ≤ 1 |x|, |y|, |z| ≤ 0.2: ρν = 3 3 3 ; (b) 0 ≤ ρ ≤ 0.1, 0 ≤ φ ≤ π , 2 ≤ z ≤ 4; ρν = x y z ρ 2 z 2 sin 0.6φ; (c) universe: ρν = e−2r /r 2 .
Ans. 0; 1.018 mC; 6.28 C

2.4 FIELD OF A LINE CHARGE
Up to this point we have considered two types of charge distribution, the point charge and charge distributed throughout a volume with a density ρν C/m3 . If we now consider a ﬁlamentlike distribution of volume charge density, such as a charged conductor of very small radius, we ﬁnd it convenient to treat the charge as a line charge of density ρ L C/m. We assume a straight-line charge extending along the z axis in a cylindrical coordinate system from −∞ to ∞, as shown in Figure 2.6. We desire the electric ﬁeld intensity E at any and every point resulting from a uniform line charge density ρ L .

36

ENGINEERING ELECTROMAGNETICS

Figure 2.6 The contribution dE = dE ρ aρ + dE z az to the electric field intensity produced by an element of charge dQ = ρ L dz located a distance z from the origin. The linear charge density is uniform and extends along the entire z axis.

Symmetry should always be considered ﬁrst in order to determine two speciﬁc factors: (1) with which coordinates the ﬁeld does not vary, and (2) which components of the ﬁeld are not present. The answers to these questions then tell us which components are present and with which coordinates they do vary. Referring to Figure 2.6, we realize that as we move around the line charge, varying φ while keeping ρ and z constant, the line charge appears the same from every angle. In other words, azimuthal symmetry is present, and no ﬁeld component may vary with φ. Again, if we maintain ρ and φ constant while moving up and down the line charge by changing z, the line charge still recedes into inﬁnite distance in both directions and the problem is unchanged. This is axial symmetry and leads to ﬁelds that are not functions of z. If we maintain φ and z constant and vary ρ, the problem changes, and Coulomb’s law leads us to expect the ﬁeld to become weaker as ρ increases. Hence, by a process of elimination we are led to the fact that the ﬁeld varies only with ρ. Now, which components are present? Each incremental length of line charge acts as a point charge and produces an incremental contribution to the electric ﬁeld intensity which is directed away from the bit of charge (assuming a positive line charge). No element of charge produces a φ component of electric intensity; E φ is zero. However, each element does produce an E ρ and an E z component, but the contribution to E z by elements of charge that are equal distances above and below the point at which we are determining the ﬁeld will cancel.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

37

We therefore have found that we have only an E ρ component and it varies only with ρ. Now to ﬁnd this component. We choose a point P(0, y, 0) on the y axis at which to determine the ﬁeld. This is a perfectly general point in view of the lack of variation of the ﬁeld with φ and z. Applying (10) to ﬁnd the incremental ﬁeld at P due to the incremental charge d Q = ρ L dz , we have dE = where ρ L dz (r − r ) 4π 0 |r − r |3

r = z az and

r = ya y = ρaρ

r − r = ρaρ − z az Therefore, dE = ρ L dz (ρaρ − z az ) 4π 0 (ρ 2 + z 2 )3/2

Because only the Eρ component is present, we may simplify: d Eρ = and Eρ = ρ L ρdz 4π 0 (ρ 2 + z 2 )3/2 ρ L ρdz 4π 0 (ρ 2 + z 2 )3/2 1 ρ2 z

∞ −∞

Integrating by integral tables or change of variable, z = ρ cot θ , we have Eρ = and Eρ = or ﬁnally, E= ρL aρ 2π 0 ρ (16) ρL 2π 0 ρ ρL ρ 4π 0
∞

ρ2 + z 2

−∞

We note that the ﬁeld falls off inversely with the distance to the charged line, as compared with the point charge, where the ﬁeld decreased with the square of the distance. Moving ten times as far from a point charge leads to a ﬁeld only 1 percent the previous strength, but moving ten times as far from a line charge only reduces the ﬁeld to 10 percent of its former value. An analogy can be drawn with a source of

38

ENGINEERING ELECTROMAGNETICS

Figure 2.7 A point P(x, y, z) is identified near an infinite uniform line charge located at x = 6, y = 8.

illumination, for the light intensity from a point source of light also falls off inversely as the square of the distance to the source. The ﬁeld of an inﬁnitely long ﬂuorescent tube thus decays inversely as the ﬁrst power of the radial distance to the tube, and we should expect the light intensity about a ﬁnite-length tube to obey this law near the tube. As our point recedes farther and farther from a ﬁnite-length tube, however, it eventually looks like a point source, and the ﬁeld obeys the inverse-square relationship. Before leaving this introductory look at the ﬁeld of the inﬁnite line charge, we should recognize the fact that not all line charges are located along the z axis. As an example, let us consider an inﬁnite line charge parallel to the z axis at x = 6, y = 8, shown in Figure 2.7. We wish to ﬁnd E at the general ﬁeld point P(x, y, z). We replace ρ in (16) by the radial distance between the line charge and point, P, R = (x − 6)2 + (y − 8)2 , and let aρ be a R . Thus, E= where aR = Therefore, E= ρ L (x − 6)ax + (y − 8)a y 2π 0 (x − 6)2 + (y − 8)2 (x − 6)ax + (y − 8)a y R = |R| (x − 6)2 + (y − 8)2 ρL 2π
0

(x − 6)2 + (y − 8)2

aR

We again note that the ﬁeld is not a function of z.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

39

In Section 2.6, we describe how ﬁelds may be sketched, and we use the ﬁeld of the line charge as one example. D2.5. Inﬁnite uniform line charges of 5 nC/m lie along the (positive and negative) x and y axes in free space. Find E at: (a) PA (0, 0, 4); (b) PB (0, 3, 4).
Ans. 45az V/m; 10.8a y + 36.9az V/m

2.5 FIELD OF A SHEET OF CHARGE
Another basic charge conﬁguration is the inﬁnite sheet of charge having a uniform density of ρ S C/m2 . Such a charge distribution may often be used to approximate that found on the conductors of a strip transmission line or a parallel-plate capacitor. As we shall see in Chapter 5, static charge resides on conductor surfaces and not in their interiors; for this reason, ρ S is commonly known as surface charge density. The charge-distribution family now is complete—point, line, surface, and volume, or Q, ρ L , ρ S , and ρν . Let us place a sheet of charge in the yz plane and again consider symmetry (Figure 2.8). We see ﬁrst that the ﬁeld cannot vary with y or with z, and then we see that the y and z components arising from differential elements of charge symmetrically located with respect to the point at which we evaluate the ﬁeld will cancel. Hence only E x is present, and this component is a function of x alone. We are again faced with a choice of many methods by which to evaluate this component, and this time we use only one method and leave the others as exercises for a quiet Sunday afternoon. Let us use the ﬁeld of the inﬁnite line charge (16) by dividing the inﬁnite sheet into differential-width strips. One such strip is shown in Figure 2.8. The line charge

Figure 2.8 An infinite sheet of charge in the yz plane, a general point P on the x axis, and the differential-width line charge used as the element in determining the field at P by dE = ρ S dy a R /(2πε0 R).

40

ENGINEERING ELECTROMAGNETICS

density, or charge per unit length, is ρ L = ρ S dy , and the distance from this line charge to our general point P on the x axis is R = x 2 + y 2 . The contribution to E x at P from this differential-width strip is then d Ex = ρS 2π 0 ρ S dy 2π
0

x2

Adding the effects of all the strips, Ex =
∞ −∞

+y

2

cos θ =

xdy ρS 2π 0 x 2 + y 2
∞ −∞

If the point P were chosen on the negative x axis, then ρS Ex = − 2 0 for the ﬁeld is always directed away from the positive charge. This difﬁculty in sign is usually overcome by specifying a unit vector a N , which is normal to the sheet and directed outward, or away from it. Then E= ρS aN 2 0 (17)

ρS x dy y = tan−1 x2 + y 2 2π 0 x

=

ρS 2 0

This is a startling answer, for the ﬁeld is constant in magnitude and direction. It is just as strong a million miles away from the sheet as it is right off the surface. Returning to our light analogy, we see that a uniform source of light on the ceiling of a very large room leads to just as much illumination on a square foot on the ﬂoor as it does on a square foot a few inches below the ceiling. If you desire greater illumination on this subject, it will do you no good to hold the book closer to such a light source. If a second inﬁnite sheet of charge, having a negative charge density −ρ S , is located in the plane x = a, we may ﬁnd the total ﬁeld by adding the contribution of each sheet. In the region x > a, ρS ρS ax E− = − ax E = E+ + E− = 0 E+ = 2 0 2 0 and for x < 0, ρS ρS E+ = − ax E− = ax E = E+ + E− = 0 2 0 2 0 and when 0 < x < a, ρS ρS E+ = ax E− = ax 2 0 2 0 and E = E+ + E− = ρS
0

ax

(18)

This is an important practical answer, for it is the ﬁeld between the parallel plates of an air capacitor, provided the linear dimensions of the plates are very much greater than their separation and provided also that we are considering a point well removed

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

41

from the edges. The ﬁeld outside the capacitor, while not zero, as we found for the preceding ideal case, is usually negligible. D2.6. Three inﬁnite uniform sheets of charge are located in free space as follows: 3 nC/m2 at z = −4, 6 nC/m2 at z = 1, and −8 nC/m2 at z = 4. Find E at the point: (a) PA (2, 5, −5); (b) PB (4, 2, −3); (c) PC (−1, −5, 2); (d) PD (−2, 4, 5).
Ans. −56.5az ; 283az ; 961az ; 56.5az all V/m

2.6 STREAMLINES AND SKETCHES OF FIELDS
We now have vector equations for the electric ﬁeld intensity resulting from several different charge conﬁgurations, and we have had little difﬁculty in interpreting the magnitude and direction of the ﬁeld from the equations. Unfortunately, this simplicity cannot last much longer, for we have solved most of the simple cases and our new charge distributions must lead to more complicated expressions for the ﬁelds and more difﬁculty in visualizing the ﬁelds through the equations. However, it is true that one picture would be worth about a thousand words, if we just knew what picture to draw. Consider the ﬁeld about the line charge, ρL E= aρ 2π 0 ρ Figure 2.9a shows a cross-sectional view of the line charge and presents what might be our ﬁrst effort at picturing the ﬁeld—short line segments drawn here and there having lengths proportional to the magnitude of E and pointing in the direction of E. The ﬁgure fails to show the symmetry with respect to φ, so we try again in Figure 2.9b with a symmetrical location of the line segments. The real trouble now appears—the longest lines must be drawn in the most crowded region, and this also plagues us if we use line segments of equal length but of a thickness that is proportional to E (Figure 2.9c). Other schemes include drawing shorter lines to represent stronger ﬁelds (inherently misleading) and using intensity of color or different colors to represent stronger ﬁelds. For the present, let us be content to show only the direction of E by drawing continuous lines, which are everywhere tangent to E, from the charge. Figure 2.9d shows this compromise. A symmetrical distribution of lines (one every 45◦ ) indicates azimuthal symmetry, and arrowheads should be used to show direction. These lines are usually called streamlines, although other terms such as ﬂux lines and direction lines are also used. A small positive test charge placed at any point in this ﬁeld and free to move would accelerate in the direction of the streamline passing through that point. If the ﬁeld represented the velocity of a liquid or a gas (which, incidentally, would have to have a source at ρ = 0), small suspended particles in the liquid or gas would trace out the streamlines.

42

ENGINEERING ELECTROMAGNETICS

Figure 2.9 (a) One very poor sketch, (b) and (c) two fair sketches, and (d ) the usual form of a streamline sketch. In the last form, the arrows show the direction of the field at every point along the line, and the spacing of the lines is inversely proportional to the strength of the field.

We will ﬁnd out later that a bonus accompanies this streamline sketch, for the magnitude of the ﬁeld can be shown to be inversely proportional to the spacing of the streamlines for some important special cases. The closer they are together, the stronger is the ﬁeld. At that time we will also ﬁnd an easier, more accurate method of making that type of streamline sketch. If we attempted to sketch the ﬁeld of the point charge, the variation of the ﬁeld into and away from the page would cause essentially insurmountable difﬁculties; for this reason sketching is usually limited to two-dimensional ﬁelds. In the case of the two-dimensional ﬁeld, let us arbitrarily set E z = 0. The streamlines are thus conﬁned to planes for which z is constant, and the sketch is the same for any such plane. Several streamlines are shown in Figure 2.10, and the E x and E y components are indicated at a general point. It is apparent from the geometry that Ey dy = Ex dx (19)

A knowledge of the functional form of E x and E y (and the ability to solve the resultant differential equation) will enable us to obtain the equations of the streamlines.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

43

Figure 2.10 The equation of a streamline is obtained by solving the differential equation E y /E x = dy/dx.

As an illustration of this method, consider the ﬁeld of the uniform line charge with ρ L = 2π 0 , E= In rectangular coordinates, E= x y ax + 2 ay x 2 + y2 x + y2 dy dx = y x 1 aρ ρ

Thus we form the differential equation Ey y dy = = dx Ex x Therefore, ln y = ln x + C1 or ln y = ln x + ln C or

from which the equations of the streamlines are obtained, y = Cx If we want to ﬁnd the equation of one particular streamline, say one passing through P(−2, 7, 10), we merely substitute the coordinates of that point into our equation and evaluate C. Here, 7 = C(−2), and C = −3.5, so y = −3.5x. Each streamline is associated with a speciﬁc value of C, and the radial lines shown in Figure 2.9d are obtained when C = 0, 1, −1, and 1/C = 0. The equations of streamlines may also be obtained directly in cylindrical or spherical coordinates. A spherical coordinate example will be examined in Section 4.7.

44

ENGINEERING ELECTROMAGNETICS

D2.7. Find the equation of that streamline that passes through the point 4x 2 −8x ax + 2 a y ; (b) 2e5x [y(5x + 1)ax + xa y ]. P(1, 4, −2) in the ﬁeld E = (a) y y
Ans. x 2 + 2y 2 = 33; y 2 = 15.7 + 0.4x − 0.08 ln(5x + 1)

REFERENCES
1. Boast, W. B. Vector Fields. New York: Harper and Row, 1964. This book contains numerous examples and sketches of ﬁelds. 2. Della Torre, E., and Longo, C. L. The Electromagnetic Field. Boston: Allyn and Bacon, 1969. The authors introduce all of electromagnetic theory with a careful and rigorous development based on a single experimental law—that of Coulomb. It begins in Chapter 1. 3. Schelkunoff, S. A. Electromagnetic Fields. New York: Blaisdell Publishing Company, 1963. Many of the physical aspects of ﬁelds are discussed early in this text without advanced mathematics.

CHAPTER 2 PROBLEMS
2.1 Three point charges are positioned in the x-y plane as follows: 5 nC at y = 5 cm, −10 nC at y = −5 cm, and 15 nC at x = −5 cm. Find the required x-y coordinates of a 20-nC fourth charge that will produce a zero electric ﬁeld at the origin. Point charges of 1 nC and −2 nC are located at (0, 0, 0) and (1, 1, 1), respectively, in free space. Determine the vector force acting on each charge. Point charges of 50 nC each are located at A(1, 0, 0), B(−1, 0, 0), C(0, 1, 0), and D(0, −1, 0) in free space. Find the total force on the charge at A. Eight identical point charges of Q C each are located at the corners of a cube of side length a, with one charge at the origin, and with the three nearest charges at (a, 0, 0), (0, a, 0), and (0, 0, a). Find an expression for the total vector force on the charge at P(a, a, a), assuming free space. Let a point charge Q 1 = 25 nC be located at P1 (4, −2, 7) and a charge Q 2 = 60 nC be at P2 (−3, 4, −2). (a) If = 0 , ﬁnd E at P3 (1, 2, 3). (b) At what point on the y axis is E x = 0? Two point charges of equal magnitude q are positioned at z = ±d/2. (a) Find the electric ﬁeld everywhere on the z axis; (b) ﬁnd the electric ﬁeld everywhere on the x axis; (c) repeat parts (a) and (b) if the charge at z = −d/2 is −q instead of +q.

2.2 2.3 2.4

2.5

2.6

2.7

A 2-µC point charge is located at A(4, 3, 5) in free space. Find E ρ , E φ , and E z at P(8, 12, 2).

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

45

2.8

A crude device for measuring charge consists of two small insulating spheres of radius a, one of which is ﬁxed in position. The other is movable along the x axis and is subject to a restraining force kx, where k is a spring constant. The uncharged spheres are centered at x = 0 and x = d, the latter ﬁxed. If the spheres are given equal and opposite charges of Q/C, obtain the expression by which Q may be found as a function of x. Determine the maximum charge that can be measured in terms of 0 , k, and d, and state the separation of the spheres then. What happens if a larger charge is applied? A 100-nC point charge is located at A(−1, 1, 3) in free space. (a) Find the locus of all points P(x, y, z) at which E x = 500 V/m. (b) Find y1 if P(−2, y1 , 3) lies on that locus.

2.9

2.10 A charge of −1 nC is located at the origin in free space. What charge must be located at (2, 0, 0) to cause E x to be zero at (3, 1, 1)? 2.11 A charge Q 0 located at the origin in free space produces a ﬁeld for which E z = 1 kV/m at point P(−2, 1, −1). (a) Find Q 0 . Find E at M(1, 6, 5) in (b) rectangular coordinates; (c) cylindrical coordinates; (d) spherical coordinates. 2.12 Electrons are in random motion in a ﬁxed region in space. During any 1 µs interval, the probability of ﬁnding an electron in a subregion of volume 10−15 m2 is 0.27. What volume charge density, appropriate for such time durations, should be assigned to that subregion? 2.13 A uniform volume charge density of 0.2 µC/m3 is present throughout the spherical shell extending from r = 3 cm to r = 5 cm. If ρν = 0 elsewhere, ﬁnd (a) the total charge present throughout the shell, and (b) r1 if half the total charge is located in the region 3 cm < r < r1 . 2.14 The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge density is represented by ρv = −0.1/(ρ 2 + 10−8 ) pC/m3 for 0 < ρ < 3 × 10−4 m, and ρv = 0 for ρ > 3 × 10−4 m. (a) Find the total charge per meter along the length of the beam; (b) if the electron velocity is 5 × 107 m/s, and with one ampere deﬁned as 1C/s, ﬁnd the beam current. 2.15 A spherical volume having a 2-µm radius contains a uniform volume charge density of 1015 C/m3 . (a) What total charge is enclosed in the spherical volume? (b) Now assume that a large region contains one of these little spheres at every corner of a cubical grid 3 mm on a side and that there is no charge between the spheres. What is the average volume charge density throughout this large region? cosθ 2.16 Within a region of free space, charge density is given as ρν = ρ0 ra C/m3 , where ρ0 and a are constants. Find the total charge lying within (a) the sphere, r ≤ a; (b) the cone, r ≤ a, 0 ≤ θ ≤ 0.1π ; (c) the region, r ≤ a, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤ 0.2π .

46

ENGINEERING ELECTROMAGNETICS

2.17 A uniform line charge of 16 nC/m is located along the line deﬁned by y = −2, z = 5. If = 0 : (a) ﬁnd E at P(1, 2, 3). (b) ﬁnd E at that point in the z = 0 plane where the direction of E is given by (1/3)a y − (2/3)az .

2.19 A uniform line charge of 2 µC/m is located on the z axis. Find E in rectangular coordinates at P(1, 2, 3) if the charge exists from (a) −∞ < z < ∞; (b) −4 ≤ z ≤ 4.

2.18 (a) Find E in the plane z = 0 that is produced by a uniform line charge, ρ L , extending along the z axis over the range −L < z < L in a cylindrical coordinate system. (b) If the ﬁnite line charge is approximated by an inﬁnite line charge (L → ∞), by what percentage is E ρ in error if ρ = 0.5L? (c) Repeat (b) with ρ = 0.1L.

2.20 A line charge of uniform charge density ρ0 C/m and of length is oriented along the z axis at − /2 < z < /2. (a) Find the electric ﬁeld strength, E, in magnitude and direction at any position along the x axis. (b) With the given line charge in position, ﬁnd the force acting on an identical line charge that is oriented along the x axis at /2 < x < 3 /2. 2.21 Two identical uniform line charges, with ρl = 75 nC/m, are located in free space at x = 0, y = ±0.4 m. What force per unit length does each line charge exert on the other?

2.22 Two identical uniform sheet charges with ρs = 100 nC/m2 are located in free space at z = ±2.0 cm. What force per unit area does each sheet exert on the other? 2.23 Given the surface charge density, ρs = 2 µC/m2 , existing in the region ρ < 0.2 m, z = 0, ﬁnd E at (a) PA (ρ = 0, z = 0.5); (b) PB (ρ = 0, z = −0.5). Show that (c) the ﬁeld along the z axis reduces to that of an inﬁnite sheet charge at small values of z; (d) the z axis ﬁeld reduces to that of a point charge at large values of z. 2.24 (a) Find the electric ﬁeld on the z axis produced by an annular ring of uniform surface charge density ρs in free space. The ring occupies the region z = 0, a ≤ ρ ≤ b, 0 ≤ φ ≤ 2π in cylindrical coordinates. (b) From your part (a) result, obtain the ﬁeld of an inﬁnite uniform sheet charge by taking appropriate limits. 2.25 Find E at the origin if the following charge distributions are present in free space: point charge, 12 nC, at P(2, 0, 6); uniform line charge density, 3 nC/m, at x = −2, y = 3; uniform surface charge density, 0.2 nC/m2 at x = 2.

2.26 A radially dependent surface charge is distributed on an inﬁnite ﬂat sheet in the x-y plane and is characterized in cylindrical coordinates by surface density ρs = ρ0 /ρ, where ρ0 is a constant. Determine the electric ﬁeld strength, E, everywhere on the z axis.

CHAPTER 2

Coulomb’s Law and Electric Field Intensity

47

2.27 Given the electric ﬁeld E = (4x − 2y)ax − (2x + 4y)a y , ﬁnd (a) the equation of the streamline that passes through the point P(2, 3, −4); (b) a unit vector specifying the direction of E at Q(3, −2, 5).

2.28 An electric dipole (discussed in detail in Section 4.7) consists of two point charges of equal and opposite magnitude ±Q spaced by distance d. With the charges along the z axis at positions z = ±d/2 (with the positive charge at the positive z location), the electric ﬁeld in spherical coordinates is given by E(r, θ ) = [Qd/(4π 0r 3 )][2 cos θ ar + sin θaθ ], where r >> d. Using rectangular coordinates, determine expressions for the vector force on a point charge of magnitude q (a) at (0, 0, z); (b) at (0, y, 0).

2.29 If E = 20e−5y (cos 5xax − sin 5xa y ), ﬁnd (a) |E| at P(π/6, 0.1, 2); (b) a unit vector in the direction of E at P; (c) the equation of the direction line passing through P. 2.30 For ﬁelds that do not vary with z in cylindrical coordinates, the equations of the streamlines are obtained by solving the differential equation E p /E φ = dρ/(ρdφ). Find the equation of the line passing through the point (2, 30◦ , 0) for the ﬁeld E = ρ cos 2φaρ − ρ sin 2φaφ .

3

C H A P T E R

Electric Flux Density, Gauss’s Law, and Divergence fter drawing a few of the ﬁelds described in the previous chapter and becoming familiar with the concept of the streamlines that show the direction of the force on a test charge at every point, it is difﬁcult to avoid giving these lines a physical signiﬁcance and thinking of them as flu lines. No physical particle is projected radially outward from the point charge, and there are no steel tentacles reaching out to attract or repel an unwary test charge, but as soon as the streamlines are drawn on paper there seems to be a picture showing “something” is present. It is very helpful to invent an electric flu that streams away symmetrically from a point charge and is coincident with the streamlines and to visualize this ﬂux wherever an electric ﬁeld is present. This chapter introduces and uses the concept of electric ﬂux and electric ﬂux density to again solve several of the problems presented in Chapter 2. The work here turns out to be much easier, and this is due to the extremely symmetrical problems that we are solving. ■

A

3.1 ELECTRIC FLUX DENSITY
About 1837, the director of the Royal Society in London, Michael Faraday, became very interested in static electric ﬁelds and the effect of various insulating materials on these ﬁelds. This problem had been bothering him during the past ten years when he was experimenting in his now-famous work on induced electromotive force, which we will discuss in Chapter 10. With that subject completed, he had a pair of concentric metallic spheres constructed, the outer one consisting of two hemispheres that could be ﬁrmly clamped together. He also prepared shells of insulating material (or dielectric material, or simply dielectric) that would occupy the entire volume between the concentric spheres. We will immediately use his ﬁndings about dielectric materials,
48

CHAPTER 3

Electric Flux Density, Gauss’s Law, and Divergence

49

for we are restricting our attention to ﬁelds in free space until Chapter 6. At that time we will see that the materials he used will be classiﬁed as ideal dielectrics. His experiment, then, consisted essentially of the following steps: 1. With the equipment dismantled, the inner sphere was given a known positive charge. 2. The hemispheres were then clamped together around the charged sphere with about 2 cm of dielectric material between them. 3. The outer sphere was discharged by connecting it momentarily to ground. 4. The outer space was separated carefully, using tools made of insulating material in order not to disturb the induced charge on it, and the negative induced charge on each hemisphere was measured. Faraday found that the total charge on the outer sphere was equal in magnitude to the original charge placed on the inner sphere and that this was true regardless of the dielectric material separating the two spheres. He concluded that there was some sort of “displacement” from the inner sphere to the outer which was independent of the medium, and we now refer to this ﬂux as displacement, displacement flux or simply electric flu . Faraday’s experiments also showed, of course, that a larger positive charge on the inner sphere induced a correspondingly larger negative charge on the outer sphere, leading to a direct proportionality between the electric ﬂux and the charge on the inner sphere. The constant of proportionality is dependent on the system of units involved, and we are fortunate in our use of SI units, because the constant is unity. If electric ﬂux is denoted by (psi) and the total charge on the inner sphere by Q, then for Faraday’s experiment =Q and the electric ﬂux is measured in coulombs. We can obtain more quantitative information by considering an inner sphere of radius a and an outer sphere of radius b, with charges of Q and −Q, respectively (Figure 3.1). The paths of electric ﬂux extending from the inner sphere to the outer sphere are indicated by the symmetrically distributed streamlines drawn radially from one sphere to the other. At the surface of the inner sphere, coulombs of electric ﬂux are produced by the charge Q(= ) Cs distributed uniformly over a surface having an area of 4π a 2 m2 . The density of the ﬂux at this surface is /4πa 2 or Q/4πa 2 C/m2 , and this is an important new quantity. Electric flu density, measured in coulombs per square meter (sometimes described as “lines per square meter,” for each line is due to one coulomb), is given the letter D, which was originally chosen because of the alternate names of displacement flu density or displacement density. Electric ﬂux density is more descriptive, however, and we will use the term consistently. The electric ﬂux density D is a vector ﬁeld and is a member of the “ﬂux density” class of vector ﬁelds, as opposed to the “force ﬁelds” class, which includes the electric

50

ENGINEERING ELECTROMAGNETICS

Figure 3.1 The electric flux in the region between a pair of charged concentric spheres. The direction and magnitude of D are not functions of the dielectric between the spheres.

ﬁeld intensity E. The direction of D at a point is the direction of the ﬂux lines at that point, and the magnitude is given by the number of ﬂux lines crossing a surface normal to the lines divided by the surface area. Referring again to Figure 3.1, the electric ﬂux density is in the radial direction and has a value of D D r =a r =b

Q ar 4πa 2 Q = ar 4π b2 =

(inner sphere) (outer sphere)

and at a radial distance r , where a ≤ r ≤ b, D= Q ar 4πr 2

If we now let the inner sphere become smaller and smaller, while still retaining a charge of Q, it becomes a point charge in the limit, but the electric ﬂux density at a point r meters from the point charge is still given by D= Q ar 4πr 2 (1)

for Q lines of ﬂux are symmetrically directed outward from the point and pass through an imaginary spherical surface of area 4πr 2 . This result should be compared with Section 2.2, Eq. (9), the radial electric ﬁeld intensity of a point charge in free space, E= Q ar 4π 0r 2

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Electric Flux Density, Gauss’s Law, and Divergence

51

In free space, therefore, D=
0E

(free space only)

(2)

Although (2) is applicable only to a vacuum, it is not restricted solely to the ﬁeld of a point charge. For a general volume charge distribution in free space, ρν dv aR 4π 0 R 2

E=

(free space only)

(3)

vol

where this relationship was developed from the ﬁeld of a single point charge. In a similar manner, (1) leads to D= ρν dv aR 4πR 2 (4)

vol

and (2) is therefore true for any free-space charge conﬁguration; we will consider (2) as deﬁning D in free space. As a preparation for the study of dielectrics later, it might be well to point out now that, for a point charge embedded in an inﬁnite ideal dielectric medium, Faraday’s results show that (1) is still applicable, and thus so is (4). Equation (3) is not applicable, however, and so the relationship between D and E will be slightly more complicated than (2). Because D is directly proportional to E in free space, it does not seem that it should really be necessary to introduce a new symbol. We do so for a few reasons. First, D is associated with the ﬂux concept, which is an important new idea. Second, the D ﬁelds we obtain will be a little simpler than the corresponding E ﬁelds, because 0 does not appear. D3.1. Given a 60-µC point charge located at the origin, ﬁnd the total electric ﬂux passing through: (a) that portion of the sphere r = 26 cm bounded by π π 0 < θ < and 0 < φ < ; (b) the closed surface deﬁned by ρ = 26 cm and 2 2 z = ±26 cm; (c) the plane z = 26 cm.
Ans. 7.5 µC; 60 µC; 30 µC

D3.2. Calculate D in rectangular coordinates at point P(2, −3, 6) produced by: (a) a point charge Q A = 55 mC at Q(−2, 3, −6); (b) a uniform line charge ρ L B = 20 mC/m on the x axis; (c) a uniform surface charge density ρ SC = 120 µC/m2 on the plane z = −5 m.
Ans. 6.38ax − 9.57a y + 19.14az µC/m2 ; −212a y + 424az µC/m2 ; 60az µC/m2

52

ENGINEERING ELECTROMAGNETICS

3.2 GAUSS’S LAW
The results of Faraday’s experiments with the concentric spheres could be summed up as an experimental law by stating that the electric ﬂux passing through any imaginary spherical surface lying between the two conducting spheres is equal to the charge enclosed within that imaginary surface. This enclosed charge is distributed on the surface of the inner sphere, or it might be concentrated as a point charge at the center of the imaginary sphere. However, because one coulomb of electric ﬂux is produced by one coulomb of charge, the inner conductor might just as well have been a cube or a brass door key and the total induced charge on the outer sphere would still be the same. Certainly the ﬂux density would change from its previous symmetrical distribution to some unknown conﬁguration, but +Q coulombs on any inner conductor would produce an induced charge of −Q coulombs on the surrounding sphere. Going one step further, we could now replace the two outer hemispheres by an empty (but completely closed) soup can. Q coulombs on the brass door key would produce = Q lines of electric ﬂux and would induce −Q coulombs on the tin can.1 These generalizations of Faraday’s experiment lead to the following statement, which is known as Gauss’s law:
The electric flu passing through any closed surface is equal to the total charge enclosed by that surface.

The contribution of Gauss, one of the greatest mathematicians the world has ever produced, was actually not in stating the law as we have, but in providing a mathematical form for this statement, which we will now obtain. Let us imagine a distribution of charge, shown as a cloud of point charges in Figure 3.2, surrounded by a closed surface of any shape. The closed surface may be the surface of some real material, but more generally it is any closed surface we wish to visualize. If the total charge is Q, then Q coulombs of electric ﬂux will pass through the enclosing surface. At every point on the surface the electric-ﬂux-density vector D will have some value D S , where the subscript S merely reminds us that D must be evaluated at the surface, and D S will in general vary in magnitude and direction from one point on the surface to another. We must now consider the nature of an incremental element of the surface. An incremental element of area S is very nearly a portion of a plane surface, and the complete description of this surface element requires not only a statement of its magnitude S but also of its orientation in space. In other words, the incremental surface element is a vector quantity. The only unique direction that may be associated with S is the direction of the normal to that plane which is tangent to the surface at the point in question. There are, of course, two such normals, and the ambiguity is removed by specifying the outward normal whenever the surface is closed and “outward” has a speciﬁc meaning.

1

If it were a perfect insulator, the soup could even be left in the can without any difference in the results.

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Electric Flux Density, Gauss’s Law, and Divergence

53

Figure 3.2 The electric flux density D S at P arising from charge Q. The total flux passing through S is D S · S.

At any point P, consider an incremental element of surface S and let D S make an angle θ with S, as shown in Figure 3.2. The ﬂux crossing S is then the product of the normal component of D S and S, = ﬂux crossing S = D S,norm S = D S cos θ S = D S · S

where we are able to apply the deﬁnition of the dot product developed in Chapter 1. The total ﬂux passing through the closed surface is obtained by adding the differential contributions crossing each surface element S, = d = closed surface

D S · dS

The resultant integral is a closed surface integral, and since the surface element dS always involves the differentials of two coordinates, such as d x d y, ρ dφ dρ, or r 2 sin θ dθ dφ, the integral is a double integral. Usually only one integral sign is used for brevity, and we will always place an S below the integral sign to indicate a surface integral, although this is not actually necessary, as the differential dS is automatically the signal for a surface integral. One last convention is to place a small circle on the integral sign itself to indicate that the integration is to be performed over a closed surface. Such a surface is often called a gaussian surface. We then have the mathematical formulation of Gauss’s law, = D S · dS = charge enclosed = Q (5)

S

The charge enclosed might be several point charges, in which case Q = Qn or a line charge, Q= ρ L dL

54

ENGINEERING ELECTROMAGNETICS

or a surface charge, Q=
S

ρ S dS

(not necessarily a closed surface)

or a volume charge distribution, Q= ρν dv vol The last form is usually used, and we should agree now that it represents any or all of the other forms. With this understanding, Gauss’s law may be written in terms of the charge distribution as D S · dS = ρν dv vol S

(6)

a mathematical statement meaning simply that the total electric ﬂux through any closed surface is equal to the charge enclosed.
E X A M P L E 3.1

To illustrate the application of Gauss’s law, let us check the results of Faraday’s experiment by placing a point charge Q at the origin of a spherical coordinate system (Figure 3.3) and by choosing our closed surface as a sphere of radius a.
Solution. We have, as before,

D= At the surface of the sphere, DS =

Q ar 4πr 2 Q ar 4πa 2

The differential element of area on a spherical surface is, in spherical coordinates from Chapter 1, d S = r 2 sin θ dθ dφ = a 2 sin θ dθ dφ or dS = a 2 sin θ dθ dφ ar The integrand is Q 2 Q sin θ dθ dφ a sin θ dθ dφar · ar = 4πa 2 4π leading to the closed surface integral D S · dS = φ=2π φ=0 θ =π θ =φ

Q sin θ dθ dφ 4π

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Electric Flux Density, Gauss’s Law, and Divergence

55

Figure 3.3 Applying Gauss’s law to the field of a point charge Q on a spherical closed surface of radius a. The electric flux density D is everywhere normal to the spherical surface and has a constant magnitude at every point on it.

where the limits on the integrals have been chosen so that the integration is carried over the entire surface of the sphere once.2 Integrating gives 2π 2π Q Q π −cos θ 0 dφ = dφ = Q 4π 0 0 2π and we obtain a result showing that Q coulombs of electric ﬂux are crossing the surface, as we should since the enclosed charge is Q coulombs. D3.3. Given the electric ﬂux density, D = 0.3r 2 ar nC/m2 in free space: (a) ﬁnd E at point P(r = 2, θ = 25◦ , φ = 90◦ ); (b) ﬁnd the total charge within the sphere r = 3; (c) ﬁnd the total electric ﬂux leaving the sphere r = 4.
Ans. 135.5ar V/m; 305 nC; 965 nC

D3.4. Calculate the total electric ﬂux leaving the cubical surface formed by the six planes x, y, z = ±5 if the charge distribution is: (a) two point charges, 0.1 µC at (1, −2, 3) and 1 µC at (−1, 2, −2); (b) a uniform line charge of π µC/m at 7 x = −2, y = 3; (c) a uniform surface charge of 0.1 µC/m2 on the plane y = 3x.
Ans. 0.243 µC; 31.4 µC; 10.54 µC

2

Note that if θ and φ both cover the range from 0 to 2π, the spherical surface is covered twice.

56

ENGINEERING ELECTROMAGNETICS

3.3 APPLICATION OF GAUSS’S LAW: SOME SYMMETRICAL CHARGE DISTRIBUTIONS
We now consider how we may use Gauss’s law, Q= D S · dS

S

to determine D S if the charge distribution is known. This is an example of an integral equation in which the unknown quantity to be determined appears inside the integral. The solution is easy if we are able to choose a closed surface which satisﬁes two conditions: 1. D S is everywhere either normal or tangential to the closed surface, so that D S · dS becomes either D S dS or zero, respectively. 2. On that portion of the closed surface for which D S · dS is not zero, D S = constant. This allows us to replace the dot product with the product of the scalars D S and d S and then to bring D S outside the integral sign. The remaining integral is then S d S over that portion of the closed surface which D S crosses normally, and this is simply the area of this section of that surface. Only a knowledge of the symmetry of the problem enables us to choose such a closed surface. Let us again consider a point charge Q at the origin of a spherical coordinate system and decide on a suitable closed surface which will meet the two requirements previously listed. The surface in question is obviously a spherical surface, centered at the origin and of any radius r . D S is everywhere normal to the surface; D S has the same value at all points on the surface. Then we have, in order, Q=
S

D S · dS =

D S dS sph φ=2π φ=0 θ =π θ =0

= DS

sph

d S = DS

r 2 sin θ dθ dφ

= 4πr 2 D S and hence Q 4πr 2 Because r may have any value and because D S is directed radially outward, DS = Q ar 4πr 2 Q ar 4π 0r 2

D=

E=

CHAPTER 3

Electric Flux Density, Gauss’s Law, and Divergence

57

which agrees with the results of Chapter 2. The example is a trivial one, and the objection could be raised that we had to know that the ﬁeld was symmetrical and directed radially outward before we could obtain an answer. This is true, and that leaves the inverse-square-law relationship as the only check obtained from Gauss’s law. The example does, however, serve to illustrate a method which we may apply to other problems, including several to which Coulomb’s law is almost incapable of supplying an answer. Are there any other surfaces which would have satisﬁed our two conditions? The student should determine that such simple surfaces as a cube or a cylinder do not meet the requirements. As a second example, let us reconsider the uniform line charge distribution ρ L lying along the z axis and extending from −∞ to +∞. We must ﬁrst know the symmetry of the ﬁeld, and we may consider this knowledge complete when the answers to these two questions are known: 1. With which coodinates does the ﬁeld vary (or of what variables is D a function)? 2. Which components of D are present? In using Gauss’s law, it is not a question of using symmetry to simplify the solution, for the application of Gauss’s law depends on symmetry, and if we cannot show that symmetry exists then we cannot use Gauss’s law to obtain a solution. The preceding two questions now become “musts.” From our previous discussion of the uniform line charge, it is evident that only the radial component of D is present, or D = Dρ aρ and this component is a function of ρ only. Dρ = f (ρ) The choice of a closed surface is now simple, for a cylindrical surface is the only surface to which Dρ is everywhere normal, and it may be closed by plane surfaces normal to the z axis. A closed right circular cylinder of radius ρ extending from z = 0 to z = L is shown in Figure 3.4. We apply Gauss’s law, Q= cyl D S · dS = D S
L z=0 2π φ=0

sides

dS + 0

top

dS + 0

dS bottom = DS and obtain

ρ dφ dz = D S 2πρ L

D S = Dρ =

Q 2πρ L Q = ρL L

In terms of the charge density ρ L , the total charge enclosed is

58

ENGINEERING ELECTROMAGNETICS

Figure 3.4 The gaussian surface for an infinite uniform line charge is a right circular cylinder of length L and radius ρ. D is constant in magnitude and everywhere perpendicular to the cylindrical surface; D is parallel to the end faces.

giving Dρ = or Eρ = ρL 2π 0 ρ ρL 2πρ

Comparing with Section 2.4, Eq. (16), shows that the correct result has been obtained and with much less work. Once the appropriate surface has been chosen, the integration usually amounts only to writing down the area of the surface at which D is normal. The problem of a coaxial cable is almost identical with that of the line charge and is an example that is extremely difﬁcult to solve from the standpoint of Coulomb’s law. Suppose that we have two coaxial cylindrical conductors, the inner of radius a and the outer of radius b, each inﬁnite in extent (Figure 3.5). We will assume a charge distribution of ρ S on the outer surface of the inner conductor. Symmetry considerations show us that only the Dρ component is present and that it can be a function only of ρ. A right circular cylinder of length L and radius ρ, where a < ρ < b, is necessarily chosen as the gaussian surface, and we quickly have Q = D S 2πρ L

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Electric Flux Density, Gauss’s Law, and Divergence

59

Figure 3.5 The two coaxial cylindrical conductors forming a coaxial cable provide an electric flux density within the cylinders, given by D ρ = aρ S/ρ.

The total charge on a length L of the inner conductor is Q= from which we have DS =
L z=0 2π φ=0

ρ S a dφ dz = 2πa Lρ S

aρ S aρ S (a < ρ < b) D= aρ ρ ρ This result might be expressed in terms of charge per unit length because the inner conductor has 2πaρ S coulombs on a meter length, and hence, letting ρ L = 2πaρ S , D= ρL aρ 2πρ

and the solution has a form identical with that of the inﬁnite line charge. Because every line of electric ﬂux starting from the charge on the inner cylinder must terminate on a negative charge on the inner surface of the outer cylinder, the total charge on that surface must be Q outer cyl = −2πa Lρ S,inner cyl and the surface charge on the outer cylinder is found as 2πbLρ S,outer cyl = −2πa Lρ S,inner cyl or a ρ S,outer cyl = − ρ S,inner cyl b What would happen if we should use a cylinder of radius ρ, ρ > b, for the gaussian surface? The total charge enclosed would then be zero, for there are equal and opposite charges on each conducting cylinder. Hence DS = 0 0 = D S 2πρ L (ρ > b) (ρ > b)

60

ENGINEERING ELECTROMAGNETICS

An identical result would be obtained for ρ < a. Thus the coaxial cable or capacitor has no external ﬁeld (we have proved that the outer conductor is a “shield”), and there is no ﬁeld within the center conductor. Our result is also useful for a finit length of coaxial cable, open at both ends, provided the length L is many times greater than the radius b so that the nonsymmetrical conditions at the two ends do not appreciably affect the solution. Such a device is also termed a coaxial capacitor. Both the coaxial cable and the coaxial capacitor will appear frequently in the work that follows.
E X A M P L E 3.2

Let us select a 50-cm length of coaxial cable having an inner radius of 1 mm and an outer radius of 4 mm. The space between conductors is assumed to be ﬁlled with air. The total charge on the inner conductor is 30 nC. We wish to know the charge density on each conductor, and the E and D ﬁelds.
Solution. We begin by ﬁnding the surface charge density on the inner cylinder,

ρ S,inner cyl =

Q inner cyl 30 × 10−9 = = 9.55 µC/m2 2πa L 2π (10−3 )(0.5)

The negative charge density on the inner surface of the outer cylinder is ρ S,outer cyl = Q outer cyl −30 × 10−9 = = −2.39 µC/m2 2π bL 2π (4 × 10−3 )(0.5) aρ S 10−3 (9.55 × 10−6 ) 9.55 = = nC/m2 ρ ρ ρ

The internal ﬁelds may therefore be calculated easily: Dρ = and Eρ = Dρ
0

=

1079 9.55 × 10−9 = V/m −12 ρ 8.854 × 10 ρ

Both of these expressions apply to the region where 1 < ρ < 4 mm. For ρ < 1 mm or ρ > 4 mm, E and D are zero. D3.5. A point charge of 0.25 µC is located at r = 0, and uniform surface charge densities are located as follows: 2 mC/m2 at r = 1 cm, and −0.6 mC/m2 at r = 1.8 cm. Calculate D at: (a) r = 0.5 cm; (b) r = 1.5 cm; (c) r = 2.5 cm. (d) What uniform surface charge density should be established at r = 3 cm to cause D = 0 at r = 3.5 cm?
Ans. 796ar µC/m2 ; 977ar µC/m2 ; 40.8ar µC/m2 ; −28.3 µC/m2

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3.4 APPLICATION OF GAUSS’S LAW: DIFFERENTIAL VOLUME ELEMENT
We are now going to apply the methods of Gauss’s law to a slightly different type of problem—one that does not possess any symmetry at all. At ﬁrst glance, it might seem that our case is hopeless, for without symmetry, a simple gaussian surface cannot be chosen such that the normal component of D is constant or zero everywhere on the surface. Without such a surface, the integral cannot be evaluated. There is only one way to circumvent these difﬁculties and that is to choose such a very small closed surface that D is almost constant over the surface, and the small change in D may be adequately represented by using the ﬁrst two terms of the Taylor’s-series expansion for D. The result will become more nearly correct as the volume enclosed by the gaussian surface decreases, and we intend eventually to allow this volume to approach zero. This example also differs from the preceding ones in that we will not obtain the value of D as our answer but will instead receive some extremely valuable information about the way D varies in the region of our small surface. This leads directly to one of Maxwell’s four equations, which are basic to all electromagnetic theory. Let us consider any point P, shown in Figure 3.6, located by a rectangular coordinate system. The value of D at the point P may be expressed in rectangular components, D0 = Dx0 ax + D y0 a y + Dz0 az . We choose as our closed surface the small rectangular box, centered at P, having sides of lengths x, y, and z, and apply Gauss’s law, D · dS = Q

S

In order to evaluate the integral over the closed surface, the integral must be broken up into six integrals, one over each face,
S

D · dS =

front

+

back

+

left

+

right

+

top

+

bottom

Consider the ﬁrst of these in detail. Because the surface element is very small, D is essentially constant (over this portion of the entire closed surface) and front = Dfront · Sfront ˙ = Dfront · y ˙ z ax z = Dx,front y ˙

where we have only to approximate the value of Dx at this front face. The front face is at a distance of x/2 from P, and hence x Dx,front = Dx0 + × rate of change of Dx with x ˙ 2 x ∂Dx = Dx0 + ˙ 2 ∂x

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ENGINEERING ELECTROMAGNETICS

Figure 3.6 A differential-sized gaussian surface about the point P is used to investigate the space rate of change of D in the neighborhood of P.

where Dx0 is the value of Dx at P, and where a partial derivative must be used to express the rate of change of Dx with x, as Dx in general also varies with y and z. This expression could have been obtained more formally by using the constant term and the term involving the ﬁrst derivative in the Taylor’s-series expansion for Dx in the neighborhood of P. We now have x ∂Dx = Dx0 + ˙ y z 2 ∂x front Consider now the integral over the back surface, = Dback · Sback ˙ = Dback · (− y ˙ = −Dx,back y ˙ z ax ) z

back

and Dx,back = Dx0 − ˙ giving = ˙ −Dx0 + ∂Dx ∂x x ∂Dx 2 ∂x y z x ∂Dx 2 ∂x

back

If we combine these two integrals, we have + = ˙ x y z

front

back

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By exactly the same process we ﬁnd that + = ˙ ∂D y ∂y x y z

right

left

and + = ˙ ∂Dz ∂z x y z

top

bottom

and these results may be collected to yield D · dS = ˙ ∂D y ∂Dx ∂Dz + + ∂x ∂y ∂z x y z

S

or D · dS = Q = ˙ ∂D y ∂Dz ∂Dx + + ∂x ∂y ∂z ν (7)

S

The expression is an approximation which becomes better as ν becomes smaller, and in the following section we shall let the volume ν approach zero. For the moment, we have applied Gauss’s law to the closed surface surrounding the volume element ν and have as a result the approximation (7) stating that Charge enclosed in volume ν= ˙ ∂D y ∂Dx ∂Dz + + ∂x ∂y ∂z × volume ν (8)

E X A M P L E 3.3

Find an approximate value for the total charge enclosed in an incremental volume of 10−9 m3 located at the origin, if D = e−x sin y ax − e−x cos y a y + 2zaz C/m2 .
Solution. We ﬁrst evaluate the three partial derivatives in (8):

∂Dx = −e−x sin y ∂x ∂D y = e−x sin y ∂y ∂Dz =2 ∂z At the origin, the ﬁrst two expressions are zero, and the last is 2. Thus, we ﬁnd that the charge enclosed in a small volume element there must be approximately 2 ν. If ν is 10−9 m3 , then we have enclosed about 2 nC.

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ENGINEERING ELECTROMAGNETICS

D3.6. In free space, let D = 8x yz 4 ax +4x 2 z 4 a y +16x 2 yz 3 az pC/m2 . (a) Find the total electric ﬂux passing through the rectangular surface z = 2, 0 < x < 2, 1 < y < 3, in the az direction. (b) Find E at P(2, −1, 3). (c) Find an approximate value for the total charge contained in an incremental sphere located at P(2, −1, 3) and having a volume of 10−12 m3 .
Ans. 1365 pC; −146.4ax + 146.4a y − 195.2az V/m; −2.38 × 10−21 C

3.5 DIVERGENCE AND MAXWELL’S FIRST EQUATION
We will now obtain an exact relationship from (7), by allowing the volume element ν to shrink to zero. We write this equation as ∂D y ∂Dz ∂Dx + + ∂x ∂y ∂z = lim
S ν→0

D · dS Q = lim = ρν ν→0 ν ν

(9)

in which the charge density, ρν , is identiﬁed in the second equality. The methods of the previous section could have been used on any vector A to ﬁnd S A · dS for a small closed surface, leading to ∂A y ∂A z ∂A x + + ∂x ∂y ∂z = lim
S ν→0

A · dS ν

(10)

where A could represent velocity, temperature gradient, force, or any other vector ﬁeld. This operation appeared so many times in physical investigations in the last century that it received a descriptive name, divergence. The divergence of A is deﬁned as Divergence of A = div A = lim
S

ν→0

A · dS ν

(11)

and is usually abbreviated div A. The physical interpretation of the divergence of a vector is obtained by describing carefully the operations implied by the right-hand side of (11), where we shall consider A to be a member of the ﬂux-density family of vectors in order to aid the physical interpretation.
The divergence of the vector flu density A is the outflo of flu from a small closed surface per unit volume as the volume shrinks to zero.

The physical interpretation of divergence afforded by this statement is often useful in obtaining qualitative information about the divergence of a vector ﬁeld without resorting to a mathematical investigation. For instance, let us consider the divergence of the velocity of water in a bathtub after the drain has been opened. The net outﬂow of water through any closed surface lying entirely within the water must be zero, for water is essentially incompressible, and the water entering and leaving

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different regions of the closed surface must be equal. Hence the divergence of this velocity is zero. If, however, we consider the velocity of the air in a tire that has just been punctured by a nail, we realize that the air is expanding as the pressure drops, and that consequently there is a net outﬂow from any closed surface lying within the tire. The divergence of this velocity is therefore greater than zero. A positive divergence for any vector quantity indicates a source of that vector quantity at that point. Similarly, a negative divergence indicates a sink. Because the divergence of the water velocity above is zero, no source or sink exists.3 The expanding air, however, produces a positive divergence of the velocity, and each interior point may be considered a source. Writing (9) with our new term, we have div D = ∂D y ∂Dx ∂Dz + + ∂x ∂y ∂z (rectangular) (12)

This expression is again of a form that does not involve the charge density. It is the result of applying the deﬁnition of divergence (11) to a differential volume element in rectangular coordinates. If a differential volume unit ρ dρ dφ dz in cylindrical coordinates, or r 2 sin θ dr dθ dφ in spherical coordinates, had been chosen, expressions for divergence involving the components of the vector in the particular coordinate system and involving partial derivatives with respect to the variables of that system would have been obtained. These expressions are obtained in Appendix A and are given here for convenience: div D = 1 ∂ 1 ∂Dφ ∂Dz (ρ Dρ ) + + ρ ∂ρ ρ ∂φ ∂z (cylindrical) (13)

div D =

1 ∂ 2 1 1 ∂Dφ ∂ (r Dr ) + (sin θ Dθ ) + r 2 ∂r r sin θ ∂θ r sin θ ∂φ

(spherical) (14)

These relationships are also shown inside the back cover for easy reference. It should be noted that the divergence is an operation which is performed on a vector, but that the result is a scalar. We should recall that, in a somewhat similar way, the dot or scalar product was a multiplication of two vectors which yielded a scalar. For some reason, it is a common mistake on meeting divergence for the ﬁrst time to impart a vector quality to the operation by scattering unit vectors around in

3

Having chosen a differential element of volume within the water, the gradual decrease in water level with time will eventually cause the volume element to lie above the surface of the water. At the instant the surface of the water intersects the volume element, the divergence is positive and the small volume is a source. This complication is avoided above by specifying an integral point.

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ENGINEERING ELECTROMAGNETICS

the partial derivatives. Divergence merely tells us how much ﬂux is leaving a small volume on a per-unit-volume basis; no direction is associated with it. We can illustrate the concept of divergence by continuing with the example at the end of Section 3.4.
E X A M P L E 3.4

Find div D at the origin if D = e−x sin y ax − e−x cos y a y + 2zaz .
Solution. We use (10) to obtain

div D =

∂D y ∂Dz ∂Dx + + ∂x ∂y ∂z −x −x = −e sin y + e sin y + 2 = 2

The value is the constant 2, regardless of location. If the units of D are C/m2 , then the units of div D are C/m3 . This is a volume charge density, a concept discussed in the next section. D3.7. In each of the following parts, ﬁnd a numerical value for div D at the point speciﬁed: (a) D = (2x yz − y 2 )ax + (x 2 z − 2x y)a y + x 2 yaz C/m2 at PA (2, 3, −1); (b) D = 2ρz 2 sin2 φ aρ + ρz 2 sin 2φ aφ + 2ρ 2 z sin2 φ az C/m2 at PB (ρ = 2, φ = 110◦ , z = −1); (c) D = 2r sin θ cos φ ar + r cos θ cos φ aθ − r sin φ aφ C/m2 at PC (r = 1.5, θ = 30◦ , φ = 50◦ ).
Ans. −10.00; 9.06; 1.29

Finally, we can combine Eqs. (9) and (12) and form the relation between electric ﬂux density and charge density: div D = ρν (15)

This is the ﬁrst of Maxwell’s four equations as they apply to electrostatics and steady magnetic ﬁelds, and it states that the electric ﬂux per unit volume leaving a vanishingly small volume unit is exactly equal to the volume charge density there. This equation is aptly called the point form of Gauss’s law. Gauss’s law relates the ﬂux leaving any closed surface to the charge enclosed, and Maxwell’s ﬁrst equation makes an identical statement on a per-unit-volume basis for a vanishingly small volume, or at a point. Because the divergence may be expressed as the sum of three partial derivatives, Maxwell’s ﬁrst equation is also described as the differential-equation form of Gauss’s law, and conversely, Gauss’s law is recognized as the integral form of Maxwell’s ﬁrst equation. As a speciﬁc illustration, let us consider the divergence of D in the region about a point charge Q located at the origin. We have the ﬁeld D= Q ar 4πr 2

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67

and use (14), the expression for divergence in spherical coordinates: div D = 1 1 ∂Dφ ∂ 1 ∂ 2 (r Dr ) + (Dθ sin θ ) + r 2 ∂r r sin θ ∂θ r sin θ ∂φ 1 d r 2 dr Q 4πr 2

Because Dθ and Dφ are zero, we have div D = r2 =0 (if r = 0)

Thus, ρν = 0 everywhere except at the origin, where it is inﬁnite. The divergence operation is not limited to electric ﬂux density; it can be applied to any vector ﬁeld. We will apply it to several other electromagnetic ﬁelds in the coming chapters. D3.8. Determine an expression for the volume charge density associated with 2x 2 2x 2 y 4x y ax + a y − 2 az ; (b) D = z sin φ aρ + each D ﬁeld: (a) D = z z z z cos φ aφ + ρ sin φ az ; (c) D = sin θ sin φ ar + cos θ sin φ aθ + cos φ aφ .
Ans. 4y 2 (x + z 2 ); 0; 0. z3

3.6 THE VECTOR OPERATOR ∇ AND THE DIVERGENCE THEOREM
If we remind ourselves again that divergence is an operation on a vector yielding a scalar result, just as the dot product of two vectors gives a scalar result, it seems possible that we can ﬁnd something that may be dotted formally with D to yield the scalar ∂D y ∂Dx ∂Dz + + ∂x ∂y ∂z Obviously, this cannot be accomplished by using a dot product; the process must be a dot operation. With this in mind, we deﬁne the del operator ∇ as a vector operator, ∇= ∂ ∂ ∂ ax + a y + az ∂x ∂y ∂z (16)

Similar scalar operators appear in several methods of solving differential equations where we often let D replace d/d x, D 2 replace d 2 /d x 2 , and so forth.4 We agree on deﬁning ∇ that it shall be treated in every way as an ordinary vector with the one important exception that partial derivatives result instead of products of scalars. Consider ∇ · D, signifying ∇ ·D = ∂ ∂ ∂ ax + a y + az ∂x ∂y ∂z · (Dx ax + D y a y + Dz az )

4

This scalar operator D, which will not appear again, is not to be confused with the electric ﬂux density.

68

ENGINEERING ELECTROMAGNETICS

We ﬁrst consider the dot products of the unit vectors, discarding the six zero terms, and obtain the result that we recognize as the divergence of D: ∇ ·D = ∂D y ∂Dz ∂Dx + + = div(D) ∂x ∂y ∂z

The use of ∇ · D is much more prevalent than that of div D, although both usages have their advantages. Writing ∇ · D allows us to obtain simply and quickly the correct partial derivatives, but only in rectangular coordinates, as we will see. On the other hand, div D is an excellent reminder of the physical interpretation of divergence. We shall use the operator notation ∇ · D from now on to indicate the divergence operation. The vector operator ∇ is used not only with divergence, but also with several other very important operations that appear later. One of these is ∇u, where u is any scalar ﬁeld, and leads to ∂u ∂ ∂ ∂u ∂u ∂ ax + a y + az u = ax + ay + az ∇u = ∂x ∂y ∂z ∂x ∂y ∂z The ∇ operator does not have a speciﬁc form in other coordinate systems. If we are considering D in cylindrical coordinates, then ∇ · D still indicates the divergence of D, or 1 ∂ 1 ∂Dφ ∂Dz ∇ ·D = (ρ Dρ ) + + ρ ∂ρ ρ ∂φ ∂z where this expression has been taken from Section 3.5. We have no form for ∇ itself to help us obtain this sum of partial derivatives. This means that ∇u, as yet unnamed but easily written in rectangular coordinates, cannot be expressed by us at this time in cylindrical coordinates. Such an expression will be obtained when ∇u is deﬁned in Chapter 4. We close our discussion of divergence by presenting a theorem that will be needed several times in later chapters, the divergence theorem. This theorem applies to any vector ﬁeld for which the appropriate partial derivatives exist, although it is easiest for us to develop it for the electric ﬂux density. We have actually obtained it already and now have little more to do than point it out and name it, for starting from Gauss’s law, we have
S

D · dS = Q =

vol

ρν dv =

vol

∇ · D dv

The ﬁrst and last expressions constitute the divergence theorem, D · dS = ∇ · D dv (17)

S

vol

which may be stated as follows:
The integral of the normal component of any vector fiel over a closed surface is equal to the integral of the divergence of this vector fiel throughout the volume enclosed by the closed surface.

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Figure 3.7 The divergence theorem states that the total flux crossing the closed surface is equal to the integral of the divergence of the flux density throughout the enclosed volume. The volume is shown here in cross section.

Again, we emphasize that the divergence theorem is true for any vector ﬁeld, although we have obtained it speciﬁcally for the electric ﬂux density D, and we will have occasion later to apply it to several different ﬁelds. Its beneﬁts derive from the fact that it relates a triple integration throughout some volume to a double integration over the surface of that volume. For example, it is much easier to look for leaks in a bottle full of some agitated liquid by inspecting the surface than by calculating the velocity at every internal point. The divergence theorem becomes obvious physically if we consider a volume ν, shown in cross section in Figure 3.7, which is surrounded by a closed surface S. Division of the volume into a number of small compartments of differential size and consideration of one cell show that the ﬂux diverging from such a cell enters, or converges on, the adjacent cells unless the cell contains a portion of the outer surface. In summary, the divergence of the ﬂux density throughout a volume leads, then, to the same result as determining the net ﬂux crossing the enclosing surface.
E X A M P L E 3.5

Evaluate both sides of the divergence theorem for the ﬁeld D = 2x yax + x a y C/m and the rectangular parellelepiped formed by the planes x = 0 and 1, y = 0 and 2, and z = 0 and 3.
2 2

Solution. Evaluating the surface integral ﬁrst, we note that D is parallel to the sur-

faces at z = 0 and z = 3, so D · dS = 0 there. For the remaining four surfaces we have
3 0 0 3 2 3 0 0 3 0 0 1 2

S

D · dS =

(D)x=0 · (−dy dz ax ) +
1 0

(D)x=1 · (dy dz ax ) (D) y=2 · (d x dz a y )

+

0

(D) y=0 · (−d x dz a y ) +

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ENGINEERING ELECTROMAGNETICS

3

2 0

=− −

0 3 0

(Dx )x=0 dy dz + (D y ) y=0 d x dz +
3 2

3 0 3 0 0 0

2

(Dx )x=1 dy dz (D y ) y=2 d x dz

1 0

1

However, (Dx )x=0 = 0, and (D y ) y=0 = (D y ) y=2 , which leaves only
S

D · dS = =

0 3 0

0

(Dx )x=1 dy dz =

3

2

2y dy dz

0

0

4 dz = 12

Since ∇ ·D = the volume integral becomes ∇ · D dv = =
3 0 3 0 0 2 0 1

∂ 2 ∂ (2x y) + (x ) = 2y ∂x ∂y
3 0 0 2

vol

2y d x dy dz =

2y dy dz

4 dz = 12

and the check is accomplished. Remembering Gauss’s law, we see that we have also determined that a total charge of 12 C lies within this parallelepiped. D3.9. Given the ﬁeld D = 6ρ sin 1 φ aρ + 1.5ρ cos 1 φ aφ C/m2 , evaluate both 2 2 sides of the divergence theorem for the region bounded by ρ = 2, φ = 0, φ = π, z = 0, and z = 5.
Ans. 225; 225

REFERENCES
1. Kraus, J. D., and D. A. Fleisch. Electromagnetics. 5th ed. New York: McGraw-Hill, 1999. The static electric ﬁeld in free space is introduced in Chapter 2. 2. Plonsey, R., and R. E. Collin. Principles and Applications of Electromagnetic Fields. New York: McGraw-Hill, 1961. The level of this text is somewhat higher than the one we are reading now, but it is an excellent text to read next. Gauss’s law appears in the second chapter. 3. Plonus, M. A. Applied Electromagnetics. New York: McGraw-Hill, 1978. This book contains rather detailed descriptions of many practical devices that illustrate electromagnetic applications. For example, see the discussion of xerography on pp. 95–98 as an electrostatics application. 4. Skilling, H. H. Fundamentals of Electric Waves. 2d ed. New York: John Wiley & Sons, 1948. The operations of vector calculus are well illustrated. Divergence is discussed on pp. 22 and 38. Chapter 1 is interesting reading.

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5. Thomas, G. B., Jr., and R. L. Finney. (see Suggested References for Chapter 1). The divergence theorem is developed and illustrated from several different points of view on pp. 976–980.

CHAPTER 3 PROBLEMS
3.1 Suppose that the Faraday concentric sphere experiment is performed in free space using a central charge at the origin, Q 1 , and with hemispheres of radius a. A second charge Q 2 (this time a point charge) is located at distance R from Q 1 , where R >> a. (a) What is the force on the point charge before the hemispheres are assembled around Q 1 ? (b) What is the force on the point charge after the hemispheres are assembled but before they are discharged? (c) What is the force on the point charge after the hemispheres are assembled and after they are discharged? (d) Qualitatively, describe what happens as Q 2 is moved toward the sphere assembly to the extent that the condition R >> a is no longer valid. An electric ﬁeld in free space is E = (5z 2 / 0 ) az V/m. Find the total charge ˆ contained within a cube, centered at the origin, of 4-m side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at ±2). The cylindrical surface ρ = 8 cm contains the surface charge density, ρ S = 5e−20|z| nC/m2 . (a) What is the total amount of charge present? (b) How much electric ﬂux leaves the surface ρ = 8 cm, 1 cm < z < 5 cm, 30◦ < φ < 90◦ ?

3.2

3.3

3.4 3.5

An electric ﬁeld in free space is E = (5z 3 / 0 ) az V/m. Find the total charge ˆ contained within a sphere of 3-m radius, centered at the origin. Let D = 4x yax + 2(x 2 + z 2 )a y + 4yzaz nC/m2 and evaluate surface integrals to ﬁnd the total charge enclosed in the rectangular parallelepiped 0 < x < 2, 0 < y < 3, 0 < z < 5 m. In free space, a volume charge of constant density ρν = ρ0 exists within the region −∞ < x < ∞, −∞ < y < ∞, and −d/2 < z < d/2. Find D and E everywhere. Volume charge density is located in free space as ρν = 2e−1000r nC/m3 for 0 < r < 1 mm, and ρν = 0 elsewhere. (a) Find the total charge enclosed by the spherical surface r = 1 mm. (b) By using Gauss’s law, calculate the value of Dr on the surface r = 1 mm. Use Gauss’s law in integral form to show that an inverse distance ﬁeld in spherical coordinates, D = Aar /r , where A is a constant, requires every spherical shell of 1 m thickness to contain 4πA coulombs of charge. Does this indicate a continuous charge distribution? If so, ﬁnd the charge density variation with r .

3.6

3.7

3.8

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ENGINEERING ELECTROMAGNETICS

3.9

3.10 An inﬁnitely long cylindrical dielectric of radius b contains charge within its volume of density ρv = aρ 2 , where a is a constant. Find the electric ﬁeld strength, E, both inside and outside the cylinder. 3.11 In cylindrical coordinates, let ρν = 0 for ρ < 1 mm, ρν = 2 sin(2000 πρ) nC/m3 for 1 mm < ρ < 1.5 mm, and ρν = 0 for ρ > 1.5 mm. Find D everywhere.

A uniform volume charge density of 80 µC/m3 is present throughout the region 8 mm < r < 10 mm. Let ρν = 0 for 0 < r < 8 mm. (a) Find the total charge inside the spherical surface r = 10 mm. (b) Find Dr at r = 10 mm. (c) If there is no charge for r > 10 mm, ﬁnd Dr at r = 20 mm.

3.12 The sun radiates a total power of about 3.86 × 1026 watts (W). If we imagine the sun’s surface to be marked off in latitude and longitude and assume uniform radiation, (a) what power is radiated by the region lying between latitude 50◦ N and 60◦ N and longitude 12◦ W and 27◦ W? (b) What is the power density on a spherical surface 93,000,000 miles from the sun in W/m2 ? 3.13 Spherical surfaces at r = 2, 4, and 6 m carry uniform surface charge densities of 20 nC/m2 , −4 nC/m2 , and ρ S0 , respectively. (a) Find D at r = 1, 3, and 5 m. (b) Determine ρ S0 such that D = 0 at r = 7 m.

3.14 A certain light-emitting diode (LED) is centered at the origin with its surface in the x y plane. At far distances, the LED appears as a point, but the glowing surface geometry produces a far-ﬁeld radiation pattern that follows a raised cosine law: that is, the optical power (ﬂux) density in watts/m2 is given in spherical coordinates by cos2 θ ar watts/m2 2πr 2 where θ is the angle measured with respect to the direction that is normal to the LED surface (in this case, the z axis), and r is the radial distance from the origin at which the power is detected. (a) In terms of P0 , ﬁnd the total power in watts emitted in the upper half-space by the LED; (b) Find the cone angle, θ1 , within which half the total power is radiated, that is, within the range 0 < θ < θ1 ; (c) An optical detector, having a 1-mm2 cross-sectional area, is positioned at r = 1 m and at θ = 45◦ , such that it faces the LED. If one milliwatt is measured by the detector, what (to a very good estimate) is the value of P0 ? Pd = P0

3.15 Volume charge density is located as follows: ρν = 0 for ρ < 1 mm and for ρ > 2 mm, ρν = 4ρ µC/m3 for 1 < ρ < 2 mm. (a) Calculate the total charge in the region 0 < ρ < ρ1 , 0 < z < L, where 1 < ρ1 < 2 mm. (b) Use Gauss’s law to determine Dρ at ρ = ρ1 . (c) Evaluate Dρ at ρ = 0.8 mm, 1.6 mm, and 2.4 mm. 3.16 An electric ﬂux density is given by D = D0 aρ , where D0 is a given constant. (a) What charge density generates this ﬁeld? (b) For the speciﬁed ﬁeld, what

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total charge is contained within a cylinder of radius a and height b, where the cylinder axis is the z axis? 3.17 A cube is deﬁned by 1 < x, y, z < 1.2. If D = 2x 2yax + 3x 2y 2 a y C/m2 (a) Apply Gauss’s law to ﬁnd the total ﬂux leaving the closed surface of the cube. (b) Evaluate ∇ · D at the center of the cube. (c) Estimate the total charge enclosed within the cube by using Eq. (8). 3.18 State whether the divergence of the following vector ﬁelds is positive, negative, or zero: (a) the thermal energy ﬂow in J/(m2 − s) at any point in a freezing ice cube; (b) the current density in A/m2 in a bus bar carrying direct current; (c) the mass ﬂow rate in kg/(m2 − s) below the surface of water in a basin, in which the water is circulating clockwise as viewed from above. 3.19 A spherical surface of radius 3 mm is centered at P(4, 1, 5) in free space. Let D = xax C/m2 . Use the results of Section 3.4 to estimate the net electric ﬂux leaving the spherical surface. 3.20 A radial electric ﬁeld distribution in free space is given in spherical coordinates as: rρ0 E1 = ar (r ≤ a) 3 0 (2a 3 − r 3 )ρ0 ar (a ≤ r ≤ b) E2 = 3 0 r2 (2a 3 − b3 )ρ0 ar (r ≥ b) E3 = 3 0 r2 where ρ0 , a, and b are constants. (a) Determine the volume charge density in the entire region (0 ≤ r ≤ ∞) by the appropriate use of ∇ · D = ρv . (b) In terms of given parameters, ﬁnd the total charge, Q, within a sphere of radius r where r > b. 3.21 Calculate ∇ · D at the point speciﬁed if (a) D = (1/z 2 )[10x yz ax + 5x 2 z a y + (2z 3 − 5x 2 y) az ] at P(−2, 3, 5); (b) D = 5z 2 aρ + 10ρz az at P(3, −45◦ , 5); (c) D = 2r sin θ sin φ ar + r cos θ sin φ aθ + r cos φ aφ at P(3, 45◦ , −45◦ ).

3.22 (a) A ﬂux density ﬁeld is given as F1 = 5az . Evaluate the outward ﬂux of F1 through the hemispherical surface, r = a, 0 < θ < π/2, 0 < φ < 2π. (b) What simple observation would have saved a lot of work in part a? (c) Now suppose the ﬁeld is given by F2 = 5zaz . Using the appropriate surface integrals, evaluate the net outward ﬂux of F2 through the closed surface consisting of the hemisphere of part a and its circular base in the x y plane. (d) Repeat part c by using the divergence theorem and an appropriate volume integral. 3.23 (a) A point charge Q lies at the origin. Show that div D is zero everywhere except at the origin. (b) Replace the point charge with a uniform volume charge density ρv0 for 0 < r < a. Relate ρv0 to Q and a so that the total charge is the same. Find div D everywhere.

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3.24 In a region in free space, electric ﬂux density is found to be D= ρ0 (z + 2d) az C/m2 −ρ0 (z − 2d) az C/m2 (−2d ≤ z ≤ 0) (0 ≤ z ≤ 2d)

3.26 If we have a perfect gas of mass density ρm kg/m3 , and we assign a velocity U m/s to each differential element, then the mass ﬂow rate is ρm U kg/(m2 − s). Physical reasoning then leads to the continuity equation, ∇ · (ρm U) = −∂ρm /∂t. (a) Explain in words the physical interpretation of this equation. (b) Show that s ρm U · dS = −d M/dt, where M is the total mass of the gas within the constant closed surface S, and explain the physical signiﬁcance of the equation. 3.27 Let D = 5.00r 2 ar mC/m2 for r ≤ 0.08 m and D = 0.205 ar /r 2 µC/m2 for r ≥ 0.08 m. (a) Find ρν for r = 0.06 m. (b) Find ρν for r = 0.1 m. (c) What surface charge density could be located at r = 0.08 m to cause D = 0 for r > 0.08 m? 3.28 Repeat Problem 3.8, but use ∇ · D = ρν and take an appropriate volume integral. 3.29 In the region of free space that includes the volume 2 < x, y, z < 3, D = 2 (yz ax + x z a y − 2x y az ) C/m2 . (a) Evaluate the volume integral side of z2 the divergence theorem for the volume deﬁned here. (b) Evaluate the surface integral side for the corresponding closed surface. 3.30 (a) Use Maxwell’s ﬁrst equation, ∇ · D = ρv , to describe the variation of the electric ﬁeld intensity with x in a region in which no charge density exists and in which a nonhomogeneous dielectric has a permittivity that increases exponentially with x. The ﬁeld has an x component only; (b) repeat part (a), but with a radially directed electric ﬁeld (spherical coordinates), in which again ρv = 0, but in which the permittivity decreases exponentially with r .

3.25 Within the spherical shell, 3 < r < 4 m, the electric ﬂux density is given as D = 5(r − 3)3 ar C/m2 . (a) What is the volume charge density at r = 4? (b) What is the electric ﬂux density at r = 4? (c) How much electric ﬂux leaves the sphere r = 4? (d) How much charge is contained within the sphere r = 4?

Everywhere else, D = 0. (a) Using ∇ · D = ρv , ﬁnd the volume charge density as a function of position everywhere. (b) Determine the electric ﬂux that passes through the surface deﬁned by z = 0, −a ≤ x ≤ a, −b ≤ y ≤ b. (c) Determine the total charge contained within the region −a ≤ x ≤ a, −b ≤ y ≤ b, −d ≤ z ≤ d. (d) Determine the total charge contained within the region −a ≤ x ≤ a, −b ≤ y ≤ b, 0 ≤ z ≤ 2d.

3.31 Given the ﬂux density D = 16 cos(2θ ) aθ C/m2 , use two different methods to r ﬁnd the total charge within the region 1 < r < 2 m, 1 < θ < 2 rad, 1 < φ < 2 rad.

C H A P T E R

4

Energy and Potential

I

n Chapters 2 and 3 we became acquainted with Coulomb’s law and its use in ﬁnding the electric ﬁeld about several simple distributions of charge, and also with Gauss’s law and its application in determining the ﬁeld about some symmetrical charge arrangements. The use of Gauss’s law was invariably easier for these highly symmetrical distributions because the problem of integration always disappeared when the proper closed surface was chosen. However, if we had attempted to ﬁnd a slightly more complicated ﬁeld, such as that of two unlike point charges separated by a small distance, we would have found it impossible to choose a suitable gaussian surface and obtain an answer. Coulomb’s law, however, is more powerful and enables us to solve problems for which Gauss’s law is not applicable. The application of Coulomb’s law is laborious, detailed, and often quite complex, the reason for this being precisely the fact that the electric ﬁeld intensity, a vector ﬁeld, must be found directly from the charge distribution. Three different integrations are needed in general, one for each component, and the resolution of the vector into components usually adds to the complexity of the integrals. Certainly it would be desirable if we could ﬁnd some as yet undeﬁned scalar function with a single integration and then determine the electric ﬁeld from this scalar by some simple straightforward procedure, such as differentiation. This scalar function does exist and is known as the potential or potential fiel . We shall ﬁnd that it has a very real physical interpretation and is more familiar to most of us than is the electric ﬁeld which it will be used to ﬁnd. We should expect, then, to be equipped soon with a third method of ﬁnding electric ﬁelds—a single scalar integration, although not always as simple as we might wish, followed by a pleasant differentiation.

75

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4.1 ENERGY EXPENDED IN MOVING A POINT CHARGE IN AN ELECTRIC FIELD
The electric ﬁeld intensity was deﬁned as the force on a unit test charge at that point at which we wish to ﬁnd the value of this vector ﬁeld. If we attempt to move the test charge against the electric ﬁeld, we have to exert a force equal and opposite to that exerted by the ﬁeld, and this requires us to expend energy or do work. If we wish to move the charge in the direction of the ﬁeld, our energy expenditure turns out to be negative; we do not do the work, the ﬁeld does. Suppose we wish to move a charge Q a distance dL in an electric ﬁeld E. The force on Q arising from the electric ﬁeld is F E = QE (1)

where the subscript reminds us that this force arises from the ﬁeld. The component of this force in the direction dL which we must overcome is FE L = F · a L = QE · a L where a L = a unit vector in the direction of dL. The force that we must apply is equal and opposite to the force associated with the ﬁeld, Fappl = −QE · a L and the expenditure of energy is the product of the force and distance. That is, the differential work done by an external source moving charge Q is d W = −QE · a L d L, or d W = −QE · dL (2) where we have replaced a L dL by the simpler expression dL. This differential amount of work required may be zero under several conditions determined easily from Eq. (2). There are the trivial conditions for which E, Q, or dL is zero, and a much more important case in which E and dL are perpendicular. Here the charge is moved always in a direction at right angles to the electric ﬁeld. We can draw on a good analogy between the electric ﬁeld and the gravitational ﬁeld, where, again, energy must be expended to move against the ﬁeld. Sliding a mass around with constant velocity on a frictionless surface is an effortless process if the mass is moved along a constant elevation contour; positive or negative work must be done in moving it to a higher or lower elevation, respectively. Returning to the charge in the electric ﬁeld, the work required to move the charge a ﬁnite distance must be determined by integrating, W = −Q ﬁnal init

E · dL

(3)

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where the path must be speciﬁed before the integral can be evaluated. The charge is assumed to be at rest at both its initial and ﬁnal positions. This deﬁnite integral is basic to ﬁeld theory, and we shall devote the following section to its interpretation and evaluation. 1 D4.1. Given the electric ﬁeld E = 2 (8x yzax + 4x 2 za y − 4x 2 yaz ) V/m, ﬁnd z the differential amount of work done in moving a 6-nC charge a distance of 2 µm, starting at P(2, −2, 3) and proceeding in the direction a L = (a) − 6 ax + 7 3 a + 2 az ; (b) 6 ax − 3 a y − 2 az ; (c) 3 ax + 6 a y . 7 y 7 7 7 7 7 7
Ans. −149.3 fJ; 149.3 fJ; 0

4.2 THE LINE INTEGRAL
The integral expression for the work done in moving a point charge Q from one position to another, Eq. (3), is an example of a line integral, which in vector-analysis notation always takes the form of the integral along some prescribed path of the dot product of a vector ﬁeld and a differential vector path length dL. Without using vector analysis we should have to write W = −Q ﬁnal init

E L dL

where E L = component of E along dL. A line integral is like many other integrals which appear in advanced analysis, including the surface integral appearing in Gauss’s law, in that it is essentially descriptive. We like to look at it much more than we like to work it out. It tells us to choose a path, break it up into a large number of very small segments, multiply the component of the ﬁeld along each segment by the length of the segment, and then add the results for all the segments. This is a summation, of course, and the integral is obtained exactly only when the number of segments becomes inﬁnite. This procedure is indicated in Figure 4.1, where a path has been chosen from an initial position B to a ﬁnal position1 A and a uniform electric fiel is selected for simplicity. The path is divided into six segments, L1 , L2 , . . . , L6 , and the components of E along each segment are denoted by E L1 , E L2 , . . . , E L6 . The work involved in moving a charge Q from B to A is then approximately W = −Q(E L1 L 1 + E L2 L 2 + · · · + E L6 L 6 ) or, using vector notation, W = −Q(E1 · L1 + E2 · L2 + · · · + E6 · L6 )
The ﬁnal position is given the designation A to correspond with the convention for potential difference, as discussed in the following section.
1

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Figure 4.1 A graphical interpretation of a line integral in a uniform field. The line integral of E between points B and A is independent of the path selected, even in a nonuniform field; this result is not, in general, true for time-varying fields.

and because we have assumed a uniform ﬁeld, E1 = E2 = · · · = E6 W = −QE · ( L1 + L2 + · · · + L6 )

What is this sum of vector segments in the preceding parentheses? Vectors add by the parallelogram law, and the sum is just the vector directed from the initial point B to the ﬁnal point A, L B A . Therefore W = −QE · L B A (uniform E) (4)

Remembering the summation interpretation of the line integral, this result for the uniform ﬁeld can be obtained rapidly now from the integral expression W = −Q as applied to a uniform ﬁeld W = −QE · where the last integral becomes L B A and W = −QE · L B A (uniform E)
A B A B

E · dL

(5)

dL

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79

For this special case of a uniform electric ﬁeld intensity, we should note that the work involved in moving the charge depends only on Q, E, and L B A , a vector drawn from the initial to the ﬁnal point of the path chosen. It does not depend on the particular path we have selected along which to carry the charge. We may proceed from B to A on a straight line or via the Old Chisholm Trail; the answer is the same. We show in Section 4.5 that an identical statement may be made for any nonuniform (static) E ﬁeld. Let us use several examples to illustrate the mechanics of setting up the line integral appearing in Eq. (5).
E X A M P L E 4.1

We are given the nonuniform ﬁeld E = yax + xa y + 2az and we are asked to determine the work expended in carrying 2C from B(1, 0, 1) to A(0.8, 0.6, 1) along the shorter arc of the circle x 2 + y2 = 1
Solution. We use W = −Q
A B

z=1

E · dL, where E is not necessarily constant. Working in rectangular coordinates, the differential path dL is dxax + dya y + dzaz , and the integral becomes W = −Q = −2 = −2
A B A B

E · dL

(yax + xa y + 2az ) · (d x ax + dy a y + dz az ) y dx − 2
0.6 0

0.8 1

x dy − 4

1 1

dz

where the limits on the integrals have been chosen to agree with the initial and ﬁnal values of the appropriate variable of integration. Using the equation of the circular path (and selecting the sign of the radical which is correct for the quadrant involved), we have W = −2
0.8 1

1 − x2 dx − 2
0.8 1

0.6 0

1 − y 2 dy − 0
0.6 0

= − x 1 − x 2 + sin−1 x = −0.96 J

− y 1 − y 2 + sin−1 y

= −(0.48 + 0.927 − 0 − 1.571) − (0.48 + 0.644 − 0 − 0)

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E X A M P L E 4.2

Again ﬁnd the work required to carry 2C from B to A in the same ﬁeld, but this time use the straight-line path from B to A.
Solution. We start by determining the equations of the straight line. Any two of the

following three equations for planes passing through the line are sufﬁcient to deﬁne the line: y A − yB y − yB = (x − x B ) xA − xB zA − zB z − zB = (y − y B ) y A − yB xA − xB x − xB = (z − z B ) zA − zB From the ﬁrst equation we have y = −3(x − 1) and from the second we obtain z=1 Thus, W = −2 =6
0.8 1 0.8 1

y dx − 2

0.6 0

x dy − 4
0.6 0

1 1

dz dy

(x − 1) d x − 2

1−

y 3

= −0.96 J This is the same answer we found using the circular path between the same two points, and it again demonstrates the statement (unproved) that the work done is independent of the path taken in any electrostatic ﬁeld. It should be noted that the equations of the straight line show that dy = −3 d x and d x = − 1 dy. These substitutions may be made in the ﬁrst two integrals, along with 3 a change in limits, and the answer may be obtained by evaluating the new integrals. This method is often simpler if the integrand is a function of only one variable. Note that the expressions for dL in our three coordinate systems use the differential lengths obtained in Chapter 1 (rectangular in Section 1.3, cylindrical in Section 1.8, and spherical in Section 1.9): dL = d x ax + dy a y + dz az (rectangular) (cylindrical) (spherical) (6) (7) (8)

dL = dr ar + r dθ aθ + r sin θ dφ aφ

dL = dρ aρ + ρ dφaφ + dz az

The interrelationships among the several variables in each expression are determined from the speciﬁc equations for the path.

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Figure 4.2 (a) A circular path and (b) a radial path along which a charge of Q is carried in the field of an infinite line charge. No work is expected in the former case.

As a ﬁnal example illustrating the evaluation of the line integral, we investigate several paths that we might take near an inﬁnite line charge. The ﬁeld has been obtained several times and is entirely in the radial direction, ρL E = E ρ aρ = aρ 2π 0 ρ First we ﬁnd the work done in carrying the positive charge Q about a circular path of radius ρb centered at the line charge, as illustrated in Figure 4.2a. Without lifting a pencil, we see that the work must be nil, for the path is always perpendicular to the electric ﬁeld intensity, or the force on the charge is always exerted at right angles to the direction in which we are moving it. For practice, however, we will set up the integral and obtain the answer. The differential element dL is chosen in cylindrical coordinates, and the circular path selected demands that dρ and dz be zero, so dL = ρ1 dφ aφ . The work is then W = −Q = −Q ﬁnal init 2π

ρL aρ · ρ1 dφ aφ 2π 0 ρ1

ρL dφ aρ · aφ = 0 2π 0 0 We will now carry the charge from ρ = a to ρ = b along a radial path (Figure 4.2b). Here dL = dρ aρ and W = −Q or ﬁnal init

ρL aρ · dρ aρ = −Q 2π 0 ρ W =−

b

a

ρL d ρ 2π 0 ρ

b Qρ L ln 2π 0 a Because b is larger than a, ln (b/a) is positive, and the work done is negative, indicating that the external source that is moving the charge receives energy.

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One of the pitfalls in evaluating line integrals is a tendency to use too many minus signs when a charge is moved in the direction of a decreasing coordinate value. This is taken care of completely by the limits on the integral, and no misguided attempt should be made to change the sign of dL. Suppose we carry Q from b to a (Figure 4.2b). We still have dL = dρ aρ and show the different direction by recognizing ρ = b as the initial point and ρ = a as the ﬁnal point, W = −Q a b

ρL d ρ b Qρ L ln = 2π 0 ρ 2π 0 a

This is the negative of the previous answer and is obviously correct. D4.2. Calculate the work done in moving a 4-C charge from B(1, 0, 0) to A(0, 2, 0) along the path y = 2 − 2x, z = 0 in the ﬁeld E = (a) 5ax V/m; (b) 5xax V/m; (c) 5xax + 5ya y V/m.
Ans. 20 J; 10 J; −30 J

D4.3. We will see later that a time-varying E ﬁeld need not be conservative. (If it is not conservative, the work expressed by Eq. (3) may be a function of the path used.) Let E = yax V/m at a certain instant of time, and calculate the work required to move a 3-C charge from (1, 3, 5) to (2, 0, 3) along the straight-line segments joining: (a) (1, 3, 5) to (2, 3, 5) to (2, 0, 5) to (2, 0, 3); (b) (1, 3, 5) to (1, 3, 3) to (1, 0, 3) to (2, 0, 3).
Ans. −9 J; 0

4.3 DEFINITION OF POTENTIAL DIFFERENCE AND POTENTIAL
We are now ready to deﬁne a new concept from the expression for the work done by an external source in moving a charge Q from one point to another in an electric ﬁeld E, “Potential difference and work.” W = −Q ﬁnal init

E · dL

In much the same way as we deﬁned the electric ﬁeld intensity as the force on a unit test charge, we now deﬁne potential difference V as the work done (by an external source) in moving a unit positive charge from one point to another in an electric ﬁeld, Potential difference = V = − ﬁnal init

E · dL

(9)

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We have to agree on the direction of movement, and we do this by stating that V AB signiﬁes the potential difference between points A and B and is the work done in moving the unit charge from B (last named) to A (ﬁrst named). Thus, in determining V AB , B is the initial point and A is the ﬁnal point. The reason for this somewhat peculiar deﬁnition will become clearer shortly, when it is seen that the initial point B is often taken at inﬁnity, whereas the ﬁnal point A represents the ﬁxed position of the charge; point A is thus inherently more signiﬁcant. Potential difference is measured in joules per coulomb, for which the volt is deﬁned as a more common unit, abbreviated as V. Hence the potential difference between points A and B is V AB = −
A B

E · dL V

(10)

and V AB is positive if work is done in carrying the positive charge from B to A. From the line-charge example of Section 4.2 we found that the work done in taking a charge Q from ρ = b to ρ = a was Qρ L b ln 2π 0 a Thus, the potential difference between points at ρ = a and ρ = b is W = Vab = W b ρL ln = Q 2π 0 a

(11)

We can try out this deﬁnition by ﬁnding the potential difference between points A and B at radial distances r A and r B from a point charge Q. Choosing an origin at Q, Q E = Er ar = ar 4π 0r 2 and dL = dr ar we have V AB = −
A B

E · dL = −

rA rB

1 Q Q 1 dr = − 4π 0r 2 4π 0 r A rB

(12)

If r B > r A , the potential difference V AB is positive, indicating that energy is expended by the external source in bringing the positive charge from r B to r A . This agrees with the physical picture showing the two like charges repelling each other. It is often convenient to speak of the potential, or absolute potential, of a point, rather than the potential difference between two points, but this means only that we agree to measure every potential difference with respect to a speciﬁed reference point that we consider to have zero potential. Common agreement must be reached on the zero reference before a statement of the potential has any signiﬁcance. A person having one hand on the deﬂection plates of a cathode-ray tube that are “at a potential of 50 V” and the other hand on the cathode terminal would probably be too shaken up

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ENGINEERING ELECTROMAGNETICS

to understand that the cathode is not the zero reference, but that all potentials in that circuit are customarily measured with respect to the metallic shield about the tube. The cathode may be several thousands of volts negative with respect to the shield. Perhaps the most universal zero reference point in experimental or physical potential measurements is “ground,” by which we mean the potential of the surface region of the earth itself. Theoretically, we usually represent this surface by an inﬁnite plane at zero potential, although some large-scale problems, such as those involving propagation across the Atlantic Ocean, require a spherical surface at zero potential. Another widely used reference “point” is inﬁnity. This usually appears in theoretical problems approximating a physical situation in which the earth is relatively far removed from the region in which we are interested, such as the static ﬁeld near the wing tip of an airplane that has acquired a charge in ﬂying through a thunderhead, or the ﬁeld inside an atom. Working with the gravitational potential ﬁeld on earth, the zero reference is normally taken at sea level; for an interplanetary mission, however, the zero reference is more conveniently selected at inﬁnity. A cylindrical surface of some deﬁnite radius may occasionally be used as a zero reference when cylindrical symmetry is present and inﬁnity proves inconvenient. In a coaxial cable the outer conductor is selected as the zero reference for potential. And, of course, there are numerous special problems, such as those for which a two-sheeted hyperboloid or an oblate spheroid must be selected as the zero-potential reference, but these need not concern us immediately. If the potential at point A is V A and that at B is VB , then V AB = V A − VB (13)

where we necessarily agree that V A and VB shall have the same zero reference point. D4.4. An electric ﬁeld is expressed in rectangular coordinates by E = 6x 2 ax + 6ya y + 4az V/m. Find: (a) VM N if points M and N are speciﬁed by M(2, 6, −1) and N (−3, −3, 2); (b) VM if V = 0 at Q(4, −2, −35); (c) VN if V = 2 at P(1, 2, −4).
Ans. −139.0 V; −120.0 V; 19.0 V

4.4 THE POTENTIAL FIELD OF A POINT CHARGE
In Section 4.3 we found an expression Eq. (12) for the potential difference between two points located at r = r A and r = r B in the ﬁeld of a point charge Q placed at the origin. How might we conveniently deﬁne a zero reference for potential? The simplest possibility is to let V = 0 at inﬁnity. If we let the point at r = r B recede to inﬁnity, the potential at r A becomes Q VA = 4π 0r A

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85

or, as there is no reason to identify this point with the A subscript, V = Q 4π 0r

(14)

This expression deﬁnes the potential at any point distant r from a point charge Q at the origin, the potential at inﬁnite radius being taken as the zero reference. Returning to a physical interpretation, we may say that Q/4π 0r joules of work must be done in carrying a unit charge from inﬁnity to any point r meters from the charge Q. A convenient method to express the potential without selecting a speciﬁc zero reference entails identifying r A as r once again and letting Q/4π 0r B be a constant. Then V = Q + C1 4π 0r (15)

and C1 may be selected so that V = 0 at any desired value of r . We could also select the zero reference indirectly by electing to let V be V0 at r = r0 . It should be noted that the potential difference between two points is not a function of C1 . Equations (14) and (15) represent the potential ﬁeld of a point charge. The potential is a scalar ﬁeld and does not involve any unit vectors. We now deﬁne an equipotential surface as a surface composed of all those points having the same value of potential. All ﬁeld lines would be perpendicular to such a surface at the points where they intersect it. Therefore, no work is involved in moving a unit charge around on an equipotential surface. The equipotential surfaces in the potential ﬁeld of a point charge are spheres centered at the point charge. An inspection of the form of the potential ﬁeld of a point charge shows that it is an inverse-distance ﬁeld, whereas the electric ﬁeld intensity was found to be an inverse-square-law function. A similar result occurs for the gravitational force ﬁeld of a point mass (inverse-square law) and the gravitational potential ﬁeld (inverse distance). The gravitational force exerted by the earth on an object one million miles from it is four times that exerted on the same object two million miles away. The kinetic energy given to a freely falling object starting from the end of the universe with zero velocity, however, is only twice as much at one million miles as it is at two million miles. D4.5. A 15-nC point charge is at the origin in free space. Calculate V1 if point P1 is located at P1 (−2, 3, −1) and (a) V = 0 at (6, 5, 4); (b) V = 0 at inﬁnity; (c) V = 5 V at (2, 0, 4).
Ans. 20.67 V; 36.0 V; 10.89 V

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ENGINEERING ELECTROMAGNETICS

4.5 THE POTENTIAL FIELD OF A SYSTEM OF CHARGES: CONSERVATIVE PROPERTY
The potential at a point has been deﬁned as the work done in bringing a unit positive charge from the zero reference to the point, and we have suspected that this work, and hence the potential, is independent of the path taken. If it were not, potential would not be a very useful concept. Let us now prove our assertion. We do so by beginning with the potential ﬁeld of the single point charge for which we showed, in Section 4.4, the independence with regard to the path, noting that the ﬁeld is linear with respect to charge so that superposition is applicable. It will then follow that the potential of a system of charges has a value at any point which is independent of the path taken in carrying the test charge to that point. Thus the potential ﬁeld of a single point charge, which we shall identify as Q 1 and locate at r1 , involves only the distance |r − r1 | from Q 1 to the point at r where we are establishing the value of the potential. For a zero reference at inﬁnity, we have V (r) = Q1 4π 0 |r − r1 |

The potential arising from two charges, Q 1 at r1 and Q 2 at r2 , is a function only of |r − r1 | and |r − r2 |, the distances from Q 1 and Q 2 to the ﬁeld point, respectively. V (r) = Q2 Q1 + 4π 0 |r − r1 | 4π 0 |r − r2 | Qm 4π 0 |r − rm | m=1 n Continuing to add charges, we ﬁnd that the potential arising from n point charges is V (r) = (16)

If each point charge is now represented as a small element of a continuous volume charge distribution ρν ν, then V (r) = ρν (r2 ) ν2 ρν (rn ) νn ρν (r1 ) ν1 + + ··· + 4π 0 |r − r1 | 4π 0 |r − r2 | 4π 0 |r − rn |

As we allow the number of elements to become inﬁnite, we obtain the integral expression V (r) = ρν (r ) dv 4π 0 |r − r | (17)

vol

We have come quite a distance from the potential ﬁeld of the single point charge, and it might be helpful to examine Eq. (17) and refresh ourselves as to the meaning of each term. The potential V (r) is determined with respect to a zero reference potential at inﬁnity and is an exact measure of the work done in bringing a unit charge from

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87

inﬁnity to the ﬁeld point at r where we are ﬁnding the potential. The volume charge density ρv (r ) and differential volume element dv combine to represent a differential amount of charge ρν (r ) dv located at r . The distance |r − r | is that distance from the source point to the ﬁeld point. The integral is a multiple (volume) integral. If the charge distribution takes the form of a line charge or a surface charge, the integration is along the line or over the surface: V (r) = V (r) = ρ L (r ) d L 4π 0 |r − r |
S

(18) (19)

ρ S (r ) d S 4π 0 |r − r |

The most general expression for potential is obtained by combining Eqs. (16)–(19). These integral expressions for potential in terms of the charge distribution should be compared with similar expressions for the electric ﬁeld intensity, such as Eq. (15) in Section 2.3: E(r) = ρν (r ) dv r−r 2 |r − r | 4π 0 |r − r |

vol

The potential again is inverse distance, and the electric ﬁeld intensity, inversesquare law. The latter, of course, is also a vector ﬁeld.
E X A M P L E 4.3

To illustrate the use of one of these potential integrals, we will ﬁnd V on the z axis for a uniform line charge ρ L in the form of a ring, ρ = a, in the z = 0 plane, as shown in Figure 4.3.
Solution. Working with Eq. (18), we have d L = adφ , r = zaz , r = aaρ , |r−r | = √

a 2 + z 2 , and

V =

2π 0

ρ L a dφ ρL a = √ √ 4π 0 a 2 + z 2 2 0 a2 + z2

For a zero reference at inﬁnity, then: 1. The potential arising from a single point charge is the work done in carrying a unit positive charge from inﬁnity to the point at which we desire the potential, and the work is independent of the path chosen between those two points. 2. The potential ﬁeld in the presence of a number of point charges is the sum of the individual potential ﬁelds arising from each charge. 3. The potential arising from a number of point charges or any continuous charge distribution may therefore be found by carrying a unit charge from inﬁnity to the point in question along any path we choose.

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Figure 4.3 The potential field of a ring of uniform line charge density is easily obtained from V = ρ L (r ) dL / (4π 0 |r − r |).

In other words, the expression for potential (zero reference at inﬁnity), VA = − or potential difference,
A ∞

E · dL
A B

V AB = V A − VB = −

E · dL

is not dependent on the path chosen for the line integral, regardless of the source of the E ﬁeld. This result is often stated concisely by recognizing that no work is done in carrying the unit charge around any closed path, or E · dL = 0 (20)

A small circle is placed on the integral sign to indicate the closed nature of the path. This symbol also appeared in the formulation of Gauss’s law, where a closed surface integral was used. Equation (20) is true for static ﬁelds, but we will see in Chapter 9 that Faraday demonstrated it was incomplete when time-varying magnetic ﬁelds were present. One of Maxwell’s greatest contributions to electromagnetic theory was in showing that a time-varying electric ﬁeld produces a magnetic ﬁeld, and therefore we should expect to ﬁnd later that Eq. (20) is not correct when either E or the magnetic ﬁeld varies with time. Restricting our attention to the static case where E does not change with time, consider the dc circuit shown in Figure 4.4. Two points, A and B, are marked, and

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Figure 4.4 A simple dc-circuit problem that must be solved by applying E · dL = 0 in the form of Kirchhoff’s voltage law.

(20) states that no work is involved in carrying a unit charge from A through R2 and R3 to B and back to A through R1 , or that the sum of the potential differences around any closed path is zero. Equation (20) is therefore just a more general form of Kirchhoff’s circuital law for voltages, more general in that we can apply it to any region where an electric ﬁeld exists and we are not restricted to a conventional circuit composed of wires, resistances, and batteries. Equation (20) must be amended before we can apply it to time-varying ﬁelds. Any ﬁeld that satisﬁes an equation of the form of Eq. (20), (i.e., where the closed line integral of the ﬁeld is zero) is said to be a conservative fiel . The name arises from the fact that no work is done (or that energy is conserved) around a closed path. The gravitational ﬁeld is also conservative, for any energy expended in moving (raising) an object against the ﬁeld is recovered exactly when the object is returned (lowered) to its original position. A nonconservative gravitational ﬁeld could solve our energy problems forever. Given a nonconservative ﬁeld, it is of course possible that the line integral may be zero for certain closed paths. For example, consider the force ﬁeld, F = sin πρ aφ . Around a circular path of radius ρ = ρ1 , we have dL = ρ dφ aφ , and F · dL =
2π 0

sin πρ1 aφ · ρ1 dφ aφ =

2π 0

ρ1 sin πρ1 dφ

= 2πρ1 sin πρ1 The integral is zero if ρ1 = 1, 2, 3, . . . , etc., but it is not zero for other values of ρ1 , or for most other closed paths, and the given ﬁeld is not conservative. A conservative ﬁeld must yield a zero value for the line integral around every possible closed path. D4.6. If we take the zero reference for potential at inﬁnity, ﬁnd the potential at (0, 0, 2) caused by this charge conﬁguration in free space (a) 12 nC/m on the line ρ = 2.5 m, z = 0; (b) point charge of 18 nC at (1, 2, −1); (c) 12 nC/m on the line y = 2.5, z = 0, −1.0 < x < 1.0.
Ans. 529 V; 43.2 V; 66.3 V

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ENGINEERING ELECTROMAGNETICS

4.6 POTENTIAL GRADIENT
We now have two methods of determining potential, one directly from the electric ﬁeld intensity by means of a line integral, and another from the basic charge distribution itself by a volume integral. Neither method is very helpful in determining the ﬁelds in most practical problems, however, for as we will see later, neither the electric ﬁeld intensity nor the charge distribution is very often known. Preliminary information is much more apt to consist of a description of two equipotential surfaces, such as the statement that we have two parallel conductors of circular cross section at potentials of 100 and −100 V. Perhaps we wish to ﬁnd the capacitance between the conductors, or the charge and current distribution on the conductors from which losses may be calculated. These quantities may be easily obtained from the potential ﬁeld, and our immediate goal will be a simple method of ﬁnding the electric ﬁeld intensity from the potential. We already have the general line-integral relationship between these quantities, V =− E · dL (21)

but this is much easier to use in the reverse direction: given E, ﬁnd V. However, Eq. (21) may be applied to a very short element of length L along which E is essentially constant, leading to an incremental potential difference V, V = −E · L ˙ (22)

Now consider a general region of space, as shown in Figure 4.5, in which E and V both change as we move from point to point. Equation (22) tells us to choose an incremental vector element of length L = L a L and multiply its magnitude by

Figure 4.5 A vector incremental element of length L is shown making an angle of θ with an E field, indicated by its streamlines. The sources of the field are not shown.

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the component of E in the direction of a L (one interpretation of the dot product) to obtain the small potential difference between the ﬁnal and initial points of L. If we designate the angle between L and E as θ , then V = − E L cos θ ˙ We now pass to the limit and consider the derivative d V /d L. To do this, we need to show that V may be interpreted as a function V (x, y, z). So far, V is merely the result of the line integral (21). If we assume a speciﬁed starting point or zero reference and then let our end point be (x, y, z), we know that the result of the integration is a unique function of the end point (x, y, z) because E is a conservative ﬁeld. Therefore V is a single-valued function V (x, y, z). We may then pass to the limit and obtain dV = −E cos θ dL In which direction should L be placed to obtain a maximum value of V ? Remember that E is a deﬁnite value at the point at which we are working and is independent of the direction of L. The magnitude L is also constant, and our variable is a L , the unit vector showing the direction of L. It is obvious that the maximum positive increment of potential, Vmax , will occur when cos θ is −1, or L points in the direction opposite to E. For this condition, dV dL =E

max

This little exercise shows us two characteristics of the relationship between E and V at any point: 1. The magnitude of the electric ﬁeld intensity is given by the maximum value of the rate of change of potential with distance. 2. This maximum value is obtained when the direction of the distance increment is opposite to E or, in other words, the direction of E is opposite to the direction in which the potential is increasing the most rapidly. We now illustrate these relationships in terms of potential. Figure 4.6 is intended to show the information we have been given about some potential ﬁeld. It does this by showing the equipotential surfaces (shown as lines in the two-dimensional sketch). We desire information about the electric ﬁeld intensity at point P. Starting at P, we lay off a small incremental distance L in various directions, hunting for that direction in which the potential is changing (increasing) the most rapidly. From the sketch, this direction appears to be left and slightly upward. From our second characteristic above, the electric ﬁeld intensity is therefore oppositely directed, or to the right and slightly downward at P. Its magnitude is given by dividing the small increase in potential by the small element of length. It seems likely that the direction in which the potential is increasing the most rapidly is perpendicular to the equipotentials (in the direction of increasing potential), and this is correct, for if L is directed along an equipotential, V = 0 by our

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Figure 4.6 A potential field is shown by its equipotential surfaces. At any point the E field is normal to the equipotential surface passing through that point and is directed toward the more negative surfaces.

deﬁnition of an equipotential surface. But then V = −E · L = 0

and as neither E nor L is zero, E must be perpendicular to this L or perpendicular to the equipotentials. Because the potential ﬁeld information is more likely to be determined ﬁrst, let us describe the direction of L, which leads to a maximum increase in potential mathematically in terms of the potential ﬁeld rather than the electric ﬁeld intensity. We do this by letting a N be a unit vector normal to the equipotential surface and directed toward the higher potentials. The electric ﬁeld intensity is then expressed in terms of the potential, E=− dV dL aN max (23)

which shows that the magnitude of E is given by the maximum space rate of change of V and the direction of E is normal to the equipotential surface (in the direction of decreasing potential). Because d V /d L|max occurs when L is in the direction of a N , we may remind ourselves of this fact by letting dV dL and dV (24) aN dN Either Eq. (23) or Eq. (24) provides a physical interpretation of the process of ﬁnding the electric ﬁeld intensity from the potential. Both are descriptive of a general procedure, and we do not intend to use them directly to obtain quantitative information. E=− = dV dN

max

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This procedure leading from V to E is not unique to this pair of quantities, however, but has appeared as the relationship between a scalar and a vector ﬁeld in hydraulics, thermodynamics, and magnetics, and indeed in almost every ﬁeld to which vector analysis has been applied. The operation on V by which −E is obtained is known as the gradient, and the gradient of a scalar ﬁeld T is deﬁned as Gradient of T = grad T = dT aN dN (25)

where a N is a unit vector normal to the equipotential surfaces, and that normal is chosen, which points in the direction of increasing values of T. Using this new term, we now may write the relationship between V and E as E = −grad V (26)

Because we have shown that V is a unique function of x, y, and z, we may take its total differential ∂V ∂V ∂V dV = dx + dy + dz ∂x ∂y ∂z But we also have Because both expressions are true for any d x, dy, and dz, then ∂V Ex = − ∂x ∂V Ey = − ∂y ∂V ∂z These results may be combined vectorially to yield Ez = − E=− ∂V ∂V ∂V ax + ay + az ∂x ∂y ∂z (27) d V = −E · dL = −E x d x − E y dy − E z dz

and comparing Eqs. (26) and (27) provides us with an expression which may be used to evaluate the gradient in rectangular coordinates, grad V = ∂V ∂V ∂V ax + ay + az ∂x ∂y ∂z (28)

The gradient of a scalar is a vector, and old quizzes show that the unit vectors that are often incorrectly added to the divergence expression appear to be those that

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ENGINEERING ELECTROMAGNETICS

were incorrectly removed from the gradient. Once the physical interpretation of the gradient, expressed by Eq. (25), is grasped as showing the maximum space rate of change of a scalar quantity and the direction in which this maximum occurs, the vector nature of the gradient should be self-evident. The vector operator ∂ ∂ ∂ ∇= ax + a y + az ∂x ∂y ∂z may be used formally as an operator on a scalar, T , ∇T , producing ∂T ∂T ∂T ∇T = ax + ay + az ∂x ∂y ∂z from which we see that ∇T = grad T This allows us to use a very compact expression to relate E and V, E = −∇V (29)

The gradient may be expressed in terms of partial derivatives in other coordinate systems through the application of its deﬁnition Eq. (25). These expressions are derived in Appendix A and repeated here for convenience when dealing with problems having cylindrical or spherical symmetry. They also appear inside the back cover. ∇V = ∇V = ∇V = ∂V ∂V ∂V ax + ay + az ∂x ∂y ∂z (rectangular) (30)

∂V 1 ∂V ∂V aρ + aφ + az ∂ρ ρ ∂φ ∂z

(cylindrical)

(31)

∂V 1 ∂V 1 ∂V ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ

(spherical)

(32)

Note that the denominator of each term has the form of one of the components of dL in that coordinate system, except that partial differentials replace ordinary differentials; for example, r sin θ dφ becomes r sin θ ∂φ. We now illustrate the gradient concept with an example.
E X A M P L E 4.4

Given the potential ﬁeld, V = 2x 2 y − 5z, and a point P(−4, 3, 6), we wish to ﬁnd several numerical values at point P: the potential V , the electric ﬁeld intensity E, the direction of E, the electric ﬂux density D, and the volume charge density ρν .
Solution. The potential at P(−4, 5, 6) is

V P = 2(−4)2 (3) − 5(6) = 66 V

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95

Next, we may use the gradient operation to obtain the electric ﬁeld intensity, The value of E at point P is E = −∇V = −4x yax − 2x 2 a y + 5az V/m E P = 48ax − 32a y + 5az V/m and The direction of E at P is given by the unit vector |E P | = 482 + (−32)2 + 52 = 57.9 V/m

a E,P = (48ax − 32a y + 5az )/57.9 If we assume these ﬁelds exist in free space, then D=
0E

= 0.829ax − 0.553a y + 0.086az

= −35.4x y ax − 17.71x 2 a y + 44.3 az pC/m3

Finally, we may use the divergence relationship to ﬁnd the volume charge density that is the source of the given potential ﬁeld, At P, ρν = −106.2 pC/m3 . ρν = ∇ · D = −35.4y pC/m3

D4.7. A portion of a two-dimensional (E z = 0) potential ﬁeld is shown in Figure 4.7. The grid lines are 1 mm apart in the actual ﬁeld. Determine approximate values for E in rectangular coordinates at: (a) a; (b) b; (c) c.
Ans. −1075a y V/m; −600ax − 700a y V/m; −500ax − 650a y V/m

100 D4.8. Given the potential ﬁeld in cylindrical coordinates, V = 2 ρ cos φV, z +1 ◦ and point P at ρ = 3 m, φ = 60 , z = 2 m, ﬁnd values at P for (a) V ; (b) E; (c) E; (d) d V /d N ; (e) a N ; ( f ) ρν in free space.
Ans. 30.0 V; −10.00aρ + 17.3aφ + 24.0az V/m; 31.2 V/m; 31.2 V/m; 0.32aρ − 0.55aφ − 0.77az ; −234 pC/m3

4.7 THE ELECTRIC DIPOLE
The dipole ﬁelds that we develop in this section are quite important because they form the basis for the behavior of dielectric materials in electric ﬁelds, as discussed in Chapter 6, as well as justifying the use of images, as described in Section 5.5 of Chapter 5. Moreover, this development will serve to illustrate the importance of the potential concept presented in this chapter. An electric dipole, or simply a dipole, is the name given to two point charges of equal magnitude and opposite sign, separated by a distance that is small compared to

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ENGINEERING ELECTROMAGNETICS

Figure 4.7 See Problem D4.7.

the distance to the point P at which we want to know the electric and potential ﬁelds. The dipole is shown in Figure 4.8a. The distant point P is described by the spherical coordinates r, θ, and φ = 90◦ , in view of the azimuthal symmetry. The positive and negative point charges have separation d and rectangular coordinates (0, 0, 1 d) and 2 (0, 0, − 1 d), respectively. 2 So much for the geometry. What would we do next? Should we ﬁnd the total electric ﬁeld intensity by adding the known ﬁelds of each point charge? Would it be easier to ﬁnd the total potential ﬁeld ﬁrst? In either case, having found one, we will ﬁnd the other from it before calling the problem solved. If we choose to ﬁnd E ﬁrst, we will have two components to keep track of in spherical coordinates (symmetry shows E φ is zero), and then the only way to ﬁnd V from E is by use of the line integral. This last step includes establishing a suitable zero reference for potential, since the line integral gives us only the potential difference between the two points at the ends of the integral path. On the other hand, the determination of V ﬁrst is a much simpler problem. This is because we ﬁnd the potential as a function of position by simply adding the scalar potentials from the two charges. The position-dependent vector magnitude and direction of E are subsequently evaluated with relative ease by taking the negative gradient of V. Choosing this simpler method, we let the distances from Q and −Q to P be R1 and R2 , respectively, and write the total potential as V = Q 4π 0 1 1 − R1 R2 = Q R2 − R1 4π 0 R1 R2

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97

Figure 4.8 (a) The geometry of the problem of an electric dipole. The dipole moment p = Qd is in the az direction. (b) For a distant point P, R1 is essentially parallel to R2 , and we find that R2 − R1 = d cos θ.

Note that the plane z = 0, midway between the two point charges, is the locus of points for which R1 = R2 , and is therefore at zero potential, as are all points at inﬁnity. For a distant point, R1 = R2 , and the R1 R2 product in the denominator may be ˙ replaced by r 2 . The approximation may not be made in the numerator, however, without obtaining the trivial answer that the potential ﬁeld approaches zero as we go very far away from the dipole. Coming back a little closer to the dipole, we see from Figure 4.8b that R2 − R1 may be approximated very easily if R1 and R2 are assumed to be parallel, R2 − R1 = d cos θ ˙

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The ﬁnal result is then V = Qd cos θ 4π 0r 2 (33)

Again, we note that the plane z = 0 (θ = 90◦ ) is at zero potential. Using the gradient relationship in spherical coordinates, E = −∇V = − we obtain E=− − or E= Qd (2 cos θ ar + sin θ aθ ) 4π 0r 3 (35) Qd sin θ Qd cos θ ar − aθ 3 2π 0r 4π 0r 3 (34) 1 ∂V 1 ∂V ∂V ar + aθ + aφ ∂r r ∂θ r sin θ ∂φ

These are the desired distant ﬁelds of the dipole, obtained with a very small amount of work. Any student who has several hours to spend may try to work the problem in the reverse direction—the authors consider the process too long and detailed to include here, even for effect. To obtain a plot of the potential ﬁeld, we choose a dipole such that Qd/(4π 0 ) = 1, and then cos θ = V r 2 . The colored lines in Figure 4.9 indicate equipotentials for which V = 0, +0.2, +0.4, +0.6, +0.8, and +1, as indicated. The dipole axis is vertical, with the positive charge on the top. The streamlines for the electric ﬁeld are obtained by applying the methods of Section 2.6 in spherical coordinates, Eθ r dθ sin θ = = Er dr 2 cos θ or dr = 2 cot θ dθ r from which we obtain r = C1 sin2 θ The black streamlines shown in Figure 4.9 are for C1 = 1, 1.5, 2, and 2.5. The potential ﬁeld of the dipole, Eq. (33), may be simpliﬁed by making use of the dipole moment. We ﬁrst identify the vector length directed from −Q to +Q as d

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99

Figure 4.9 The electrostatic field of a point dipole with its moment in the az direction. Six equipotential surfaces are labeled with relative values of V .

and then deﬁne the dipole moment as Qd and assign it the symbol p. Thus p = Qd The units of p are C · m. Because d · ar = d cos θ, we then have p · ar V = 4π 0r 2 This result may be generalized as V = r−r 1 p· 2 4π 0 |r − r | |r − r | (36)

(37)

(38)

where r locates the ﬁeld point P, and r determines the dipole center. Equation (38) is independent of any coordinate system.

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ENGINEERING ELECTROMAGNETICS

The dipole moment p will appear again when we discuss dielectric materials. Since it is equal to the product of the charge and the separation, neither the dipole moment nor the potential will change as Q increases and d decreases, provided the product remains constant. The limiting case of a point dipole is achieved when we let d approach zero and Q approach inﬁnity such that the product p is ﬁnite. Turning our attention to the resultant ﬁelds, it is interesting to note that the potential ﬁeld is now proportional to the inverse square of the distance, and the electric ﬁeld intensity is proportional to the inverse cube of the distance from the dipole. Each ﬁeld falls off faster than the corresponding ﬁeld for the point charge, but this is no more than we should expect because the opposite charges appear to be closer together at greater distances and to act more like a single point charge of zero Coulombs. Symmetrical arrangements of larger numbers of point charges produce ﬁelds proportional to the inverse of higher and higher powers of r . These charge distributions are called multipoles, and they are used in inﬁnite series to approximate more unwieldy charge conﬁgurations. D4.9. An electric dipole located at the origin in free space has a moment p = 3ax − 2a y + az nC · m. (a) Find V at PA (2, 3, 4). (b) Find V at r = 2.5, θ = 30◦ , φ = 40◦ .
Ans. 0.23 V; 1.97 V

D4.10. A dipole of moment p = 6az nC · m is located at the origin in free space. (a) Find V at P(r = 4, θ = 20◦ , φ = 0◦ ). (b) Find E at P.
Ans. 3.17 V; 1.58ar + 0.29aθ V/m

4.8 ENERGY DENSITY IN THE ELECTROSTATIC FIELD
We have introduced the potential concept by considering the work done, or energy expended, in moving a point charge around in an electric ﬁeld, and now we must tie up the loose ends of that discussion by tracing the energy ﬂow one step further. Bringing a positive charge from inﬁnity into the ﬁeld of another positive charge requires work, the work being done by the external source moving the charge. Let us imagine that the external source carries the charge up to a point near the ﬁxed charge and then holds it there. Energy must be conserved, and the energy expended in bringing this charge into position now represents potential energy, for if the external source released its hold on the charge, it would accelerate away from the ﬁxed charge, acquiring kinetic energy of its own and the capability of doing work. In order to ﬁnd the potential energy present in a system of charges, we must ﬁnd the work done by an external source in positioning the charges.

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We may start by visualizing an empty universe. Bringing a charge Q 1 from inﬁnity to any position requires no work, for there is no ﬁeld present.2 The positioning of Q 2 at a point in the ﬁeld of Q 1 requires an amount of work given by the product of the charge Q 2 and the potential at that point due to Q 1 . We represent this potential as V2,1 , where the ﬁrst subscript indicates the location and the second subscript the source. That is, V2,1 is the potential at the location of Q 2 due to Q 1 . Then Work to position Q 2 = Q 2 V2,1 Similarly, we may express the work required to position each additional charge in the ﬁeld of all those already present: Work to position Q 4 = Q 4 V4,1 + Q 4 V4,2 + Q 4 V4,3 and so forth. The total work is obtained by adding each contribution: Total positioning work = potential energy of ﬁeld = W E = Q 2 V2,1 + Q 3 V3,1 + Q 3 V3,2 + Q 4 V4,1 +Q 4 V4,2 + Q 4 V4,3 + · · · (39) Work to position Q 3 = Q 3 V3,1 + Q 3 V3,2

Noting the form of a representative term in the preceding equation, Q1 Q3 Q 3 V3,1 = Q 3 = Q1 4π 0 R13 4π 0 R31 where R13 and R31 each represent the scalar distance between Q 1 and Q 3 , we see that it might equally well have been written as Q 1 V1,3 . If each term of the total energy expression is replaced by its equal, we have W E = Q 1 V1,2 + Q 1 V1,3 + Q 2 V2,3 + Q 1 V1,4 + Q 2 V2,4 + Q 3 V3,4 + · · · (40)

Adding the two energy expressions (39) and (40) gives us a chance to simplify the result a little: 2W E = Q 1 (V1,2 + V1,3 + V1,4 + · · ·) + Q 2 (V2,1 + V2,3 + V2,4 + · · ·) + Q 3 (V3,1 + V3,2 + V3,4 + · · ·) + ···

Each sum of potentials in parentheses is the combined potential due to all the charges except for the charge at the point where this combined potential is being found. In other words, V1,2 + V1,3 + V1,4 + · · · = V1
2 However, somebody in the workshop at inﬁnity had to do an inﬁnite amount of work to create the point charge in the ﬁrst place! How much energy is required to bring two half-charges into coincidence to make a unit charge?

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V1 is the potential at the location of Q 1 due to the presence of Q 2 , Q 3 , . . . . We therefore have W E = 1 (Q 1 V1 + Q 2 V2 + Q 3 V3 + · · ·) = 2 m=N 1 2 m=1

Q m Vm

(41)

In order to obtain an expression for the energy stored in a region of continuous charge distribution, each charge is replaced by ρν dv, and the summation becomes an integral, WE =
1 2

ρν V dv vol (42)

Equations (41) and (42) allow us to ﬁnd the total potential energy present in a system of point charges or distributed volume charge density. Similar expressions may be easily written in terms of line or surface charge density. Usually we prefer to use Eq. (42) and let it represent all the various types of charge which may have to be considered. This may always be done by considering point charges, line charge density, or surface charge density to be continuous distributions of volume charge density over very small regions. We will illustrate such a procedure with an example shortly. Before we undertake any interpretation of this result, we should consider a few lines of more difﬁcult vector analysis and obtain an expression equivalent to Eq. (42) but written in terms of E and D. We begin by making the expression a little bit longer. Using Maxwell’s ﬁrst equation, replace ρν by its equal ∇ · D and make use of a vector identity which is true for any scalar function V and any vector function D, ∇ · (V D) ≡ V (∇ · D) + D · (∇V ) (43)

This may be proved readily by expansion in rectangular coordinates. We then have, successively, WE = =
1 2 1 2 vol

ρν V dv =

1 2

(∇ · D)V dv vol vol

[∇ · (V D) − D · (∇V )] dv

Using the divergence theorem from Chapter 3, the ﬁrst volume integral of the last equation is changed into a closed surface integral, where the closed surface surrounds the volume considered. This volume, ﬁrst appearing in Eq. (42), must contain every charge, and there can then be no charges outside of the volume. We may therefore consider the volume as infinit in extent if we wish. We have WE =
1 2 S

(V D) · dS −

1 2

D · (∇V ) dv vol The surface integral is equal to zero, for over this closed surface surrounding the universe we see that V is approaching zero at least as rapidly as 1/r (the charges look like point charges from there), and D is approaching zero at least as rapidly as 1/r 2 . The integrand therefore approaches zero at least as rapidly as 1/r 3 , while the

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differential area of the surface, looking more and more like a portion of a sphere, is increasing only as r 2 . Consequently, in the limit as r → ∞, the integrand and the integral both approach zero. Substituting E = −∇V in the remaining volume integral, we have our answer, WE =
1 2

vol

D · E dv =

1 2

0E vol

2

dv

(44)

We may now use this last expression to calculate the energy stored in the electrostatic ﬁeld of a section of a coaxial cable or capacitor of length L. We found in Section 3.3 that aρ S Dρ = ρ Hence, aρ S E= aρ 0ρ where ρ S is the surface charge density on the inner conductor, whose radius is a. Thus, WE =
1 2 L 0 0 2π a b 2 a 2 ρS 0 2 2ρ 0ρ

dρ dφ dz =

2 π L a 2 ρS 0

ln

b a

This same result may be obtained from Eq. (42). We choose the outer conductor as our zero-potential reference, and the potential of the inner cylinder is then Va = − a b

E ρ dρ = −

a b

aρ S aρ S b ln dρ = ρ a 0 0

The surface charge density ρ S at ρ = a can be interpreted as a volume charge density ρν = ρ S /t, extending from ρ = a − 1 t to ρ = a + 1 t, where t a. The integrand 2 2 in Eq. (42) is therefore zero everywhere between the cylinders (where the volume charge density is zero), as well as at the outer cylinder (where the potential is zero). The integration is therefore performed only within the thin cylindrical shell at ρ = a, WE = from which WE =
2 a 2 ρ S ln(b/a) 0 1 2 vol

ρν V d V =

1 2

L

2π

a+t/2

0

0

a−t/2

ρS ρS b a ln ρ dρ dφ dz t a 0

πL

once again. This expression takes on a more familiar form if we recognize the total charge on the inner conductor as Q = 2πa Lρ S . Combining this with the potential difference between the cylinders, Va , we see that W E = 1 QVa 2 which should be familiar as the energy stored in a capacitor.

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The question of where the energy is stored in an electric ﬁeld has not yet been answered. Potential energy can never be pinned down precisely in terms of physical location. Someone lifts a pencil, and the pencil acquires potential energy. Is the energy stored in the molecules of the pencil, in the gravitational ﬁeld between the pencil and the earth, or in some obscure place? Is the energy in a capacitor stored in the charges themselves, in the ﬁeld, or where? No one can offer any proof for his or her own private opinion, and the matter of deciding may be left to the philosophers. Electromagnetic ﬁeld theory makes it easy to believe that the energy of an electric ﬁeld or a charge distribution is stored in the ﬁeld itself, for if we take Eq. (44), an exact and rigorously correct expression, WE = and write it on a differential basis, d W E = 1 D · E dv 2 or d WE = 1D · E 2 dv (45)
1 2

D · E dv vol we obtain a quantity 1 D · E, which has the dimensions of an energy density, or joules 2 per cubic meter. We know that if we integrate this energy density over the entire ﬁeldcontaining volume, the result is truly the total energy present, but we have no more justiﬁcation for saying that the energy stored in each differential volume element dv is 1 D · E dv than we have for looking at Eq. (42) and saying that the stored energy is 2 1 ρ V dv. The interpretation afforded by Eq. (45), however, is a convenient one, and 2 ν we will use it until proved wrong. D4.11. Find the energy stored in free space for the region 2 mm < r < 3 200 V; mm, 0 < θ < 90◦ , 0 < φ < 90◦ , given the potential ﬁeld V = : (a) r 300 cos θ V. (b) r2
Ans. 46.4 µJ; 36.7 J

REFERENCES
1. Attwood, S. S. Electric and Magnetic Fields. 3d ed. New York: John Wiley & Sons, 1949. There are a large number of well-drawn ﬁeld maps of various charge distributions, including the dipole ﬁeld. Vector analysis is not used. 2. Skilling, H. H. (See Suggested References for Chapter 3.) Gradient is described on pp. 19–21. 3. Thomas, G. B., Jr., and R. L. Finney. (See Suggested References for Chapter 1.) The directional derivative and the gradient are presented on pp. 823–30.

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105

CHAPTER 4 PROBLEMS
4.1 The value of E at P(ρ = 2, φ = 40◦ , z = 3) is given as E = 100aρ − 200aφ + 300az V/m. Determine the incremental work required to move a 20 µC charge a distance of 6 µm: (a) in the direction of aρ ; (b) in the direction of aφ ; (c) in the direction of az ; (d) in the direction of E; (e) in the direction of G = 2ax − 3a y + 4az .

4.2

4.3

A positive point charge of magnitude q1 lies at the origin. Derive an expression for the incremental work done in moving a second point charge q2 through a distance d x from the starting position (x, y, z), in the direction of −ax .

If E = 120aρ V/m, ﬁnd the incremental amount of work done in moving a 50-µC charge a distance of 2 mm from (a) P(1, 2, 3) toward Q(2, 1, 4); (b) Q(2, 1, 4) toward P(1, 2, 3). An electric ﬁeld in free space is given by E = xax + ya y + zaz V/m. Find the work done in moving a 1-µC charge through this ﬁeld (a) from (1, 1, 1) to (0, 0, 0); (b) from (ρ = 2, φ = 0) to (ρ = 2, φ = 90◦ ); (c) from (r = 10, θ = θ0 ) to (r = 10, θ = θ0 + 180◦ ).
P

4.4

4.5

Compute the value of A G · dL for G = 2yax with A(1, −1, 2) and P(2, 1, 2) using the path (a) straight-line segments A(1, −1, 2) to B(1, 1, 2) to P(2, 1, 2); (b) straight-line segments A(1, −1, 2) to C(2, −1, 2) to P(2, 1, 2). An electric ﬁeld in free space is given as E = x ax + 4z a y + 4y az . Given ˆ ˆ ˆ V (1, 1, 1) = 10 V, determine V (3, 3, 3). Let G = 3x y 2 ax + 2za y Given an initial point P(2, 1, 1) and a ﬁnal point Q(4, 3, 1), ﬁnd G · dL using the path (a) straight line: y = x − 1, z = 1; (b) parabola: 6y = x 2 + 2, z = 1.

4.6 4.7

4.8

4.9

Given E = −xax + ya y , (a) ﬁnd the work involved in moving a unit positive charge on a √ circular arc, the circle centered at the origin, from x = a to x = y = a/ 2; (b) verify that the work done in moving the charge around the full circle from x = a is zero. A uniform surface charge density of 20 nC/m2 is present on the spherical surface r = 0.6 cm in free space. (a) Find the absolute potential at P(r = 1 cm, θ = 25◦ , φ = 50◦ ). (b) Find V AB , given points A(r = 2 cm, θ = 30◦ , φ = 60◦ ) and B(r = 3 cm, θ = 45◦ , φ = 90◦ ).

4.10 A sphere of radius a carries a surface charge density of ρs0 C/m2 . (a) Find the absolute potential at the sphere surface. (b) A grounded conducting shell of radius b where b > a is now positioned around the charged sphere. What is the potential at the inner sphere surface in this case? 4.11 Let a uniform surface charge density of 5 nC/m2 be present at the z = 0 plane, a uniform line charge density of 8 nC/m be located at x = 0, z = 4,

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and a point charge of 2 µC be present at P(2, 0, 0). If V = 0 at M(0, 0, 5), ﬁnd V at N (1, 2, 3). 4.12 In spherical coordinates, E = 2r/(r 2 + a 2 )2 ar V/m. Find the potential at any point, using the reference (a)V = 0 at inﬁnity; (b) V = 0 at r = 0; (c)V = 100 V at r = a.

4.13 Three identical point charges of 4 pC each are located at the corners of an equilateral triangle 0.5 mm on a side in free space. How much work must be done to move one charge to a point equidistant from the other two and on the line joining them?

4.14 Given the electric ﬁeld E = (y + 1)ax + (x − 1)a y + 2az ﬁnd the potential difference between the points (a) (2, −2, −1) and (0, 0, 0); (b) (3, 2, −1) and (−2, −3, 4). 4.15 Two uniform line charges, 8 nC/m each, are located at x = 1, z = 2, and at x = −1, y = 2 in free space. If the potential at the origin is 100 V, ﬁnd V at P(4, 1, 3).

4.16 A spherically symmetric charge distribution in free space (with 0 < r < ∞) is known to have a potential function V (r ) = V0 a 2 /r 2 , where V0 and a are constants. (a) Find the electric ﬁeld intensity. (b) Find the volume charge density. (c) Find the charge contained inside radius a. (d) Find the total energy stored in the charge (or equivalently, in its electric ﬁeld). 4.17 Uniform surface charge densities of 6 and 2 nC/m2 are present at ρ = 2 and 6 cm, respectively, in free space. Assume V = 0 at ρ = 4 cm, and calculate V at (a) ρ = 5 cm; (b) ρ = 7 cm.

4.18 Find the potential at the origin produced by a line charge ρ L = kx/(x 2 + a 2 ) extending along the x axis from x = a to +∞, where a > 0. Assume a zero reference at inﬁnity.

4.20 In a certain medium, the electric potential is given by V (x) = ρ0 1 − e−ax a 0

4.19 The annular surface 1 cm < ρ < 3 cm, z = 0, carries the nonuniform surface charge density ρs = 5ρ nC/m2 . Find V at P(0, 0, 2 cm) if V = 0 at inﬁnity.

where ρ0 and a are constants. (a) Find the electric ﬁeld intensity, E. (b) Find the potential difference between the points x = d and x = 0. (c) If the medium permittivity is given by (x) = 0 eax , ﬁnd the electric ﬂux density, D, and the volume charge density, ρv , in the region. (d) Find the stored energy in the region (0 < x < d), (0 < y < 1), (0 < z < 1). 4.21 Let V = 2x y 2 z 3 + 3 ln(x 2 + 2y 2 + 3z 2 ) V in free space. Evaluate each of the following quantities at P(3, 2, −1) (a) V ; (b) |V |; (c) E; (d) |E|; (e) a N ; ( f ) D.

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4.22 A line charge of inﬁnite length lies along the z axis and carries a uniform linear charge density of ρ C/m. A perfectly conducting cylindrical shell, whose axis is the z axis, surrounds the line charge. The cylinder (of radius b), is at ground potential. Under these conditions, the potential function inside the cylinder (ρ < b) is given by ρ ln(ρ) V (ρ) = k − 2π 0 where k is a constant. (a) Find k in terms of given or known parameters. (b) Find the electric ﬁeld strength, E, for ρ < b. (c) Find the electric ﬁeld strength, E, for ρ > b. (d) Find the stored energy in the electric ﬁeld per unit length in the z direction within the volume deﬁned by ρ > a, where a < b. 4.23 It is known that the potential is given as V = 80ρ 0.6 V. Assuming free space conditions, ﬁnd. (a) E; (b) the volume charge density at ρ = 0.5 m; (c) the total charge lying within the closed surface ρ = 0.6, 0 < z < 1. 4.24 A certain spherically symmetric charge conﬁguration in free space produces an electric ﬁeld given in spherical coordinates by E(r ) = (ρ0r 2 )/(100 0 ) ar V/m (100ρ0 )/( 0r 2 ) ar V/m (r ≤ 10) (r ≥ 10)

where ρ0 is a constant. (a) Find the charge density as a function of position. (b) Find the absolute potential as a function of position in the two regions, r ≤ 10 and r ≥ 10. (c) Check your result of part b by using the gradient. (d) Find the stored energy in the charge by an integral of the form of Eq. (43). (e) Find the stored energy in the ﬁeld by an integral of the form of Eq. (45). 4.25 Within the cylinder ρ = 2, 0 < z < 1, the potential is given by V = 100 + 50ρ + 150ρ sin φV. (a) Find V, E, D, and ρν at P(1, 60◦ , 0.5) in free space. (b) How much charge lies within the cylinder? 4.26 Let us assume that we have a very thin, square, imperfectly conducting plate 2 m on a side, located in the plane z = 0 with one corner at the origin such that it lies entirely within the ﬁrst quadrant. The potential at any point in the plate is given as V = −e−x sin y. (a) An electron enters the plate at x = 0, y = π/3 with zero initial velocity; in what direction is its initial movement? (b) Because of collisions with the particles in the plate, the electron achieves a relatively low velocity and little acceleration (the work that the ﬁeld does on it is converted largely into heat). The electron therefore moves approximately along a streamline. Where does it leave the plate and in what direction is it moving at the time? 4.27 Two point charges, 1 nC at (0, 0, 0.1) and −1 nC at (0, 0, −0.1), are in free space. (a) Calculate V at P(0.3, 0, 0.4). (b) Calculate |E| at P. (c) Now treat the two charges as a dipole at the origin and ﬁnd V at P. 4.28 Use the electric ﬁeld intensity of the dipole [Section 4.7, Eq. (35)] to ﬁnd the difference in potential between points at θa and θb , each point having the

108

ENGINEERING ELECTROMAGNETICS

same r and φ coordinates. Under what conditions does the answer agree with Eq. (33), for the potential at θa ? 4.29 A dipole having a moment p = 3ax − 5a y + 10az nC · m is located at Q(1, 2, −4) in free space. Find V at P(2, 3, 4).

4.31 A potential ﬁeld in free space is expressed as V = 20/(x yz) V. (a) Find the total energy stored within the cube 1 < x, y, z < 2. (b) What value would be obtained by assuming a uniform energy density equal to the value at the center of the cube? 4.32 (a) Using Eq. (35), ﬁnd the energy stored in the dipole ﬁeld in the region r > a. (b) Why can we not let a approach zero as a limit? 4.33 A copper sphere of radius 4 cm carries a uniformly distributed total charge of 5 µC in free space. (a) Use Gauss’s law to ﬁnd D external to the sphere. (b) Calculate the total energy stored in the electrostatic ﬁeld. (c) Use W E = Q 2 /(2C) to calculate the capacitance of the isolated sphere. 4.34 A sphere of radius a contains volume charge of uniform density ρ0 C/m3 . Find the total stored energy by applying (a) Eq. (42); (b) Eq. (44). 4.35 Four 0.8 nC point charges are located in free space at the corners of a square 4 cm on a side. (a) Find the total potential energy stored. (b) A ﬁfth 0.8 nC charge is installed at the center of the square. Again ﬁnd the total stored energy. 4.36 Surface charge of uniform density ρs lies on a spherical shell of radius b, centered at the origin in free space. (a) Find the absolute potential everywhere, with zero reference at inﬁnity. (b) Find the stored energy in the sphere by considering the charge density and the potential in a two-dimensional version of Eq. (42). (c) Find the stored energy in the electric ﬁeld and show that the results of parts (b) and (c) are identical.

4.30 A dipole for which p = 10 0 az C · m is located at the origin. What is the equation of the surface on which E z = 0 but E = 0?

C H A P T E R

5

Conductors and Dielectrics

I

n this chapter, we apply the methods we have learned to some of the materials with which an engineer must work. In the ﬁrst part of the chapter, we consider conducting materials by describing the parameters that relate current to an applied electric ﬁeld. This leads to a general deﬁnition of Ohm’s law. We then develop methods of evaluating resistances of conductors in a few simple geometric forms. Conditions that must be met at a conducting boundary are obtained next, and this knowledge leads to a discussion of the method of images. The properties of semiconductors are described to conclude the discussion of conducting media. In the second part of the chapter, we consider insulating materials, or dielectrics. Such materials differ from conductors in that ideally, there is no free charge that can be transported within them to produce conduction current. Instead, all charge is conﬁned to molecular or lattice sites by coulomb forces. An applied electric ﬁeld has the effect of displacing the charges slightly, leading to the formation of ensembles of electric dipoles. The extent to which this occurs is measured by the relative permittivity, or dielectric constant. Polarization of the medium may modify the electric ﬁeld, whose magnitude and direction may differ from the values it would have in a different medium or in free space. Boundary conditions for the ﬁelds at interfaces between dielectrics are developed to evaluate these differences. It should be noted that most materials will possess both dielectric and conductive properties; that is, a material considered a dielectric may be slightly conductive, and a material that is mostly conductive may be slightly polarizable. These departures from the ideal cases lead to some interesting behavior, particularly as to the effects on electromagnetic wave propagation, as we will see later. ■

109

110

ENGINEERING ELECTROMAGNETICS

5.1 CURRENT AND CURRENT DENSITY
Electric charges in motion constitute a current. The unit of current is the ampere (A), deﬁned as a rate of movement of charge passing a given reference point (or crossing a given reference plane) of one coulomb per second. Current is symbolized by I , and therefore dQ I = (1) dt Current is thus deﬁned as the motion of positive charges, even though conduction in metals takes place through the motion of electrons, as we will see shortly. In ﬁeld theory, we are usually interested in events occurring at a point rather than within a large region, and we ﬁnd the concept of current density, measured in amperes per square meter (A/m2 ), more useful. Current density is a vector1 represented by J. The increment of current I crossing an incremental surface S normal to the current density is I = JN S and in the case where the current density is not perpendicular to the surface, I =J· Total current is obtained by integrating, I = J · dS (2) S

S

Current density may be related to the velocity of volume charge density at a point. Consider the element of charge Q = ρν ν = ρν S L, as shown in Figure 5.1a. To simplify the explanation, assume that the charge element is oriented with its edges parallel to the coordinate axes and that it has only an x component of velocity. In the time interval t, the element of charge has moved a distance x, as indicated in Figure 5.1b. We have therefore moved a charge Q = ρν S x through a reference plane perpendicular to the direction of motion in a time increment t, and the resulting current is Q x I = = ρν S t t As we take the limit with respect to time, we have I = ρν
1

S vx

Current is not a vector, for it is easy to visualize a problem in which a total current I in a conductor of nonuniform cross section (such as a sphere) may have a different direction at each point of a given cross section. Current in an exceedingly ﬁne wire, or a filamentar current, is occasionally deﬁned as a vector, but we usually prefer to be consistent and give the direction to the ﬁlament, or path, and not to the current.

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111

where νx represents the x component of the velocity v.2 In terms of current density, we ﬁnd Jx = ρν νx and in general J = ρν v (3) This last result shows clearly that charge in motion constitutes a current. We call this type of current a convection current, and J or ρν v is the convection current density. Note that the convection current density is related linearly to charge density as well as to velocity. The mass rate of ﬂow of cars (cars per square foot per second) in the Holland Tunnel could be increased either by raising the density of cars per cubic foot, or by going to higher speeds, if the drivers were capable of doing so. D5.1. Given the vector current density J = 10ρ 2 zaρ − 4ρ cos2 φ aφ mA/m2 : (a) ﬁnd the current density at P(ρ = 3, φ = 30◦ , z = 2); (b) determine the total current ﬂowing outward through the circular band ρ = 3, 0 < φ < 2π, 2 < z < 2.8.
Ans. 180aρ − 9aφ mA/m2 ; 3.26 A

Figure 5.1 An increment of charge, Q = ρν S L, which moves a distance a time t, produces a component of current density in the limit of Jx = ρν νx .

x in

5.2 CONTINUITY OF CURRENT
The introduction of the concept of current is logically followed by a discussion of the conservation of charge and the continuity equation. The principle of conservation of charge states simply that charges can be neither created nor destroyed, although equal

lowercase ν is used both for volume and velocity. Note, however, that velocity always appears as a vector v, a component νx , or a magnitude |v|, whereas volume appears only in differential form as dν or ν.

2 The

112

ENGINEERING ELECTROMAGNETICS

amounts of positive and negative charge may be simultaneously created, obtained by separation, or lost by recombination. The continuity equation follows from this principle when we consider any region bounded by a closed surface. The current through the closed surface is I =
S

J · dS

and this outward flo of positive charge must be balanced by a decrease of positive charge (or perhaps an increase of negative charge) within the closed surface. If the charge inside the closed surface is denoted by Q i , then the rate of decrease is −d Q i /dt and the principle of conservation of charge requires d Qi I = J · dS = − (4) dt S It might be well to answer here an often-asked question. “Isn’t there a sign error? I thought I = dQ/dt.” The presence or absence of a negative sign depends on what current and charge we consider. In circuit theory we usually associate the current ﬂow into one terminal of a capacitor with the time rate of increase of charge on that plate. The current of (4), however, is an outward-flowin current. Equation (4) is the integral form of the continuity equation; the differential, or point, form is obtained by using the divergence theorem to change the surface integral into a volume integral:
S

J · dS =

(∇ · J) dv vol We next represent the enclosed charge Q i by the volume integral of the charge density, d (∇ · J) dv = − ρν dv dt vol vol If we agree to keep the surface constant, the derivative becomes a partial derivative and may appear within the integral, ∂ρν (∇ · J) dv = − dv ∂t vol vol from which we have our point form of the continuity equation, (∇ · J) = − ∂ρν ∂t (5)

Remembering the physical interpretation of divergence, this equation indicates that the current, or charge per second, diverging from a small volume per unit volume is equal to the time rate of decrease of charge per unit volume at every point. As a numerical example illustrating some of the concepts from the last two sections, let us consider a current density that is directed radially outward and decreases exponentially with time, 1 J = e−t ar A/m2 r

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113

Selecting an instant of time t = 1 s, we may calculate the total outward current at r = 5 m: I = Jr S = I = Jr S =
1 −1 e 5

(4π52 ) = 23.1 A

At the same instant, but for a slightly larger radius, r = 6 m, we have
1 −1 e 6

4π62 = 27.7 A

Thus, the total current is larger at r = 6 than it is at r = 5. To see why this happens, we need to look at the volume charge density and the velocity. We use the continuity equation ﬁrst: − ∂ρν =∇ ·J=∇ · ∂t 1 −t e ar r = 1 1 ∂ r 2 e−t 2 ∂r r r = 1 −t e r2

We next seek the volume charge density by integrating with respect to t. Because ρν is given by a partial derivative with respect to time, the “constant” of integration may be a function of r : ρν = − 1 −t 1 e dt + K(r ) = 2 e−t + K(r ) r2 r

If we assume that ρν → 0 as t → ∞, then K(r ) = 0, and ρν = 1 −t e C/m3 r2

We may now use J = ρν v to ﬁnd the velocity, 1 −t e Jr r νr = = = r m/s 1 −t ρν e r2 The velocity is greater at r = 6 than it is at r = 5, and we see that some (unspeciﬁed) force is accelerating the charge density in an outward direction. In summary, we have a current density that is inversely proportional to r , a charge density that is inversely proportional to r 2 , and a velocity and total current that are proportional to r . All quantities vary as e−t. D5.2. Current density is given in cylindrical coordinates as J = −106 z 1.5 az A/m2 in the region 0 ≤ ρ ≤ 20 µm; for ρ ≥ 20 µm, J = 0. (a) Find the total current crossing the surface z = 0.1 m in the az direction. (b) If the charge velocity is 2 × 106 m/s at z = 0.1 m, ﬁnd ρν there. (c) If the volume charge density at z = 0.15 m is −2000 C/m3 , ﬁnd the charge velocity there.
Ans. −39.7 µA; −15.8 mC/m3 ; 29.0 m/s

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ENGINEERING ELECTROMAGNETICS

5.3 METALLIC CONDUCTORS
Physicists describe the behavior of the electrons surrounding the positive atomic nucleus in terms of the total energy of the electron with respect to a zero reference level for an electron at an inﬁnite distance from the nucleus. The total energy is the sum of the kinetic and potential energies, and because energy must be given to an electron to pull it away from the nucleus, the energy of every electron in the atom is a negative quantity. Even though this picture has some limitations, it is convenient to associate these energy values with orbits surrounding the nucleus, the more negative energies corresponding to orbits of smaller radius. According to the quantum theory, only certain discrete energy levels, or energy states, are permissible in a given atom, and an electron must therefore absorb or emit discrete amounts of energy, or quanta, in passing from one level to another. A normal atom at absolute zero temperature has an electron occupying every one of the lower energy shells, starting outward from the nucleus and continuing until the supply of electrons is exhausted. In a crystalline solid, such as a metal or a diamond, atoms are packed closely together, many more electrons are present, and many more permissible energy levels are available because of the interaction forces between adjacent atoms. We ﬁnd that the allowed energies of electrons are grouped into broad ranges, or “bands,” each band consisting of very numerous, closely spaced, discrete levels. At a temperature of absolute zero, the normal solid also has every level occupied, starting with the lowest and proceeding in order until all the electrons are located. The electrons with the highest (least negative) energy levels, the valence electrons, are located in the valence band. If there are permissible higher-energy levels in the valence band, or if the valence band merges smoothly into a conduction band, then additional kinetic energy may be given to the valence electrons by an external ﬁeld, resulting in an electron ﬂow. The solid is called a metallic conductor. The ﬁlled valence band and the unﬁlled conduction band for a conductor at absolute zero temperature are suggested by the sketch in Figure 5.2a. If, however, the electron with the greatest energy occupies the top level in the valence band and a gap exists between the valence band and the conduction band, then

Figure 5.2 The energy-band structure in three different types of materials at 0 K. (a) The conductor exhibits no energy gap between the valence and conduction bands. (b) The insulator shows a large energy gap. (c) The semiconductor has only a small energy gap.

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115

the electron cannot accept additional energy in small amounts, and the material is an insulator. This band structure is indicated in Figure 5.2b. Note that if a relatively large amount of energy can be transferred to the electron, it may be sufﬁciently excited to jump the gap into the next band where conduction can occur easily. Here the insulator breaks down. An intermediate condition occurs when only a small “forbidden region” separates the two bands, as illustrated by Figure 5.2c. Small amounts of energy in the form of heat, light, or an electric ﬁeld may raise the energy of the electrons at the top of the ﬁlled band and provide the basis for conduction. These materials are insulators which display many of the properties of conductors and are called semiconductors. Let us ﬁrst consider the conductor. Here the valence electrons, or conduction, or free, electrons, move under the inﬂuence of an electric ﬁeld. With a ﬁeld E, an electron having a charge Q = −e will experience a force F = −eE In free space, the electron would accelerate and continuously increase its velocity (and energy); in the crystalline material, the progress of the electron is impeded by continual collisions with the thermally excited crystalline lattice structure, and a constant average velocity is soon attained. This velocity vd is termed the drift velocity, and it is linearly related to the electric ﬁeld intensity by the mobility of the electron in the given material. We designate mobility by the symbol µ (mu), so that vd = −µe E (6)

where µ is the mobility of an electron and is positive by deﬁnition. Note that the electron velocity is in a direction opposite to the direction of E. Equation (6) also shows that mobility is measured in the units of square meters per volt-second; typical values3 are 0.0012 for aluminum, 0.0032 for copper, and 0.0056 for silver. For these good conductors, a drift velocity of a few centimeters per second is sufﬁcient to produce a noticeable temperature rise and can cause the wire to melt if the heat cannot be quickly removed by thermal conduction or radiation. Substituting (6) into Eq. (3) of Section 5.1, we obtain J = −ρe µe E (7)

where ρe is the free-electron charge density, a negative value. The total charge density ρν is zero because equal positive and negative charges are present in the neutral material. The negative value of ρe and the minus sign lead to a current density J that is in the same direction as the electric ﬁeld intensity E. The relationship between J and E for a metallic conductor, however, is also speciﬁed by the conductivity σ (sigma), J = σE
3

(8)

Wert and Thomson, p. 238, listed in the References at the end of this chapter.

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where σ is measured is siemens4 per meter (S/m). One siemens (1 S) is the basic unit of conductance in the SI system and is deﬁned as one ampere per volt. Formerly, the unit of conductance was called the mho and was symbolized by an inverted . Just as the siemens honors the Siemens brothers, the reciprocal unit of resistance that we call the ohm (1 is one volt per ampere) honors Georg Simon Ohm, a German physicist who ﬁrst described the current-voltage relationship implied by Eq. (8). We call this equation the point form of Ohm’s law; we will look at the more common form of Ohm’s law shortly. First, however, it is informative to note the conductivity of several metallic conductors; typical values (in siemens per meter) are 3.82×107 for aluminum, 5.80×107 for copper, and 6.17 × 107 for silver. Data for other conductors may be found in Appendix C. On seeing data such as these, it is only natural to assume that we are being presented with constant values; this is essentially true. Metallic conductors obey Ohm’s law quite faithfully, and it is a linear relationship; the conductivity is constant over wide ranges of current density and electric ﬁeld intensity. Ohm’s law and the metallic conductors are also described as isotropic, or having the same properties in every direction. A material which is not isotropic is called anisotropic, and we shall mention such a material in Chapter 6. The conductivity is a function of temperature, however. The resistivity, which is the reciprocal of the conductivity, varies almost linearly with temperature in the region of room temperature, and for aluminum, copper, and silver it increases about 0.4 percent for a 1-K rise in temperature.5 For several metals the resistivity drops abruptly to zero at a temperature of a few kelvin; this property is termed superconductivity. Copper and silver are not superconductors, although aluminum is (for temperatures below 1.14 K). If we now combine Equations (7) and (8), conductivity may be expressed in terms of the charge density and the electron mobility, σ = −ρe µe (9)

From the deﬁnition of mobility (6), it is now satisfying to note that a higher temperature infers a greater crystalline lattice vibration, more impeded electron progress for a given electric ﬁeld strength, lower drift velocity, lower mobility, lower conductivity from Eq. (9), and higher resistivity as stated. The application of Ohm’s law in point form to a macroscopic (visible to the naked eye) region leads to a more familiar form. Initially, assume that J and E are uniform, as they are in the cylindrical region shown in Figure 5.3. Because they are uniform, I =
4

S

J · dS = JS

(10)

This is the family name of two German-born brothers, Karl Wilhelm and Werner von Siemens, who were famous engineer-inventors in the nineteenth century. Karl became a British subject and was knighted, becoming Sir William Siemens. Copious temperature data for conducting materials are available in the Standard Handbook for Electrical Engineers, listed among the References at the end of this chapter.

5

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117

Figure 5.3 Uniform current density J and electric field intensity E in a cylindrical region of length L and crosssectional area S. Here V = I R, where R = L/σ S.

and Vab = − b a

E · dL = −E ·

a b

dL = −E · Lba (11)

= E · Lab or V = EL Thus J= or V I = σE = σ S L V =

L I σS The ratio of the potential difference between the two ends of the cylinder to the current entering the more positive end, however, is recognized from elementary circuit theory as the resistance of the cylinder, and therefore V = IR where R= L σS (13) (12)

Equation (12) is, of course, known as Ohm’s law, and Eq. (13) enables us to compute the resistance R, measured in ohms (abbreviated as ), of conducting objects which possess uniform ﬁelds. If the ﬁelds are not uniform, the resistance may still be deﬁned as the ratio of V to I , where V is the potential difference between two speciﬁed equipotential surfaces in the material and I is the total current crossing the more positive surface into the material. From the general integral relationships in Eqs. (10) and (11), and from Ohm’s law (8), we may write this general expression for resistance

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ENGINEERING ELECTROMAGNETICS

when the ﬁelds are nonuniform, R= − b E · dL Vab = I S σ E · dS a (14)

The line integral is taken between two equipotential surfaces in the conductor, and the surface integral is evaluated over the more positive of these two equipotentials. We cannot solve these nonuniform problems at this time, but we should be able to solve several of them after reading Chapter 6.
E X A M P L E 5.1

As an example of the determination of the resistance of a cylinder, we ﬁnd the resistance of a 1-mile length of #16 copper wire, which has a diameter of 0.0508 in. the cross section is π(1.291 × 10−3 /2)2 = 1.308 × 10−6 m2 , and the length is 1609 m. Using a conductivity of 5.80 × 107 S/m, the resistance of the wire is, therefore, R= 1609 (5.80 × 107 )(1.308 × 10−6 ) = 21.2
Solution. The diameter of the wire is 0.0508×0.0254 = 1.291×10−3 m, the area of

This wire can safely carry about 10 A dc, corresponding to a current density of 10/(1.308×10−6 ) = 7.65×106 A/m2 , or 7.65 A/mm2 . With this current, the potential difference between the two ends of the wire is 212 V, the electric ﬁeld intensity is 0.312 V/m, the drift velocity is 0.000 422 m/s, or a little more than one furlong a week, and the free-electron charge density is −1.81 × 1010 C/m3 , or about one electron within a cube two angstroms on a side. D5.3. Find the magnitude of the current density in a sample of silver for which σ = 6.17 × 107 S/m and µe = 0.0056 m2 /V · s if (a) the drift velocity is 1.5 µm/s ; (b) the electric ﬁeld intensity is 1 mV/m; (c) the sample is a cube 2.5 mm on a side having a voltage of 0.4 mV between opposite faces; (d) the sample is a cube 2.5 mm on a side carrying a total current of 0.5 A.
Ans. 16.5 kA/m2 ; 61.7 kA/m2 ; 9.9 MA/m2 ; 80.0 kA/m2

D5.4. A copper conductor has a diameter of 0.6 in. and it is 1200 ft long. Assume that it carries a total dc current of 50 A. (a) Find the total resistance of the conductor. (b) What current density exists in it? (c) What is the dc voltage between the conductor ends? (d) How much power is dissipated in the wire?
Ans. 0.035 ; 2.74 × 105 A/m2 ; 1.73 V; 86.4 W

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119

5.4 CONDUCTOR PROPERTIES AND BOUNDARY CONDITIONS
Once again, we must temporarily depart from our assumed static conditions and let time vary for a few microseconds to see what happens when the charge distribution is suddenly unbalanced within a conducting material. Suppose, for the sake of argument, that there suddenly appear a number of electrons in the interior of a conductor. The electric ﬁelds set up by these electrons are not counteracted by any positive charges, and the electrons therefore begin to accelerate away from each other. This continues until the electrons reach the surface of the conductor or until a number of electrons equal to the number injected have reached the surface. Here, the outward progress of the electrons is stopped, for the material surrounding the conductor is an insulator not possessing a convenient conduction band. No charge may remain within the conductor. If it did, the resulting electric ﬁeld would force the charges to the surface. Hence the ﬁnal result within a conductor is zero charge density, and a surface charge density resides on the exterior surface. This is one of the two characteristics of a good conductor. The other characteristic, stated for static conditions in which no current may ﬂow, follows directly from Ohm’s law: the electric ﬁeld intensity within the conductor is zero. Physically, we see that if an electric ﬁeld were present, the conduction electrons would move and produce a current, thus leading to a nonstatic condition. Summarizing for electrostatics, no charge and no electric ﬁeld may exist at any point within a conducting material. Charge may, however, appear on the surface as a surface charge density, and our next investigation concerns the ﬁelds external to the conductor. We wish to relate these external ﬁelds to the charge on the surface of the conductor. The problem is a simple one, and we may ﬁrst talk our way to the solution with a little mathematics. If the external electric ﬁeld intensity is decomposed into two components, one tangential and one normal to the conductor surface, the tangential component is seen to be zero. If it were not zero, a tangential force would be applied to the elements of the surface charge, resulting in their motion and nonstatic conditions. Because static conditions are assumed, the tangential electric ﬁeld intensity and electric ﬂux density are zero. Gauss’s law answers our questions concerning the normal component. The electric ﬂux leaving a small increment of surface must be equal to the charge residing on that incremental surface. The ﬂux cannot penetrate into the conductor, for the total ﬁeld there is zero. It must then leave the surface normally. Quantitatively, we may say that the electric ﬂux density in coulombs per square meter leaving the surface normally is equal to the surface charge density in coulombs per square meter, or D N = ρS . If we use some of our previously derived results in making a more careful analysis (and incidentally introducing a general method which we must use later), we should set up a boundary between a conductor and free space (Figure 5.4) showing tangential

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n

Figure 5.4 An appropriate closed path and gaussian surface are used to determine boundary conditions at a boundary between a conductor and free space; E t = 0 and D N = ρ S.

and normal components of D and E on the free-space side of the boundary. Both ﬁelds are zero in the conductor. The tangential ﬁeld may be determined by applying Section 4.5, Eq. (21), E · dL = 0 around the small closed path abcda. The integral must be broken up into four parts b a c d a

+

b

+

c

+

d

=0

Remembering that E = 0 within the conductor, we let the length from a to b or c to d be w and from b to c or d to a be h, and obtain As we allow h to approach zero, keeping w small but ﬁnite, it makes no difference whether or not the normal ﬁelds are equal at a and b, for h causes these products to become negligibly small. Hence, E t w = 0 and, therefore, E t = 0. The condition on the normal ﬁeld is found most readily by considering D N rather than E N and choosing a small cylinder as the gaussian surface. Let the height be h and the area of the top and bottom faces be S. Again, we let h approach zero. Using Gauss’s law, D · dS = Q
1 1 E t w − E N ,at b 2 h + E N ,at a 2 h = 0

S

we integrate over the three distinct surfaces + + =Q

top

bottom

sides

and ﬁnd that the last two are zero (for different reasons). Then D N S = Q = ρS S

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121

or These are the desired boundary conditions for the conductor-to-free-space boundary in electrostatics, Dt = E t = 0 DN =
0 EN

D N = ρS

(15) (16)

= ρS

The electric ﬂux leaves the conductor in a direction normal to the surface, and the value of the electric ﬂux density is numerically equal to the surface charge density. Equations (15) and (16) can be more formally expressed using the vector ﬁelds E×n s =0 D · n s = ρs (17) (18)

where n is the unit normal vector at the surface that points away from the conductor, as shown in Figure 5.4, and where both operations are evaluated at the conductor surface, s. Taking the cross product or the dot product of either ﬁeld quantity with n gives the tangential or the normal component of the ﬁeld, respectively. An immediate and important consequence of a zero tangential electric ﬁeld intensity is the fact that a conductor surface is an equipotential surface. The evaluation of the potential difference between any two points on the surface by the line integral leads to a zero result, because the path may be chosen on the surface itself where E · dL = 0. To summarize the principles which apply to conductors in electrostatic ﬁelds, we may state that 1. The static electric ﬁeld intensity inside a conductor is zero. 2. The static electric ﬁeld intensity at the surface of a conductor is everywhere directed normal to that surface. 3. The conductor surface is an equipotential surface. Using these three principles, there are a number of quantities that may be calculated at a conductor boundary, given a knowledge of the potential ﬁeld.
E X A M P L E 5.2

Given the potential, and a point P(2, −1, 3) that is stipulated to lie on a conductor-to-free-space boundary, ﬁnd V , E, D, and ρ S at P, and also the equation of the conductor surface.
Solution. The potential at point P is

V = 100(x 2 − y 2 )

V P = 100[22 − (−1)2 ] = 300 V

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ENGINEERING ELECTROMAGNETICS

Figure 5.5 Given point P(2, −1, 3) and the potential field, V = 100(x 2 − y 2 ), we find the equipotential surface through P is x 2 − y 2 = 3, and the streamline through P is xy = −2.

Because the conductor is an equipotential surface, the potential at the entire surface must be 300 V. Moreover, if the conductor is a solid object, then the potential everywhere in and on the conductor is 300 V, for E = 0 within the conductor. The equation representing the locus of all points having a potential of 300 V is 300 = 100(x 2 − y 2 ) or x 2 − y2 = 3 This is therefore the equation of the conductor surface; it happens to be a hyperbolic cylinder, as shown in Figure 5.5. Let us assume arbitrarily that the solid conductor lies above and to the right of the equipotential surface at point P, whereas free space is down and to the left. Next, we ﬁnd E by the gradient operation, At point P, E = −100∇(x 2 − y 2 ) = −200xax + 200ya y E p = −400ax − 200a y V/m

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123

Because D =

0 E,

we have

The ﬁeld is directed downward and to the left at P; it is normal to the equipotential surface. Therefore, Thus, the surface charge density at P is D N = |D P | = 3.96 nC/m2 ρ S,P = D N = 3.96 nC/m2 Note that if we had taken the region to the left of the equipotential surface as the conductor, the E ﬁeld would terminate on the surface charge and we would let ρ S = −3.96 nC/m2 .

D P = 8.854 × 10−12 E P = −3.54ax − 1.771a y nC/m2

E X A M P L E 5.3

Finally, let us determine the equation of the streamline passing through P.
Solution. We see that

Thus,

Ey 200y y dy = =− = Ex −200x x dx dx dy + =0 y x

and ln y + ln x = C1 Therefore, The line (or surface) through P is obtained when C2 = (2)(−1) = −2. Thus, the streamline is the trace of another hyperbolic cylinder, x y = −2 This is also shown on Figure 5.5. D5.5. Given the potential ﬁeld in free space, V = 100 sinh 5x sin 5y V , and a point P(0.1, 0.2, 0.3), ﬁnd at P: (a) V ; (b) E; (c) |E|; (d) |ρ S | if it is known that P lies on a conductor surface.
Ans. 43.8 V; −474ax − 140.8a y V/m; 495 V/m; 4.38 nC/m2

x y = C2

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ENGINEERING ELECTROMAGNETICS

5.5 THE METHOD OF IMAGES
One important characteristic of the dipole ﬁeld that we developed in Chapter 4 is the inﬁnite plane at zero potential that exists midway between the two charges. Such a plane may be represented by a vanishingly thin conducting plane that is inﬁnite in extent. The conductor is an equipotential surface at a potential V = 0, and the electric ﬁeld intensity is therefore normal to the surface. Thus, if we replace the dipole conﬁguration shown in Figure 5.6a with the single charge and conducting plane shown in Figure 5.6b, the ﬁelds in the upper half of each ﬁgure are the same. Below the conducting plane, all ﬁelds are zero, as we have not provided any charges in that region. Of course, we might also substitute a single negative charge below a conducting plane for the dipole arrangement and obtain equivalence for the ﬁelds in the lower half of each region. If we approach this equivalence from the opposite point of view, we begin with a single charge above a perfectly conducting plane and then see that we may maintain the same ﬁelds above the plane by removing the plane and locating a negative charge at a symmetrical location below the plane. This charge is called the image of the original charge, and it is the negative of that value. If we can do this once, linearity allows us to do it again and again, and thus any charge conﬁguration above an inﬁnite ground plane may be replaced by an arrangement composed of the given charge conﬁguration, its image, and no conducting plane. This is suggested by the two illustrations of Figure 5.7. In many cases, the potential ﬁeld of the new system is much easier to ﬁnd since it does not contain the conducting plane with its unknown surface charge distribution. As an example of the use of images, let us ﬁnd the surface charge density at P(2, 5, 0) on the conducting plane z = 0 if there is a line charge of 30 nC/m located at x = 0, z = 3, as shown in Figure 5.8a. We remove the plane and install an image line charge of −30 nC/m at x = 0, z = −3, as illustrated in Figure 5.8b. The ﬁeld at P may now be obtained by superposition of the known ﬁelds of the line

Figure 5.6 (a) Two equal but opposite charges may be replaced by (b) a single charge and a conducting plane without affecting the fields above the V = 0 surface.

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125

Figure 5.7 (a) A given charge configuration above an infinite conducting plane may be replaced by (b) the given charge configuration plus the image configuration, without the conducting plane.

charges. The radial vector from the positive line charge to P is R+ = 2ax − 3az , while R− = 2ax + 3az . Thus, the individual ﬁelds are E+ = and E− = Adding these results, we have −180 × 10−9 az = −249az V/m 2π 0 (13) This then is the ﬁeld at (or just above) P in both the conﬁgurations of Figure 5.8, and it is certainly satisfying to note that the ﬁeld is normal to the conducting plane, as it must be. Thus, D = 0 E = −2.20az nC/m2 , and because this is directed toward the conducting plane, ρ S is negative and has a value of −2.20 nC/m2 at P. E= ρL 2π
0 R+

a R+ =

30 × 10−9 2ax − 3az √ √ 2π 0 13 13

30 × 10−9 2ax + 3az √ √ 2π 0 13 13

Figure 5.8 (a) A line charge above a conducting plane. (b) The conductor is removed, and the image of the line charge is added.

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ENGINEERING ELECTROMAGNETICS

D5.6. A perfectly conducting plane is located in free space at x = 4, and a uniform inﬁnite line charge of 40 nC/m lies along the line x = 6, y = 3. Let V = 0 at the conducting plane. At P(7, −1, 5) ﬁnd: (a) V ; (b) E.
Ans. 317 V; −45.3ax − 99.2a y V/m

5.6 SEMICONDUCTORS
If we now turn our attention to an intrinsic semiconductor material, such as pure germanium or silicon, two types of current carriers are present, electrons and holes. The electrons are those from the top of the ﬁlled valence band that have received sufﬁcient energy (usually thermal) to cross the relatively small forbidden band into the conduction band. The forbidden-band energy gap in typical semiconductors is of the order of one electronvolt. The vacancies left by these electrons represent unﬁlled energy states in the valence band which may also move from atom to atom in the crystal. The vacancy is called a hole, and many semiconductor properties may be described by treating the hole as if it had a positive charge of e, a mobility, µh , and an effective mass comparable to that of the electron. Both carriers move in an electric ﬁeld, and they move in opposite directions; hence each contributes a component of the total current which is in the same direction as that provided by the other. The conductivity is therefore a function of both hole and electron concentrations and mobilities, σ = −ρe µe + ρh µh (19)

For pure, or intrinsic, silicon, the electron and hole mobilities are 0.12 and 0.025, respectively, whereas for germanium, the mobilities are, respectively, 0.36 and 0.17. These values are given in square meters per volt-second and range from 10 to 100 times as large as those for aluminum, copper, silver, and other metallic conductors.6 These mobilities are given for a temperature of 300 K. The electron and hole concentrations depend strongly on temperature. At 300 K the electron and hole volume charge densities are both 0.0024 C/m3 in magnitude in intrinsic silicon and 3.0 C/m3 in intrinsic germanium. These values lead to conductivities of 0.000 35 S/m in silicon and 1.6 S/m in germanium. As temperature increases, the mobilities decrease, but the charge densities increase very rapidly. As a result, the conductivity of silicon increases by a factor of 10 as the temperature increases from 300 to about 330 K and decreases by a factor of 10 as the temperature drops from 300 to about 275 K. Note that the conductivity of the intrinsic semiconductor increases with temperature, whereas that of a metallic conductor decreases with temperature; this is one of the characteristic differences between the metallic conductors and the intrinsic semiconductors.

6

Mobility values for semiconductors are given in References 2, 3, and 5 listed at the end of this chapter.

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127

Intrinsic semiconductors also satisfy the point form of Ohm’s law; that is, the conductivity is reasonably constant with current density and with the direction of the current density. The number of charge carriers and the conductivity may both be increased dramatically by adding very small amounts of impurities. Donor materials provide additional electrons and form n-type semiconductors, whereas acceptors furnish extra holes and form p-type materials. The process is known as doping, and a donor concentration in silicon as low as one part in 107 causes an increase in conductivity by a factor of 105 . The range of value of the conductivity is extreme as we go from the best insulating materials to semiconductors and the ﬁnest conductors. In siemens per meter, σ ranges from 10−17 for fused quartz, 10−7 for poor plastic insulators, and roughly unity for semiconductors to almost 108 for metallic conductors at room temperature. These values cover the remarkably large range of some 25 orders of magnitude. D5.7. Using the values given in this section for the electron and hole mobilities in silicon at 300 K, and assuming hole and electron charge densities are 0.0029 C/m3 and −0.0029 C/m3 , respectively, ﬁnd: (a) the component of the conductivity due to holes; (b) the component of the conductivity due to electrons; (c) the conductivity.
Ans. 72.5 µS/m; 348 µS/m; 421 µS/m

5.7 THE NATURE OF DIELECTRIC MATERIALS
A dielectric in an electric ﬁeld can be viewed as a free-space arrangement of microscopic electric dipoles, each of which is composed of a positive and a negative charge whose centers do not quite coincide.These are not free charges, and they cannot contribute to the conduction process. Rather, they are bound in place by atomic and molecular forces and can only shift positions slightly in response to external ﬁelds. They are called bound charges, in contrast to the free charges that determine conductivity. The bound charges can be treated as any other sources of the electrostatic ﬁeld. Therefore, we would not need to introduce the dielectric constant as a new parameter or to deal with permittivities different from the permittivity of free space; however, the alternative would be to consider every charge within a piece of dielectric material. This is too great a price to pay for using all our previous equations in an unmodiﬁed form, and we shall therefore spend some time theorizing about dielectrics in a qualitative way; introducing polarization P, permittivity , and relative permittivity r; and developing some quantitative relationships involving these new parameters. The characteristic that all dielectric materials have in common, whether they are solid, liquid, or gas, and whether or not they are crystalline in nature, is their ability to store electric energy. This storage takes place by means of a shift in the relative positions of the internal, bound positive and negative charges against the normal molecular and atomic forces.

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ENGINEERING ELECTROMAGNETICS

This displacement against a restraining force is analogous to lifting a weight or stretching a spring and represents potential energy. The source of the energy is the external ﬁeld, the motion of the shifting charges resulting perhaps in a transient current through a battery that is producing the ﬁeld. The actual mechanism of the charge displacement differs in the various dielectric materials. Some molecules, termed polar molecules, have a permanent displacement existing between the centers of “gravity” of the positive and negative charges, and each pair of charges acts as a dipole. Normally the dipoles are oriented in a random way throughout the interior of the material, and the action of the external ﬁeld is to align these molecules, to some extent, in the same direction. A sufﬁciently strong ﬁeld may even produce an additional displacement between the positive and negative charges. A nonpolar molecule does not have this dipole arrangement until after a ﬁeld is applied. The negative and positive charges shift in opposite directions against their mutual attraction and produce a dipole that is aligned with the electric ﬁeld. Either type of dipole may be described by its dipole moment p, as developed in Section 4.7, Eq. (36), where Q is the positive one of the two bound charges composing the dipole, and d is the vector from the negative to the positive charge. We note again that the units of p are coulomb-meters. If there are n dipoles per unit volume and we deal with a volume ν, then there are n ν dipoles, and the total dipole moment is obtained by the vector sum, ptotal = n ν

p = Qd

(20)

pi

i=1

If the dipoles are aligned in the same general direction, ptotal may have a signiﬁcant value. However, a random orientation may cause ptotal to be essentially zero. We now deﬁne the polarization P as the dipole moment per unit volume, P = lim ν→0 1 ν

n

ν

pi

(21)

i=1

with units of coulombs per square meter. We will treat P as a typical continuous ﬁeld, even though it is obvious that it is essentially undeﬁned at points within an atom or molecule. Instead, we should think of its value at any point as an average value taken over a sample volume ν—large enough to contain many molecules (n ν in number), but yet sufﬁciently small to be considered incremental in concept. Our immediate goal is to show that the bound-volume charge density acts like the free-volume charge density in producing an external ﬁeld; we will obtain a result similar to Gauss’s law. To be speciﬁc, assume that we have a dielectric containing nonpolar molecules. No molecule has a dipole moment, and P = 0 throughout the material. Somewhere in the interior of the dielectric we select an incremental surface element S, as shown in Figure 5.9a, and apply an electric ﬁeld E. The electric ﬁeld produces a moment

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129

Figure 5.9 (a) An incremental surface element S is shown in the interior of a dielectric in which an electric field E is present. (b) The nonpolar molecules form dipole moments p and a polarization P. There is a net transfer of bound charge across S.

p = Qd in each molecule, such that p and d make an angle θ with S, as indicated in Figure 5.9b. The bound charges will now move across S. Each of the charges associated with the creation of a dipole must have moved a distance 1 d cos θ in the direction 2 perpendicular to S. Thus, any positive charges initially lying below the surface S and within the distance 1 d cos θ of the surface must have crossed S going upward. 2 Also, any negative charges initially lying above the surface and within that distance ( 1 d cos θ ) from S must have crossed S going downward. Therefore, because there 2 are n molecules/m3 , the net total charge that crosses the elemental surface in an upward direction is equal to n Qd cos θ S, or where the subscript on Q b reminds us that we are dealing with a bound charge and not a free charge. In terms of the polarization, we have If we interpret S as an element of a closed surface inside the dielectric material, then the direction of S is outward, and the net increase in the bound charge within the closed surface is obtained through the integral Qb = −
S

Q b = n Qd · S

Qb = P · S

P · dS

(22)

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ENGINEERING ELECTROMAGNETICS

This last relationship has some resemblance to Gauss’s law, and we may now generalize our deﬁnition of electric ﬂux density so that it applies to media other than free space. We ﬁrst write Gauss’s law in terms of 0 E and Q T , the total enclosed charge, bound plus free: QT = where and Q is the total free charge enclosed by the surface S. Note that the free charge appears without a subscript because it is the most important type of charge and will appear in Maxwell’s equations. Combining these last three equations, we obtain an expression for the free charge enclosed, Q = QT − Qb =
S S 0 E · dS

(23)

QT = Qb + Q

( 0 E + P) · dS

(24)

D is now deﬁned in more general terms than was done in Chapter 3, D=
0E

+P

(25)

There is thus an added term to D that appears when polarizable material is present. Thus, Q=
S

D · dS

(26)

where Q is the free charge enclosed. Utilizing the several volume charge densities, we have Qb = Q= QT = ρb dv ν ρν dv ν ρT dv ν With the help of the divergence theorem, we may therefore transform Eqs. (22), (23), and (26) into the equivalent divergence relationships, ∇· ∇ · P = −ρb
0E

= ρT

∇ · D = ρν

(27)

We will emphasize only Eq. (26) and (27), the two expressions involving the free charge, in the work that follows.

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131

In order to make any real use of these new concepts, it is necessary to know the relationship between the electric ﬁeld intensity E and the polarization P that results. This relationship will, of course, be a function of the type of material, and we will essentially limit our discussion to those isotropic materials for which E and P are linearly related. In an isotropic material, the vectors E and P are always parallel, regardless of the orientation of the ﬁeld. Although most engineering dielectrics are linear for moderate-to-large ﬁeld strengths and are also isotropic, single crystals may be anisotropic. The periodic nature of crystalline materials causes dipole moments to be formed most easily along the crystal axes, and not necessarily in the direction of the applied ﬁeld. In ferroelectric materials, the relationship between P and E not only is nonlinear, but also shows hysteresis effects; that is, the polarization produced by a given electric ﬁeld intensity depends on the past history of the sample. Important examples of this type of dielectric are barium titanate, often used in ceramic capacitors, and Rochelle salt. The linear relationship between P and E is P = χe 0 E (28)

where χe (chi) is a dimensionless quantity called the electric susceptibility of the material. Using this relationship in Eq. (25), we have The expression within the parentheses is now deﬁned as r = χe + 1 D=
0E

+ χe 0 E = (χe + 1) 0 E (29)

This is another dimensionless quantity, and it is known as the relative permittivity, or dielectric constant of the material. Thus, D= where =
0 r 0 rE

= E

(30)

(31)

and is the permittivity. The dielectric constants are given for some representative materials in Appendix C. Anisotropic dielectric materials cannot be described in terms of a simple susceptibility or permittivity parameter. Instead, we ﬁnd that each component of D may be a function of every component of E, and D = E becomes a matrix equation where D and E are each 3 × 1 column matrices and is a 3 × 3 square matrix. Expanding the matrix equation gives Dy = Dz = Dx = x x Ex yx E x zx E x

+ + +

xy Ey yy E y zy E y

+ + +

x z Ez yz E z zz E z

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ENGINEERING ELECTROMAGNETICS

Note that the elements of the matrix depend on the selection of the coordinate axes in the anisotropic material. Certain choices of axis directions lead to simpler matrices.7 D and E (and P) are no longer parallel, and although D = 0 E + P remains a valid equation for anisotropic materials, we may continue to use D = E only by interpreting it as a matrix equation. We will concentrate our attention on linear isotropic materials and reserve the general case for a more advanced text. In summary, then, we now have a relationship between D and E that depends on the dielectric material present, D= E =
0 r

(30) (31)

where

This electric ﬂux density is still related to the free charge by either the point or integral form of Gauss’s law: ∇ · D = ρν D · dS = Q (27)

S

(26)

Use of the relative permittivity, as indicated by Eq. (31), makes consideration of the polarization, dipole moments, and bound charge unnecessary. However, when anisotropic or nonlinear materials must be considered, the relative permittivity, in the simple scalar form that we have discussed, is no longer applicable.
E X A M P L E 5.4

We locate a slab of Teﬂon in the region 0 ≤ x ≤ a, and assume free space where x < 0 and x > a. Outside the Teﬂon there is a uniform ﬁeld Eout = E 0 ax V/m. We seek values for D, E, and P everywhere.
Solution. The dielectric constant of the Teﬂon is 2.1, and thus the electric suscepti-

bility is 1.1. Outside the slab, we have immediately Dout = 0 E 0 ax. Also, as there is no dielectric material there, Pout = 0. Now, any of the last four or ﬁve equations will enable us to relate the several ﬁelds inside the material to each other. Thus Din = 2.1 0 Ein Pin = 1.1 0 Ein (0 ≤ x ≤ a) (0 ≤ x ≤ a)

7

A more complete discussion of this matrix may be found in the Ramo, Whinnery, and Van Duzer reference listed at the end of this chapter.

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133

As soon as we establish a value for any of these three ﬁelds within the dielectric, the other two can be found immediately. The difﬁculty lies in crossing over the boundary from the known ﬁelds external to the dielectric to the unknown ones within it. To do this we need a boundary condition, and this is the subject of the next section. We will complete this example then. In the remainder of this text we will describe polarizable materials in terms of D and rather than P and χe . We will limit our discussion to isotropic materials. D5.8. A slab of dielectric material has a relative dielectric constant of 3.8 and contains a uniform electric ﬂux density of 8 nC/m2 . If the material is lossless, ﬁnd: (a) E; (b) P; (c) the average number of dipoles per cubic meter if the average dipole moment is 10−29 C · m.
Ans. 238 V/m; 5.89 nC/m2 ; 5.89 × 1020 m−3

5.8 BOUNDARY CONDITIONS FOR PERFECT DIELECTRIC MATERIALS
How do we attack a problem in which there are two different dielectrics, or a dielectric and a conductor? This is another example of a boundary condition, such as the condition at the surface of a conductor whereby the tangential ﬁelds are zero and the normal electric ﬂux density is equal to the surface charge density on the conductor. Now we take the ﬁrst step in solving a two-dielectric problem, or a dielectric-conductor problem, by determining the behavior of the ﬁelds at the dielectric interface. Let us ﬁrst consider the interface between two dielectrics having permittivities 1 and 2 and occupying regions 1 and 2, as shown in Figure 5.10. We ﬁrst examine

n

Figure 5.10 The boundary between perfect dielectrics of permittivities 1 and 2 . The continuity of D N is shown by the gaussian surface on the right, and the continuity of E tan is shown by the line integral about the closed path at the left.

134

ENGINEERING ELECTROMAGNETICS

the tangential components by using E · dL = 0 around the small closed path on the left, obtaining E tan 1 w − E tan 2 w =0

The small contribution to the line integral by the normal component of E along the sections of length h becomes negligible as h decreases and the closed path crowds the surface. Immediately, then, E tan 1 = E tan 2 (32)

Evidently, Kirchhoff’s voltage law is still applicable to this case. Certainly we have shown that the potential difference between any two points on the boundary that are separated by a distance w is the same immediately above or below the boundary. If the tangential electric ﬁeld intensity is continuous across the boundary, then tangential D is discontinuous, for Dtan 1
1

= E tan 1 = E tan 2 = Dtan 1 = Dtan 2
1 2

Dtan 2
2

or (33)

The boundary conditions on the normal components are found by applying Gauss’s law to the small “pillbox” shown at the right in Figure 5.10. The sides are again very short, and the ﬂux leaving the top and bottom surfaces is the difference D N 1 S − D N 2 S = Q = ρS S from which D N 1 − D N 2 = ρS (34)

What is this surface charge density? It cannot be a bound surface charge density, because we are taking the polarization of the dielectric into effect by using a dielectric constant different from unity; that is, instead of considering bound charges in free space, we are using an increased permittivity. Also, it is extremely unlikely that any free charge is on the interface, for no free charge is available in the perfect dielectrics we are considering. This charge must then have been placed there deliberately, thus unbalancing the total charge in and on this dielectric body. Except for this special case, then, we may assume ρ S is zero on the interface and DN 1 = DN 2 (35)

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Conductors and Dielectrics

135

or the normal component of D is continuous. It follows that
1 EN1

=

2 EN2

(36)

and normal E is discontinuous. Equations (32) and (34) can be written in terms of ﬁeld vectors in any direction, along with the unit normal to the surface as shown in Figure 5.10. Formally stated, the boundary conditions for the electric ﬂux density and the electric ﬁeld strength at the surface of a perfect dielectric are (D1 − D2 ) · n = ρs which is the general statement of Eq. (32), and (E1 − E2 ) × n = 0 (38) (37)

generally states Eq. (34). This construction was used previously in Eqs. (17) and (18) for a conducting surface, in which the normal or tangential components of the ﬁelds are obtained through the dot product or cross product with the normal, respectively. These conditions may be used to show the change in the vectors D and E at the surface. Let D1 (and E1 ) make an angle θ1 with a normal to the surface (Figure 5.11). Because the normal components of D are continuous, D N 1 = D1 cos θ1 = D2 cos θ2 = D N 2 The ratio of the tangential components is given by (33) as D1 sin θ1 Dtan 1 = = Dtan 2 D2 sin θ2 or
2 D1 1 2

(39)

sin θ1 =

1 D2

sin θ2

(40)

n

Figure 5.11 The refraction of D at a dielectric interface. For the case shown, 1 > 2 ; E1 and E2 are directed along D1 and D2 , with D 1 > D 2 and E 1 < E 2 .

136

ENGINEERING ELECTROMAGNETICS

and the division of this equation by (39) gives tan θ1 = tan θ2
1 2

(41)

In Figure 5.11 we have assumed that 1 > 2 , and therefore θ1 > θ2 . The direction of E on each side of the boundary is identical with the direction of D, because D = E. The magnitude of D in region 2 may be found from Eq. (39) and (40), D2 = D1 cos2 θ1 + and the magnitude of E2 is E 2 = E 1 sin2 θ1 +
1 2 2 2 1 2

sin2 θ1

(42)

cos2 θ1

(43)

An inspection of these equations shows that D is larger in the region of larger permittivity (unless θ1 = θ2 = 0◦ where the magnitude is unchanged) and that E is larger in the region of smaller permittivity (unless θ1 = θ2 = 90◦ , where its magnitude is unchanged).
E X A M P L E 5.5

Solution. We recall that we had a slab of Teﬂon extending from x = 0 to x = a, as shown in Figure 5.12, with free space on both sides of it and an external ﬁeld Eout = E 0 ax . We also have Dout = 0 E 0 ax and Pout = 0. Inside, the continuity of D N at the boundary allows us to ﬁnd that Din = Dout = 0 E 0 ax . This gives us Ein = Din / = 0 E 0 ax /( r 0 ) = 0.476E 0 ax . To get the polarization ﬁeld in the dielectric, we use D = 0 E + P and obtain

Complete Example 5.4 by ﬁnding the ﬁelds within the Teﬂon ( uniform external ﬁeld Eout = E 0 ax in free space.

r

= 2.1), given the

Pin = Din − Summarizing then gives

0 Ein

=

0 E 0 ax

− 0.476 0 E 0 ax = 0.524 0 E 0 ax (0 ≤ x ≤ a) (0 ≤ x ≤ a)

Din =

0 E 0 ax

Ein = 0.476E 0 ax

Pin = 0.524 0 E 0 ax

(0 ≤ x ≤ a)

A practical problem most often does not provide us with a direct knowledge of the ﬁeld on either side of the boundary. The boundary conditions must be used to help us determine the ﬁelds on both sides of the boundary from the other information that is given.

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137

Figure 5.12 A knowledge of the electric field external to the dielectric enables us to find the remaining external fields first and then to use the continuity of normal D to begin finding the internal fields.

D5.9. Let Region 1 (z < 0) be composed of a uniform dielectric material for which r = 3.2, while Region 2 (z > 0) is characterized by r = 2. Let D1 = −30ax + 50a y + 70az nC/m2 and ﬁnd: (a) D N 1 ; (b) Dt1 ; (c) Dt1 ; (d) D1 ; (e) θ1 ; ( f ) P1 .
Ans. 70 nC/m2 ; −30ax + 50a y nC/m2 ; 58.3 nC/m2 ; 91.1 nC/m2 ; 39.8◦ ; −20.6ax + 34.4a y + 48.1az nC/m2

D5.10. Continue Problem D5.9 by ﬁnding: (a) D N 2 ; (b) Dt2 ; (c) D2 ; (d) P2 ; (e) θ2 .
Ans. 70az nC/m2 ; −18.75ax + 31.25a y nC/m2 ; −18.75ax + 31.25a y + 70az nC/m2 ; −9.38ax + 15.63a y + 35az nC/m2 ; 27.5◦

REFERENCES
1. Fano, R. M., L. J. Chu, and R. B. Adler. Electromagnetic Fields, Energy, and Forces. Cambridge, MA: MIT Press, 1968. Polarization in dielectrics is discussed in Chapter 5. This junior-level text presupposes a full-term physics course in electricity and magnetism, and it is therefore slightly more advanced in level. The introduction beginning on p. 1 should be read. 2. Dekker, A. J. Electrical Engineering Materials. Englewood Cliffs, NJ: Prentice-Hall, 1963. This admirable little book covers dielectrics, conductors, semiconductors, and magnetic materials.

138

ENGINEERING ELECTROMAGNETICS

3. Fink, D. G., and H. W. Beaty. Standard Handbook for Electrical Engineers. 15th ed. New York: McGraw-Hill, 2006. 4. Maxwell, J. C. A Treatise on Electricity and Magnetism. New York: Cambridge University Press, 2010. 5. Wert, C. A., and R. M. Thomson. Physics of Solids. 2nd ed. New York: McGraw-Hill, 1970. This is an advanced undergraduate-level text that covers metals, semiconductors, and dielectrics.

CHAPTER 5 PROBLEMS
5.1 Given the current density J = −104 [sin(2x)e−2y ax + cos(2x)e−2y a y ] kA/m2 (a) Find the total current crossing the plane y = 1 in the a y direction in the region 0 < x < 1, 0 < z < 2. (b) Find the total current leaving the region 0 < x, y < 1, 2 < z < 3 by integrating J · d S over the surface of the cube. (c) Repeat part (b), but use the divergence theorem. Given J = −10−4 (yax + xa y ) A/m2 , ﬁnd the current crossing the y = 0 plane in the −a y direction between z = 0 and 1, and x = 0 and 2.

5.2 5.3

Let J = 400 sin θ/(r 2 + 4) ar A/m2 . (a) Find the total current ﬂowing through that portion of the spherical surface r = 0.8, bounded by 0.1π < θ < 0.3π, 0 < φ < 2π. (b) Find the average value of J over the deﬁned area.

5.4 5.5

If volume charge density is given as ρv = (cos ωt)/r 2 C/m2 in spherical coordinates, ﬁnd J. It is reasonable to assume that J is not a function of θ or φ. Let J = 25/ρaρ − 20/(ρ 2 + 0.01) az A/m2 . (a) Find the total current crossing the plane z = 0.2 in the az direction for ρ < 0.4. (b) Calculate ∂ρν /∂t. (c) Find the outward current crossing the closed surface deﬁned by ρ = 0.01, ρ = 0.4, z = 0, and z = 0.2. (d) Show that the divergence theorem is satisiﬁed for J and the surface speciﬁed in part (c). In spherical coordinates, a current density J = −k/(r sin θ ) aθ A/m2 exists in a conducting medium, where k is a constant. Determine the total current in the az direction that crosses a circular disk of radius R, centered on the z axis and located at (a) z = 0; (b) z = h. Assuming that there is no transformation of mass to energy or vice versa, it is possible to write a continuity equation for mass. (a) If we use the continuity equation for charge as our model, what quantities correspond to J and ρν? (b) Given a cube 1 cm on a side, experimental data show that the rates at which mass is leaving each of the six faces are 10.25, −9.85, 1.75, −2.00, −4.05, and 4.45 mg/s. If we assume that the cube is an incremental volume element, determine an approximate value for the time rate of change of density at its center. A truncated cone has a height of 16 cm. The circular faces on the top and bottom have radii of 2 mm and 0.1 mm, respectively. If the material from

5.6

5.7

5.8

CHAPTER 5

Conductors and Dielectrics

139

which this solid cone is constructed has a conductivity of 2 × 106 S/m, use some good approximations to determine the resistance between the two circular faces. 5.9 (a) Using data tabulated in Appendix C, calculate the required diameter for a 2-m-long nichrome wire that will dissipate an average power of 450 W when 120 V rms at 60 Hz is applied to it. (b) Calculate the rms current density in the wire.

5.10 A large brass washer has a 2-cm inside diameter, a 5-cm outside diameter, and is 0.5 cm thick. Its conductivity is σ = 1.5 × 107 S/m. The washer is cut in half along a diameter, and a voltage is applied between the two rectangular faces of one part. The resultant electric ﬁeld in the interior of the half-washer is E = (0.5/ρ) aφ V/m in cylindrical coordinates, where the z axis is the axis of the washer. (a) What potential difference exists between the two rectangular faces? (b) What total current is ﬂowing? (c) What is the resistance between the two faces? 5.11 Two perfectly conducting cylindrical surfaces of length are located at ρ = 3 and ρ = 5 cm. The total current passing radially outward through the medium between the cylinders is 3 A dc. (a) Find the voltage and resistance between the cylinders, and E in the region between the cylinders, if a conducting material having σ = 0.05 S/m is present for 3 < ρ < 5 cm. (b) Show that integrating the power dissipated per unit volume over the volume gives the total dissipated power. 5.12 Two identical conducting plates, each having area A, are located at z = 0 and z = d. The region between plates is ﬁlled with a material having z-dependent conductivity, σ (z) = σ0 e−z/d , where σ0 is a constant. Voltage V0 is applied to the plate at z = d; the plate at z = 0 is at zero potential. Find, in terms of the given parameters, (a) the resistance of the material; (b) the total current ﬂowing between plates; (c) the electric ﬁeld intensity E within the material. 5.13 A hollow cylindrical tube with a rectangular cross section has external dimensions of 0.5 in. by 1 in. and a wall thickness of 0.05 in. Assume that the material is brass, for which σ = 1.5 × 107 S/m. A current of 200 A dc is ﬂowing down the tube. (a) What voltage drop is present across a 1 m length of the tube? (b) Find the voltage drop if the interior of the tube is ﬁlled with a conducting material for which σ = 1.5 × 105 S/m.

5.14 A rectangular conducting plate lies in the x y plane, occupying the region 0 < x < a, 0 < y < b. An identical conducting plate is positioned directly above and parallel to the ﬁrst, at z = d. The region between plates is ﬁlled with material having conductivity σ (x) = σ0 e−x/a , where σ0 is a constant. Voltage V0 is applied to the plate at z = d; the plate at z = 0 is at zero potential. Find, in terms of the given parameters, (a) the electric ﬁeld intensity E within the material; (b) the total current ﬂowing between plates; (c) the resistance of the material.

140

ENGINEERING ELECTROMAGNETICS

5.16 A coaxial transmission line has inner and outer conductor radii a and b. Between conductors (a < ρ < b) lies a conductive medium whose conductivity is σ (ρ) = σ0 /ρ, where σ0 is a constant. The inner conductor is charged to potential V0 , and the outer conductor is grounded. (a) Assuming dc radial current I per unit length in z, determine the radial current density ﬁeld J in A/m2 . (b) Determine the electric ﬁeld intensity E in terms of I and other parameters, given or known. (c) By taking an appropriate line integral of E as found in part (b), ﬁnd an expression that relates V0 to I . (d) Find an expression for the conductance of the line per unit length, G. 5.17 Given the potential ﬁeld V = 100x z/(x 2 + 4) V in free space: (a) Find D at the surface z = 0. (b) Show that the z = 0 surface is an equipotential surface. (c) Assume that the z = 0 surface is a conductor and ﬁnd the total charge on that portion of the conductor deﬁned by 0 < x < 2, −3 < y < 0.

5.15 Let V = 10(ρ + 1)z 2 cos φ V in free space. (a) Let the equipotential surface V = 20 V deﬁne a conductor surface. Find the equation of the conductor surface. (b) Find ρ and E at that point on the conductor surface where φ = 0.2π and z = 1.5. (c) Find |ρ S | at that point.

5.18 Two parallel circular plates of radius a are located at z = 0 and z = d. The top plate (z = d) is raised to potential V0 ; the bottom plate is grounded. Between the plates is a conducting material having radial-dependent conductivity, σ (ρ) = σ0 ρ, where σ0 is a constant. (a) Find the ρ-independent electric ﬁeld strength, E, between plates. (b) Find the current density, J between plates. (c) Find the total current, I , in the structure. (d) Find the resistance between plates.

5.19 Let V = 20x 2 yz − 10z 2 V in free space. (a) Determine the equations of the equipotential surfaces on which V = 0 and 60 V. (b) Assume these are conducting surfaces and ﬁnd the surface charge density at that point on the V = 60 V surface where x = 2 and z = 1. It is known that 0 ≤ V ≤ 60 V is the ﬁeld-containing region. (c) Give the unit vector at this point that is normal to the conducting surface and directed toward the V = 0 surface. 5.20 Two point charges of −100π µC are located at (2, −1, 0) and (2, 1, 0). The surface x = 0 is a conducting plane. (a) Determine the surface charge density at the origin. (b) Determine ρ S at P(0, h, 0).

5.21 Let the surface y = 0 be a perfect conductor in free space. Two uniform inﬁnite line charges of 30 nC/m each are located at x = 0, y = 1, and x = 0, y = 2. (a) Let V = 0 at the plane y = 0, and ﬁnd V at P(1, 2, 0). (b) Find E at P. 5.22 The line segment x = 0, −1 ≤ y ≤ 1, z = 1, carries a linear charge density ρ L = π |y| µC/m. Let z = 0 be a conducting plane and determine the surface charge density at: (a) (0, 0, 0); (b) (0, 1, 0).

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Conductors and Dielectrics

141

5.23 A dipole with p = 0.1az µC · m is located at A(1, 0, 0) in free space, and the x = 0 plane is perfectly conducting. (a) Find V at P(2, 0, 1). (b) Find the equation of the 200 V equipotential surface in rectangular coordinates. 5.24 At a certain temperature, the electron and hole mobilities in intrinsic germanium are given as 0.43 and 0.21 m2 /V · s, respectively. If the electron and hole concentrations are both 2.3 × 1019 m−3 , ﬁnd the conductivity at this temperature. 5.25 Electron and hole concentrations increase with temperature. For pure silicon, suitable expressions are ρh = −ρe = 6200T 1.5 e−7000/T C/m3 . The functional dependence of the mobilities on temperature is given by µh = 2.3 × 105 T −2.7 m2 /V · s and µe = 2.1 × 105 T −2.5 m2 /V · s, where the temperature, T , is in degrees Kelvin. Find σ at: (a) 0◦ C; (b) 40◦ C; (c) 80◦ C. 5.26 A semiconductor sample has a rectangular cross section 1.5 by 2.0 mm, and a length of 11.0 mm. The material has electron and hole densities of 1.8 × 1018 and 3.0 × 1015 m−3 , respectively. If µe = 0.082 m2 /V · s and µh = 0.0021 m2 / V · s, ﬁnd the resistance offered between the end faces of the sample. 5.27 Atomic hydrogen contains 5.5 × 1025 atoms/m3 at a certain temperature and pressure. When an electric ﬁeld of 4 kV/m is applied, each dipole formed by the electron and positive nucleus has an effective length of 7.1 × 10−19 m. (a) Find P. (b) Find r .

5.28 Find the dielectric constant of a material in which the electric ﬂux density is four times the polarization. 5.29 A coaxial conductor has radii a = 0.8 mm and b = 3 mm and a polystyrene dielectric for which r = 2.56. If P = (2/ρ)aρ nC/m2 in the dielectric, ﬁnd (a) D and E as functions of ρ; (b) Vab and χe . (c) If there are 4 × 1019 molecules per cubic meter in the dielectric, ﬁnd p(ρ). 5.30 Consider a composite material made up of two species, having number densities N1 and N2 molecules/m3, respectively. The two materials are uniformly mixed, yielding a total number density of N = N1 + N2 . The presence of an electric ﬁeld E induces molecular dipole moments p1 and p2 within the individual species, whether mixed or not. Show that the dielectric constant of the composite material is given by r = f r 1 + (1 − f ) r 2 , where f is the number fraction of species 1 dipoles in the composite, and where r 1 and r 2 are the dielectric constants that the unmixed species would have if each had number density N . 5.31 The surface x = 0 separates two perfect dielectrics. For x > 0, let r = r 1 = 3, while r 2 = 5 where x < 0. If E1 = 80ax − 60a y − 30az V/m, ﬁnd (a) E N 1 ; (b) ET 1 ; (c) E1 ; (d) the angle θ1 between E1 and a normal to the surface; (e) D N 2 ; ( f ) DT 2 ; (g) D2 ; (h) P2 ; (i) the angle θ2 between E2 and a normal to the surface.

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ENGINEERING ELECTROMAGNETICS

5.32 Two equal but opposite-sign point charges of 3 µC are held x meters apart by a spring that provides a repulsive force given by Fsp = 12(0.5 − x) N. Without any force of attraction, the spring would be fully extended to 0.5 m. (a) Determine the charge separation. (b) What is the dipole moment? 5.33 Two perfect dielectrics have relative permittivities r 1 = 2 and r 2 = 8. The planar interface between them is the surface x − y + 2z = 5. The origin lies in region 1. If E1 = 100ax + 200a y − 50az V/m, ﬁnd E2 .

5.34 Region 1 (x ≥ 0) is a dielectric with r 1 = 2, while region 2(x < 0) has r 2 = 5. Let E1 = 20ax − 10a y + 50az V/m. (a) Find D2 . (b) Find the energy density in both regions. 5.35 Let the cylindrical surfaces ρ = 4 cm and ρ = 9 cm enclose two wedges of perfect dielectrics, r 1 = 2 for 0 < φ < π/2 and r 2 = 5 for π/2 < φ < 2π. If E1 = (2000/ρ)aρ V/m, ﬁnd (a) E2 ; (b) the total electrostatic energy stored in a 1 m length of each region.

C H A P T E R

6

Capacitance

C

apacitance measures the capability of energy storage in electrical devices. It can be deliberately designed for a speciﬁc purpose, or it may exist as an unavoidable by-product of the device structure that one must live with. Understanding capacitance and its impact on device or system operation is critical in every aspect of electrical engineering. A capacitor is a device that stores energy; energy thus stored can either be associated with accumulated charge or it can be related to the stored electric ﬁeld, as was discussed in Section 4.8. In fact, one can think of a capacitor as a device that stores electric flu , in a similar way that an inductor — an analogous device — stores magnetic ﬂux (or ultimately magnetic ﬁeld energy). We will explore this in Chapter 8. A primary goal in this chapter is to present the methods for calculating capacitance for a number of cases, including transmission line geometries, and to be able to make judgments on how capacitance will be altered by changes in materials or their conﬁguration. ■

6.1 CAPACIT ANCE DEFINED
Consider two conductors embedded in a homogeneous dielectric (Figure 6.1). Conductor M2 carries a total positive charge Q, and M1 carries an equal negative charge. There are no other charges present, and the total charge of the system is zero. We now know that the charge is carried on the surface as a surface charge density and also that the electric ﬁeld is normal to the conductor surface. Each conductor is, moreover, an equipotential surface. Because M2 carries the positive charge, the electric ﬂux is directed from M2 to M1, and M2 is at the more positive potential. In other words, work must be done to carry a positive charge from M1 to M2 . Let us designate the potential difference between M2 and M1 as V0 . We may now deﬁne the capacitance of this two-conductor system as the ratio of the magnitude

143

144

ENGINEERING ELECTROMAGNETICS

Figure 6.1 Two oppositely charged conductors M1 and M2 surrounded by a uniform dielectric. The ratio of the magnitude of the charge on either conductor to the magnitude of the potential difference between them is the capacitance C.

of the total charge on either conductor to the magnitude of the potential difference between conductors, C= Q V0 (1)

In general terms, we determine Q by a surface integral over the positive conductors, and we ﬁnd V0 by carrying a unit positive charge from the negative to the positive surface, C=
S

E · dS
+ −

−

E · dL

(2)

The capacitance is independent of the potential and total charge, for their ratio is constant. If the charge density is increased by a factor of N, Gauss’s law indicates that the electric ﬂux density or electric ﬁeld intensity also increases by N, as does the potential difference. The capacitance is a function only of the physical dimensions of the system of conductors and of the permittivity of the homogeneous dielectric. Capacitance is measured in farads (F), where a farad is deﬁned as one coulomb per volt. Common values of capacitance are apt to be very small fractions of a farad, and consequently more practical units are the microfarad (µF), the nanofarad (nF), and the picofarad (pF).

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6.2 PARALLEL-PLATE CAPACITOR
We can apply the deﬁnition of capacitance to a simple two-conductor system in which the conductors are identical, inﬁnite parallel planes with separation d (Figure 6.2). Choosing the lower conducting plane at z = 0 and the upper one at z = d, a uniform sheet of surface charge ±ρ S on each conductor leads to the uniform ﬁeld [Section 2.5, Eq. (18)] ρS E= az where the permittivity of the homogeneous dielectric is , and D = ρ S az Note that this result could be obtained by applying the boundary condition at a conducting surface (Eq. (18), Chapter 5) at either one of the plate surfaces. Referring to the surfaces and their unit normal vectors in Fig. 6.2, where n = az and nu = −az , we ﬁnd on the lower plane: D·n D · nu z=0 = D · az = ρs ⇒ D = ρs az

On the upper plane, we get the same result z=d = D · (−az ) = −ρs ⇒ D = ρs az

This is a key advantage of the conductor boundary condition, in that we need to apply it only to a single boundary to obtain the total ﬁeld there (arising from all other sources). The potential difference between lower and upper planes is V0 = − lower upper

E · dL = −

0 d

ρS

dz =

ρS

d

Since the total charge on either plane is inﬁnite, the capacitance is inﬁnite. A more practical answer is obtained by considering planes, each of area S, whose linear dimensions are much greater than their separation d. The electric ﬁeld and charge

nu

nl

Figure 6.2 The problem of the parallel-plate capacitor. The capacitance per square meter of surface area is /d.

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ENGINEERING ELECTROMAGNETICS

distribution are then almost uniform at all points not adjacent to the edges, and this latter region contributes only a small percentage of the total capacitance, allowing us to write the familiar result Q = ρS S ρS V0 = d C= Q S = V0 d (3)

More rigorously, we might consider Eq. (3) as the capacitance of a portion of the inﬁnite-plane arrangement having a surface area S. Methods of calculating the effect of the unknown and nonuniform distribution near the edges must wait until we are able to solve more complicated potential problems.
E X A M P L E 6.1

Calculate the capacitance of a parallel-plate capacitor having a mica dielectric, a plate area of 10 in.2 , and a separation of 0.01 in.
Solution. We may ﬁnd that

r

= 6,

d = 0.01 × 0.0254 = 2.54 × 10−4 m and therefore C= 6 × 8.854 × 10−12 × 6.45 × 10−3 = 1.349 nF 2.54 × 10−4

S = 10 × 0.02542 = 6.45 × 10−3 m2

A large plate area is obtained in capacitors of small physical dimensions by stacking smaller plates in 50- or 100-decker sandwiches, or by rolling up foil plates separated by a ﬂexible dielectric. Table C.1 in Appendix C also indicates that materials are available having dielectric constants greater than 1000. Finally, the total energy stored in the capacitor is WE = or
2 W E = 1 C V 0 = 1 Q V0 = 2 2 1 2 1 2

vol

E 2 dv =

1 2

S 0 0

d

2 ρS 2

dz dS =

1 2

2 ρS

Sd =

1 2

2 S ρS d 2 2 d

Q2 C

(4)

which are all familiar expressions. Equation (4) also indicates that the energy stored in a capacitor with a ﬁxed potential difference across it increases as the dielectric constant of the medium increases.

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D6.1. Find the relative permittivity of the dielectric material present in a parallel-plate capacitor if: (a) S = 0.12 m2 , d = 80 µm, V0 = 12 V, and the capacitor contains 1 µJ of energy; (b) the stored energy density is 100 J/m3 , V0 = 200 V, and d = 45 µm; (c) E = 200 kV/m and ρ S = 20 µC/m2 .
Ans. 1.05; 1.14; 11.3

6.3 SEVERAL CAPACIT ANCE EXAMPLES
As a ﬁrst brief example, we choose a coaxial cable or coaxial capacitor of inner radius a, outer radius b, and length L. No great derivational struggle is required, because the potential difference is given as Eq. (11) in Section 4.3, and we ﬁnd the capacitance very simply by dividing this by the total charge ρ L L in the length L. Thus, C= 2π L ln(b/a) (5)

Next we consider a spherical capacitor formed of two concentric spherical conducting shells of radius a and b, b > a. The expression for the electric ﬁeld was obtained previously by Gauss’s law, Q Er = 4π r 2 where the region between the spheres is a dielectric with permittivity . The expression for potential difference was found from this by the line integral [Section 4.3, Eq. (12)]. Thus, Q 1 1 − Vab = 4π a b Here Q represents the total charge on the inner sphere, and the capacitance becomes C= Q 4π = 1 1 Vab − a b (6)

If we allow the outer sphere to become inﬁnitely large, we obtain the capacitance of an isolated spherical conductor, C = 4π a For a diameter of 1 cm, or a sphere about the size of a marble, C = 0.556 pF in free space. (7)

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ENGINEERING ELECTROMAGNETICS

C =

1 d d1 + 2 e1S e 2 S e2 e1

Area, S d2 d1

Conducting plates

d

Figure 6.3 A parallel-plate capacitor containing two dielectrics with the dielectric interface parallel to the conducting plates.

Coating this sphere with a different dielectric layer, for which from r = a to r = r1, Dr = Q 4πr 2 Q Er = 4π 1r 2 = and the potential difference is Va − V∞ = − = Therefore, C= 1
1 a r1

=

1,

extending

(a < r < r1 ) (r1 < r )

Q 4π 0r 2

Q dr − 4π 1r 2 1 1 − a r1

r1 ∞

Q dr 4π 0r 2 1 0 r1

Q 4π

1
1

+

4π 1 1 1 + − a r1 0 r1

(8)

In order to look at the problem of multiple dielectrics a little more thoroughly, let us consider a parallel-plate capacitor of area S and spacing d, with the usual assumption that d is small compared to the linear dimensions of the plates. The capacitance is 1 S/d, using a dielectric of permittivity 1 . Now replace a part of this dielectric by another of permittivity 2 , placing the boundary between the two dielectrics parallel to the plates (Figure 6.3). Some of us may immediately suspect that this combination is effectively two capacitors in series, yielding a total capacitance of C= 1 1 1 + C1 C2

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149

where C1 = 1 S/d1 and C2 = 2 S/d2 . This is the correct result, but we can obtain it using less intuition and a more basic approach. Because the capacitance deﬁnition, C = Q/V , involves a charge and a voltage, we may assume either and then ﬁnd the other in terms of it. The capacitance is not a function of either, but only of the dielectrics and the geometry. Suppose we assume a potential difference V0 between the plates. The electric ﬁeld intensities in the two regions, E 2 and E 1 , are both uniform, and V0 = E 1 d1 + E 2 d2 . At the dielectric interface, E is normal, and our boundary condition, Eq. (35) Chapter 5, tells us that D N 1 = D N 2 , or 1 E 1 = 2 E 2 . This assumes (correctly) that there is no surface charge at the interface. Eliminating E 2 in our V0 relation, we have V0 E1 = d1 + d 2 ( 1 / 2 ) and the surface charge density on the lower plate therefore has the magnitude V0 ρ S1 = D1 = 1 E 1 = d1 d2 +
1 2

Because D1 = D2 , the magnitude of the surface charge is the same on each plate. The capacitance is then Q ρS S 1 1 C= = = = d1 d2 1 1 V0 V0 + + S S C1 C2 1 2 As an alternate (and slightly simpler) solution, we might assume a charge Q on one plate, leading to a charge density Q/S and a value of D that is also Q/S. This is true in both regions, as D N 1 = D N 2 and D is normal. Then E 1 = D/ 1 = Q/( 1 S), E 2 = D/ 2 = Q/( 2 S), and the potential differences across the regions are V1 = E 1 d1 = Qd1 /( 1 S), and V2 = E 2 d2 = Qd2 /( 2 S). The capacitance is C= Q 1 Q = = d1 d2 V V1 + V2 + 1S 2S (9)

How would the method of solution or the answer change if there were a third conducting plane along the interface? We would now expect to ﬁnd surface charge on each side of this conductor, and the magnitudes of these charges should be equal. In other words, we think of the electric lines not as passing directly from one outer plate to the other, but as terminating on one side of this interior plane and then continuing on the other side. The capacitance is unchanged, provided, of course, that the added conductor is of negligible thickness. The addition of a thick conducting plate will increase the capacitance if the separation of the outer plates is kept constant, and this is an example of a more general theorem which states that the replacement of any portion of the dielectric by a conducting body will cause an increase in the capacitance. If the dielectric boundary were placed normal to the two conducting plates and the dielectrics occupied areas of S1 and S2 , then an assumed potential difference V0 would produce ﬁeld strengths E 1 = E 2 = V0 /d. These are tangential ﬁelds at the

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ENGINEERING ELECTROMAGNETICS

interface, and they must be equal. Then we may ﬁnd in succession D1 , D2 , ρ S1 , ρ S2 , and Q, obtaining a capacitance C=
1 S1

+ d

2 S2

= C1 + C2

(10)

as we should expect. At this time we can do very little with a capacitor in which two dielectrics are used in such a way that the interface is not everywhere normal or parallel to the ﬁelds. Certainly we know the boundary conditions at each conductor and at the dielectric interface; however, we do not know the ﬁelds to which to apply the boundary conditions. Such a problem must be put aside until our knowledge of ﬁeld theory has increased and we are willing and able to use more advanced mathematical techniques. D6.2. Determine the capacitance of: (a) a 1-ft length of 35B/U coaxial cable, which has an inner conductor 0.1045 in. in diameter, a polyethylene dielectric ( r = 2.26 from Table C.1), and an outer conductor that has an inner diameter of 0.680 in.; (b) a conducting sphere of radius 2.5 mm, covered with a polyethylene layer 2 mm thick, surrounded by a conducting sphere of radius 4.5 mm; (c) two rectangular conducting plates, 1 cm by 4 cm, with negligible thickness, between which are three sheets of dielectric, each 1 cm by 4 cm, and 0.1 mm thick, having dielectric constants of 1.5, 2.5, and 6.
Ans. 20.5 pF; 1.41 pF; 28.7 pF

6.4 CAPACIT ANCE OF A TWO-WIRE LINE
We next consider the problem of the two-wire line. The conﬁguration consists of two parallel conducting cylinders, each of circular cross section, and we will ﬁnd complete information about the electric ﬁeld intensity, the potential ﬁeld, the surface-chargedensity distribution, and the capacitance. This arrangement is an important type of transmission line, as is the coaxial cable. We begin by investigating the potential ﬁeld of two inﬁnite line charges. Figure 6.4 shows a positive line charge in the x z plane at x = a and a negative line charge at x = −a. The potential of a single line charge with zero reference at a radius of R0 is V = R0 ρL ln 2π R

We now write the expression for the combined potential ﬁeld in terms of the radial distances from the positive and negative lines, R1 and R2 , respectively, V = ρL 2π ln R10 R20 − ln R1 R2 = ρ L R10 R2 ln 2π R20 R1

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151

y

R2 (−a, 0, 0) (a, 0, 0) 2a R1

P(x, y, 0) x

−rL

z

+rL

Figure 6.4 Two parallel infinite line charges carrying opposite charge. The positive line is at x = a, y = 0, and the negative line is at x = −a, y = 0. A general point P(x, y, 0) in the xy plane is radially distant R1 and R2 from the positive and negative lines, respectively. The equipotential surfaces are circular cylinders.

We choose R10 = R20 , thus placing the zero reference at equal distances from each line. This surface is the x = 0 plane. Expressing R1 and R2 in terms of x and y, V = (x + a)2 + y 2 ρ L (x + a)2 + y 2 ρL = ln ln 2π (x − a)2 + y 2 4π (x − a)2 + y 2 (11)

In order to recognize the equipotential surfaces and adequately understand the problem we are going to solve, some algebraic manipulations are necessary. Choosing an equipotential surface V = V1, we deﬁne K 1 as a dimensionless parameter that is a function of the potential V1, K 1 = e4π so that (x + a)2 + y 2 (x − a)2 + y 2 After multiplying and collecting like powers, we obtain K1 + 1 + y2 + a2 = 0 x 2 − 2ax K1 − 1 We next work through a couple of lines of algebra and complete the square, √ 2 2a K 1 K1 + 1 2 + y2 = x −a K1 − 1 K1 − 1 This shows that the V = V1 equipotential surface is independent of z (or is a cylinder) and intersects the x y plane in a circle of radius b, √ 2a K 1 b= K1 − 1 K1 =
V1 /ρ L

(12)

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ENGINEERING ELECTROMAGNETICS

which is centered at x = h, y = 0, where h=a

K1 + 1 K1 − 1 Now let us attack a physical problem by considering a zero-potential conducting plane located at x = 0, and a conducting cylinder of radius b and potential V0 with its axis located a distance h from the plane. We solve the last two equations for a and K 1 in terms of the dimensions b and h, a= h 2 − b2 h+ √ (13)

and K1 =

h 2 − b2 b But the potential of the cylinder is V0 , so Eq. (12) leads to K 1 = e2π Therefore, ρL =
V0 /ρ L

(14)

The solid line in Figure 6.5 shows the cross section of a cylinder of 5 m radius at a potential of 100 V in free space, with its axis 13 m distant from a plane at zero potential. Thus, b = 5, h = 13, V0 = 100, and we rapidly ﬁnd the location of the equivalent line charge from Eq. (13), the value of the potential parameter K 1 from Eq. (14), √ 13 + 12 h + h 2 − b2 = =5 K1 = b 5 the strength of the equivalent line charge from Eq. (15), ρL = a= h 2 − b2 = 132 − 52 = 12 m K 1 = 25

4π V0 (15) lnK 1 Thus, given h, b, and V0 , we may determine a, ρ L , and the parameter K 1 . The capacitance between the cylinder and plane is now available. For a length L in the z direction, we have ρL L 4π L 2π L C= = = √ V0 lnK 1 ln K 1 or 2π L 2π L C= = (16) √ 2 − b2 )/b] cosh−1 (h/b) ln[(h + h

4π V0 4π × 8.854 × 10−12 × 100 = = 3.46 nC/m lnK 1 ln 25 and the capacitance between cylinder and plane from Eq. (16), C= 2π × 8.854 × 10−12 2π = = 34.6 pF/m −1 cosh (h/b) cosh−1 (13/5)

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153

y V=0

Equivalent line charge x

h = 13

b=5

Center, x = 13, y = 0, V = 100

Center, x = 18, y = 0 radius = 13.42 V = 50

Figure 6.5 A numerical example of the capacitance, linear charge density, position of an equivalent line charge, and characteristics of the mid-equipotential surface for a cylindrical conductor of 5 m radius at a potential of 100 V, parallel to and 13 m from a conducting plane at zero potential.

We may also identify the cylinder representing the 50 V equipotential surface by ﬁnding new values for K 1 , h, and b. We ﬁrst use Eq. (12) to obtain K 1 = e4π Then the new radius is
V1 /ρ L

= e4π ×8.854×10

−12

×50/3.46×10−9

= 5.00

√ √ 2a K 1 2 × 12 5 b= = = 13.42 m K1 − 1 5−1 and the corresponding value of h becomes 5+1 K1 + 1 = 12 = 18 m K1 − 1 5−1 This cylinder is shown in color in Figure 6.5. The electric ﬁeld intensity can be found by taking the gradient of the potential ﬁeld, as given by Eq. (11), h=a E = −∇ Thus, E=− and ρL 4π ρ L (x + a)2 + y 2 ln 4π (x − a)2 + y 2

2(x − a)ax + 2ya y 2(x + a)ax + 2ya y − (x + a)2 + y 2 (x − a)2 + y 2 (x − a)ax + ya y ρ L (x + a)ax + ya y − 2π (x + a)2 + y 2 (x − a)2 + y 2

D= E=−

154

ENGINEERING ELECTROMAGNETICS

If we evaluate Dx at x = h − b, y = 0, we may obtain ρ S,max ρ S,max = −Dx,x=h−b,y=0 = For our example, ρ S,max = 3.46 × 10−9 2π

ρL h − b + a h−b−a − 2 2π (h − b + a) (h − b − a)2

Similarly, ρ S,min = Dx,x=h+b,y=0, and ρ S,min = Thus, 3.46 × 10−9 2π

13 − 5 − 12 13 − 5 + 12 = 0.165 nC/m2 − (13 − 5 + 12)2 (13 − 5 − 12)2 13 + 5 + 12 13 + 5 − 12 = 0.073 nC/m2 − 302 62 ρ S,max = 2.25ρ S,min

If we apply Eq. (16) to the case of a conductor for which b ln h + and

h, then

˙ ˙ h 2 − b2 /b = ln[(h + h)/b] = ln(2h/b) C=

2π L (b h) (17) ln(2h/b) The capacitance between two circular conductors separated by a distance 2h is one-half the capacitance given by Eqs. (16) or (17). This last answer is of interest because it gives us an expression for the capacitance of a section of two-wire transmission line, one of the types of transmission lines studied later in Chapter 13. D6.3. A conducting cylinder with a radius of 1 cm and at a potential of 20 V is parallel to a conducting plane which is at zero potential. The plane is 5 cm distant from the cylinder axis. If the conductors are embedded in a perfect dielectric for which r = 4.5, ﬁnd: (a) the capacitance per unit length between cylinder and plane; (b) ρ S,max on the cylinder.
Ans. 109.2 pF/m; 42.6 nC/m2

6.5 USING FIELD SKETCHES TO ESTIMATE CAPACIT ANCE IN TWO-DIMENSIONAL PROBLEMS
In capacitance problems in which the conductor conﬁgurations cannot be easily described using a single coordinate system, other analysis techniques are usually applied. Such methods typically involve a numerical determination of ﬁeld or potential values over a grid within the region of interest. In this section, another approach is described that involves making sketches of ﬁeld lines and equipotential surfaces in a manner that follows a few simple rules. This approach, although lacking the accuracy of more

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elegant methods, allows fairly quick estimates of capacitance while providing a useful visualization of the ﬁeld conﬁguration. The method, requiring only pencil and paper, is capable of yielding good accuracy if used skillfully and patiently. Fair accuracy (5 to 10 percent on a capacitance determination) may be obtained by a beginner who does no more than follow the few rules and hints of the art. The method to be described is applicable only to ﬁelds in which no variation exists in the direction normal to the plane of the sketch. The procedure is based on several facts that we have already demonstrated: 1. A conductor boundary is an equipotential surface. 2. The electric ﬁeld intensity and electric ﬂux density are both perpendicular to the equipotential surfaces. 3. E and D are therefore perpendicular to the conductor boundaries and possess zero tangential values. 4. The lines of electric ﬂux, or streamlines, begin and terminate on charge and hence, in a charge-free, homogeneous dielectric, begin and terminate only on the conductor boundaries. We consider the implications of these statements by drawing the streamlines on a sketch that already shows the equipotential surfaces. In Figure 6.6a, two conductor boundaries are shown, and equipotentials are drawn with a constant potential difference between lines. We should remember that these lines are only the cross sections of the equipotential surfaces, which are cylinders (although not circular). No variation in the direction normal to the surface of the paper is permitted. We arbitrarily choose to begin a streamline, or ﬂux line, at A on the surface of the more positive conductor. It leaves the surface normally and must cross at right angles the undrawn but very real equipotential surfaces between the conductor and the ﬁrst surface shown. The line is continued to the other conductor, obeying the single rule that the intersection with each equipotential must be square. In a similar manner, we may start at B and sketch another streamline ending at B . We need to understand the meaning of this pair of streamlines. The streamline,

Figure 6.6 (a) Sketch of the equipotential surfaces between two conductors. The increment of potential between each of the two adjacent equipotentials is the same. (b) One flux line has been drawn from A to A , and a second from B to B .

156

ENGINEERING ELECTROMAGNETICS

by deﬁnition, is everywhere tangent to the electric ﬁeld intensity or to the electric ﬂux density. Because the streamline is tangent to the electric ﬂux density, the ﬂux density is tangent to the streamline, and no electric ﬂux may cross any streamline. In other words, if there is a charge of 5 µC on the surface between A and B (and extending 1 m into the paper), then 5 µC of ﬂux begins in this region, and all must terminate between A and B . Such a pair of lines is sometimes called a ﬂux tube, because it physically seems to carry ﬂux from one conductor to another without losing any. We next construct a third streamline, and both the mathematical and visual interpretations we may make from the sketch will be greatly simpliﬁed if we draw this line starting from some point C chosen so that the same amount of ﬂux is carried in the tube BC as is contained in AB. How do we choose the position of C? The electric ﬁeld intensity at the midpoint of the line joining A to B may be found approximately by assuming a value for the ﬂux in the tube AB, say , which allows us to express the electric ﬂux density by / L t , where the depth of the tube into the paper is 1 m and L t is the length of the line joining A to B. The magnitude of E is then 1 E= Lt We may also ﬁnd the magnitude of the electric ﬁeld intensity by dividing the potential difference between points A and A1, lying on two adjacent equipotential surfaces, by the distance from A to A1 . If this distance is designated L N and an increment of potential between equipotentials of V is assumed, then V E= LN This value applies most accurately to the point at the middle of the line segment from A to A1 , while the previous value was most accurate at the midpoint of the line segment from A to B. If, however, the equipotentials are close together ( V small) and the two streamlines are close together ( small), the two values found for the electric ﬁeld intensity must be approximately equal, V 1 = (18) Lt LN Throughout our sketch we have assumed a homogeneous medium ( constant), a constant increment of potential between equipotentials ( V constant), and a constant amount of ﬂux per tube ( constant). To satisfy all these conditions, Eq. (18) shows that Lt 1 = constant = LN V (19)

A similar argument might be made at any point in our sketch, and we are therefore led to the conclusion that a constant ratio must be maintained between the distance between streamlines as measured along an equipotential, and the distance between equipotentials as measured along a streamline. It is this ratio that must have the same value at every point, not the individual lengths. Each length must decrease in regions of greater ﬁeld strength, because V is constant.

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Figure 6.7 The remaining of the streamlines have been added to Fig. 6.6b by beginning each new line normally to the conductor and maintaining curvilinear squares throughout the sketch.

The simplest ratio we can use is unity, and the streamline from B to B shown in Figure 6.6b was started at a point for which L t = L N . Because the ratio of these distances is kept at unity, the streamlines and equipotentials divide the ﬁeld-containing region into curvilinear squares, a term implying a planar geometric ﬁgure that differs from a true square in having slightly curved and slightly unequal sides but which approaches a square as its dimensions decrease. Those incremental surface elements in our three coordinate systems which are planar may also be drawn as curvilinear squares. We may now sketch in the remainder of the streamlines by keeping each small box as square as possible. One streamline is begun, an equipotential line is roughed in, another streamline is added, forming a curvilinear square, and the map is gradually extended throughout the desired region. The complete sketch is shown in Figure 6.7. The construction of a useful ﬁeld map is an art; the science merely furnishes the rules. Proﬁciency in any art requires practice. A good problem for beginners is the coaxial cable or coaxial capacitor, since all the equipotentials are circles while the ﬂux lines are straight lines. The next sketch attempted should be two parallel circular conductors, where the equipotentials are again circles but with different centers. Each of these is given as a problem at the end of the chapter. Figure 6.8 shows a completed map for a cable containing a square inner conductor surrounded by a circular conductor. The capacitance is found from C = Q/V0 by replacing Q by NQ Q = NQ , where NQ is the number of ﬂux tubes joining the two conductors, and letting V0 = N V V, where N V is the number of potential increments between conductors, C= and then using Eq. (19), C= NQ NV Lt NQ = LN NV (20) NQ Q NV V

158

ENGINEERING ELECTROMAGNETICS

Figure 6.8 An example of a curvilinear-square field map. The side of the square is two-thirds the radius of the circle. NV = 4 and NQ = 8 × 3.25 × 26, and therefore C = 0 NQ /NV = 57.6 pF/m.

since L t / L N = 1. The determination of the capacitance from a ﬂux plot merely consists of counting squares in two directions, between conductors and around either conductor. From Figure 6.8 we obtain 8 × 3.25 = 57.6 pF/m 4 Ramo, Whinnery, and Van Duzer have an excellent discussion with examples of the construction of ﬁeld maps by curvilinear squares. They offer the following suggestions:1 C=
0

1. Plan on making a number of rough sketches, taking only a minute or so apiece, before starting any plot to be made with care. The use of transparent paper over the basic boundary will speed up this preliminary sketching. 2. Divide the known potential difference between electrodes into an equal number of divisions, say four or eight to begin with. 3. Begin the sketch of equipotentials in the region where the ﬁeld is known best, for example, in some region where it approaches a uniform ﬁeld. Extend the equipotentials according to your best guess throughout the plot. Note that they should tend to hug acute angles of the conducting boundary and be spread out in the vicinity of obtuse angles of the boundary.

By permission from S. Ramo, J. R. Whinnery, and T. Van Duzer, pp. 51–52. See References at the end of this chapter. Curvilinear maps are discussed on pp. 50–52.

1

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159

4. Draw in the orthogonal set of ﬁeld lines. As these are started, they should form curvilinear squares, but, as they are extended, the condition of orthogonality should be kept paramount, even though this will result in some rectangles with ratios other than unity. 5. Look at the regions with poor side ratios and try to see what was wrong with the ﬁrst guess of equipotentials. Correct them and repeat the procedure until reasonable curvilinear squares exist throughout the plot. 6. In regions of low ﬁeld intensity, there will be large ﬁgures, often of ﬁve or six sides. To judge the correctness of the plot in this region, these large units should be subdivided. The subdivisions should be started back away from the region needing subdivision, and each time a ﬂux tube is divided in half, the potential divisions in this region must be divided by the same factor. D6.4. Figure 6.9 shows the cross section of two circular cylinders at potentials of 0 and 60 V. The axes are parallel and the region between the cylinders is airﬁlled. Equipotentials at 20 V and 40 V are also shown. Prepare a curvilinearsquare map on the ﬁgure and use it to establish suitable values for: (a) the capacitance per meter length; (b) E at the left side of the 60 V conductor if its true radius is 2 mm; (c) ρ S at that point.
Ans. 69 pF/m; 60 kV/m; 550 nC/m2

Figure 6.9 See Problem D6.4.

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ENGINEERING ELECTROMAGNETICS

6.6 POISSON’S AND LAPLACE’S EQUATIONS
In preceding sections, we have found capacitance by ﬁrst assuming a known charge distribution on the conductors and then ﬁnding the potential difference in terms of the assumed charge. An alternate approach would be to start with known potentials on each conductor, and then work backward to ﬁnd the charge in terms of the known potential difference. The capacitance in either case is found by the ratio Q/V . The ﬁrst objective in the latter approach is thus to ﬁnd the potential function between conductors, given values of potential on the boundaries, along with possible volume charge densities in the region of interest. The mathematical tools that enable this to happen are Poisson’s and Laplace’s equations, to be explored in the remainder of this chapter. Problems involving one to three dimensions can be solved either analytically or numerically. Laplace’s and Poisson’s equations, when compared to other methods, are probably the most widely useful because many problems in engineering practice involve devices in which applied potential differences are known, and in which constant potentials occur at the boundaries. Obtaining Poisson’s equation is exceedingly simple, for from the point form of Gauss’s law, ∇ · D = ρν the deﬁnition of D, D= E and the gradient relationship, E = −∇V by substitution we have ∇ · D = ∇ · ( E) = −∇ · ( ∇V ) = ρν or ∇ · ∇V = − ρν (24) (23) (22) (21)

for a homogeneous region in which is constant. Equation (24) is Poisson’s equation, but the “double ∇” operation must be interpreted and expanded, at least in rectangular coordinates, before the equation can be useful. In rectangular coordinates, ∇ ·A = ∇V = ∂A y ∂A z ∂A x + + ∂x ∂y ∂z ∂V ∂V ∂V ax + ay + az ∂x ∂y ∂z

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and therefore ∇ · ∇V = = ∂ ∂x ∂V ∂x + ∂ ∂V ∂y ∂y + ∂ ∂V ∂z ∂z (25)

∂2V ∂2V ∂2V + + ∂x2 ∂ y2 ∂z 2

Usually the operation ∇ · ∇ is abbreviated ∇ 2 (and pronounced “del squared”), a good reminder of the second-order partial derivatives appearing in Eq. (5), and we have ∇2V = ∂2V ∂2V ∂2V ρν + + =− ∂x2 ∂ y2 ∂z 2 (26)

in rectangular coordinates. If ρν = 0, indicating zero volume charge density, but allowing point charges, line charge, and surface charge density to exist at singular locations as sources of the ﬁeld, then ∇2V = 0 which is Laplace’s equation. The ∇ 2 operation is called the Laplacian of V. In rectangular coordinates Laplace’s equation is ∇2V = ∂2V ∂2V ∂2V + + =0 2 2 ∂x ∂y ∂z 2 (rectangular) (28) (27)

and the form of ∇ 2 V in cylindrical and spherical coordinates may be obtained by using the expressions for the divergence and gradient already obtained in those coordinate systems. For reference, the Laplacian in cylindrical coordinates is ∇2V = 1 ∂ ∂V ρ ρ ∂ρ ∂ρ + 1 ∂2V ρ 2 ∂φ 2 + ∂2V ∂z 2 (cylindrical) (29)

and in spherical coordinates is ∇2V = ∂V 1 ∂ r2 r 2 ∂r ∂r + 1 ∂ r 2 sin θ ∂θ sin θ ∂V ∂θ + 1 ∂2V r 2 sin2 θ ∂φ 2 (spherical) (30) These equations may be expanded by taking the indicated partial derivatives, but it is usually more helpful to have them in the forms just given; furthermore, it is much easier to expand them later if necessary than it is to put the broken pieces back together again. Laplace’s equation is all-embracing, for, applying as it does wherever volume charge density is zero, it states that every conceivable conﬁguration of electrodes

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or conductors produces a ﬁeld for which ∇ 2 V = 0. All these ﬁelds are different, with different potential values and different spatial rates of change, yet for each of them ∇ 2 V = 0. Because every ﬁeld (if ρν = 0) satisﬁes Laplace’s equation, how can we expect to reverse the procedure and use Laplace’s equation to ﬁnd one speciﬁc ﬁeld in which we happen to have an interest? Obviously, more information is required, and we shall ﬁnd that we must solve Laplace’s equation subject to certain boundary conditions. Every physical problem must contain at least one conducting boundary and usually contains two or more. The potentials on these boundaries are assigned values, perhaps V0 , V1 , . . . , or perhaps numerical values. These deﬁnite equipotential surfaces will provide the boundary conditions for the type of problem to be solved. In other types of problems, the boundary conditions take the form of speciﬁed values of E (alternatively, a surface charge density, ρ S ) on an enclosing surface, or a mixture of known values of V and E. Before using Laplace’s equation or Poisson’s equation in several examples, we must state that if our answer satisﬁes Laplace’s equation and also satisﬁes the boundary conditions, then it is the only possible answer. This is a statement of the Uniqueness Theorem, the proof of which is presented in Appendix D. D6.5. Calculate numerical values for V and ρν at point P in free space if: π 4yz , at P(1, 2, 3); (b) V = 5ρ 2 cos 2φ, at P(ρ = 3, φ = , (a) V = 2 x +1 3 2 cos φ , at P(r = 0.5, θ = 45◦ , φ = 60◦ ). z = 2); (c) V = r2
Ans. 12 V, −106.2 pC/m3 ; −22.5 V, 0; 4 V, 0

6.7 EXAMPLES OF THE SOLUTION OF LAPLACE’S EQUATION
Several methods have been developed for solving Laplace’s equation. The simplest method is that of direct integration. We will use this technique to work several examples involving one-dimensional potential variation in various coordinate systems in this section. The method of direct integration is applicable only to problems that are “onedimensional,” or in which the potential ﬁeld is a function of only one of the three coordinates. Since we are working with only three coordinate systems, it might seem, then, that there are nine problems to be solved, but a little reﬂection will show that a ﬁeld that varies only with x is fundamentally the same as a ﬁeld that varies only with y. Rotating the physical problem a quarter turn is no change. Actually, there are only ﬁve problems to be solved, one in rectangular coordinates, two in cylindrical, and two in spherical. We will solve them all. First, let us assume that V is a function only of x and worry later about which physical problem we are solving when we have a need for boundary conditions. Laplace’s equation reduces to ∂2V =0 ∂x2

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and the partial derivative may be replaced by an ordinary derivative, since V is not a function of y or z, d2V =0 dx2 We integrate twice, obtaining dV =A dx and where A and B are constants of integration. Equation (31) contains two such constants, as we would expect for a second-order differential equation. These constants can be determined only from the boundary conditions. Since the ﬁeld varies only with x and is not a function of y and z, then V is a constant if x is a constant or, in other words, the equipotential surfaces are parallel planes normal to the x axis. The ﬁeld is thus that of a parallel-plate capacitor, and as soon as we specify the potential on any two planes, we may evaluate our constants of integration.
E X A M P L E 6.2

V = Ax + B

(31)

Start with the potential function, Eq. (31), and ﬁnd the capacitance of a parallel-plate capacitor of plate area S, plate separation d, and potential difference V0 between plates.
Solution. Take V = 0 at x = 0 and V = V0 at x = d. Then from Eq. (31),

A= and

V0 d

B=0 V0 x d

V =

(32)

We still need the total charge on either plate before the capacitance can be found. We should remember that when we ﬁrst solved this capacitor problem, the sheet of charge provided our starting point. We did not have to work very hard to ﬁnd the charge, for all the ﬁelds were expressed in terms of it. The work then was spent in ﬁnding potential difference. Now the problem is reversed (and simpliﬁed). The necessary steps are these, after the choice of boundary conditions has been made: 1. 2. 3. 4. 5. Given V, use E = −∇V to ﬁnd E. Use D = E to ﬁnd D. Evaluate D at either capacitor plate, D = D S = D N a N. Recognize that ρ S = D N. Find Q by a surface integration over the capacitor plate, Q =

S

ρ S dS.

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Here we have

V = V0

x d V0 E = − ax d V0 ax D=− d x=0 DS = D

a N = ax

=−

V0 ax d

DN = − Q= and the capacitance is
S

V0 = ρS d − V0 V0 S dS = − d d

C=

|Q| S = V0 d

(33)

We will use this procedure several times in the examples to follow.

E X A M P L E 6.3

Because no new problems are solved by choosing ﬁelds which vary only with y or with z in rectangular coordinates, we pass on to cylindrical coordinates for our next example. Variations with respect to z are again nothing new, and we next assume variation with respect to ρ only. Laplace’s equation becomes ∂V 1 ∂ ρ ρ ∂ρ ∂ρ =0

Noting the ρ in the denominator, we exclude ρ = 0 from our solution and then multiply by ρ and integrate, ρ dV =A dρ

where a total derivative replaces the partial derivative because V varies only with ρ. Next, rearrange, and integrate again, V = A ln ρ + B (34)

The equipotential surfaces are given by ρ = constant and are cylinders, and the problem is that of the coaxial capacitor or coaxial transmission line. We choose a

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potential difference of V0 by letting V = V0 at ρ = a, V = 0 at ρ = b, b > a, and obtain V = V0 from which E= D N (ρ=a) = Q= 1 V0 aρ ρ ln(b/a) V0 a ln(b/a) V0 2πaL a ln(b/a) 2π L ln(b/a) ln(b/ρ) ln(b/a) (35)

C=

(36)

which agrees with our result in Section 6.3 (Eq. (5)).

E X A M P L E 6.4

Now assume that V is a function only of φ in cylindrical coordinates. We might look at the physical problem ﬁrst for a change and see that equipotential surfaces are given by φ = constant. These are radial planes. Boundary conditions might be V = 0 at φ = 0 and V = V0 at φ = α, leading to the physical problem detailed in Figure 6.10.

Figure 6.10 Two infinite radial planes with an interior angle α. An infinitesimal insulating gap exists at ρ = 0. The potential field may be found by applying Laplace’s equation in cylindrical coordinates.

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Laplace’s equation is now 1 ∂2V =0 ρ 2 ∂φ 2 We exclude ρ = 0 and have d2V =0 dφ 2 V = Aφ + B φ α

The solution is The boundary conditions determine A and B, and V = V0 (37)

Taking the gradient of Eq. (37) produces the electric ﬁeld intensity, E=− V0 aφ αρ (38)

and it is interesting to note that E is a function of ρ and not of φ. This does not contradict our original assumptions, which were restrictions only on the potential ﬁeld. Note, however, that the vector ﬁeld E is in the φ direction. A problem involving the capacitance of these two radial planes is included at the end of the chapter.
E X A M P L E 6.5

We now turn to spherical coordinates, dispose immediately of variations with respect to φ only as having just been solved, and treat ﬁrst V = V (r ). The details are left for a problem later, but the ﬁnal potential ﬁeld is given by 1 1 − b V = V0 r 1 1 − a b

(39)

where the boundary conditions are evidently V = 0 at r = b and V = V0 at r = a, b > a. The problem is that of concentric spheres. The capacitance was found previously in Section 6.3 (by a somewhat different method) and is C= 4π 1 1 − a b

(40)

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E X A M P L E 6.6

In spherical coordinates we now restrict the potential function to V = V (θ ), obtaining r2 d 1 sin θ dθ sin θ dV dθ =0

We exclude r = 0 and θ = 0 or π and have dV sin θ =A dθ The second integral is then A dθ V = +B sin θ which is not as obvious as the previous ones. From integral tables (or a good memory) we have θ +B (41) V = A ln tan 2 The equipotential surfaces of Eq. (41) are cones. Figure 6.11 illustrates the case where V = 0 at θ = π/2 and V = V0 at θ = α, α < π/2. We obtain θ 2 V = V0 α ln tan 2 ln tan

(42)

Figure 6.11 For the cone θ = α at V0 and the plane θ = π/2 at V = 0, the potential field is given by V = V0 [ln(tan θ/2)]/[ln(tan α/2)].

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In order to ﬁnd the capacitance between a conducting cone with its vertex separated from a conducting plane by an inﬁnitesimal insulating gap and its axis normal to the plane, we ﬁrst ﬁnd the ﬁeld strength: E = −∇V = −1 ∂V aθ = − r ∂θ V0 α r sin θ ln tan 2 aθ

The surface charge density on the cone is then ρS = producing a total charge Q, Q= − V0 sin α ln tan −2π 0 V0 α ln tan 2
∞ 2π 0

− V0 r sin α ln tan

α 2

α 2
∞ 0

0

r sin α dφ dr r

=

dr

This leads to an inﬁnite value of charge and capacitance, and it becomes necessary to consider a cone of ﬁnite size. Our answer will now be only an approximation because the theoretical equipotential surface is θ = α, a conical surface extending from r = 0 to r = ∞, whereas our physical conical surface extends only from r = 0 to, say, r = r1 . The approximate capacitance is C = ˙ 2π r1 α ln cot 2 (43)

If we desire a more accurate answer, we may make an estimate of the capacitance of the base of the cone to the zero-potential plane and add this amount to our answer. Fringing, or nonuniform, ﬁelds in this region have been neglected and introduce an additional source of error. D6.6. Find |E| at P(3, 1, 2) in rectangular coordinates for the ﬁeld of: (a) two coaxial conducting cylinders, V = 50 V at ρ = 2 m, and V = 20 V at ρ = 3 m; (b) two radial conducting planes, V = 50 V at φ = 10◦ , and V = 20 V at φ = 30◦ .
Ans. 23.4 V/m; 27.2 V/m

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6.8 EXAMPLE OF THE SOLUTION OF POISSON’S EQUATION: THE P-N JUNCTION CAPACITANCE
To select a reasonably simple problem that might illustrate the application of Poisson’s equation, we must assume that the volume charge density is speciﬁed. This is not usually the case, however; in fact, it is often the quantity about which we are seeking further information. The type of problem which we might encounter later would begin with a knowledge only of the boundary values of the potential, the electric ﬁeld intensity, and the current density. From these we would have to apply Poisson’s equation, the continuity equation, and some relationship expressing the forces on the charged particles, such as the Lorentz force equation or the diffusion equation, and solve the whole system of equations simultaneously. Such an ordeal is beyond the scope of this text, and we will therefore assume a reasonably large amount of information. As an example, let us select a pn junction between two halves of a semiconductor bar extending in the x direction. We will assume that the region for x < 0 is doped p type and that the region for x > 0 is n type. The degree of doping is identical on each side of the junction. To review some of the facts about the semiconductor junction, we note that initially there are excess holes to the left of the junction and excess electrons to the right. Each diffuses across the junction until an electric ﬁeld is built up in such a direction that the diffusion current drops to zero. Thus, to prevent more holes from moving to the right, the electric ﬁeld in the neighborhood of the junction must be directed to the left; E x is negative there. This ﬁeld must be produced by a net positive charge to the right of the junction and a net negative charge to the left. Note that the layer of positive charge consists of two parts—the holes which have crossed the junction and the positive donor ions from which the electrons have departed. The negative layer of charge is constituted in the opposite manner by electrons and negative acceptor ions. The type of charge distribution that results is shown in Figure 6.12a, and the negative ﬁeld which it produces is shown in Figure 6.12b. After looking at these two ﬁgures, one might proﬁtably read the previous paragraph again. A charge distribution of this form may be approximated by many different expressions. One of the simpler expressions is x x ρν = 2ρν0 sech tanh (44) a a which has a maximum charge density ρv,max = ρv0 that occurs at x = 0.881a. The maximum charge density ρv0 is related to the acceptor and donor concentrations Na and Nd by noting that all the donor and acceptor ions in this region (the depletion layer) have been stripped of an electron or a hole, and thus ρv0 = eNa = eNd We now solve Poisson’s equation, ∇2V = − ρν

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Figure 6.12 (a) The charge density, (b) the electric field intensity, and (c) the potential are plotted for a pn junction as functions of distance from the center of the junction. The p-type material is on the left, and the n-type is on the right.

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subject to the charge distribution assumed above, d2V x x 2ρv0 sech tanh =− 2 dx a a in this one-dimensional problem in which variations with y and z are not present. We integrate once, dV 2ρv0 a x = sech + C1 dx a and obtain the electric ﬁeld intensity, 2ρv0 a x Ex = − sech − C1 a To evaluate the constant of integration C1 , we note that no net charge density and no ﬁelds can exist far from the junction. Thus, as x → ±∞, E x must approach zero. Therefore C1 = 0, and 2ρv0 a x Ex = − sech (45) a Integrating again, V = 4ρv0 a 2 tan−1 e x/a + C2

Let us arbitrarily select our zero reference of potential at the center of the junction, x = 0, 4ρv0 a 2 π 0= + C2 4 and ﬁnally, π (46) 4 Figure 6.12 shows the charge distribution (a), electric ﬁeld intensity (b), and the potential (c), as given by Eqs. (44), (45), and (46), respectively. The potential is constant once we are a distance of about 4a or 5a from the junction. The total potential difference V0 across the junction is obtained from Eq. (46), V = tan−1 e x/a − V0 = Vx→∞ − Vx→−∞ = 2πρv0 a 2 (47) 4ρv0 a 2

This expression suggests the possibility of determining the total charge on one side of the junction and then using Eq. (47) to ﬁnd a junction capacitance. The total positive charge is ∞ x x Q=S 2ρν0 sech tanh d x = 2ρν0 aS a a 0 where S is the area of the junction cross section. If we make use of Eq. (47) to eliminate the distance parameter a, the charge becomes Q=S 2ρν0 V0 π (48)

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Because the total charge is a function of the potential difference, we have to be careful in deﬁning a capacitance. Thinking in “circuit” terms for a moment, dQ d V0 I = =C dt dt and thus dQ C= d V0 By differentiating Eq. (48), we therefore have the capacitance C= ρν0 S S= 2πV0 2πa (49)

The ﬁrst form of Eq. (49) shows that the capacitance varies inversely as the square root of the voltage. That is, a higher voltage causes a greater separation of the charge layers and a smaller capacitance. The second form is interesting in that it indicates that we may think of the junction as a parallel-plate capacitor with a “plate” separation of 2πa. In view of the dimensions of the region in which the charge is concentrated, this is a logical result. Poisson’s equation enters into any problem involving volume charge density. Besides semiconductor diode and transistor models, we ﬁnd that vacuum tubes, magnetohydrodynamic energy conversion, and ion propulsion require its use in constructing satisfactory theories. D6.7. In the neighborhood of a certain semiconductor junction, the volume charge density is given by ρν = 750 sech 106 πx tanh 106 πx C/m3 . The dielectric constant of the semiconductor material is 10 and the junction area is 2 × 10−7 m2 . Find: (a) V0 ; (b) C; (c) E at the junction.
Ans. 2.70 V; 8.85 pF; 2.70 MV/m

√ D6.8. Given the volume charge density ρν = −2 × 107 0 x C/m3 in free space, let V = 0 at x = 0 and let V = 2 V at x = 2.5 mm. At x = 1 mm, ﬁnd: (a) V ; (b) E x .
Ans. 0.302 V; −555 V/m

REFERENCES
1. Matsch, L. W. Capacitors, Magnetic Circuits, and Transformers. Englewood Cliffs, NJ: Prentice-Hall, 1964. Many of the practical aspects of capacitors are discussed in Chapter 2. 2. Ramo, S., J. R. Whinnery, and T. Van Duzer. Fields and Waves in Communication Electronics. 3rd ed. New York: John Wiley and Sons, 1994. This classic text is primarily directed toward beginning graduate students, but it may be read by anyone familiar with basic electromagnetics concepts. Curvilinear square plotting is described on pp. 50–52. A more advanced discussion of methods of solving Laplace’s equation is given in Chapter 7.

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3. Dekker, A. J. See references for Chapter 5. 4. Hayt, W. H. Jr., and J. E. Kemmerly. Engineering Circuit Analysis. 5th ed. New York: McGraw-Hill, 1993. 5. Collin, R. E., and R. E. Plonsey. Principles and Applications of Electromagnetic Fields. New York: McGraw-Hill, 1961. Provides an excellent treatment of methods of solving Laplace’s and Poisson’s equations. 6. Smythe, W. R. Static and Dynamic Electricity. 3rd ed. Taylor and Francis, 1989. An advanced treatment of potential theory is given in Chapter 4.

CHAPTER 6 PROBLEMS
6.1 Consider a coaxial capacitor having inner radius a, outer radius b, unit length, and ﬁlled with a material with dielectric constant, r . Compare this to a parallel-plate capacitor having plate width w, plate separation d, ﬁlled with the same dielectric, and having unit length. Express the ratio b/a in terms of the ratio d/w, such that the two structures will store the same energy for a given applied voltage. Let S = 100 mm2 , d = 3 mm, and r = 12 for a parallel-plate capacitor. (a) Calculate the capacitance. (b) After connecting a 6-V battery across the capacitor, calculate E, D, Q, and the total stored electrostatic energy. (c) With the source still connected, the dielectric is carefully withdrawn from between the plates. With the dielectric gone, recalculate E, D, Q, and the energy stored in the capacitor. (d) If the charge and energy found in part (c) are less than the values found in part (b) (which you should have discovered), what became of the missing charge and energy? Capacitors tend to be more expensive as their capacitance and maximum voltage Vmax increase. The voltage Vmax is limited by the ﬁeld strength at which the dielectric breaks down, E BD . Which of these dielectrics will give the largest CVmax product for equal plate areas? (a) Air: r = 1, E BD = 3 MV/m. (b) Barium titanate: r = 1200, E BD = 3 MV/m. (c) Silicon dioxide: r = 3.78, E BD = 16 MV/m. (d) Polyethylene: r = 2.26, E BD = 4.7 MV/m. An air-ﬁlled parallel-plate capacitor with plate separation d and plate area A is connected to a battery that applies a voltage V0 between plates. With the battery left connected, the plates are moved apart to a distance of 10d. Determine by what factor each of the following quantities changes: (a) V0 ; (b) C; (c) E; (d) D; (e) Q; ( f ) ρ S ; (g) W E . A parallel-plate capacitor is ﬁlled with a nonuniform dielectric characterized by r = 2 + 2 × 106 x 2 , where x is the distance from one plate in meters. If S = 0.02 m2 and d = 1 mm, ﬁnd C. Repeat Problem 6.4, assuming the battery is disconnected before the plate separation is increased. Let r 1 = 2.5 for 0 < y < 1 mm, r 2 = 4 for 1 < y < 3 mm, and r 3 for 3 < y < 5 mm (region 3). Conducting surfaces are present at y = 0 and

6.2

6.3

6.4

6.5

6.6 6.7

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6.8

6.9

A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the x y plane, centered at the origin. The top plate is located at z = d, with its center on the z axis. Potential V0 is on the top plate; the bottom plate is grounded. Dielectric having radially dependent permittivity ﬁlls the region between plates. The permittivity is given by (ρ) = 0 (1 + ρ 2 /a 2 ). Find (a) E; (b) D; (c) Q; (d) C. Two coaxial conducting cylinders of radius 2 cm and 4 cm have a length of 1 m. The region between the cylinders contains a layer of dielectric from ρ = c to ρ = d with r = 4. Find the capacitance if (a) c = 2 cm, d = 3 cm; (b) d = 4 cm, and the volume of the dielectric is the same as in part (a).

y = 5 mm. Calculate the capacitance per square meter of surface area if (a) region 3 is air; (b) r 3 = r 1 ; (c) r 3 = r 2 ; (d) region 3 is silver.

6.10 A coaxial cable has conductor dimensions of a = 1.0 mm and b = 2.7 mm. The inner conductor is supported by dielectric spacers ( r = 5) in the form of washers with a hole radius of 1 mm and an outer radius of 2.7 mm, and with a thickness of 3.0 mm. The spacers are located every 2 cm down the cable. (a) By what factor do the spacers increase the capacitance per unit length? (b) If 100 V is maintained across the cable, ﬁnd E at all points. 6.11 Two conducting spherical shells have radii a = 3 cm and b = 6 cm. The interior is a perfect dielectric for which r = 8. (a) Find C. (b) A portion of the dielectric is now removed so that r = 1.0, 0 < φ < π/2, and r = 8, π/2 < φ < 2π. Again ﬁnd C.

6.13 With reference to Figure 6.5, let b = 6 m, h = 15 m, and the conductor potential be 250 V. Take = 0 . Find values for K 1 , ρ L , a, and C.

6.12 (a) Determine the capacitance of an isolated conducting sphere of radius a in free space (consider an outer conductor existing at r → ∞). (b) The sphere is to be covered with a dielectric layer of thickness d and dielectric contant r . If r = 3, ﬁnd d in terms of a such that the capacitance is twice that of part (a).

6.14 Two #16 copper conductors (1.29 mm diameter) are parallel with a separation d between axes. Determine d so that the capacitance between wires in air is 30 pF/m.

6.15 A 2-cm-diameter conductor is suspended in air with its axis 5 cm from a conducting plane. Let the potential of the cylinder be 100 V and that of the plane be 0 V. (a) Find the surface charge density on the cylinder at a point nearest the plane. (b) Plane at a point nearest the cylinder; (c) ﬁnd the capacitance per unit length. 6.16 Consider an arrangement of two isolated conducting surfaces of any shape that form a capacitor. Use the deﬁnitions of capacitance (Eq. (2) in this chapter) and resistance (Eq. (14) in Chapter 5) to show that when the region between the conductors is ﬁlled with either conductive material (conductivity σ ) or a perfect dielectric (permittivity ), the resulting

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resistance and capacitance of the structures are related through the simple formula RC = /σ . What basic properties must be true about both the dielectric and the conducting medium for this condition to hold for certain? 6.17 Construct a curvilinear-square map for a coaxial capacitor of 3 cm inner radius and 8 cm outer radius. These dimensions are suitable for the drawing. (a) Use your sketch to calculate the capacitance per meter length, assuming r = 1. (b) Calculate an exact value for the capacitance per unit length.

6.18 Construct a curvilinear-square map of the potential ﬁeld about two parallel circular cylinders, each of 2.5 cm radius, separated by a centerto-center distance of 13 cm. These dimensions are suitable for the actual sketch if symmetry is considered. As a check, compute the capacitance per meter both from your sketch and from the exact formula. Assume r = 1. 6.19 Construct a curvilinear-square map of the potential ﬁeld between two parallel circular cylinders, one of 4 cm radius inside another of 8 cm radius. The two axes are displaced by 2.5 cm. These dimensions are suitable for the drawing. As a check on the accuracy, compute the capacitance per meter from the sketch and from the exact expression: C= cosh
−1

[(a 2

where a and b are the conductor radii and D is the axis separation. 6.20 A solid conducting cylinder of 4 cm radius is centered within a rectangular conducting cylinder with a 12 cm by 20 cm cross section. (a) Make a full-size sketch of one quadrant of this conﬁguration and construct a curvilinear-square map for its interior. (b) Assume = 0 and estimate C per meter length.

2π + b2 − D 2 )/(2ab)]

6.21 The inner conductor of the transmission line shown in Figure 6.13 has a square cross section 2a × 2a, whereas the outer square is 4a × 5a. The axes are displaced as shown. (a) Construct a good-sized drawing of this transmission line, say with a = 2.5 cm, and then prepare a curvilinear-square plot of the electrostatic ﬁeld between the conductors. (b) Use the map to calculate the capacitance per meter length if = 1.6 0 . (c) How would your result to part (b) change if a = 0.6 cm? 6.22 Two conducting plates, each 3 × 6 cm, and three slabs of dielectric, each 1 × 3 × 6 cm, and having dielectric constants of 1, 2, and 3, are assembled into a capacitor with d = 3 cm. Determine the two values of capacitance obtained by the two possible methods of assembling the capacitor.

6.23 A two-wire transmission line consists of two parallel perfectly conducting cylinders, each having a radius of 0.2 mm, separated by a center-to-center distance of 2 mm. The medium surrounding the wires has r = 3 and σ = 1.5 mS/m. A 100-V battery is connected between the wires. (a) Calculate the magnitude of the charge per meter length on each wire. (b) Using the result of Problem 6.16, ﬁnd the battery current.

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Figure 6.13 See Problem 6.21.

6.24 A potential ﬁeld in free space is given in spherical coordinates as V (r ) = [ρ0 /(6 0 )] [3a 2 − r 2 ] (r ≤ a) (a 3 ρ0 )/(3 0r ) (r ≥ a)

where ρ0 and a are constants. (a) Use Poisson’s equation to ﬁnd the volume charge density everywhere. (b) Find the total charge present. 6.25 Let V = 2x y 2 z 3 and = 0 . Given point P(1, 2, −1), ﬁnd. (a) V at P; (b) E at P; (c) ρν at P; (d) the equation of the equipotential surface passing through P; (e) the equation of the streamline passing through P. ( f ) Does V satisfy Laplace’s equation? 6.26 Given the spherically symmetric potential ﬁeld in free space, V = V0 e−r/a , ﬁnd. (a) ρν at r = a; (b) the electric ﬁeld at r = a; (c) the total charge.

6.27 Let V (x, y) = 4e2x + f (x) − 3y 2 in a region of free space where ρν = 0. It is known that both E x and V are zero at the origin. Find f (x) and V (x, y).

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6.28 Show that in a homogeneous medium of conductivity σ , the potential ﬁeld V satisﬁes Laplace’s equation if any volume charge density present does not vary with time. 6.29 Given the potential ﬁeld V = (Aρ 4 + Bρ −4 ) sin 4φ: (a) Show that ∇ 2 V = 0. (b) Select A and B so that V = 100 V and |E| = 500 V/m at P(ρ = 1, φ = 22.5◦ , z = 2).

6.30 A parallel-plate capacitor has plates located at z = 0 and z = d. The region between plates is ﬁlled with a material that contains volume charge of uniform density ρ0 C/m3 and has permittivity . Both plates are held at ground potential. (a) Determine the potential ﬁeld between plates. (b) Determine the electric ﬁeld intensity E between plates. (c) Repeat parts (a) and (b) for the case of the plate at z = d raised to potential V0 , with the z = 0 plate grounded. 6.31 Let V = (cos 2φ)/ρ in free space. (a) Find the volume charge density at point A(0.5, 60◦ , 1). (b) Find the surface charge density on a conductor surface passing through the point B(2, 30◦ , 1).

6.32 A uniform volume charge has constant density ρν = ρ0 C/m3 and ﬁlls the region r < a, in which permittivity is assumed. A conducting spherical shell is located at r = a and is held at ground potential. Find (a) the potential everywhere; (b) the electric ﬁeld intensity, E, everywhere. 6.33 The functions V1 (ρ, φ, z) and V2 (ρ, φ, z) both satisfy Laplace’s equation in the region a < ρ < b, 0 ≤ φ < 2π, −L < z < L; each is zero on the surfaces ρ = b for −L < z < L; z = −L for a < ρ < b; and z = L for a < ρ < b; and each is 100 V on the surface ρ = a for −L < z < L . (a) In the region speciﬁed, is Laplace’s equation satisﬁed by the functions V1 + V2 , V1 − V2 , V1 + 3, and V1 V2 ? (b) On the boundary surfaces speciﬁed, are the potential values given in this problem obtained from the functions V1 + V2 , V1 − V2 , V1 + 3, and V1 V2 ? (c) Are the functions V1 + V2 , V1 − V2 , V1 + 3, and V1 V2 identical with V1 ? 6.34 Consider the parallel-plate capacitor of Problem 6.30, but this time the charged dielectric exists only between z = 0 and z = b, where b < d. Free space ﬁlls the region b < z < d. Both plates are at ground potential. By solving Laplace’s and Poisson’s equations, ﬁnd (a) V (z) for 0 < z < d; (b) the electric ﬁeld intensity for 0 < z < d. No surface charge exists at z = b, so both V and D are continuous there.

6.35 The conducting planes 2x + 3y = 12 and 2x + 3y = 18 are at potentials of 100 V and 0, respectively. Let = 0 and ﬁnd (a) V at P(5, 2, 6); (b) E at P.

6.36 The derivation of Laplace’s and Poisson’s equations assumed constant permittivity, but there are cases of spatially varying permittivity in which the equations will still apply. Consider the vector identity, ∇ · (ψG) = G · ∇ψ + ψ∇ · G, where ψ and G are scalar and vector functions, respectively.

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Figure 6.14 See Problem 6.39.

Determine a general rule on the allowed directions in which with respect to the local electric ﬁeld.

may vary

6.37 Coaxial conducting cylinders are located at ρ = 0.5 cm and ρ = 1.2 cm. The region between the cylinders is ﬁlled with a homogeneous perfect dielectric. If the inner cylinder is at 100 V and the outer at 0 V, ﬁnd (a) the location of the 20 V equipotential surface; (b) E ρ max ; (c) r if the charge per meter length on the inner cylinder is 20 nC/m. 6.38 Repeat Problem 6.37, but with the dielectric only partially ﬁlling the volume, within 0 < φ < π, and with free space in the remaining volume. 6.39 The two conducting planes illustrated in Figure 6.14 are deﬁned by 0.001 < ρ < 0.120 m, 0 < z < 0.1 m, φ = 0.179 and 0.188 rad. The medium surrounding the planes is air. For Region 1, 0.179 < φ < 0.188; neglect fringing and ﬁnd (a) V (φ); (b) E(ρ); (c) D(ρ); (d) ρs on the upper surface of the lower plane; (e) Q on the upper surface of the lower plane. ( f ) Repeat parts (a) through (c) for Region 2 by letting the location of the upper plane be φ = .188 − 2π , and then ﬁnd ρs and Q on the lower surface of the lower plane. (g) Find the total charge on the lower plane and the capacitance between the planes. 6.40 A parallel-plate capacitor is made using two circular plates of radius a, with the bottom plate on the x y plane, centered at the origin. The top plate is located at z = d, with its center on the z axis. Potential V0 is on the top plate; the bottom plate is grounded. Dielectric having radially dependent permittivity ﬁlls the region between plates. The permittivity is given by (ρ) = 0 (1 + ρ 2 /a 2 ). Find (a)V (z); (b) E; (c) Q; (d) C. This is a reprise of Problem 6.8, but it starts with Laplace’s equation. 6.41 Concentric conducting spheres are located at r = 5 mm and r = 20 mm. The region between the spheres is ﬁlled with a perfect dielectric. If the inner sphere is at 100 V and the outer sphere is at 0 V (a) Find the

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location of the 20 V equipotential surface. (b) Find Er,max . (c) Find the surface charge density on the inner sphere is 1.0 µC/m2 .

r

if

6.42 The hemisphere 0 < r < a, 0 < θ < π/2, is composed of homogeneous conducting material of conductivity σ . The ﬂat side of the hemisphere rests on a perfectly conducting plane. Now, the material within the conical region 0 < θ < α, 0 < r < a is drilled out and replaced with material that is perfectly conducting. An air gap is maintained between the r = 0 tip of this new material and the plane. What resistance is measured between the two perfect conductors? Neglect fringing ﬁelds. 6.43 Two coaxial conducting cones have their vertices at the origin and the z axis as their axis. Cone A has the point A(1, 0, 2) on its surface, while cone B has the point B(0, 3, 2) on its surface. Let V A = 100 V and VB = 20 V. Find (a) α for each cone; (b) V at P(1, 1, 1). 6.44 A potential ﬁeld in free space is given as V = 100 ln tan(θ/2) + 50 V. (a) Find the maximum value of |Eθ | on the surface θ = 40◦ for 0.1 < r < 0.8 m, 60◦ < φ < 90◦ . (b) Describe the surface V = 80 V.

6.45 In free space, let ρν = 200 0 /r 2.4 . (a) Use Poisson’s equation to ﬁnd V (r ) if it is assumed that r 2 Er → 0 when r → 0, and also that V → 0 as r → ∞. (b) Now ﬁnd V (r ) by using Gauss’s law and a line integral. 6.46 By appropriate solution of Laplace’s and Poisson’s equations, determine the absolute potential at the center of a sphere of radius a, containing uniform volume charge of density ρ0 . Assume permittivity 0 everywhere. Hint: What must be true about the potential and the electric ﬁeld at r = 0 and at r = a?

7

C H A P T E R

The Steady Magnetic Field t this point, the concept of a ﬁeld should be a familiar one. Since we ﬁrst accepted the experimental law of forces existing between two point charges and deﬁned electric ﬁeld intensity as the force per unit charge on a test charge in the presence of a second charge, we have discussed numerous ﬁelds. These ﬁelds possess no real physical basis, for physical measurements must always be in terms of the forces on the charges in the detection equipment. Those charges that are the source cause measurable forces to be exerted on other charges, which we may think of as detector charges. The fact that we attribute a ﬁeld to the source charges and then determine the effect of this ﬁeld on the detector charges amounts merely to a division of the basic problem into two parts for convenience. We will begin our study of the magnetic ﬁeld with a deﬁnition of the magnetic ﬁeld itself and show how it arises from a current distribution. The effect of this ﬁeld on other currents, or the second half of the physical problem, will be discussed in Chapter 8. As we did with the electric ﬁeld, we conﬁne our initial discussion to freespace conditions, and the effect of material media will also be saved for discussion in Chapter 8. The relation of the steady magnetic ﬁeld to its source is more complicated than is the relation of the electrostatic ﬁeld to its source. We will ﬁnd it necessary to accept several laws temporarily on faith alone. The proof of the laws does exist and is available on the Web site for the disbelievers or the more advanced student. ■

A

7.1 BIOT-SAVART LAW
The source of the steady magnetic ﬁeld may be a permanent magnet, an electric ﬁeld changing linearly with time, or a direct current. We will largely ignore the permanent magnet and save the time-varying electric ﬁeld for a later discussion. Our present study will concern the magnetic ﬁeld produced by a differential dc element in free space. We may think of this differential current element as a vanishingly small section of a current-carrying ﬁlamentary conductor, where a ﬁlamentary conductor is the limiting
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181

Figure 7.1 The law of Biot-Savart expresses the magnetic field intensity dH2 produced by a differential current element I 1 dL1 . The direction of dH2 is into the page.

case of a cylindrical conductor of circular cross section as the radius approaches zero. We assume a current I ﬂowing in a differential vector length of the ﬁlament dL. The law of Biot-Savart1 then states that at any point P the magnitude of the magnetic ﬁeld intensity produced by the differential element is proportional to the product of the current, the magnitude of the differential length, and the sine of the angle lying between the ﬁlament and a line connecting the ﬁlament to the point P at which the ﬁeld is desired; also, the magnitude of the magnetic ﬁeld intensity is inversely proportional to the square of the distance from the differential element to the point P. The direction of the magnetic ﬁeld intensity is normal to the plane containing the differential ﬁlament and the line drawn from the ﬁlament to the point P. Of the two possible normals, that one to be chosen is the one which is in the direction of progress of a right-handed screw turned from dL through the smaller angle to the line from the ﬁlament to P. Using rationalized mks units, the constant of proportionality is 1/4π . The Biot-Savart law, just described in some 150 words, may be written concisely using vector notation as I dL × R I dL × a R = (1) dH = 2 4πR 4πR 3 The units of the magnetic fiel intensity H are evidently amperes per meter (A/m). The geometry is illustrated in Figure 7.1. Subscripts may be used to indicate the point to which each of the quantities in (1) refers. If we locate the current element at point 1 and describe the point P at which the ﬁeld is to be determined as point 2, then dH2 = I1 dL1 × a R12 2 4πR12 (2)

Biot and Savart were colleagues of Amp` re, and all three were professors of physics at the Coll` ge de e e France at one time or another. The Biot-Savart law was proposed in 1820.

1

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The law of Biot-Savart is sometimes called Amp` re’s law for the current element, e but we will retain the former name because of possible confusion with Amp` re’s e circuital law, to be discussed later. In some aspects, the Biot-Savart law is reminiscent of Coulomb’s law when that law is written for a differential element of charge, dE2 = dQ 1 a R12 2 4π 0 R12

Both show an inverse-square-law dependence on distance, and both show a linear relationship between source and ﬁeld. The chief difference appears in the direction of the ﬁeld. It is impossible to check experimentally the law of Biot-Savart as expressed by (1) or (2) because the differential current element cannot be isolated. We have restricted our attention to direct currents only, so the charge density is not a function of time. The continuity equation in Section 5.2, Eq. (5), ∇ ·J = − therefore shows that ∇ ·J = 0 or upon applying the divergence theorem, s ∂ρν ∂t

J · dS = 0

The total current crossing any closed surface is zero, and this condition may be satisﬁed only by assuming a current ﬂow around a closed path. It is this current ﬂowing in a closed circuit that must be our experimental source, not the differential element. It follows that only the integral form of the Biot-Savart law can be veriﬁed experimentally, H= I dL × a R 4πR 2 (3)

Equation (1) or (2), of course, leads directly to the integral form (3), but other differential expressions also yield the same integral formulation. Any term may be added to (1) whose integral around a closed path is zero. That is, any conservative ﬁeld could be added to (1). The gradient of any scalar ﬁeld always yields a conservative ﬁeld, and we could therefore add a term ∇G to (1), where G is a general scalar ﬁeld, without changing (3) in the slightest. This qualiﬁcation on (1) or (2) is mentioned to show that if we later ask some foolish questions, not subject to any experimental check, concerning the force exerted by one differential current element on another, we should expect foolish answers. The Biot-Savart law may also be expressed in terms of distributed sources, such as current density J and surface current density K. Surface current ﬂows in a sheet of vanishingly small thickness, and the current density J, measured in amperes per square

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Figure 7.2 The total current I within a transverse width b, in which there is a uniform surface current density K, is K b.

meter, is therefore inﬁnite. Surface current density, however, is measured in amperes per meter width and designated by K. If the surface current density is uniform, the total current I in any width b is I = Kb where we assume that the width b is measured perpendicularly to the direction in which the current is ﬂowing. The geometry is illustrated by Figure 7.2. For a nonuniform surface current density, integration is necessary: I = KdN (4)

where dN is a differential element of the path across which the current is ﬂowing. Thus the differential current element I dL, where dL is in the direction of the current, may be expressed in terms of surface current density K or current density J, I dL = K d S = J dν and alternate forms of the Biot-Savart law obtained, H= and H= J × a R dν 4πR 2 (7) K × aR d S 4πR 2 (6) (5)

s

vol

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Figure 7.3 An infinitely long straight filament carrying a direct current I. The field at point 2 is H = (I/2πρ)aφ .

so that

We illustrate the application of the Biot-Savart law by considering an inﬁnitely long straight ﬁlament. We apply (2) ﬁrst and then integrate. This, of course, is the same as using the integral form (3) in the ﬁrst place.2 Referring to Figure 7.3, we should recognize the symmetry of this ﬁeld. No variation with z or with φ can exist. Point 2, at which we will determine the ﬁeld, is therefore chosen in the z = 0 plane. The ﬁeld point r is therefore r = ρaρ . The source point r is given by r = z az , and therefore R12 = r − r = ρaρ − z az a R12 = ρaρ − z az ρ2 + z 2

We take dL = dz az and (2) becomes I dz az × (ρaρ − z az ) dH2 = 4π (ρ 2 + z 2 )3/2 Because the current is directed toward increasing values of z , the limits are −∞ and ∞ on the integral, and we have ∞ I dz az × (ρaρ − z az ) H2 = 4π (ρ 2 + z 2 )3/2 −∞ = I 4π
∞ −∞

(ρ 2

ρdz aφ + z 2 )3/2

2

The closed path for the current may be considered to include a return ﬁlament parallel to the ﬁrst ﬁlament and inﬁnitely far removed. An outer coaxial conductor of inﬁnite radius is another theoretical possibility. Practically, the problem is an impossible one, but we should realize that our answer will be quite accurate near a very long, straight wire having a distant return path for the current.

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Figure 7.4 The streamlines of the magnetic field intensity about an infinitely long straight filament carrying a direct current I. The direction of I is into the page.

At this point the unit vector aφ under the integral sign should be investigated, for it is not always a constant, as are the unit vectors of the rectangular coordinate system. A vector is constant when its magnitude and direction are both constant. The unit vector certainly has constant magnitude, but its direction may change. Here aφ changes with the coordinate φ but not with ρ or z. Fortunately, the integration here is with respect to z , and aφ is a constant and may be removed from under the integral sign, H2 = Iρaφ 4π
∞ −∞

(ρ 2

dz + z 2 )3/2
∞ 2

Iρaφ z = 2 ρ2 + z 4π ρ and H2 = I aφ 2πρ

−∞

(8)

The magnitude of the ﬁeld is not a function of φ or z, and it varies inversely with the distance from the ﬁlament. The direction of the magnetic-ﬁeld-intensity vector is circumferential. The streamlines are therefore circles about the ﬁlament, and the ﬁeld may be mapped in cross section as in Figure 7.4. The separation of the streamlines is proportional to the radius, or inversely proportional to the magnitude of H. To be speciﬁc, the streamlines have been drawn with curvilinear squares in mind. As yet, we have no name for the family of lines3 that are perpendicular to these circular streamlines, but the spacing of the streamlines has

3

If you can’t wait, see Section 7.6.

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been adjusted so that the addition of this second set of lines will produce an array of curvilinear squares. A comparison of Figure 7.4 with the map of the electric ﬁeld about an inﬁnite line charge shows that the streamlines of the magnetic ﬁeld correspond exactly to the equipotentials of the electric ﬁeld, and the unnamed (and undrawn) perpendicular family of lines in the magnetic ﬁeld corresponds to the streamlines of the electric ﬁeld. This correspondence is not an accident, but there are several other concepts which must be mastered before the analogy between electric and magnetic ﬁelds can be explored more thoroughly. Using the Biot-Savart law to ﬁnd H is in many respects similar to the use of Coulomb’s law to ﬁnd E. Each requires the determination of a moderately complicated integrand containing vector quantities, followed by an integration. When we were concerned with Coulomb’s law we solved a number of examples, including the ﬁelds of the point charge, line charge, and sheet of charge. The law of Biot-Savart can be used to solve analogous problems in magnetic ﬁelds, and some of these problems appear as exercises at the end of the chapter rather than as examples here. One useful result is the ﬁeld of the ﬁnite-length current element, shown in Figure 7.5. It turns out (see Problem 7.8 at the end of the chapter) that H is most easily expressed in terms of the angles α1 and α2 , as identiﬁed in the ﬁgure. The result is I H= (sin α2 − sin α1 )aφ (9) 4πρ If one or both ends are below point 2, then α1 is or both α1 and α2 are negative.

Figure 7.5 The magnetic field intensity caused by a finite-length current filament on the z axis is (I/4πρ)(sin α2 − sin α1 )aφ .

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Equation (9) may be used to ﬁnd the magnetic ﬁeld intensity caused by current ﬁlaments arranged as a sequence of straight-line segments.
E X A M P L E 7.1

As a numerical example illustrating the use of (9), we determine H at P2 (0.4, 0.3, 0) in the ﬁeld of an 8. A ﬁlamentary current is directed inward from inﬁnity to the origin on the positive x axis, and then outward to inﬁnity along the y axis. This arrangement is shown in Figure 7.6. two angles, α1x = −90◦ and α2x = tan−1 (0.4/0.3) = 53.1◦ . The radial distance ρ is measured from the x axis, and we have ρx = 0.3. Thus, this contribution to H2 is H2(x) = 8 2 12 (sin 53.1◦ + 1)aφ = (1.8)aφ = aφ 4π(0.3) 0.3π π
Solution. We ﬁrst consider the semi-inﬁnite current on the x axis, identifying the

The unit vector aφ must also be referred to the x axis. We see that it becomes −az . Therefore, H2(x) = − 12 az A/m π

For the current on the y axis, we have α1y = − tan−1 (0.3/0.4) = −36.9◦ , α2y = 90◦ , and ρ y = 0.4. It follows that H2(y) = 8 8 (1 + sin 36.9◦ )(−az ) = − az A/m 4π(0.4) π

Figure 7.6 The individual fields of two semi-infinite current segments are found by (9) and added to obtain H2 at P2 .

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Adding these results, we have H2 = H2(x) + H2(y) = − 20 az = −6.37az A/m π

D7.1. Given the following values for P1 , P2 , and I1 L 1 , calculate H2 : (a) P1 (0, 0, 2), P2 (4, 2, 0), 2π az µA·m; (b) P1 (0, 2, 0), P2 (4, 2, 3), 2π az µA·m; (c) P1 (1, 2, 3), P2 (−3, −1, 2), 2π (−ax + a y + 2az )µA·m.
Ans. −8.51ax + 17.01a y nA/m; 16a y nA/m; 18.9ax − 33.9a y + 26.4az nA/m

D7.2. A current ﬁlament carrying 15 A in the az direction lies along the entire √ z axis. Find H in rectangular coordinates at: (a) PA ( 20, 0, 4); (b) PB (2, −4, 4).
Ans. 0.534a y A/m; 0.477ax + 0.239a y A/m

` 7.2 AMP ERE’S CIRCUITAL LAW
After solving a number of simple electrostatic problems with Coulomb’s law, we found that the same problems could be solved much more easily by using Gauss’s law whenever a high degree of symmetry was present. Again, an analogous procedure exists in magnetic ﬁelds. Here, the law that helps us solve problems more easily is known as Amp` re’s circuital4 law, sometimes called Amp` re’s work law. This law e e may be derived from the Biot-Savart law (see Section 7.7). Amp` re’s circuital law states that the line integral of H about any closed path is e exactly equal to the direct current enclosed by that path, H · dL = I (10)

We deﬁne positive current as ﬂowing in the direction of advance of a right-handed screw turned in the direction in which the closed path is traversed. Referring to Figure 7.7, which shows a circular wire carrying a direct current I, the line integral of H about the closed paths lettered a and b results in an answer of I; the integral about the closed path c which passes through the conductor gives an answer less than I and is exactly that portion of the total current that is enclosed by the path c. Although paths a and b give the same answer, the integrands are, of course, different. The line integral directs us to multiply the component of H in the direction of the path by a small increment of path length at one point of the path, move along the path to the next incremental length, and repeat the process, continuing until the path is completely traversed. Because H will generally vary from point to point, and because paths a and b are not alike, the contributions to the integral made by, say,

4

The preferred pronunciation puts the accent on “circ-.”

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Figure 7.7 A conductor has a total current I. The line integral of H about the closed paths a and b is equal to I, and the integral around path c is less than I, since the entire current is not enclosed by the path.

each micrometer of path length are quite different. Only the ﬁnal answers are the same. We should also consider exactly what is meant by the expression “current enclosed by the path.” Suppose we solder a circuit together after passing the conductor once through a rubber band, which we use to represent the closed path. Some strange and formidable paths can be constructed by twisting and knotting the rubber band, but if neither the rubber band nor the conducting circuit is broken, the current enclosed by the path is that carried by the conductor. Now replace the rubber band by a circular ring of spring steel across which is stretched a rubber sheet. The steel loop forms the closed path, and the current-carrying conductor must pierce the rubber sheet if the current is to be enclosed by the path. Again, we may twist the steel loop, and we may also deform the rubber sheet by pushing our ﬁst into it or folding it in any way we wish. A single current-carrying conductor still pierces the sheet once, and this is the true measure of the current enclosed by the path. If we should thread the conductor once through the sheet from front to back and once from back to front, the total current enclosed by the path is the algebraic sum, which is zero. In more general language, given a closed path, we recognize this path as the perimeter of an inﬁnite number of surfaces (not closed surfaces). Any current-carrying conductor enclosed by the path must pass through every one of these surfaces once. Certainly some of the surfaces may be chosen in such a way that the conductor pierces them twice in one direction and once in the other direction, but the algebraic total current is still the same. We will ﬁnd that the nature of the closed path is usually extremely simple and can be drawn on a plane. The simplest surface is, then, that portion of the plane enclosed by the path. We need merely ﬁnd the total current passing through this region of the plane. The application of Gauss’s law involves ﬁnding the total charge enclosed by a closed surface; the application of Amp` re’s circuital law involves ﬁnding the total e current enclosed by a closed path.

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Let us again ﬁnd the magnetic ﬁeld intensity produced by an inﬁnitely long ﬁlament carrying a current I . The ﬁlament lies on the z axis in free space (as in Figure 7.3), and the current ﬂows in the direction given by az . Symmetry inspection comes ﬁrst, showing that there is no variation with z or φ. Next we determine which components of H are present by using the Biot-Savart law. Without speciﬁcally using the cross product, we may say that the direction of dH is perpendicular to the plane conaining dL and R and therefore is in the direction of aφ . Hence the only component of H is Hφ , and it is a function only of ρ. We therefore choose a path, to any section of which H is either perpendicular or tangential, and along which H is constant. The ﬁrst requirement (perpendicularity or tangency) allows us to replace the dot product of Amp` re’s circuital law with the e product of the scalar magnitudes, except along that portion of the path where H is normal to the path and the dot product is zero; the second requirement (constancy) then permits us to remove the magnetic ﬁeld intensity from the integral sign. The integration required is usually trivial and consists of ﬁnding the length of that portion of the path to which H is parallel. In our example, the path must be a circle of radius ρ, and Amp` re’s circuital law e becomes H · dL = or Hφ = I 2πρ
2π 0

Hφ ρdφ = Hφ ρ

2π 0

dφ = Hφ 2πρ = I

as before. As a second example of the application of Amp` re’s circuital law, consider an e inﬁnitely long coaxial transmission line carrying a uniformly distributed total current I in the center conductor and −I in the outer conductor. The line is shown in Figure 7.8a. Symmetry shows that H is not a function of φ or z. In order to determine the components present, we may use the results of the previous example by considering the solid conductors as being composed of a large number of ﬁlaments. No ﬁlament has a z component of H. Furthermore, the Hρ component at φ = 0◦, produced by one ﬁlament located at ρ = ρ1 , φ = φ1 , is canceled by the Hρ component produced by a symmetrically located ﬁlament at ρ = ρ1 , φ = −φ1 . This symmetry is illustrated by Figure 7.8b. Again we ﬁnd only an Hφ component which varies with ρ. A circular path of radius ρ, where ρ is larger than the radius of the inner conductor but less than the inner radius of the outer conductor, then leads immediately to Hφ = I 2πρ (a < ρ < b)

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191

Figure 7.8 (a) Cross section of a coaxial cable carrying a uniformly distributed current I in the inner conductor and −I in the outer conductor. The ` magnetic field at any point is most easily determined by applying Ampere’s circuital law about a circular path. (b) Current filaments at ρ = ρ1 , φ = ±φ1 , produces Hρ components which cancel. For the total field, H = Hφ aφ .

If we choose ρ smaller than the radius of the inner conductor, the current enclosed is Iencl = I and 2πρHφ = I or Iρ (ρ < a) 2πa 2 If the radius ρ is larger than the outer radius of the outer conductor, no current is enclosed and Hφ = Hφ = 0 (ρ > c) Finally, if the path lies within the outer conductor, we have 2πρHφ = I − I Hφ = ρ 2 − b2 c2 − b2 (b < ρ < c) ρ2 a2 ρ2 a2

The magnetic-ﬁeld-strength variation with radius is shown in Figure 7.9 for a coaxial cable in which b = 3a, c = 4a. It should be noted that the magnetic ﬁeld intensity H is continuous at all the conductor boundaries. In other words, a slight increase in the radius of the closed path does not result in the enclosure of a tremendously different current. The value of Hφ shows no sudden jumps.

I c2 − ρ 2 2πρ c2 − b2

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Figure 7.9 The magnetic field intensity as a function of radius in an infinitely long coaxial transmission line with the dimensions shown.

The external ﬁeld is zero. This, we see, results from equal positive and negative currents enclosed by the path. Each produces an external ﬁeld of magnitude I /2πρ, but complete cancellation occurs. This is another example of “shielding”; such a coaxial cable carrying large currents would, in principle, not produce any noticeable effect in an adjacent circuit. As a ﬁnal example, let us consider a sheet of current ﬂowing in the positive y direction and located in the z = 0 plane. We may think of the return current as equally divided between two distant sheets on either side of the sheet we are considering. A sheet of uniform surface current density K = K y a y is shown in Figure 7.10. H cannot vary with x or y. If the sheet is subdivided into a number of ﬁlaments, it is evident that no ﬁlament can produce an Hy component. Moreover, the Biot-Savart law shows that the contributions to Hz produced by a symmetrically located pair of ﬁlaments cancel. Thus, Hz is zero also; only an Hx component is present. We therefore choose the path 1-1 -2 -2-1 composed of straight-line segments that are either parallel or

Figure 7.10 A uniform sheet of surface current K = K y a y in the z = 0 plane. H may be found by applying ` Ampere’s circuital law about the paths 1-1 -2 -2-1 and 3-3 -2 -2-3.

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perpendicular to Hx . Amp` re’s circuital law gives e Hx1 L + Hx2 (−L) = K y L or Hx1 − Hx2 = K y If the path 3-3 -2 -2-3 is now chosen, the same current is enclosed, and Hx3 − Hx2 = K y and therefore Hx3 = Hx1

It follows that Hx is the same for all positive z. Similarly, Hx is the same for all negative z. Because of the symmetry, then, the magnetic ﬁeld intensity on one side of the current sheet is the negative of that on the other. Above the sheet, Hx = 1 K y 2 while below it Hx = − 1 K y 2 (z < 0) (z > 0)

Letting a N be a unit vector normal (outward) to the current sheet, the result may be written in a form correct for all z as H = 1 K × aN 2 (11)

If a second sheet of current ﬂowing in the opposite direction, K = −K y a y , is placed at z = h, (11) shows that the ﬁeld in the region between the current sheets is H = K × aN and is zero elsewhere, H = 0 (z < 0, z > h) (13) (0 < z < h) (12)

The most difﬁcult part of the application of Amp` re’s circuital law is the detere mination of the components of the ﬁeld that are present. The surest method is the logical application of the Biot-Savart law and a knowledge of the magnetic ﬁelds of simple form. Problem 7.13 at the end of this chapter outlines the steps involved in applying Amp` re’s circuital law to an inﬁnitely long solenoid of radius a and uniform current e density K a aφ , as shown in Figure 7.11a. For reference, the result is H = K a az H=0 (ρ < a) (ρ > a) (14a) (14b)

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Figure 7.11 (a) An ideal solenoid of infinite length with a circular current sheet K = K a aφ . (b) An N-turn solenoid of finite length d.

If the solenoid has a ﬁnite length d and consists of N closely wound turns of a ﬁlament that carries a current I (Figure 7.11b), then the ﬁeld at points well within the solenoid is given closely by NI H= (15) az (well within the solenoid) d The approximation is useful it if is not applied closer than two radii to the open ends, nor closer to the solenoid surface than twice the separation between turns. For the toroids shown in Figure 7.12, it can be shown that the magnetic ﬁeld intensity for the ideal case, Figure 7.12a, is ρ0 − a H = Ka (16a) aφ (inside toroid) ρ H=0 (outside) (16b) For the N -turn toroid of Figure 7.12b, we have the good approximations, H= NI aφ 2πρ (inside toroid) (outside) (17a) (17b)

H=0

as long as we consider points removed from the toroidal surface by several times the separation between turns. Toroids having rectangular cross sections are also treated quite readily, as you can see for yourself by trying Problem 7.14. Accurate formulas for solenoids, toroids, and coils of other shapes are available in Section 2 of the Standard Handbook for Electrical Engineers (see References for Chapter 5).

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Figure 7.12 (a) An ideal toroid carrying a surface current K in the direction shown. (b) An N-turn toroid carrying a filamentary current I.

D7.3. Express the value of H in rectangular components at P(0, 0.2, 0) in the ﬁeld of: (a) a current ﬁlament, 2.5 A in the az direction at x = 0.1, y = 0.3; (b) a coax, centered on the z axis, with a = 0.3, b = 0.5, c = 0.6, I = 2.5 A in the az direction in the center conductor; (c) three current sheets, 2.7ax A/m at y = 0.1, −1.4ax A/m at y = 0.15, and −1.3ax A/m at y = 0.25.
Ans. 1.989ax − 1.989a y A/m; −0.884ax A/m; 1.300az A/m

7.3 CURL
We completed our study of Gauss’s law by applying it to a differential volume element and were led to the concept of divergence. We now apply Amp` re’s circuital law to e the perimeter of a differential surface element and discuss the third and last of the special derivatives of vector analysis, the curl. Our objective is to obtain the point form of Amp` re’s circuital law. e Again we choose rectangular coordinates, and an incremental closed path of sides x and y is selected (Figure 7.13). We assume that some current, as yet unspeciﬁed, produces a reference value for H at the center of this small rectangle, H0 = Hx0 ax + Hy0 a y + Hz0 az

The closed line integral of H about this path is then approximately the sum of the four values of H · L on each side. We choose the direction of traverse as 1-2-3-4-1, which corresponds to a current in the az direction, and the ﬁrst contribution is therefore (H · L)1−2 = Hy,1−2 y

The value of Hy on this section of the path may be given in terms of the reference value Hy0 at the center of the rectangle, the rate of change of Hy with x, and the

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Figure 7.13 An incremental closed path in rectangular coordinates is selected for the ` application of Ampere’s circuital law to determine the spatial rate of change of H.

distance

x/2 from the center to the midpoint of side 1–2: ∂ Hy . Hy,1−2 = Hy0 + ∂x 1 x 2 x y

Thus . (H · L)1−2 = Hy0 + 1 ∂ Hy 2 ∂x

Along the next section of the path we have 1 ∂ Hx . . (H · L)2−3 = Hx,2−3 (− x) = − Hx0 + 2 ∂y . H · dL = ∂ Hy ∂ Hx − ∂x ∂y y x

Continuing for the remaining two segments and adding the results, x y

By Amp` re’s circuital law, this result must be equal to the current enclosed by the e path, or the current crossing any surface bounded by the path. If we assume a general . current density J, the enclosed current is then I = Jz x y, and . H · dL = or H · dL . ∂ Hy ∂ Hx . = − = Jz x y ∂x ∂y As we cause the closed path to shrink, the preceding expression becomes more nearly exact, and in the limit we have the equality x, y→0

∂ Hy ∂ Hx − ∂x ∂y

. x y = Jz x y

lim

H · dL ∂ Hy ∂ Hx = − = Jz x y ∂x ∂y

(18)

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After beginning with Amp` re’s circuital law equating the closed line integral of e H to the current enclosed, we have now arrived at a relationship involving the closed line integral of H per unit area enclosed and the current per unit area enclosed, or current density. We performed a similar analysis in passing from the integral form of Gauss’s law, involving ﬂux through a closed surface and charge enclosed, to the point form, relating ﬂux through a closed surface per unit volume enclosed and charge per unit volume enclosed, or volume charge density. In each case a limit is necessary to produce an equality. If we choose closed paths that are oriented perpendicularly to each of the remaining two coordinate axes, analogous processes lead to expressions for the x and y components of the current density, H · dL ∂ Hy ∂ Hz = − = Jx y z ∂y ∂z

y, z→0

lim

(19)

and lim H · dL ∂ Hx ∂ Hz = − = Jy z x ∂z ∂x (20)

z, x→0

Comparing (18)–(20), we see that a component of the current density is given by the limit of the quotient of the closed line integral of H about a small path in a plane normal to that component and of the area enclosed as the path shrinks to zero. This limit has its counterpart in other ﬁelds of science and long ago received the name of curl. The curl of any vector is a vector, and any component of the curl is given by the limit of the quotient of the closed line integral of the vector about a small path in a plane normal to that component desired and the area enclosed, as the path shrinks to zero. It should be noted that this deﬁnition of curl does not refer speciﬁcally to a particular coordinate system. The mathematical form of the deﬁnition is (curl H) N = lim H · dL SN (21)

S N →0

where S N is the planar area enclosed by the closed line integral. The N subscript indicates that the component of the curl is that component which is normal to the surface enclosed by the closed path. It may represent any component in any coordinate system. In rectangular coordinates, the deﬁnition (21) shows that the x, y, and z components of the curl H are given by (18)–(20), and therefore curl H = ∂ Hy ∂ Hz − ax + ∂y ∂z ∂ Hz ∂ Hx − ay + ∂z ∂x ∂ Hy ∂ Hx − az ∂x ∂y

(22)

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This result may be written in the form of a determinant, ax ∂ curl H = ∂x Hx ay ∂ ∂y Hy az ∂ ∂z Hz

(23)

and may also be written in terms of the vector operator, curl H = ∇ × H (24)

Equation (22) is the result of applying the deﬁnition (21) to the rectangular coordinate system. We obtained the z component of this expression by evaluating Amp` re’s e circuital law about an incremental path of sides x and y, and we could have obtained the other two components just as easily by choosing the appropriate paths. Equation (23) is a neat method of storing the rectangular coordinate expression for curl; the form is symmetrical and easily remembered. Equation (24) is even more concise and leads to (22) upon applying the deﬁnitions of the cross product and vector operator. The expressions for curl H in cylindrical and spherical coordinates are derived in Appendix A by applying the deﬁnition (21). Although they may be written in determinant form, as explained there, the determinants do not have one row of unit vectors on top and one row of components on the bottom, and they are not easily memorized. For this reason, the curl expansions in cylindrical and spherical coordinates that follow here and appear inside the back cover are usually referred to whenever necessary. ∇ ×H= 1 ∂Hz ∂Hρ ∂Hφ ∂Hz − aρ + − aφ ρ ∂φ ∂z ∂z ∂ρ 1 ∂Hρ 1 ∂(ρ Hφ ) − az (cylindrical) + ρ ∂ρ ρ ∂φ

(25)

∇ ×H=

1 r sin θ

1 ∂(Hφ sin θ) ∂Hθ 1 ∂Hr ∂(rHφ ) − ar + − aθ ∂θ ∂φ r sin θ ∂φ ∂r 1 ∂(rHθ ) ∂Hr + − aφ (spherical) r ∂r ∂θ

(26)

Although we have described curl as a line integral per unit area, this does not provide everyone with a satisfactory physical picture of the nature of the curl operation, for the closed line integral itself requires physical interpretation. This integral was ﬁrst met in the electrostatic ﬁeld, where we saw that E · dL = 0. Inasmuch as the integral was zero, we did not belabor the physical picture. More recently we have discussed the closed line integral of H, H · dL = I . Either of these closed line integrals is also known by the name of circulation, a term borrowed from the ﬁeld of ﬂuid dynamics.

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Figure 7.14 (a) The curl meter shows a component of the curl of the water velocity into the page. (b) The curl of the magnetic field intensity about an infinitely long filament is shown.

The circulation of H, or H · dL, is obtained by multiplying the component of H parallel to the speciﬁed closed path at each point along it by the differential path length and summing the results as the differential lengths approach zero and as their number becomes inﬁnite. We do not require a vanishingly small path. Amp` re’s e circuital law tells us that if H does possess circulation about a given path, then current passes through this path. In electrostatics we see that the circulation of E is zero about every path, a direct consequence of the fact that zero work is required to carry a charge around a closed path. We may describe curl as circulation per unit area. The closed path is vanishingly small, and curl is deﬁned at a point. The curl of E must be zero, for the circulation is zero. The curl of H is not zero, however; the circulation of H per unit area is the current density by Amp` re’s circuital law [or (18), (19), and (20)]. e Skilling5 suggests the use of a very small paddle wheel as a “curl meter.” Our vector quantity, then, must be thought of as capable of applying a force to each blade of the paddle wheel, the force being proportional to the component of the ﬁeld normal to the surface of that blade. To test a ﬁeld for curl, we dip our paddle wheel into the ﬁeld, with the axis of the paddle wheel lined up with the direction of the component of curl desired, and note the action of the ﬁeld on the paddle. No rotation means no curl; larger angular velocities mean greater values of the curl; a reversal in the direction of spin means a reversal in the sign of the curl. To ﬁnd the direction of the vector curl and not merely to establish the presence of any particular component, we should place our paddle wheel in the ﬁeld and hunt around for the orientation which produces the greatest torque. The direction of the curl is then along the axis of the paddle wheel, as given by the right-hand rule. As an example, consider the ﬂow of water in a river. Figure 7.14a shows the longitudinal section of a wide river taken at the middle of the river. The water velocity is zero at the bottom and increases linearly as the surface is approached. A paddle wheel placed in the position shown, with its axis perpendicular to the paper, will turn in a clockwise direction, showing the presence of a component of curl in the direction
5

See the References at the end of the chapter.

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ENGINEERING ELECTROMAGNETICS

of an inward normal to the surface of the page. If the velocity of water does not change as we go up- or downstream and also shows no variation as we go across the river (or even if it decreases in the same fashion toward either bank), then this component is the only component present at the center of the stream, and the curl of the water velocity has a direction into the page. In Figure 7.14b, the streamlines of the magnetic ﬁeld intensity about an inﬁnitely long ﬁlamentary conductor are shown. The curl meter placed in this ﬁeld of curved lines shows that a larger number of blades have a clockwise force exerted on them but that this force is in general smaller than the counterclockwise force exerted on the smaller number of blades closer to the wire. It seems possible that if the curvature of the streamlines is correct and also if the variation of the ﬁeld strength is just right, the net torque on the paddle wheel may be zero. Actually, the paddle wheel does not rotate in this case, for since H = (I /2πρ)aφ , we may substitute into (25) obtaining curl H = − 1 ∂(ρ Hφ ) ∂Hφ aρ + az = 0 ∂z ρ ∂ρ

E X A M P L E 7.2

As an example of the evaluation of curl H from the deﬁnition and of the evaluation of another line integral, suppose that H = 0.2z 2 ax for z > 0, and H = 0 elsewhere, as shown in Figure 7.15. Calculate H · dL about a square path with side d, centered at (0, 0, z 1 ) in the y = 0 plane where z 1 > d/2.

Figure 7.15 A square path of side d with its center on the z axis at z = z1 is used to evaluate H · dL and find curl H.

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Solution. We evaluate the line integral of H along the four segments, beginning at

the top: H · dL = 0.2 z 1 + 1 d 2 = 0.4z 1 d 2 In the limit as the area approaches zero, we ﬁnd (∇ × H) y = lim H · dL 0.4z 1 d 2 = lim = 0.4z 1 d→0 d2 d2
2

d + 0 − 0.2 z 1 − 1 d 2

2

d +0

d→0

The other components are zero, so ∇ × H = 0.4z 1 a y . To evaluate the curl without trying to illustrate the deﬁnition or the evaluation of a line integral, we simply take the partial derivative indicated by (23): ax ∂ ∇× H= ∂x 0.2z 2 ay ∂ ∂y 0 az ∂ ∂ (0.2z 2 )a y = 0.4za y = ∂z ∂z 0

which checks with the preceding result when z = z 1 . Returning now to complete our original examination of the application of Amp` re’s circuital law to a differential-sized path, we may combine (18)–(20), (22), e and (24), curl H = ∇ × H = ∂Hy ∂Hz ∂Hx ∂Hz − ax + − ay ∂y ∂z ∂z ∂x ∂Hy ∂Hx + − az = J ∂x ∂y

(27)

and write the point form of Amp` re’s circuital law, e ∇ ×H = J (28)

This is the second of Maxwell’s four equations as they apply to non-time-varying conditions. We may also write the third of these equations at this time; it is the point form of E · dL = 0, or ∇ ×E = 0 The fourth equation appears in Section 7.5. (29)

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D7.4. (a) Evaluate the closed line integral of H about the rectangular path P1 (2, 3, 4) to P2 (4, 3, 4) to P3 (4, 3, 1) to P4 (2, 3, 1) to P1 , given H = 3zax − 2x 3 az A/m. (b) Determine the quotient of the closed line integral and the area enclosed by the path as an approximation to (∇ × H) y . (c) Determine (∇ × H) y at the center of the area.
Ans. 354 A; 59 A/m2 ; 57 A/m2

D7.5. Calculate the value of the vector current density: (a) in rectangular coordinates at PA (2, 3, 4) if H = x 2 za y − y 2 xaz ; (b) in cylindrical coordi2 nates at PB (1.5, 90◦ , 0.5) if H = (cos 0.2φ)aρ ; (c) in spherical coordinates at ρ 1 aθ . PC (2, 30◦ , 20◦ ) if H = sin θ
Ans. −16ax + 9a y + 16az A/m2 ; 0.055az A/m2 ; aφ A/m2

7.4 STOKES’ THEOREM
Although Section 7.3 was devoted primarily to a discussion of the curl operation, the contribution to the subject of magnetic ﬁelds should not be overlooked. From Amp` re’s circuital law we derived one of Maxwell’s equations, ∇ × H = J. This e latter equation should be considered the point form of Amp` re’s circuital law and e applies on a “per-unit-area” basis. In this section we shall again devote a major share of the material to the mathematical theorem known as Stokes’ theorem, but in the process we will show that we may obtain Amp` re’s circuital law from ∇ × H = J. e In other words, we are then prepared to obtain the integral form from the point form or to obtain the point form from the integral form. Consider the surface S of Figure 7.16, which is broken up into incremental surfaces of area S. If we apply the deﬁnition of the curl to one of these incremental surfaces, then H · dL S . = (∇ × H) N S where the N subscript again indicates the right-hand normal to the surface. The subscript on dL S indicates that the closed path is the perimeter of an incremental area S. This result may also be written H · dL S or H · dL
S S

. = (∇ × H) · a N

. = (∇ × H) · a N S = (∇ × H) · S

where a N is a unit vector in the direction of the right-hand normal to S. Now let us determine this circulation for every S comprising S and sum the results. As we evaluate the closed line integral for each S, some cancellation will occur

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Figure 7.16 The sum of the closed line integrals about the perimeter of every S is the same as the closed line integral about the perimeter of S because of cancellation on every interior path.

because every interior wall is covered once in each direction. The only boundaries on which cancellation cannot occur form the outside boundary, the path enclosing S. Therefore we have H · dL ≡ (∇ × H) · dS (30)

S

where dL is taken only on the perimeter of S. Equation (30) is an identity, holding for any vector ﬁeld, and is known as Stokes’ theorem.
E X A M P L E 7.3

A numerical example may help to illustrate the geometry involved in Stokes’ theorem. Consider the portion of a sphere shown in Figure 7.17. The surface is speciﬁed by r = 4, 0 ≤ θ ≤ 0.1π, 0 ≤ φ ≤ 0.3π, and the closed path forming its perimeter is composed of three circular arcs. We are given the ﬁeld H = 6r sin φar + 18r sin θ cos φaφ and are asked to evaluate each side of Stokes’ theorem.
Solution. The ﬁrst path segment is described in spherical coordinates by r = 4, 0 ≤

θ ≤ 0.1π, φ = 0; the second one by r = 4, θ = 0.1π, 0 ≤ φ ≤ 0.3π ; and the third by r = 4, 0 ≤ θ ≤ 0.1π, φ = 0.3π . The differential path element dL is the vector sum of the three differential lengths of the spherical coordinate system ﬁrst discussed in Section 1.9, dL = dr ar + r dθ aθ + r sin θ dφ aφ

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Figure 7.17 A portion of a spherical cap is used as a surface and a closed path to illustrate Stokes’ theorem.

The ﬁrst term is zero on all three segments of the path since r = 4 and dr = 0, the second is zero on segment 2 as θ is constant, and the third term is zero on both segments 1 and 3. Thus, H · dL =
1

Hθ r dθ +
0.3π

2

Hφ r sin θ dφ +

Hθ r dθ
3

Because Hθ = 0, we have only the second integral to evaluate, H · dL = [18(4) sin 0.1π cos φ]4 sin 0.1πdφ
0

= 288 sin2 0.1π sin 0.3π = 22.2 A We next attack the surface integral. First, we use (26) to ﬁnd ∇ ×H = 1 1 1 (36r sin θ cos θ cos φ)ar + 6r cos φ − 36r sin θ cos φ aθ r sin θ r sin θ
0.3π 0.1π

Because dS = r 2 sin θ dθ dφ ar , the integral is
S

(∇ × H) · dS = =

(36 cos θ cos φ)16 sin θ dθ dφ
1 2

0 0.3π

0

576
0

sin2 θ

0.1π 0

cos φ dφ

= 288 sin2 0.1π sin 0.3π = 22.2 A

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Thus, the results check Stokes’ theorem, and we note in passing that a current of 22.2 A is ﬂowing upward through this section of a spherical cap. Next, let us see how easy it is to obtain Amp` re’s circuital law from ∇ × H = J. e We merely have to dot each side by dS, integrate each side over the same (open) surface S, and apply Stokes’ theorem:
S

(∇ × H) · dS =

S

J · dS =

H · dL

The integral of the current density over the surface S is the total current I passing through the surface, and therefore H · dL = I This short derivation shows clearly that the current I , described as being “enclosed by the closed path,” is also the current passing through any of the inﬁnite number of surfaces that have the closed path as a perimeter. Stokes’ theorem relates a surface integral to a closed line integral. It should be recalled that the divergence theorem relates a volume integral to a closed surface integral. Both theorems ﬁnd their greatest use in general vector proofs. As an example, let us ﬁnd another expression for ∇ · ∇ × A, where A represents any vector ﬁeld. The result must be a scalar (why?), and we may let this scalar be T , or ∇ ·∇ ×A = T Multiplying by dν and integrating throughout any volume ν, (∇ · ∇ × A) dν = T dν vol vol

we ﬁrst apply the divergence theorem to the left side, obtaining (∇ × A) · dS = T dν vol S

The left side is the surface integral of the curl of A over the closed surface surrounding the volume ν. Stokes’ theorem relates the surface integral of the curl of A over the open surface enclosed by a given closed path. If we think of the path as the opening of a laundry bag and the open surface as the surface of the bag itself, we see that as we gradually approach a closed surface by pulling on the drawstrings, the closed path becomes smaller and smaller and ﬁnally disappears as the surface becomes closed. Hence, the application of Stokes’ theorem to a closed surface produces a zero result, and we have T dν = 0

vol

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Because this is true for any volume, it is true for the differential volume dν, T dν = 0 and therefore T =0 or ∇ ·∇ ×A ≡ 0 (31)

Equation (31) is a useful identity of vector calculus.6 Of course, it may also be proven easily by direct expansion in rectangular coordinates. Let us apply the identity to the non-time-varying magnetic ﬁeld for which ∇ ×H = J This shows quickly that ∇ ·J = 0 which is the same result we obtained earlier in the chapter by using the continuity equation. Before introducing several new magnetic ﬁeld quantities in the following section, we may review our accomplishments at this point. We initially accepted the BiotSavart law as an experimental result, I dL × a R 4πR 2 and tentatively accepted Amp` re’s circuital law, subject to later proof, e H= H · dL = I From Amp` re’s circuital law the deﬁnition of curl led to the point form of this same e law, ∇ ×H = J We now see that Stokes’ theorem enables us to obtain the integral form of Amp` re’s e circuital law from the point form. D7.6. Evaluate both sides of Stokes’ theorem for the ﬁeld H = 6x yax − 3y 2 a y A/m and the rectangular path around the region, 2 ≤ x ≤ 5, −1 ≤ y ≤ 1, z = 0. Let the positive direction of dS be az .
Ans. −126 A; −126 A

6

This and other vector identities are tabulated in Appendix A.3.

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7.5 MAGNETIC FLUX AND MAGNETIC FLUX DENSITY
In free space, let us deﬁne the magnetic flu density B as B = µ0 H (free space only) (32)

where B is measured in webers per square meter (Wb/m2 ) or in a newer unit adopted in the International System of Units, tesla (T). An older unit that is often used for magnetic ﬂux density is the gauss (G), where 1 T or 1Wb/m2 is the same as 10, 000 G. The constant µ0 is not dimensionless and has the define value for free space, in henrys per meter (H/m), of µ0 = 4π × 10−7 H/m (33)

The name given to µ0 is the permeability of free space. We should note that since H is measured in amperes per meter, the weber is dimensionally equal to the product of henrys and amperes. Considering the henry as a new unit, the weber is merely a convenient abbreviation for the product of henrys and amperes. When time-varying ﬁelds are introduced, it will be shown that a weber is also equivalent to the product of volts and seconds. The magnetic-ﬂux-density vector B, as the name weber per square meter implies, is a member of the ﬂux-density family of vector ﬁelds. One of the possible analogies between electric and magnetic ﬁelds7 compares the laws of Biot-Savart and Coulomb, thus establishing an analogy between H and E. The relations B = µ0 H and D = 0 E then lead to an analogy between B and D. If B is measured in teslas or webers per square meter, then magnetic ﬂux should be measured in webers. Let us represent magnetic ﬂux by and deﬁne as the ﬂux passing through any designated area, = B · dS Wb (34)

S

Our analogy should now remind us of the electric ﬂux , measured in coulombs, and of Gauss’s law, which states that the total ﬂux passing through any closed surface is equal to the charge enclosed, =
S

D · dS = Q

The charge Q is the source of the lines of electric ﬂux and these lines begin and terminate on positive and negative charges, respectively.

7

An alternate analogy is presented in Section 9.2.

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No such source has ever been discovered for the lines of magnetic ﬂux. In the example of the inﬁnitely long straight ﬁlament carrying a direct current I , the H ﬁeld formed concentric circles about the ﬁlament. Because B = µ0 H, the B ﬁeld is of the same form. The magnetic ﬂux lines are closed and do not terminate on a “magnetic charge.” For this reason Gauss’s law for the magnetic ﬁeld is B · dS = 0 (35)

S

and application of the divergence theorem shows us that ∇ ·B = 0 (36)

Equation (36) is the last of Maxwell’s four equations as they apply to static electric ﬁelds and steady magnetic ﬁelds. Collecting these equations, we then have for static electric ﬁelds and steady magnetic ﬁelds

∇ ×H= J ∇ ·B = 0

∇ ×E = 0

∇ · D = ρν (37)

To these equations we may add the two expressions relating D to E and B to H in free space, D=
0E

(38)

B = µ0 H We have also found it helpful to deﬁne an electrostatic potential, E = −∇V

(39)

(40)

and we will discuss a potential for the steady magnetic ﬁeld in the following section. In addition, we extended our coverage of electric ﬁelds to include conducting materials and dielectrics, and we introduced the polarization P. A similar treatment will be applied to magnetic ﬁelds in the next chapter. Returning to (37), it may be noted that these four equations specify the divergence and curl of an electric and a magnetic ﬁeld. The corresponding set of four integral

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209

equations that apply to static electric ﬁelds and steady magnetic ﬁelds is D · dS = Q = E · dL = 0 H · dL = I =
S S

S

ρν dν vol (41) J · dS

B · dS = 0

Our study of electric and magnetic ﬁelds would have been much simpler if we could have begun with either set of equations, (37) or (41). With a good knowledge of vector analysis, such as we should now have, either set may be readily obtained from the other by applying the divergence theorem or Stokes’ theorem. The various experimental laws can be obtained easily from these equations. As an example of the use of ﬂux and ﬂux density in magnetic ﬁelds, let us ﬁnd the ﬂux between the conductors of the coaxial line of Figure 7.8a. The magnetic ﬁeld intensity was found to be I Hφ = (a < ρ < b) 2πρ and therefore µ0 I B = µ0 H = aφ 2πρ The magnetic ﬂux contained between the conductors in a length d is the ﬂux crossing any radial plane extending from ρ = a to ρ = b and from, say, z = 0 to z=d d b µ0 I = B · dS = aφ · dρ dz aφ 0 a 2πρ S or µ0 Id b = ln (42) 2π a This expression will be used later to obtain the inductance of the coaxial transmission line. D7.7. A solid conductor of circular cross section is made of a homogeneous nonmagnetic material. If the radius a = 1 mm, the conductor axis lies on the z axis, and the total current in the az direction is 20 A, ﬁnd: (a) Hφ at ρ = 0.5 mm; (b) Bφ at ρ = 0.8 mm; (c) the total magnetic ﬂux per unit length inside the conductor; (d) the total ﬂux for ρ < 0.5 mm; (e) the total magnetic ﬂux outside the conductor.
Ans. 1592 A/m; 3.2 mT; 2 µWb/m; 0.5 µWb; ∞

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ENGINEERING ELECTROMAGNETICS

7.6 THE SCALAR AND VECTOR MAGNETIC POTENTIALS
The solution of electrostatic ﬁeld problems is greatly simpliﬁed by the use of the scalar electrostatic potential V. Although this potential possesses a very real physical signiﬁcance for us, it is mathematically no more than a stepping-stone which allows us to solve a problem by several smaller steps. Given a charge conﬁguration, we may ﬁrst ﬁnd the potential and then from it the electric ﬁeld intensity. We should question whether or not such assistance is available in magnetic ﬁelds. Can we deﬁne a potential function which may be found from the current distribution and from which the magnetic ﬁelds may be easily determined? Can a scalar magnetic potential be deﬁned, similar to the scalar electrostatic potential? We will show in the next few pages that the answer to the ﬁrst question is yes, but the second must be answered “sometimes.” Let us attack the second question ﬁrst by assuming the existence of a scalar magnetic potential, which we designate Vm , whose negative gradient gives the magnetic ﬁeld intensity H = −∇Vm The selection of the negative gradient provides a closer analogy to the electric potential and to problems which we have already solved. This deﬁnition must not conﬂict with our previous results for the magnetic ﬁeld, and therefore ∇ × H = J = ∇ × (−∇Vm ) However, the curl of the gradient of any scalar is identically zero, a vector identity the proof of which is left for a leisure moment. Therefore, we see that if H is to be deﬁned as the gradient of a scalar magnetic potential, then current density must be zero throughout the region in which the scalar magnetic potential is so deﬁned. We then have H = −∇Vm (J = 0) (43)

Because many magnetic problems involve geometries in which the current-carrying conductors occupy a relatively small fraction of the total region of interest, it is evident that a scalar magnetic potential can be useful. The scalar magnetic potential is also applicable in the case of permanent magnets. The dimensions of Vm are obviously amperes. This scalar potential also satisﬁes Laplace’s equation. In free space, ∇ · B = µ0 ∇ · H = 0 and hence µ0 ∇ · (−∇Vm ) = 0

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211

or ∇ 2 Vm = 0 (J = 0) (44)

We will see later that Vm continues to satisfy Laplace’s equation in homogeneous magnetic materials; it is not deﬁned in any region in which current density is present. Although we shall consider the scalar magnetic potential to a much greater extent in Chapter 8, when we introduce magnetic materials and discuss the magnetic circuit, one difference between V and Vm should be pointed out now: Vm is not a single-valued function of position. The electric potential V is single-valued; once a zero reference is assigned, there is only one value of V associated with each point in space. Such is not the case with Vm . Consider the cross section of the coaxial line shown in Figure 7.18. In the region a < ρ < b, J = 0, and we may establish a scalar magnetic potential. The value of H is I H= aφ 2πρ where I is the total current ﬂowing in the az direction in the inner conductor. We ﬁnd Vm by integrating the appropriate component of the gradient. Applying (43), I = −∇Vm 2πρ or ∂Vm I =− ∂φ 2π φ =−

1 ∂Vm ρ ∂φ

Figure 7.18 The scalar magnetic potential Vm is a multivalued function of φ in the region a < ρ < b. The electrostatic potential is always single valued.

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ENGINEERING ELECTROMAGNETICS

Thus, I φ 2π where the constant of integration has been set equal to zero. What value of potential do we associate with point P, where φ = π/4? If we let Vm be zero at φ = 0 and proceed counterclockwise around the circle, the magnetic potential goes negative linearly. When we have made one circuit, the potential is −I , but that was the point at which we said the potential was zero a moment ago. At P, then, φ = π/4, 9π/4, 17π/4, . . . , or −7π/4, −15π/4, −23π/4, . . . , or I 2n − 1 π (n = 0, ±1, ±2, . . .) VmP = 4 2π or Vm = − VmP = I n −
1 8

(n = 0, ±1, ±2, . . .)

The reason for this multivaluedness may be shown by a comparison with the electrostatic case. There, we know that ∇ ×E = 0

E · dL = 0 and therefore the line integral Vab = − ∇ ×H = 0 but H · dL = I even if J is zero along the path of integration. Every time we make another complete lap around the current, the result of the integration increases by I . If no current I is enclosed by the path, then a single-valued potential function may be deﬁned. In general, however, Vm,ab = − a b a b

E · dL

is independent of the path. In the magnetostatic case, however, (wherever J = 0)

H · dL

(speciﬁed path)

(45)

where a speciﬁc path or type of path must be selected. We should remember that the electrostatic potential V is a conservative ﬁeld; the magnetic scalar potential Vm is not a conservative ﬁeld. In our coaxial problem, let us erect a barrier8 at φ = π ; we
8

This corresponds to the more precise mathematical term “branch cut.”

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213

agree not to select a path that crosses this plane. Therefore, we cannot encircle I , and a single-valued potential is possible. The result is seen to be Vm = − and VmP = − I 8 φ= π 4 I φ 2π (−π < φ < π)

The scalar magnetic potential is evidently the quantity whose equipotential surfaces will form curvilinear squares with the streamlines of H in Figure 7.4. This is one more facet of the analogy between electric and magnetic ﬁelds about which we will have more to say in the next chapter. Let us temporarily leave the scalar magnetic potential now and investigate a vector magnetic potential. This vector ﬁeld is one which is extremely useful in studying radiation from antennas (as we will ﬁnd in Chapter 14) as well as radiation leakage from transmission lines, waveguides, and microwave ovens. The vector magnetic potential may be used in regions where the current density is zero or nonzero, and we shall also be able to extend it to the time-varying case later. Our choice of a vector magnetic potential is indicated by noting that ∇ ·B = 0 Next, a vector identity that we proved in Section 7.4 shows that the divergence of the curl of any vector ﬁeld is zero. Therefore, we select B = ∇ ×A (46)

where A signiﬁes a vector magnetic potential, and we automatically satisfy the condition that the magnetic ﬂux density shall have zero divergence. The H ﬁeld is H= and ∇ ×H = J = 1 ∇ ×∇ ×A µ0 1 ∇ ×A µ0

The curl of the curl of a vector ﬁeld is not zero and is given by a fairly complicated expression,9 which we need not know now in general form. In speciﬁc cases for which the form of A is known, the curl operation may be applied twice to determine the current density.

9 ∇ × ∇ × A ≡ ∇(∇ · A) − ∇ 2 A. In rectangular coordinates, it may be shown that ∇ 2 A ≡ ∇ 2 A a + x x ∇ 2 A y a y + ∇ 2 Az az . In other coordinate systems, ∇ 2 A may be found by evaluating the second-order partial derivatives in ∇ 2 A = ∇(∇ · A) − ∇ × ∇ × A.

214

ENGINEERING ELECTROMAGNETICS

Equation (46) serves as a useful deﬁnition of the vector magnetic potential A. Because the curl operation implies differentiation with respect to a length, the units of A are webers per meter. As yet we have seen only that the deﬁnition for A does not conﬂict with any previous results. It still remains to show that this particular deﬁnition can help us to determine magnetic ﬁelds more easily. We certainly cannot identify A with any easily measured quantity or history-making experiment. We will show in Section 7.7 that, given the Biot-Savart law, the deﬁnition of B, and the deﬁnition of A, A may be determined from the differential current elements by A= µ0 I dL 4πR (47)

The signiﬁcance of the terms in (47) is the same as in the Biot-Savart law; a direct current I ﬂows along a ﬁlamentary conductor of which any differential length dL is distant R from the point at which A is to be found. Because we have deﬁned A only through speciﬁcation of its curl, it is possible to add the gradient of any scalar ﬁeld to (47) without changing B or H, for the curl of the gradient is identically zero. In steady magnetic ﬁelds, it is customary to set this possible added term equal to zero. The fact that A is a vector magnetic potential is more apparent when (47) is compared with the similar expression for the electrostatic potential, V = ρ L dL 4π 0 R

Each expression is the integral along a line source, in one case line charge and in the other case line current; each integrand is inversely proportional to the distance from the source to the point of interest; and each involves a characteristic of the medium (here free space), the permeability or the permittivity. Equation (47) may be written in differential form, dA = µ0 I dL 4π R (48)

if we again agree not to attribute any physical signiﬁcance to any magnetic ﬁelds we obtain from (48) until the entire closed path in which the current flow is considered. With this reservation, let us go right ahead and consider the vector magnetic potential ﬁeld about a differential ﬁlament. We locate the ﬁlament at the origin in free space, as shown in Figure 7.19, and allow it to extend in the positive z direction so that dL = dz az . We use cylindrical coordinates to ﬁnd dA at the point (ρ, φ, z): dA = or dAz = µ0 I dz 4π ρ 2 + z 2 µ0 I dz az 4π ρ 2 + z 2 dAφ = 0 dAρ = 0 (49)

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215

Figure 7.19 The differential current element I dzaz at the origin establishes the differential vector magnetic potential field, µ0 I dzaz dA = at P(ρ, φ, z). 4π ρ 2 + z2

We note that the direction of dA is the same as that of I dL. Each small section of a current-carrying conductor produces a contribution to the total vector magnetic potential which is in the same direction as the current ﬂow in the conductor. The magnitude of the vector magnetic potential varies inversely with the distance to the current element, being strongest in the neighborhood of the current and gradually falling off to zero at distant points. Skilling10 describes the vector magnetic potential ﬁeld as “like the current distribution but fuzzy around the edges, or like a picture of the current out of focus.” In order to ﬁnd the magnetic ﬁeld intensity, we must take the curl of (49) in cylindrical coordinates, leading to dH = or I dz ρ a 2 + z 2 )3/2 φ 4π (ρ which is easily shown to be the same as the value given by the Biot-Savart law. Expressions for the vector magnetic potential A can also be obtained for a current source which is distributed. For a current sheet K, the differential current element becomes dH = In the case of current ﬂow throughout a volume with a density J, we have I dL = J dν
10

1 1 ∇ × dA = µ0 µ0

−

∂dA z aφ ∂ρ

I dL = K dS

See the References at the end of the chapter.

216

ENGINEERING ELECTROMAGNETICS

In each of these two expressions the vector character is given to the current. For the ﬁlamentary element it is customary, although not necessary, to use I dL instead of I dL. Since the magnitude of the ﬁlamentary element is constant, we have chosen the form which allows us to remove one quantity from the integral. The alternative expressions for A are then A= and A= µ0 J dν 4πR (51) µ0 K dS 4πR (50)

S

vol

Equations (47), (50), and (51) express the vector magnetic potential as an integration over all of its sources. From a comparison of the form of these integrals with those which yield the electrostatic potential, it is evident that once again the zero reference for A is at inﬁnity, for no ﬁnite current element can produce any contribution as R → ∞. We should remember that we very seldom used the similar expressions for V ; too often our theoretical problems included charge distributions that extended to inﬁnity, and the result would be an inﬁnite potential everywhere. Actually, we calculated very few potential ﬁelds until the differential form of the potential equation was obtained, ∇ 2 V = −ρν / , or better yet, ∇ 2 V = 0. We were then at liberty to select our own zero reference. The analogous expressions for A will be derived in the next section, and an example of the calculation of a vector magnetic potential ﬁeld will be completed. D7.8. A current sheet, K = 2.4az A/m, is present at the surface ρ = 1.2 in free space. (a) Find H for ρ > 1.2. Find Vm at P(ρ = 1.5, φ = 0.6π, z = 1) if: (b) Vm = 0 at φ = 0 and there is a barrier at φ = π ; (c) Vm = 0 at φ = 0 and there is a barrier at φ = π/2; (d) Vm = 0 at φ = π and there is a barrier at φ = 0; (e) Vm = 5 V at φ = π and there is a barrier at φ = 0.8π .
Ans. 2.88 aφ ; −5.43 V; 12.7 V; 3.62 V; −9.48 V ρ

D7.9. The value of A within a solid nonmagnetic conductor of radius a carrying a total current I in the az direction may be found easily. Using the known value of H or B for ρ < a, then (46) may be solved for A. Select A = (µ0 I ln 5)/2π at ρ = a (to correspond with an example in the next section) and ﬁnd A at ρ =: (a) 0; (b) 0.25a; (c) 0.75a; (d) a.
Ans. 0.422I az µWb/m; 0.416I az µWb/m; 0.366I az µWb/m; 0.322I az µWb/m

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217

7.7 DERIVATION OF THE STEADY-MAGNETIC-FIELD LAWS
We will now supply the promised proofs of the several relationships between the magnetic ﬁeld quantities. All these relationships may be obtained from the deﬁnitions of H, I dL × a R H= (3) 4πR 2 of B (in free space), and of A, B = µ0 H B = ∇ ×A (32)

(46)

Let us ﬁrst assume that we may express A by the last equation of Section 7.6, A= µ0 J dν 4πR (51)

vol

and then demonstrate the correctness of (51) by showing that (3) follows. First, we should add subscripts to indicate the point at which the current element is located (x1 , y1 , z 1 ) and the point at which A is given (x2 , y2 , z 2 ). The differential volume element dν is then written dν1 and in rectangular coordinates would be d x1 dy1 dz 1 . The variables of integration are x1 , y1 , and z 1 . Using these subscripts, then, A2 = From (32) and (46) we have H= ∇ ×A B = µ0 µ0 (53) vol µ0 J1 dν1 4πR12

(52)

To show that (3) follows from (52), it is necessary to substitute (52) into (53). This step involves taking the curl of A2 , a quantity expressed in terms of the variables x2 , y2 , and z 2 , and the curl therefore involves partial derivatives with respect to x2 , y2 , and z 2 . We do this, placing a subscript on the del operator to remind us of the variables involved in the partial differentiation process, H2 = ∇2 × A2 1 = ∇2 × µ0 µ0 vol µ0 J1 dν1 4πR12

The order of partial differentiation and integration is immaterial, and µ0 /4π is constant, allowing us to write H2 = 1 4π vol ∇2 ×

J1 dν1 R12

The curl operation within the integrand represents partial differentiation with respect to x2 , y2 , and z 2 . The differential volume element dν1 is a scalar and a function

218

ENGINEERING ELECTROMAGNETICS

only of x1 , y1 , and z 1 . Consequently, it may be factored out of the curl operation as any other constant, leaving H2 = 1 4π ∇2 × J1 dν1 R12 (54)

vol

The curl of the product of a scalar and a vector is given by an identity which may be checked by expansion in rectangular coordinates or obtained from Appendix A.3, ∇ × (SV) ≡ (∇ S) × V + S(∇ × V) This identity is used to expand the integrand of (54), H2 = 1 4π ∇2 1 R12 × J1 + 1 (∇2 × J1 ) dν1 R12 (56) (55)

vol

The second term of this integrand is zero because ∇2 × J1 indicates partial derivatives of a function of x1 , y1 , and z 1 , taken with respect to the variables x2 , y2 , and z 2 ; the ﬁrst set of variables is not a function of the second set, and all partial derivatives are zero. The ﬁrst term of the integrand may be determined by expressing R12 in terms of the coordinate values, R12 = (x2 − x1 )2 + (y2 − y1 )2 + (z 2 − z 1 )2 ∇2 1 R12 a R12 =− 3 =− 2 R12 R12 R12 1 4π a R12 × J1 dν1 2 R12

and taking the gradient of its reciprocal. Problem 7.42 shows that the result is

Substituting this result into (56), we have H2 = − or H2 = vol vol

J1 × a R12 dν1 2 4πR12

which is the equivalent of (3) in terms of current density. Replacing J1 dν1 by I1 dL1 , we may rewrite the volume integral as a closed line integral, H2 = I1 dL1 × a R12 2 4πR12

Equation (51) is therefore correct and agrees with the three deﬁnitions (3), (32), and (46). Next we will prove Amp` re’s circuital law in point form, e ∇ ×H = J Combining (28), (32), and (46), we obtain ∇ ×H = ∇ × B 1 = ∇ ×∇ ×A µ0 µ0 (57) (28)

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219

We now need the expansion in rectangular coordinates for ∇ × ∇ × A. Performing the indicated partial differentiations and collecting the resulting terms, we may write the result as ∇ × ∇ × A ≡ ∇(∇ · A) − ∇ 2 A where ∇ 2 A ≡ ∇ 2 A x a x + ∇ 2 A y a y + ∇ 2 A z az (59) (58)

Equation (59) is the deﬁnition (in rectangular coordinates) of the Laplacian of a vector. Substituting (58) into (57), we have ∇ ×H = 1 [∇(∇ · A) − ∇ 2 A] µ0 (60)

and now require expressions for the divergence and the Laplacian of A. We may ﬁnd the divergence of A by applying the divergence operation to (52), ∇2 · A2 = µ0 4π vol ∇2 ·

J1 dν1 R12

(61)

and using the vector identity (44) of Section 4.8, ∇ · (SV) ≡ V · (∇ S) + S(∇ · V) Thus, ∇2 · A2 = µ0 4π vol J1 · ∇ 2

1 R12

+

1 (∇2 · J1 ) dν1 R12

(62)

The second part of the integrand is zero because J1 is not a function of x2 , y2 , and z 2 . 3 We have already used the result that ∇2 (1/R12 ) = −R12 /R12 , and it is just as easily shown that ∇1 or that ∇1 µ0 4π 1 1 = −∇2 R12 R12 1 R12 1 R12 = 3 R12 R12

Equation (62) can therefore be written as ∇ 2 · A2 = µ0 4π − J 1 · ∇1 dν1

vol

and the vector identity applied again, ∇2 · A2 = 1 (∇1 · J1 ) − ∇1 · R12 J1 R12 dν1 (63)

vol

220

ENGINEERING ELECTROMAGNETICS

Because we are concerned only with steady magnetic ﬁelds, the continuity equation shows that the ﬁrst term of (63) is zero. Application of the divergence theorem to the second term gives ∇2 · A2 = − µ0 4π
S1

J1 · dS1 R12

where the surface S1 encloses the volume throughout which we are integrating. This volume must include all the current, for the original integral expression for A was an integration such as to include the effect of all the current. Because there is no current outside this volume (otherwise we should have had to increase the volume to include it), we may integrate over a slightly larger volume or a slightly larger enclosing surface without changing A. On this larger surface the current density J1 must be zero, and therefore the closed surface integral is zero, since the integrand is zero. Hence the divergence of A is zero. In order to ﬁnd the Laplacian of the vector A, let us compare the x component of (51) with the similar expression for electrostatic potential, µ0 Jx dν ρν dν V = vol 4πR vol 4π 0 R We note that one expression can be obtained from the other by a straightforward change of variable, Jx for ρν , µ0 for 1/ 0 , and A x for V . However, we have derived some additional information about the electrostatic potential which we shall not have to repeat now for the x component of the vector magnetic potential. This takes the form of Poisson’s equation, ρν ∇2V = − Ax =
0

which becomes, after the change of variables, ∇ 2 A x = −µ0 Jx Similarly, we have ∇ 2 A y = −µ0 Jy and ∇ 2 A z = −µ0 Jz or ∇ 2 A = −µ0 J (64)

Returning to (60), we can now substitute for the divergence and Laplacian of A and obtain the desired answer, ∇ ×H = J (28)

We have already shown the use of Stokes’ theorem in obtaining the integral form of Amp` re’s circuital law from (28) and need not repeat that labor here. e

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221

We thus have succeeded in showing that every result we have essentially pulled from thin air11 for magnetic ﬁelds follows from the basic deﬁnitions of H, B, and A. The derivations are not simple, but they should be understandable on a step-by-step basis. Finally, let us return to (64) and make use of this formidable second-order vector partial differential equation to ﬁnd the vector magnetic potential in one simple example. We select the ﬁeld between conductors of a coaxial cable, with radii of a and b as usual, and current I in the az direction in the inner conductor. Between the conductors, J = 0, and therefore ∇ 2A = 0 We have already been told (and Problem 7.44 gives us the opportunity to check the results for ourselves) that the vector Laplacian may be expanded as the vector sum of the scalar Laplacians of the three components in rectangular coordinates, ∇ 2 A = ∇ 2 A x a x + ∇ 2 A y a y + ∇ 2 A z az but such a relatively simple result is not possible in other coordinate systems. That is, in cylindrical coordinates, for example, However, it is not difﬁcult to show for cylindrical coordinates that the z component of the vector Laplacian is the scalar Laplacian of the z component of A, or ∇ 2 A = ∇ 2 Az z ∇ 2 A = ∇ 2 Aρ aρ + ∇ 2 Aφ aφ + ∇ 2 A z az

(65)

and because the current is entirely in the z direction in this problem, A has only a z component. Therefore, ∇ 2 Az = 0 or 1 ∂ ∂ 2 Az ∂ Az 1 ∂ 2 Az + =0 ρ + 2 ρ ∂ρ ∂ρ ρ ∂φ 2 ∂z 2 Thinking symmetrical thoughts about (51) shows us that A z is a function only of ρ, and thus 1 d d Az ρ =0 ρ dρ dρ We have solved this equation before, and the result is If we choose a zero reference at ρ = b, then A z = C1 ln ρ + C2 A z = C1 ln ρ b

11

Free space.

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ENGINEERING ELECTROMAGNETICS

In order to relate C1 to the sources in our problem, we may take the curl of A, ∇ ×A = − obtain H, H=− and evaluate the line integral, H · dL = I = Thus C1 = − or Az = and Hφ = I 2πρ µ0 I b ln 2π ρ (66) µ0 I 2π
2π 0

C1 ∂ Az aφ = − aφ = B ∂ρ ρ C1 aφ µ0 ρ

−

C1 2πC1 aφ · ρ dφ aφ = − µ0 ρ µ0

as before. A plot of A z versus ρ for b = 5a is shown in Figure 7.20; the decrease of |A| with distance from the concentrated current source that the inner conductor represents is evident. The results of Problem D7.9 have also been added to Figure 7.20. The extension of the curve into the outer conductor is left as Problem 7.43. It is also possible to ﬁnd A z between conductors by applying a process some of us informally call “uncurling.” That is, we know H or B for the coax, and we may

Figure 7.20 The vector magnetic potential is shown within the inner conductor and in the region between conductors for a coaxial cable with b = 5a carrying I in the az direction. Az = 0 is arbitrarily selected at ρ = b.

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223

therefore select the φ component of ∇ × A = B and integrate to obtain A z . Try it, you’ll like it! D7.10. Equation (66) is obviously also applicable to the exterior of any conductor of circular cross section carrying a current I in the az direction in free space. The zero reference is arbitrarily set at ρ = b. Now consider two conductors, each of 1 cm radius, parallel to the z axis with their axes lying in the x = 0 plane. One conductor whose axis is at (0, 4 cm, z) carries 12 A in the az direction; the other axis is at (0, −4 cm, z) and carries 12 A in the −az direction. Each current has its zero reference for A located 4 cm from its axis. Find the total A ﬁeld at: (a) (0, 0, z); (b) (0, 8 cm, z); (c) (4 cm, 4 cm, z); (d) (2 cm, 4 cm, z).
Ans. 0; 2.64 µWb/m; 1.93 µWb/m; 3.40 µWb/m

REFERENCES
1. Boast, W. B. (See References for Chapter 2.) The scalar magnetic potential is deﬁned on p. 220, and its use in mapping magnetic ﬁelds is discussed on p. 444. 2. Jordan, E. C., and K. G. Balmain. Electromagnetic Waves and Radiating Systems. 2d ed. Englewood Cliffs, N.J.: Prentice-Hall, 1968. Vector magnetic potential is discussed on pp. 90–96. 3. Paul, C. R., K. W. Whites, and S. Y. Nasar. Introduction to Electromagnetic Fields. 3d ed. New York: McGraw-Hill, 1998. The vector magnetic potential is presented on pp. 216–20. 4. Skilling, H. H. (See References for Chapter 3.) The “paddle wheel” is introduced on pp. 23–25.

CHAPTER 7
7.1

PROBLEMS

7.2

(a) Find H in rectangular components at P(2, 3, 4) if there is a current ﬁlament on the z axis carrying 8 mA in the az direction. (b) Repeat if the ﬁlament is located at x = −1, y = 2. (c) Find H if both ﬁlaments are present. A ﬁlamentary conductor is formed into an equilateral triangle with sides of length carrying current I . Find the magnetic ﬁeld intensity at the center of the triangle. Two semi-inﬁnite ﬁlaments on the z axis lie in the regions −∞ < z < −a and a < z < ∞. Each carries a current I in the az direction. (a) Calculate H as a function of ρ and φ at z = 0. (b) What value of a will cause the magnitude of H at ρ = 1, z = 0, to be one-half the value obtained for an inﬁnite ﬁlament? Two circular current loops are centered on the z axis at z = ±h. Each loop has radius a and carries current I in the aφ direction. (a) Find H on the z axis over the range −h < z < h. Take I = 1 A and plot |H| as a function of z/a if

7.3

7.4

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ENGINEERING ELECTROMAGNETICS

Figure 7.21 See Problem 7.5.

(b) h = a/4; (c) h = a/2; (d) h = a. Which choice for h gives the most uniform ﬁeld? These are called Helmholtz coils (of a single turn each in this case), and are used in providing uniform ﬁelds. 7.5 7.6 The parallel ﬁlamentary conductors shown in Figure 7.21 lie in free space. Plot |H| versus y, −4 < y < 4, along the line x = 0, z = 2.

A disk of radius a lies in the x y plane, with the z axis through its center. Surface charge of uniform density ρs lies on the disk, which rotates about the z axis at angular velocity rad/s. Find H at any point on the z axis.

7.7

A ﬁlamentary conductor carrying current I in the az direction extends along the entire negative z axis. At z = 0 it connects to a copper sheet that ﬁlls the x > 0, y > 0 quadrant of the x y plane. (a) Set up the Biot-Savart law and ﬁnd H everywhere on the z axis; (b) repeat part (a), but with the copper sheet occupying the entire x y plane (Hint: express aφ in terms of ax and a y and angle φ in the integral). For the ﬁnite-length current element on the z axis, as shown in Figure 7.5, use the Biot-Savart law to derive Eq. (9) of Section 7.1. A current sheet K = 8ax A/m ﬂows in the region −2 < y < 2 in the plane z = 0. Calculate H at P(0, 0, 3).

7.8 7.9

7.10 A hollow spherical conducting shell of radius a has ﬁlamentary connections made at the top (r = a, θ = 0) and bottom (r = a, θ = π ). A direct current I ﬂows down the upper ﬁlament, down the spherical surface, and out the lower ﬁlament. Find H in spherical coordinates (a) inside and (b) outside the sphere. 7.11 An inﬁnite ﬁlament on the z axis carries 20π mA in the az direction. Three az -directed uniform cylindrical current sheets are also present: 400 mA/m at

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Figure 7.22 See Problem 7.12.

7.12 In Figure 7.22, let the regions 0 < z < 0.3 m and 0.7 < z < 1.0 m be conducting slabs carrying uniform current densities of 10 A/m2 in opposite directions as shown. Find H at z =: (a) −0.2; (b) 0.2; (c) 0.4; (d) 0.75; (e) 1.2 m.

ρ = 1 cm, −250 mA/m at ρ = 2 cm, and −300 mA/m at ρ = 3 cm. Calculate Hφ at ρ = 0.5, 1.5, 2.5, and 3.5 cm.

7.13 A hollow cylindrical shell of radius a is centered on the z axis and carries a uniform surface current density of K a aφ . (a) Show that H is not a function of φ or z. (b) Show that Hφ and Hρ are everywhere zero. (c) Show that Hz = 0 for ρ > a. (d) Show that Hz = K a for ρ < a. (e) A second shell, ρ = b, carries a current K b aφ . Find H everywhere. 7.14 A toroid having a cross section of rectangular shape is deﬁned by the following surfaces: the cylinders ρ = 2 and ρ = 3 cm, and the planes z = 1 and z = 2.5 cm. The toroid carries a surface current density of −50az A/m on the surface ρ = 3 cm. Find H at the point P(ρ, φ, z): (a) PA (1.5 cm, 0, 2 cm); (b) PB (2.1 cm, 0, 2 cm); (c) PC (2.7 cm, π/2, 2 cm); (d) PD (3.5 cm, π/2, 2 cm). 7.15 Assume that there is a region with cylindrical symmetry in which the conductivity is given by σ = 1.5e−150ρ kS/m. An electric ﬁeld of 30az V/m is present. (a) Find J. (b) Find the total current crossing the surface ρ < ρ0 , z = 0, all φ. (c) Make use of Amp` re’s circuital law to ﬁnd H. e

7.16 A current ﬁlament carrying I in the −az direction lies along the entire positive z axis. At the origin, it connects to a conducting sheet that forms the x y plane. (a) Find K in the conducting sheet. (b) Use Ampere’s circuital law to ﬁnd H everywhere for z > 0; (c) ﬁnd H for z < 0.

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7.17 A current ﬁlament on the z axis carries a current of 7 mA in the az direction, and current sheets of 0.5 az A/m and −0.2 az A/m are located at ρ = 1 cm and ρ = 0.5 cm, respectively. Calculate H at: (a) ρ = 0.5 cm; (b) ρ = 1.5 cm; (c) ρ = 4 cm. (d) What current sheet should be located at ρ = 4 cm so that H = 0 for all ρ > 4 cm? 7.18 A wire of 3 mm radius is made up of an inner material (0 < ρ < 2 mm) for which σ = 107 S/m, and an outer material (2 mm < ρ < 3 mm) for which σ = 4 × 107 S/m. If the wire carries a total current of 100 mA dc, determine H everywhere as a function of ρ.

7.19 In spherical coordinates, the surface of a solid conducting cone is described by θ = π/4 and a conducting plane by θ = π/2. Each carries a total current I . The current ﬂows as a surface current radially inward on the plane to the vertex of the cone, and then ﬂows radially outward throughout the cross section of the conical conductor. (a) Express the surface current density as a function of r ; (b) express the volume current density inside the cone as a function of r ; (c) determine H as a function of r and θ in the region between the cone and the plane; (d) determine H as a function of r and θ inside the cone. 7.20 A solid conductor of circular cross section with a radius of 5 mm has a conductivity that varies with radius. The conductor is 20 m long, and there is a potential difference of 0.1 V dc between its two ends. Within the conductor, H = 105 ρ 2 aφ A/m. (a) Find σ as a function of ρ. (b) What is the resistance between the two ends? 7.21 A cylindrical wire of radius a is oriented with the z axis down its center line. The wire carries a nonuniform current down its length of density J = bρ az A/m2 where b is a constant. (a) What total current ﬂows in the wire? (b) Find Hin (0 < ρ < a), as a function of ρ; (c) ﬁnd Hout (ρ > a), as a function of ρ; (d) verify your results of parts (b) and (c) by using ∇ × H = J. 7.22 A solid cylinder of radius a and length L, where L a, contains volume charge of uniform density ρ0 C/m3 . The cylinder rotates about its axis (the z axis) at angular velocity rad/s. (a) Determine the current density J as a function of position within the rotating cylinder. (b) Determine H on-axis by applying the results of Problem 7.6. (c) Determine the magnetic ﬁeld intensity H inside and outside. (d) Check your result of part (c) by taking the curl of H.

7.23 Given the ﬁeld H = 20ρ 2 aφ A/m: (a) Determine the current density J. (b) Integrate J over the circular surface ρ ≤ 1, 0 < φ < 2π, z = 0, to determine the total current passing through that surface in the az direction. (c) Find the total current once more, this time by a line integral around the circular path ρ = 1, 0 < φ < 2π, z = 0.

7.24 Inﬁnitely long ﬁlamentary conductors are located in the y = 0 plane at x = n meters where n = 0, ±1, ±2, . . . Each carries 1 A in the az direction.

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227

(a) Find H on the y axis. As a help,
∞ n=1

y2

y π 1 π = − + 2π y 2 +n 2 2y e −1

7.25 When x, y, and z are positive and less than 5, a certain magnetic ﬁeld intensity may be expressed as H = [x 2 yz/(y + 1)]ax + 3x 2 z 2 a y − [x yz 2 /(y + 1)]az . Find the total current in the ax direction that crosses the strip x = 2, 1 ≤ y ≤ 4, 3 ≤ z ≤ 4, by a method utilizing: (a) a surface integral; (b) a closed line integral. 7.26 Consider a sphere of radius r = 4 centered at (0, 0, 3). Let S1 be that portion of the spherical surface that lies above the x y plane. Find S1 (∇ × H) · dS if H = 3ρ aφ in cylindrical coordinates.

(b) Compare your result of part (a) to that obtained if the ﬁlaments are replaced by a current sheet in the y = 0 plane that carries surface current density K = 1az A/m.

7.27 The magnetic ﬁeld intensity is given in a certain region of space as H = [(x + 2y)/z 2 ]a y + (2/z)az A/m. (a) Find ∇ × H. (b) Find J. (c) Use J to ﬁnd the total current passing through the surface z = 4, 1 ≤ x ≤ 2, 3 ≤ z ≤ 5, in the az direction. (d) Show that the same result is obtained using the other side of Stokes’ theorem. 7.28 Given H = (3r 2 / sin θ)aθ + 54r cos θaφ A/m in free space: (a) Find the total current in the aθ direction through the conical surface θ = 20◦ , 0 ≤ φ ≤ 2π , 0 ≤ r ≤ 5, by whatever side of Stokes’ theorem you like the best. (b) Check the result by using the other side of Stokes’ theorem. 7.29 A long, straight, nonmagnetic conductor of 0.2 mm radius carries a uniformly distributed current of 2 A dc. (a) Find J within the conductor. (b) Use Amp` re’s circuital law to ﬁnd H and B within the conductor. e (c) Show that ∇ × H = J within the conductor. (d) Find H and B outside the conductor. (e) Show that ∇ × H = J outside the conductor.

7.30 (An inversion of Problem 7.20.) A solid, nonmagnetic conductor of circular cross section has a radius of 2 mm. The conductor is inhomogeneous, with σ = 106 (1 + 106 ρ 2 ) S/m. If the conductor is 1 m in length and has a voltage of 1 mV between its ends, ﬁnd: (a) H inside; (b) the total magnetic ﬂux inside the conductor.

7.31 The cylindrical shell deﬁned by 1 cm < ρ < 1.4 cm consists of a nonmagnetic conducting material and carries a total current of 50 A in the az direction. Find the total magnetic ﬂux crossing the plane φ = 0, 0 < z < 1: (a) 0 < ρ < 1.2 cm; (b) 1.0 cm < ρ < 1.4 cm; (c) 1.4 cm < ρ < 20 cm. 7.32 The free space region deﬁned by 1 < z < 4 cm and 2 < ρ < 3 cm is a toroid of rectangular cross section. Let the surface at ρ = 3 cm carry a surface current K = 2az kA/m. (a) Specify the current densities on the surfaces at

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ρ = 2 cm, z = 1 cm, and z = 4 cm. (b) Find H everywhere. (c) Calculate the total ﬂux within the toroid. 7.33 Use an expansion in rectangular coordinates to show that the curl of the gradient of any scalar ﬁeld G is identically equal to zero. 7.34 A ﬁlamentary conductor on the z axis carries a current of 16 A in the az direction, a conducting shell at ρ = 6 carries a total current of 12 A in the −az direction, and another shell at ρ = 10 carries a total current of 4 A in the −az direction. (a) Find H for 0 < ρ < 12. (b) Plot Hφ versus ρ. (c) Find the total ﬂux crossing the surface 1 < ρ < 7, 0 < z < 1, at ﬁxed φ. 7.35 A current sheet, K = 20 az A/m, is located at ρ = 2, and a second sheet, K = −10az A/m, is located at ρ = 4. (a) Let Vm = 0 at P(ρ = 3, φ = 0, z = 5) and place a barrier at φ = π . Find Vm (ρ, φ, z) for −π < φ < π. (b) Let A = 0 at P and ﬁnd A(ρ, φ, z) for 2 < ρ < 4.

7.36 Let A = (3y − z)ax + 2x za y Wb/m in a certain region of free space. (a) Show that ∇ · A = 0. (b) At P(2, −1, 3), ﬁnd A, B, H, and J.

7.38 A square ﬁlamentary differential current loop, d L on a side, is centered at the origin in the z = 0 plane in free space. The current I ﬂows generally in the aφ direction. (a) Assuming that r >> d L, and following a method similar to that in Section 4.7, show that dA = (b) Show that dH = I (d L)2 (2 cos θ ar + sin θ aθ ) 4πr 3 µ0 I (d L)2 sin θ aφ 4πr 2

7.37 Let N = 1000, I = 0.8 A, ρ0 = 2 cm, and a = 0.8 cm for the toroid shown in Figure 7.12b. Find Vm in the interior of the toroid if Vm = 0 at ρ = 2.5 cm, φ = 0.3π . Keep φ within the range 0 < φ < 2π.

The square loop is one form of a magnetic dipole. 7.39 Planar current sheets of K = 30az A/m and −30az A/m are located in free space at x = 0.2 and x = −0.2, respectively. For the region −0.2 < x < 0.2 (a) ﬁnd H; (b) obtain an expression for Vm if Vm = 0 at P(0.1, 0.2, 0.3); (c) ﬁnd B; (d) obtain an expression for A if A = 0 at P. 7.40 Show that the line integral of the vector potential A about any closed path is equal to the magnetic ﬂux enclosed by the path, or A · dL = B · dS.

7.41 Assume that A = 50ρ 2 az Wb/m in a certain region of free space. (a) Find H and B. (b) Find J. (c) Use J to ﬁnd the total current crossing the surface 0 ≤ ρ ≤ 1, 0 ≤ φ < 2π , z = 0. (d) Use the value of Hφ at ρ = 1 to calculate H · dL for ρ = 1, z = 0.

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229

3 7.42 Show that ∇2 (1/R12 ) = −∇1 (1/R12 ) = R21 /R12 .

7.43 Compute the vector magnetic potential within the outer conductor for the coaxial line whose vector magnetic potential is shown in Figure 7.20 if the outer radius of the outer conductor is 7a. Select the proper zero reference and sketch the results on the ﬁgure.

7.44 By expanding Eq. (58), Section 7.7 in rectangular coordinates, show that (59) is correct.

8

C H A P T E R

Magnetic Forces, Materials, and Inductance

W

e are now ready to undertake the second half of the magnetic ﬁeld problem, that of determining the forces and torques exerted by the magnetic ﬁeld on other charges. The electric ﬁeld causes a force to be exerted on a charge that may be either stationary or in motion; we will see that the steady magnetic ﬁeld is capable of exerting a force only on a moving charge. This result appears reasonable; a magnetic ﬁeld may be produced by moving charges and may exert forces on moving charges; a magnetic ﬁeld cannot arise from stationary charges and cannot exert any force on a stationary charge. This chapter initially considers the forces and torques on current-carrying conductors that may either be of a ﬁlamentary nature or possess a ﬁnite cross section with a known current density distribution. The problems associated with the motion of particles in a vacuum are largely avoided. With an understanding of the fundamental effects produced by the magnetic ﬁeld, we may then consider the varied types of magnetic materials, the analysis of elementary magnetic circuits, the forces on magnetic materials, and ﬁnally, the important electrical circuit concepts of self-inductance and mutual inductance. ■

8.1 FORCE ON A MOVING CHARGE
In an electric ﬁeld, the deﬁnition of the electric ﬁeld intensity shows us that the force on a charged particle is F = QE
230

(1)

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231

The force is in the same direction as the electric ﬁeld intensity (for a positive charge) and is directly proportional to both E and Q. If the charge is in motion, the force at any point in its trajectory is then given by (1). A charged particle in motion in a magnetic ﬁeld of ﬂux density B is found experimentally to experience a force whose magnitude is proportional to the product of the magnitudes of the charge Q, its velocity v, and the ﬂux density B, and to the sine of the angle between the vectors v and B. The direction of the force is perpendicular to both v and B and is given by a unit vector in the direction of v × B. The force may therefore be expressed as F = Qv × B (2)

A fundamental difference in the effect of the electric and magnetic ﬁelds on charged particles is now apparent, for a force which is always applied in a direction at right angles to the direction in which the particle is proceeding can never change the magnitude of the particle velocity. In other words, the acceleration vector is always normal to the velocity vector. The kinetic energy of the particle remains unchanged, and it follows that the steady magnetic ﬁeld is incapable of transferring energy to the moving charge. The electric ﬁeld, on the other hand, exerts a force on the particle which is independent of the direction in which the particle is progressing and therefore effects an energy transfer between ﬁeld and particle in general. The ﬁrst two problems at the end of this chapter illustrate the different effects of electric and magnetic ﬁelds on the kinetic energy of a charged particle moving in free space. The force on a moving particle arising from combined electric and magnetic ﬁelds is obtained easily by superposition, F = Q(E + v × B) (3)

This equation is known as the Lorentz force equation, and its solution is required in determining electron orbits in the magnetron, proton paths in the cyclotron, plasma characteristics in a magnetohydrodynamic (MHD) generator, or, in general, chargedparticle motion in combined electric and magnetic ﬁelds. D8.1. The point charge Q = 18 nC has a velocity of 5×106 m/s in the direction aν = 0.60ax +0.75a y + 0.30az . Calculate the magnitude of the force exerted on the charge by the ﬁeld: (a) B = −3ax + 4a y + 6az mT; (b) E = −3ax + 4a y + 6az kV/m; (c) B and E acting together.
Ans. 660 µN; 140 µN; 670 µN

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8.2 FORCE ON A DIFFERENTIAL CURRENT ELEMENT
The force on a charged particle moving through a steady magnetic ﬁeld may be written as the differential force exerted on a differential element of charge, dF = dQ v × B (4)

Physically, the differential element of charge consists of a large number of very small, discrete charges occupying a volume which, although small, is much larger than the average separation between the charges. The differential force expressed by (4) is thus merely the sum of the forces on the individual charges. This sum, or resultant force, is not a force applied to a single object. In an analogous way, we might consider the differential gravitational force experienced by a small volume taken in a shower of falling sand. The small volume contains a large number of sand grains, and the differential force is the sum of the forces on the individual grains within the small volume. If our charges are electrons in motion in a conductor, however, we can show that the force is transferred to the conductor and that the sum of this extremely large number of extremely small forces is of practical importance. Within the conductor, electrons are in motion throughout a region of immobile positive ions which form a crystalline array, giving the conductor its solid properties. A magnetic ﬁeld which exerts forces on the electrons tends to cause them to shift position slightly and produces a small displacement between the centers of “gravity” of the positive and negative charges. The Coulomb forces between electrons and positive ions, however, tend to resist such a displacement. Any attempt to move the electrons, therefore, results in an attractive force between electrons and the positive ions of the crystalline lattice. The magnetic force is thus transferred to the crystalline lattice, or to the conductor itself. The Coulomb forces are so much greater than the magnetic forces in good conductors that the actual displacement of the electrons is almost immeasurable. The charge separation that does result, however, is disclosed by the presence of a slight potential difference across the conductor sample in a direction perpendicular to both the magnetic ﬁeld and the velocity of the charges. The voltage is known as the Hall voltage, and the effect itself is called the Hall effect. Figure 8.1 illustrates the direction of the Hall voltage for both positive and negative charges in motion. In Figure 8.1a, v is in the −ax direction, v × B is in the a y direction, and Q is positive, causing F Q to be in the a y direction; thus, the positive charges move to the right. In Figure 8.1b, v is now in the +ax direction, B is still in the az direction, v × B is in the −a y direction, and Q is negative; thus, F Q is again in the a y direction. Hence, the negative charges end up at the right edge. Equal currents provided by holes and electrons in semiconductors can therefore be differentiated by their Hall voltages. This is one method of determining whether a given semiconductor is n-type or p-type. Devices employ the Hall effect to measure the magnetic ﬂux density and, in some applications where the current through the device can be made proportional to the

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233

Figure 8.1 Equal currents directed into the material are provided by positive charges moving inward in (a) and negative charges moving outward in (b). The two cases can be distinguished by oppositely directed Hall voltages, as shown.

magnetic ﬁeld across it, to serve as electronic wattmeters, squaring elements, and so forth. Returning to (4), we may therefore say that if we are considering an element of moving charge in an electron beam, the force is merely the sum of the forces on the individual electrons in that small volume element, but if we are considering an element of moving charge within a conductor, the total force is applied to the solid conductor itself. We will now limit our attention to the forces on current-carrying conductors. In Chapter 5 we deﬁned convection current density in terms of the velocity of the volume charge density, J = ρν v The differential element of charge in (4) may also be expressed in terms of volume charge density,1 dQ = ρν dν Thus dF = ρν dν v × B or dF = J × B dν (5)

We saw in Chapter 7 that J dν may be interpreted as a differential current element; that is, J dν = K dS = I dL
1 Remember

that dν is a differential volume element and not a differential increase in velocity.

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ENGINEERING ELECTROMAGNETICS

and thus the Lorentz force equation may be applied to surface current density, dF = K × B dS or to a differential current ﬁlament, dF = I dL × B (7) (6)

Integrating (5), (6), or (7) over a volume, a surface which may be either open or closed (why?), or a closed path, respectively, leads to the integral formulations F= F= and F= I dL × B = −I B × dL (10) vol J × B dν

(8) (9)

S

K × B dS

One simple result is obtained by applying (7) or (10) to a straight conductor in a uniform magnetic ﬁeld, F = IL × B The magnitude of the force is given by the familiar equation F = BIL sin θ (12) (11)

where θ is the angle between the vectors representing the direction of the current ﬂow and the direction of the magnetic ﬂux density. Equation (11) or (12) applies only to a portion of the closed circuit, and the remainder of the circuit must be considered in any practical problem.
E X A M P L E 8.1

As a numerical example of these equations, consider Figure 8.2. We have a square loop of wire in the z = 0 plane carrying 2 mA in the ﬁeld of an inﬁnite ﬁlament on the y axis, as shown. We desire the total force on the loop.
Solution. The ﬁeld produced in the plane of the loop by the straight ﬁlament is

H= Therefore,

15 I az = az A/m 2πx 2πx 3 × 10−6 az T x

B = µ0 H = 4π × 10−7 H =

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235

Figure 8.2 A square loop of wire in the xy plane carrying 2 mA is subjected to a nonuniform B field.

We use the integral form (10), F = −I B × dL

Let us assume a rigid loop so that the total force is the sum of the forces on the four sides. Beginning with the left side: F = −2 × 10−3 × 3 × 10−6
1 3 x=1 0 y=2

az × d x ax + x

2 y=0

az × dy a y 3

+

x=3

az × d x ax + x
3

az × dy a y 1

= −6 × 10−9 ln x a y +
1

1 0 1 2 y (−ax ) + ln x a y + y (−ax ) 3 0 3 2

2 1 = −6 × 10−9 (ln 3)a y − ax + ln a y + 2ax 3 3 = −8ax nN Thus, the net force on the loop is in the −ax direction. D8.2. The ﬁeld B = −2ax + 3a y + 4az mT is present in free space. Find the vector force exerted on a straight wire carrying 12 A in the a AB direction, given A(1, 1, 1) and: (a) B(2, 1, 1); (b) B(3, 5, 6).
Ans. −48a y + 36az mN; 12ax − 216a y + 168az mN

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D8.3. The semiconductor sample shown in Figure 8.1 is n-type silicon, having a rectangular cross section of 0.9 mm by 1.1 cm and a length of 1.3 cm. Assume the electron and hole mobilities are 0.13 and 0.03 m2 /V · s, respectively, at the operating temperature. Let B = 0.07 T and the electric ﬁeld intensity in the direction of the current ﬂow be 800 V/m. Find the magnitude of: (a) the voltage across the sample length; (b) the drift velocity; (c) the transverse force per coulomb of moving charge caused by B; (d) the transverse electric ﬁeld intensity; (e) the Hall voltage.
Ans. 10.40 V; 104.0 m/s; 7.28 N/C; 7.28 V/m; 80.1 mV

8.3 FORCE BETWEEN DIFFERENTIAL CURRENT ELEMENTS
The concept of the magnetic ﬁeld was introduced to break into two parts the problem of ﬁnding the interaction of one current distribution on a second current distribution. It is possible to express the force on one current element directly in terms of a second current element without ﬁnding the magnetic ﬁeld. Because we claimed that the magnetic-ﬁeld concept simpliﬁes our work, it then behooves us to show that avoidance of this intermediate step leads to more complicated expressions. The magnetic ﬁeld at point 2 due to a current element at point 1 was found to be dH2 = I1 dL1 × a R12 2 4πR12

Now, the differential force on a differential current element is dF = I dL × B and we apply this to our problem by letting B be dB2 (the differential ﬂux density at point 2 caused by current element 1), by identifying I dL as I2 dL2 , and by symbolizing the differential amount of our differential force on element 2 as d(dF2 ): d(dF2 ) = I2 dL2 × dB2 Because dB2 = µ0 dH2 , we obtain the force between two differential current elements, d(dF2 ) = µ0
E X A M P L E 8.2

I1 I2 dL2 × (dL1 × a R12 ) 2 4πR12

(13)

As an example that illustrates the use (and misuse) of these results, consider the two differential current elements shown in Figure 8.3. We seek the differential force on dL2 .

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Figure 8.3 Given P1 (5, 2, 1), P2 (1, 8, 5), I 1 dL1 = −3a y A · m, and I 2 dL2 = −4az A · m, the force on I 2 dL2 is 8.56 nN in the a y direction.

Solution. We have I1 dL1 = −3a y A · m at P1 (5, 2, 1), and I2 dL2 = −4az A · m at

P2 (1, 8, 5). Thus, R12 = −4ax +6a y +4az , and we may substitute these data into (13), d(dF2 ) = 4π10−7 (−4az ) × [(−3a y ) × (−4ax + 6a y + 4az )] 4π (16 + 36 + 16)1.5 = 8.56a y nN

Many chapters ago, when we discussed the force exerted by one point charge on another point charge, we found that the force on the ﬁrst charge was the negative of that on the second. That is, the total force on the system was zero. This is not the case with the differential current elements, and d(dF1 ) = −12.84az nN in Example 8.2. The reason for this different behavior lies with the nonphysical nature of the current element. Whereas point charges may be approximated quite well by small charges, the continuity of current demands that a complete circuit be considered. This we shall now do. The total force between two ﬁlamentary circuits is obtained by integrating twice: F 2 = µ0 I1 I2 4π dL2 × dL1 × a R12 2 R12

(14) a R12 × dL1 × dL2 2 R12 Equation (14) is quite formidable, but the familiarity gained in Chapter 7 with the magnetic ﬁeld should enable us to recognize the inner integral as the integral necessary to ﬁnd the magnetic ﬁeld at point 2 due to the current element at point 1. Although we shall only give the result, it is not very difﬁcult to use (14) to ﬁnd the force of repulsion between two inﬁnitely long, straight, parallel, ﬁlamentary I1 I2 = µ0 4π

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Figure 8.4 Two infinite parallel filaments with separation d and equal but opposite currents I experience a repulsive force of µ0 I 2 /(2πd ) N/m.

conductors with separation d, and carrying equal but opposite currents I , as shown in Figure 8.4. The integrations are simple, and most errors are made in determining suitable expressions for a R12 , dL1 , and dL2 . However, since the magnetic ﬁeld intensity at either wire caused by the other is already known to be I/(2πd), it is readily apparent that the answer is a force of µ0 I 2 /(2πd) newtons per meter length. D8.4. Two differential current elements, I1 L1 = 3 × 10−6 a y A · m at P1 (1, 0, 0) and I2 L2 = 3 × 10−6 (−0.5ax + 0.4a y + 0.3az ) A · m at P2 (2, 2, 2), are located in free space. Find the vector force exerted on: (a) I2 L2 by I1 L1 ; (b) I1 L1 by I2 L2 .
Ans. (−1.333ax + 0.333a y − 2.67az )10−20 N; (4.67ax + 0.667az )10−20 N

8.4 FORCE AND TORQUE ON A CLOSED CIRCUIT
We have already obtained general expressions for the forces exerted on current systems. One special case is easily disposed of, for if we take our relationship for the force on a ﬁlamentary closed circuit, as given by Eq. (10), Section 8.2, F = −I B × dL dL

and assume a uniform magnetic ﬂux density, then B may be removed from the integral: F = −I B ×

However, we discovered during our investigation of closed line integrals in an electrostatic potential ﬁeld that dL = 0, and therefore the force on a closed ﬁlamentary circuit in a uniform magnetic ﬁeld is zero. If the ﬁeld is not uniform, the total force need not be zero.

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Figure 8.5 (a) Given a lever arm R extending from an origin O to a point P where force F is applied, the torque about O is T = R × F. (b) If F2 = −F1 , then the torque T = R21 × F1 is independent of the choice of origin for R1 and R2 .

This result for uniform ﬁelds does not have to be restricted to ﬁlamentary circuits only. The circuit may contain surface currents or volume current density as well. If the total current is divided into ﬁlaments, the force on each one is zero, as we have shown, and the total force is again zero. Therefore, any real closed circuit carrying direct currents experiences a total vector force of zero in a uniform magnetic ﬁeld. Although the force is zero, the torque is generally not equal to zero. In deﬁning the torque, or moment, of a force, it is necessary to consider both an origin at or about which the torque is to be calculated, and the point at which the force is applied. In Figure 8.5a, we apply a force F at point P, and we establish an origin at O with a rigid lever arm R extending from O to P. The torque about point O is a vector whose magnitude is the product of the magnitudes of R, of F, and of the sine of the angle between these two vectors. The direction of the vector torque T is normal to both the force F and the lever arm R and is in the direction of progress of a right-handed screw as the lever arm is rotated into the force vector through the smaller angle. The torque is expressible as a cross product, T=R×F Now assume that two forces, F1 at P1 and F2 at P2 , having lever arms R1 and R2 extending from a common origin O, as shown in Figure 8.5b, are applied to an object of ﬁxed shape and that the object does not undergo any translation. Then the torque about the origin is T = R1 × F1 + R2 × F2 where F1 + F 2 = 0

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and therefore T = (R1 − R2 ) × F1 = R21 × F1 The vector R21 = R1 − R2 joins the point of application of F2 to that of F1 and is independent of the choice of origin for the two vectors R1 and R2 . Therefore, the torque is also independent of the choice of origin, provided that the total force is zero. This may be extended to any number of forces. Consider the application of a vertically upward force at the end of a horizontal crank handle on an elderly automobile. This cannot be the only applied force, for if it were, the entire handle would be accelerated in an upward direction. A second force, equal in magnitude to that exerted at the end of the handle, is applied in a downward direction by the bearing surface at the axis of rotation. For a 40-N force on a crank handle 0.3 m in length, the torque is 12 N · m. This ﬁgure is obtained regardless of whether the origin is considered to be on the axis of rotation (leading to 12 N · m plus 0 N · m), at the midpoint of the handle (leading to 6 N · m plus 6 N · m), or at some point not even on the handle or an extension of the handle. We may therefore choose the most convenient origin, and this is usually on the axis of rotation and in the plane containing the applied forces if the several forces are coplanar. With this introduction to the concept of torque, let us now consider the torque on a differential current loop in a magnetic ﬁeld B. The loop lies in the x y plane (Figure 8.6); the sides of the loop are parallel to the x and y axes and are of length d x and dy. The value of the magnetic ﬁeld at the center of the loop is taken as B0 .

Figure 8.6 A differential current loop in a magnetic field B. The torque on the loop is d T = I (dx dyaz ) × B0 = I dS × B.

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Since the loop is of differential size, the value of B at all points on the loop may be taken as B0 . (Why was this not possible in the discussion of curl and divergence?) The total force on the loop is therefore zero, and we are free to choose the origin for the torque at the center of the loop. The vector force on side 1 is dF1 = I d x ax × B0 or dF1 = I d x(B0y az − B0z a y )

For this side of the loop the lever arm R extends from the origin to the midpoint of the side, R1 = − 1 dy a y , and the contribution to the total torque is 2 dT1 = R1 × dF1 = − 1 dy a y × I d x(B0y az − B0z a y ) 2 = − 1 d x d y I B0y ax 2

The torque contribution on side 3 is found to be the same, dT3 = R3 × dF3 = 1 dy a y × (−I d x ax × B0 ) 2 = − 1 d x d y IB0y ax = dT1 2 and dT1 + dT3 = −d x d y IB0y ax Evaluating the torque on sides 2 and 4, we ﬁnd dT2 + dT4 = d x d y IB0x a y and the total torque is then dT = I d x dy(B0x a y − B0y ax ) The quantity within the parentheses may be represented by a cross product, dT = I d x dy(az × B0 ) or dT = I dS × B (15)

where dS is the vector area of the differential current loop and the subscript on B0 has been dropped. We now deﬁne the product of the loop current and the vector area of the loop as the differential magnetic dipole moment dm, with units of A · m2 . Thus dm = I dS (16)

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and dT = dm × B (17)

If we extend the results we obtained in Section 4.7 for the differential electric dipole by determining the torque produced on it by an electric ﬁeld, we see a similar result, dT = dp × E Equations (15) and (17) are general results that hold for differential loops of any shape, not just rectangular ones. The torque on a circular or triangular loop is also given in terms of the vector surface or the moment by (15) or (17). Because we selected a differential current loop so that we might assume B was constant throughout it, it follows that the torque on a planar loop of any size or shape in a uniform magnetic ﬁeld is given by the same expression, T = IS × B = m × B (18)

We should note that the torque on the current loop always tends to turn the loop so as to align the magnetic ﬁeld produced by the loop with the applied magnetic ﬁeld that is causing the torque. This is perhaps the easiest way to determine the direction of the torque.
E X A M P L E 8.3

To illustrate some force and torque calculations, consider the rectangular loop shown in Figure 8.7. Calculate the torque by using T = I S × B. B0 = −0.6a y + 0.8az T. The loop current is 4 mA, a value that is sufﬁciently small to avoid causing any magnetic ﬁeld that might affect B0 . We have T = 4 × 10−3 [(1)(2)az ] × (−0.6a y + 0.8az ) = 4.8ax mN · m
Solution. The loop has dimensions of 1 m by 2 m and lies in the uniform ﬁeld

Thus, the loop tends to rotate about an axis parallel to the positive x axis. The small magnetic ﬁeld produced by the 4 mA loop current tends to line up with B0 .
E X A M P L E 8.4

Now let us ﬁnd the torque once more, this time by calculating the total force and torque contribution for each side.
Solution. On side 1 we have

F1 = I L1 × B0 = 4 × 10−3 (1ax ) × (−0.6a y + 0.8az ) = −3.2a y − 2.4az mN

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Figure 8.7 A rectangular loop is located in a uniform magnetic flux density B0 .

On side 3 we obtain the negative of this result, F3 = 3.2a y + 2.4az mN Next, we attack side 2: F2 = I L2 × B0 = 4 × 10−3 (2a y ) × (−0.6a y + 0.8az ) = 6.4ax mN with side 4 again providing the negative of this result, F4 = −6.4ax mN Because these forces are distributed uniformly along each of the sides, we treat each force as if it were applied at the center of the side. The origin for the torque may be established anywhere since the sum of the forces is zero, and we choose the center of the loop. Thus, T = T1 + T2 + T3 + T4 = R1 × F1 + R2 × F2 + R3 × F3 + R4 × F4 = (−1a y ) × (−3.2a y − 2.4az ) + (0.5ax ) × (6.4ax ) + (1a y ) × (3.2a y + 2.4az ) + (−0.5ax ) × (−6.4ax )

= 2.4ax + 2.4ax = 4.8ax mN · m

Crossing the loop moment with the magnetic ﬂux density is certainly easier.

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D8.5. A conducting ﬁlamentary triangle joins points A(3, 1, 1), B(5, 4, 2), and C(1, 2, 4). The segment AB carries a current of 0.2 A in the a AB direction. There is present a magnetic ﬁeld B = 0.2ax − 0.1a y + 0.3az T. Find: (a) the force on segment BC; (b) the force on the triangular loop; (c) the torque on the loop about an origin at A; (d) the torque on the loop about an origin at C.
Ans. −0.08ax + 0.32a y + 0.16az N; 0; −0.16ax − 0.08a y + 0.08az N · m; −0.16ax − 0.08a y + 0.08az N · m

8.5 THE NATURE OF MAGNETIC MATERIALS
We are now in a position to combine our knowledge of the action of a magnetic ﬁeld on a current loop with a simple model of an atom and obtain some appreciation of the difference in behavior of various types of materials in magnetic ﬁelds. Although accurate quantitative results can only be predicted through the use of quantum theory, the simple atomic model, which assumes that there is a central positive nucleus surrounded by electrons in various circular orbits, yields reasonable quantitative results and provides a satisfactory qualitative theory. An electron in an orbit is analogous to a small current loop (in which the current is directed oppositely to the direction of electron travel) and, as such, experiences a torque in an external magnetic ﬁeld, the torque tending to align the magnetic ﬁeld produced by the orbiting electron with the external magnetic ﬁeld. If there were no other magnetic moments to consider, we would then conclude that all the orbiting electrons in the material would shift in such a way as to add their magnetic ﬁelds to the applied ﬁeld, and thus that the resultant magnetic ﬁeld at any point in the material would be greater than it would be at that point if the material were not present. A second moment, however, is attributed to electron spin. Although it is tempting to model this phenomenon by considering the electron as spinning about its own axis and thus generating a magnetic dipole moment, satisfactory quantitative results are not obtained from such a theory. Instead, it is necessary to digest the mathematics of relativistic quantum theory to show that an electron may have a spin magnetic moment of about ±9 × 10−24 A · m2 ; the plus and minus signs indicate that alignment aiding or opposing an external magnetic ﬁeld is possible. In an atom with many electrons present, only the spins of those electrons in shells which are not completely ﬁlled will contribute to a magnetic moment for the atom. A third contribution to the moment of an atom is caused by nuclear spin. Although this factor provides a negligible effect on the overall magnetic properties of materials, it is the basis of the nuclear magnetic resonance imaging (MRI) procedure provided by many of the larger hospitals. Thus each atom contains many different component moments, and their combination determines the magnetic characteristics of the material and provides its general magnetic classiﬁcation. We describe brieﬂy six different types of material: diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, ferrimagnetic, and superparamagnetic.

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Let us ﬁrst consider atoms in which the small magnetic ﬁelds produced by the motion of the electrons in their orbits and those produced by the electron spin combine to produce a net ﬁeld of zero. Note that we are considering here the ﬁelds produced by the electron motion itself in the absence of any external magnetic ﬁeld; we might also describe this material as one in which the permanent magnetic moment m0 of each atom is zero. Such a material is termed diamagnetic. It would seem, therefore, that an external magnetic ﬁeld would produce no torque on the atom, no realignment of the dipole ﬁelds, and consequently an internal magnetic ﬁeld that is the same as the applied ﬁeld. With an error that only amounts to about one part in a hundred thousand, this is correct. Let us select an orbiting electron whose moment m is in the same direction as the applied ﬁeld B0 (Figure 8.8). The magnetic ﬁeld produces an outward force on the orbiting electron. Since the orbital radius is quantized and cannot change, the inward Coulomb force of attraction is also unchanged. The force unbalance created by the outward magnetic force must therefore be compensated for by a reduced orbital velocity. Hence, the orbital moment decreases, and a smaller internal ﬁeld results. If we had selected an atom for which m and B0 were opposed, the magnetic force would be inward, the velocity would increase, the orbital moment would increase, and greater cancellation of B0 would occur. Again a smaller internal ﬁeld would result. Metallic bismuth shows a greater diamagnetic effect than most other diamagnetic materials, among which are hydrogen, helium, the other “inert” gases, sodium chloride, copper, gold, silicon, germanium, graphite, and sulfur. We should also realize that the diamagnetic effect is present in all materials, because it arises from an interaction of the external magnetic ﬁeld with every orbiting electron; however, it is overshadowed by other effects in the materials we shall consider next. Now consider an atom in which the effects of the electron spin and orbital motion do not quite cancel. The atom as a whole has a small magnetic moment, but the random orientation of the atoms in a larger sample produces an average magnetic moment of zero. The material shows no magnetic effects in the absence of an external ﬁeld.

Figure 8.8 An orbiting electron is shown having a magnetic moment m in the same direction as an applied field B0 .

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When an external ﬁeld is applied, however, there is a small torque on each atomic moment, and these moments tend to become aligned with the external ﬁeld. This alignment acts to increase the value of B within the material over the external value. However, the diamagnetic effect is still operating on the orbiting electrons and may counteract the increase. If the net result is a decrease in B, the material is still called diamagnetic. However, if there is an increase in B, the material is termed paramagnetic. Potassium, oxygen, tungsten, and the rare earth elements and many of their salts, such as erbium chloride, neodymium oxide, and yttrium oxide, one of the materials used in masers, are examples of paramagnetic substances. The remaining four classes of material, ferromagnetic, antiferromagnetic, ferrimagnetic, and superparamagnetic, all have strong atomic moments. Moreover, the interaction of adjacent atoms causes an alignment of the magnetic moments of the atoms in either an aiding or exactly opposing manner. In ferromagnetic materials, each atom has a relatively large dipole moment, caused primarily by uncompensated electron spin moments. Interatomic forces cause these moments to line up in a parallel fashion over regions containing a large number of atoms. These regions are called domains, and they may have a variety of shapes and sizes ranging from one micrometer to several centimeters, depending on the size, shape, material, and magnetic history of the sample. Virgin ferromagnetic materials will have domains which each have a strong magnetic moment; the domain moments, however, vary in direction from domain to domain. The overall effect is therefore one of cancellation, and the material as a whole has no magnetic moment. Upon application of an external magnetic ﬁeld, however, those domains which have moments in the direction of the applied ﬁeld increase their size at the expense of their neighbors, and the internal magnetic ﬁeld increases greatly over that of the external ﬁeld alone. When the external ﬁeld is removed, a completely random domain alignment is not usually attained, and a residual, or remnant, dipole ﬁeld remains in the macroscopic structure. The fact that the magnetic moment of the material is different after the ﬁeld has been removed, or that the magnetic state of the material is a function of its magnetic history, is called hysteresis, a subject which will be discussed again when magnetic circuits are studied in Section 8.8. Ferromagnetic materials are not isotropic in single crystals, and we will therefore limit our discussion to polycrystalline materials, except for mentioning that one of the characteristics of anisotropic magnetic materials is magnetostriction, or the change in dimensions of the crystal when a magnetic ﬁeld is impressed on it. The only elements that are ferromagnetic at room temperature are iron, nickel, and cobalt, and they lose all their ferromagnetic characteristics above a temperature called the Curie temperature, which is 1043 K (770◦ C) for iron. Some alloys of these metals with each other and with other metals are also ferromagnetic, as for example alnico, an aluminum-nickel-cobalt alloy with a small amount of copper. At lower temperatures some of the rare earth elements, such as gadolinium and dysprosium, are ferromagnetic. It is also interesting that some alloys of nonferromagnetic metals are ferromagnetic, such as bismuth-manganese and copper-manganese-tin. In antiferromagnetic materials, the forces between adjacent atoms cause the atomic moments to line up in an antiparallel fashion. The net magnetic moment is

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Table 8.1

Characteristics of magnetic materials

Classificatio Diamagnetic Paramagnetic Ferromagnetic Antiferromagnetic Ferrimagnetic Superparamagnetic

Magnetic Moments morb + mspin = 0 |mspin | |morb | |morb | |morb | |morb | morb + mspin = small |mspin | |mspin | |mspin |

B Values Bint < Bappl Bint > Bappl Bint Bappl . Bint = Bappl Bint > Bappl Bint > Bappl

Domains

Comments . Bint = Bappl . Bint = Bappl

Adjacent moments oppose Unequal adjacent moments oppose; low σ Nonmagnetic matrix; recording tapes

zero, and antiferromagnetic materials are affected only slightly by the presence of an external magnetic ﬁeld. This effect was ﬁrst discovered in manganese oxide, but several hundred antiferromagnetic materials have been identiﬁed since then. Many oxides, sulﬁdes, and chlorides are included, such as nickel oxide (NiO), ferrous sulﬁde (FeS), and cobalt chloride (CoCl2 ). Antiferromagnetism is only present at relatively low temperatures, often well below room temperature. The effect is not of engineering importance at present. The ferrimagnetic substances also show an antiparallel alignment of adjacent atomic moments, but the moments are not equal. A large response to an external magnetic ﬁeld therefore occurs, although not as large as that in ferromagnetic materials. The most important group of ferrimagnetic materials are the ferrites, in which the conductivity is low, several orders of magnitude less than that of semiconductors. The fact that these substances have greater resistance than the ferromagnetic materials results in much smaller induced currents in the material when alternating ﬁelds are applied, as for example in transformer cores that operate at the higher frequencies. The reduced currents (eddy currents) lead to lower ohmic losses in the transformer core. The iron oxide magnetite (Fe3 O4 ), a nickel-zinc ferrite (Ni1/2 Zn1/2 Fe2 O4 ), and a nickel ferrite (NiFe2 O4 ) are examples of this class of materials. Ferrimagnetism also disappears above the Curie temperature. Superparamagnetic materials are composed of an assembly of ferromagnetic particles in a nonferromagnetic matrix. Although domains exist within the individual particles, the domain walls cannot penetrate the intervening matrix material to the adjacent particle. An important example is the magnetic tape used in audiotape or videotape recorders. Table 8.1 summarizes the characteristics of the six types of magnetic materials we have discussed.

8.6 MAGNETIZATION AND PERMEABILITY
To place our description of magnetic materials on a more quantitative basis, we will now devote a page or so to showing how the magnetic dipoles act as a distributed source for the magnetic ﬁeld. Our result will be an equation that looks very much like Amp` re’s circuital law, H · dL = I . The current, however, will be the movement of e

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bound charges (orbital electrons, electron spin, and nuclear spin), and the ﬁeld, which has the dimensions of H, will be called the magnetization M. The current produced by the bound charges is called a bound current or Amperian current. Let us begin by deﬁning the magnetization M in terms of the magnetic dipole moment m. The bound current Ib circulates about a path enclosing a differential area dS, establishing a dipole moment (A · m2 ), If there are n magnetic dipoles per unit volume and we consider a volume the total magnetic dipole moment is found by the vector sum mtotal = n ν i=1

m = Ib dS

ν, then

mi

(19)

Each of the mi may be different. Next, we deﬁne the magnetization M as the magnetic dipole moment per unit volume, M = lim ν→0 1 ν

n ν i=1

mi

and see that its units must be the same as for H, amperes per meter. Now let us consider the effect of some alignment of the magnetic dipoles as the result of the application of a magnetic ﬁeld. We shall investigate this alignment along a closed path, a short portion of which is shown in Figure 8.9. The ﬁgure shows several magnetic moments m that make an angle θ with the element of path dL; each moment consists of a bound current Ib circulating about an area dS. We are therefore considering a small volume, dS cos θdL, or dS · dL, within which there are ndS · dL magnetic dipoles. In changing from a random orientation to this partial alignment, the bound current crossing the surface enclosed by the path (to our left as we travel in the a L direction in Figure 8.9) has increased by Ib for each of the ndS · dL dipoles. Thus the differential change in the net bound current I B over the segment dL will be d I B = n Ib dS · dL = M · dL and within an entire closed contour, IB = M · dL (21) (20)

Figure 8.9 A section dL of a closed path along which magnetic dipoles have been partially aligned by some external magnetic field. The alignment has caused the bound current crossing the surface defined by the closed path to increase by nI bdS · dL A.

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Equation (21) merely says that if we go around a closed path and ﬁnd dipole moments going our way more often than not, there will be a corresponding current composed of, for example, orbiting electrons crossing the interior surface. This last expression has some resemblance to Amp` re’s circuital law, and we e may now generalize the relationship between B and H so that it applies to media other than free space. Our present discussion is based on the forces and torques on differential current loops in a B ﬁeld, and we therefore take B as our fundamental quantity and seek an improved deﬁnition of H. We thus write Amp` re’s circuital law e in terms of the total current, bound plus free, B · dL = IT µ0 where I T = IB + I (22)

and I is the total free current enclosed by the closed path. Note that the free current appears without subscript since it is the most important type of current and will be the only current appearing in Maxwell’s equations. Combining these last three equations, we obtain an expression for the free current enclosed, I = I T − IB = We may now deﬁne H in terms of B and M, H= B −M µ0 (24) B −M µ0 · dL (23)

and we see that B = µ0 H in free space where the magnetization is zero. This relationship is usually written in a form that avoids fractions and minus signs: B = µ0 (H + M) We may now use our newly deﬁned H ﬁeld in (23), I = H · dL (26) (25)

obtaining Amp` re’s circuital law in terms of the free currents. e Using the several current densities, we have IB = IT = I =
S

J B · dS JT · dS J · dS

S

S

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With the help of Stokes’ theorem, we may therefore transform (21), (26), and (22) into the equivalent curl relationships: ∇ × M = JB ∇× B = JT µ0

∇ ×H=J

(27)

We will emphasize only (26) and (27), the two expressions involving the free charge, in the work that follows. The relationship between B, H, and M expressed by (25) may be simpliﬁed for linear isotropic media where a magnetic susceptibility χm can be deﬁned: M = χm H Thus we have B = µ0 (H + χm H) = µ0 µr H where µr = 1 + χm µ = µ0 µr B = µH
E X A M P L E 8.5

(28)

(29)

is deﬁned as the relative permeability µr . We next deﬁne the permeability µ: (30) and this enables us to write the simple relationship between B and H, (31)

Given a ferrite material that we shall specify to be operating in a linear mode with B = 0.05 T, let us assume µr = 50, and calculate values for χm , M, and H.
Solution. Because µr = 1 + χm , we have

χm = µr − 1 = 49 Also, B = µr µ0 H and H= 0.05 = 796 A/m 50 × 4π × 10−7

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The magnetization is M = χm H, or 39, 000 A/m. The alternate ways of relating B and H are, ﬁrst, B = µ0 (H + M) or 0.05 = 4π × 10−7 (796 + 39, 000) showing that Amperian currents produce 49 times the magnetic ﬁeld intensity that the free charges do; and second, B = µr µ0 H or 0.05 = 50 × 4π × 10−7 × 796 where we use a relative permeability of 50 and let this quantity account completely for the notion of the bound charges. We shall emphasize the latter interpretation in the chapters that follow. The ﬁrst two laws that we investigated for magnetic ﬁelds were the Biot-Savart law and Amp` re’s circuital law. Both were restricted to free space in their application. e We may now extend their use to any homogeneous, linear, isotropic magnetic material that may be described in terms of a relative permeability µr . Just as we found for anisotropic dielectric materials, the permeability of an anisotropic magnetic material must be given as a 3 × 3 matrix, and B and H are both 3 × 1 matrices. We have Bx = µx x Hx + µx y Hy + µx z Hz B y = µ yx Hx + µ yy Hy + µ yz Hz Bz = µzx Hx + µzy Hy + µzz Hz

For anisotropic materials, then, B = µH is a matrix equation; however, B = µ0 (H + M) remains valid, although B, H, and M are no longer parallel in general. The most common anisotropic magnetic material is a single ferromagnetic crystal, although thin magnetic ﬁlms also exhibit anisotropy. Most applications of ferromagnetic materials, however, involve polycrystalline arrays that are much easier to make. Our deﬁnitions of susceptibility and permeability also depend on the assumption of linearity. Unfortunately, this is true only in the less interesting paramagnetic and diamagnetic materials for which the relative permeability rarely differs from unity by more than one part in a thousand. Some typical values of the susceptibility for diamagnetic materials are hydrogen, −2 × 10−5 ; copper, −0.9 × 10−5 ; germanium, −0.8 × 10−5 ; silicon, −0.3 × 10−5 ; and graphite,−12 × 10−5 . Several representative paramagnetic susceptibilities are oxygen, 2 × 10−6 ; tungsten, 6.8 × 10−5 ; ferric oxide (Fe2 O3 ), 1.4 × 10−3 ; and yttrium oxide (Y2 O3 ), 0.53 × 10−6 . If we simply take the ratio of B to µ0 H as the relative permeability of a ferromagnetic material, typical

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values of µr would range from 10 to 100, 000. Diamagnetic, paramagnetic, and antiferromagnetic materials are commonly said to be nonmagnetic. D8.6. Find the magnetization in a magnetic material where: (a) µ = 1.8 × 10−5 H/m and H = 120 A/m; (b) µr = 22, there are 8.3 × 1028 atoms/m3 , and each atom has a dipole moment of 4.5 × 10−27 A · m2 ; (c) B = 300 µT and χm = 15.
Ans. 1599 A/m; 374 A/m; 224 A/m

D8.7. The magnetization in a magnetic material for which χm = 8 is given in a certain region as 150z 2 ax A/m. At z = 4 cm, ﬁnd the magnitude of: (a) JT ; (b) J; (c) J B .
Ans. 13.5 A/m2 ; 1.5 A/m2 ; 12 A/m2

8.7 MAGNETIC BOUNDARY CONDITIONS
We should have no difﬁculty in arriving at the proper boundary conditions to apply to B, H, and M at the interface between two different magnetic materials, for we have solved similar problems for both conducting materials and dielectrics. We need no new techniques. Figure 8.10 shows a boundary between two isotropic homogeneous linear materials with permeabilities µ1 and µ2 . The boundary condition on the normal components

Figure 8.10 A gaussian surface and a closed path are constructed at the boundary between media 1 and 2, having permeabilities of µ1 and µ2 , respectively. From this we determine the boundary conditions BN1 = BN2 and Ht1 − Ht2 = K , the component of the surface current density directed into the page.

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is determined by allowing the surface to cut a small cylindrical gaussian surface. Applying Gauss’s law for the magnetic ﬁeld from Section 7.5,
S

B · dS = 0

we ﬁnd that BN 1 S − BN 2 S = 0 or BN 2 = BN 1 Thus HN 2 = µ1 HN 1 µ2 (33) (32)

The normal component of B is continuous, but the normal component of H is discontinuous by the ratio µ1 /µ2 . The relationship between the normal components of M, of course, is ﬁxed once the relationship between the normal components of H is known. For linear magnetic materials, the result is written simply as µ1 χm2 µ1 MN 1 (34) M N 2 = χm2 HN 1 = µ2 χm1 µ2 Next, Amp` re’s circuital law e H · dL = I is applied about a small closed path in a plane normal to the boundary surface, as shown to the right in Figure 8.10. Taking a clockwise trip around the path, we ﬁnd that where we assume that the boundary may carry a surface current K whose component normal to the plane of the closed path is K. Thus Ht1 − Ht2 = K (35) Ht1 L − Ht2 L = K L

The directions are speciﬁed more exactly by using the cross product to identify the tangential components, where a N 12 is the unit normal at the boundary directed from region 1 to region 2. An equivalent formulation in terms of the vector tangential components may be more convenient for H: Ht1 − Ht2 = a N 12 × K (H1 − H2 ) × a N 12 = K

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For tangential B, we have Bt2 Bt1 − =K µ1 µ2 (36)

The boundary condition on the tangential component of the magnetization for linear materials is therefore χm2 Mt1 − χm2 K (37) Mt2 = χm1 The last three boundary conditions on the tangential components are much simpler, of course, if the surface current density is zero. This is a free current density, and it must be zero if neither material is a conductor.
E X A M P L E 8.6

To illustrate these relationships with an example, let us assume that µ = µ1 = 4 µH/m in region 1 where z > 0, whereas µ2 = 7 µH/m in region 2 wherever z < 0. Moreover, let K = 80ax A/m on the surface z = 0. We establish a ﬁeld, B1 = 2ax − 3a y + az mT, in region 1 and seek the value of B2 .
Solution. The normal component of B1 is

B N 1 = (B1 · a N 12 )a N 12 = [(2ax − 3a y + az ) · (−az )](−az ) = az mT Thus, B N 2 = B N 1 = az mT We next determine the tangential components: Bt1 = B1 − B N 1 = 2ax − 3ay mT and Ht1 = Thus, Ht2 = Ht1 − a N 12 × K = 500ax − 750a y − (−az ) × 80ax = 500ax − 750a y + 80a y = 500ax − 670a y A/m and Bt2 = µ2 Ht2 = 7 × 10−6 (500ax − 670a y ) = 3.5ax − 4.69a y mT Therefore, B2 = B N 2 + Bt2 = 3.5ax − 4.69a y + az mT (2ax − 3a y )10−3 Bt1 = = 500ax − 750a y A/m µ1 4 × 10−6

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D8.8. Let the permittivity be 5 µH/m in region A where x < 0, and 20 µH/m in region B where x > 0. If there is a surface current density K = 150a y − 200az A/m at x = 0, and if H A = 300ax − 400a y + 500az A/m, ﬁnd: (a) |Ht A |; (b) |H N A |; (c) |Ht B |; (d) |H N B |.
Ans. 640 A/m; 300 A/m; 695 A/m; 75 A/m

8.8 THE MAGNETIC CIRCUIT
In this section, we digress brieﬂy to discuss the fundamental techniques involved in solving a class of magnetic problems known as magnetic circuits. As we will see shortly, the name arises from the great similarity to the dc-resistive-circuit analysis with which it is assumed we are all familiar. The only important difference lies in the nonlinear nature of the ferromagnetic portions of the magnetic circuit; the methods which must be adopted are similar to those required in nonlinear electric circuits which contain diodes, thermistors, incandescent ﬁlaments, and other nonlinear elements. As a convenient starting point, let us identify those ﬁeld equations on which resistive circuit analysis is based. At the same time we will point out or derive the analogous equations for the magnetic circuit. We begin with the electrostatic potential and its relationship to electric ﬁeld intensity, E = −∇V (38a)

The scalar magnetic potential has already been deﬁned, and its analogous relation to the magnetic ﬁeld intensity is H = −∇Vm (38b)

In dealing with magnetic circuits, it is convenient to call Vm the magnetomotive force, or mmf, and we shall acknowledge the analogy to the electromotive force, or emf, by doing so. The units of the mmf are, of course, amperes, but it is customary to recognize that coils with many turns are often employed by using the term “ampereturns.” Remember that no current may ﬂow in any region in which Vm is deﬁned. The electric potential difference between points A and B may be written as V AB =
B A

E · dL

(39a)

and the corresponding relationship between the mmf and the magnetic ﬁeld intensity, Vm AB =
B A

H · dL

(39b)

was developed in Chapter 7, where we learned that the path selected must not cross the chosen barrier surface.

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Ohm’s law for the electric circuit has the point form J = σE (40a)

and we see that the magnetic ﬂux density will be the analog of the current density, B = µH To ﬁnd the total current, we must integrate: I =
S

(40b)

J · dS

(41a)

A corresponding operation is necessary to determine the total magnetic ﬂux ﬂowing through the cross section of a magnetic circuit: = B · dS (41b)

S

We then deﬁned resistance as the ratio of potential difference and current, or V = IR (42a)

and we shall now deﬁne reluctance as the ratio of the magnetomotive force to the total ﬂux; thus Vm = (42b)

where reluctance is measured in ampere-turns per weber (A · t/Wb). In resistors that are made of a linear isotropic homogeneous material of conductivity σ and have a uniform cross section of area S and length d, the total resistance is d (43a) σS If we are fortunate enough to have such a linear isotropic homogeneous magnetic material of length d and uniform cross section S, then the total reluctance is R= = d µS (43b)

The only such material to which we shall commonly apply this relationship is air. Finally, let us consider the analog of the source voltage in an electric circuit. We know that the closed line integral of E is zero, E · dL = 0 In other words, Kirchhoff’s voltage law states that the rise in potential through the source is exactly equal to the fall in potential through the load. The expression for

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magnetic phenomena takes on a slightly different form, H · dL = Itotal for the closed line integral is not zero. Because the total current linked by the path is usually obtained by allowing a current I to ﬂow through an N -turn coil, we may express this result as H · dL = NI (44)

In an electric circuit, the voltage source is a part of the closed path; in the magnetic circuit, the current-carrying coil will surround or link the magnetic circuit. In tracing a magnetic circuit, we will not be able to identify a pair of terminals at which the magnetomotive force is applied. The analogy is closer here to a pair of coupled circuits in which induced voltages exist (and in which we will see in Chapter 9 that the closed line integral of E is also not zero). Let us try out some of these ideas on a simple magnetic circuit. In order to avoid the complications of ferromagnetic materials at this time, we will assume that we have an air-core toroid with 500 turns, a cross-sectional area of 6 cm2 , a mean radius of 15 cm, and a coil current of 4 A. As we already know, the magnetic ﬁeld is conﬁned to the interior of the toroid, and if we consider the closed path of our magnetic circuit along the mean radius, we link 2000 A · t, Vm, source = 2000 A · t Although the ﬁeld in the toroid is not quite uniform, we may assume that it is, for all practical purposes, and calculate the total reluctance of the circuit as = Thus d 2π(0.15) = = 1.25 × 109 A·t/Wb −7 × 6 × 10−4 µS 4π10 = Vm,S = 2000 = 1.6 × 10−6 Wb 1.25 × 109

This value of the total ﬂux is in error by less than 1 percent, in comparison with the 4 value obtained when the exact distribution of ﬂux over the cross section is used. Hence B= and ﬁnally, H= S = 1.6 × 10−6 = 2.67 × 10−3 T 6 × 10−4

2.67 × 10−3 B = 2120 A·t/m = µ 4π10−7 Hφ 2πr = NI

As a check, we may apply Amp` re’s circuital law directly in this symmetrical problem, e

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and obtain Hφ = NI 500 × 4 = = 2120 A/m 2πr 6.28 × 0.15

at the mean radius. Our magnetic circuit in this example does not give us any opportunity to ﬁnd the mmf across different elements in the circuit, for there is only one type of material. The analogous electric circuit is, of course, a single source and a single resistor. We could make it look just as long as the preceding analysis, however, if we found the current density, the electric ﬁeld intensity, the total current, the resistance, and the source voltage. More interesting and more practical problems arise when ferromagnetic materials are present in the circuit. Let us begin by considering the relationship between B and H in such a material. We may assume that we are establishing a curve of B versus H for a sample of ferromagnetic material which is completely demagnetized; both B and H are zero. As we begin to apply an mmf, the ﬂux density also rises, but not linearly, as the experimental data of Figure 8.11 show near the origin. After H reaches a value of about 100 A · t/m, the ﬂux density rises more slowly and begins to saturate when H is several hundred A · t/m. Having reached partial saturation, let us now turn to Figure 8.12, where we may continue our experiment at point x by reducing H. As we do so, the effects of hysteresis begin to show, and we do not retrace our original curve. Even after H is zero, B = Br , the remnant ﬂux density. As H is reversed, then brought back to zero, and the complete cycle traced several times, the hysteresis loop of Figure 8.12 is obtained. The mmf required to reduce the ﬂux density to zero is identiﬁed as Hc , the coercive “force.” For smaller maximum values of H , smaller

Figure 8.11 Magnetization curve of a sample of silicon sheet steel.

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Figure 8.12 A hysteresis loop for silicon steel. The coercive force Hc and remnant flux density Br are indicated.

hysteresis loops are obtained, and the locus of the tips is about the same as the virgin magnetization curve of Figure 8.11.
E X A M P L E 8.7

Let us use the magnetization curve for silicon steel to solve a magnetic circuit problem that is slightly different from our previous example. We use a steel core in the toroid, except for an air gap of 2 mm. Magnetic circuits with air gaps occur because gaps are deliberately introduced in some devices, such as inductors, which must carry large direct currents, because they are unavoidable in other devices such as rotating machines, or because of unavoidable problems in assembly. There are still 500 turns about the toroid, and we ask what current is required to establish a ﬂux density of 1 T everywhere in the core.
Solution. This magnetic circuit is analogous to an electric circuit containing a voltage

source and two resistors, one of which is nonlinear. Because we are given the “current,” it is easy to ﬁnd the “voltage” across each series element, and hence the total “emf.” In the air gap, air =

dair 2 × 10−3 = 2.65 × 106 A·t/Wb = −7 × 6 × 10−4 µS 4π10 = BS = 1(6 × 10−4 ) = 6 × 10−4 Wb

Knowing the total ﬂux,

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which is the same in both steel and air, we may ﬁnd the mmf required for the gap, Vm,air = (6 × 10−4 )(2.65 × 106 ) = 1590 A·t Referring to Figure 8.11, a magnetic ﬁeld strength of 200 A · t/m is required to produce a ﬂux density of 1 T in the steel. Thus, Vm,steel = Hsteel dsteel = 200 × 0.30π = 188 A·t The total mmf is therefore 1778 A·t, and a coil current of 3.56 A is required. We have made several approximations in obtaining this answer. We have already mentioned the lack of a completely uniform cross section, or cylindrical symmetry; the path of every ﬂux line is not of the same length. The choice of a “mean” path length can help compensate for this error in problems in which it may be more important than it is in our example. Fringing ﬂux in the air gap is another source of error, and formulas are available by which we may calculate an effective length and cross-sectional area for the gap which will yield more accurate results. There is also a leakage ﬂux between the turns of wire, and in devices containing coils concentrated on one section of the core, a few ﬂux lines bridge the interior of the toroid. Fringing and leakage are problems that seldom arise in the electric circuit because the ratio of the conductivities of air and the conductive or resistive materials used is so high. In contrast, the magnetization curve for silicon steel shows that the ratio of H to B in the steel is about 200 up to the “knee” of the magnetization curve; this compares with a ratio in air of about 800, 000. Thus, although ﬂux prefers steel to air by the commanding ratio of 4000 to 1, this is not very close to the ratio of conductivities of, say, 1015 for a good conductor and a fair insulator.
E X A M P L E 8.8

Hsteel = 200 A·t

As a last example, let us consider the reverse problem. Given a coil current of 4 A in the magnetic circuit of Example 8.7, what will the ﬂux density be?
Solution. First let us try to linearize the magnetization curve by a straight line from

the origin to B = 1, H = 200. We then have B = H/200 in steel and B = µ0 H in air. The two reluctances are found to be 0.314×106 for the steel path and 2.65×106 for the air gap, or 2.96×106 A · t/Wb total. Since Vm is 2000 A · t, the ﬂux is 6.76×10−4 Wb, and B = 1.13 T. A more accurate solution may be obtained by assuming several values of B and calculating the necessary mmf. Plotting the results enables us to determine the true value of B by interpolation. With this method we obtain B = 1.10 T. The good accuracy of the linear model results from the fact that the reluctance of the air gap in a magnetic circuit is often much greater than the reluctance of the ferromagnetic portion of the circuit. A relatively poor approximation for the iron or steel can thus be tolerated.

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261

Figure 8.13

See Problem D8.9.

D8.9. Given the magnetic circuit of Figure 8.13, assume B = 0.6 T at the midpoint of the left leg and ﬁnd: (a) Vm,air ; (b) Vm,steel ; (c) the current required in a 1300-turn coil linking the left leg.
Ans. 3980 A · t; 72 A · t; 3.12 A

D8.10. The magnetization curve for material X under normal operating conditions may be approximated by the expression B = (H/160)(0.25 + e−H/320 ), where H is in A/m and B is in T. If a magnetic circuit contains a 12 cm length of material X , as well as a 0.25-mm air gap, assume a uniform cross section of 2.5 cm2 and ﬁnd the total mmf required to produce a ﬂux of (a) 10 µWb; (b) 100 µWb.
Ans. 8.58 A · t; 86.7 A · t

8.9 POTENTIAL ENERGY AND FORCES ON MAGNETIC MATERIALS
In the electrostatic ﬁeld we ﬁrst introduced the point charge and the experimental law of force between point charges. After deﬁning electric ﬁeld intensity, electric ﬂux density, and electric potential, we were able to ﬁnd an expression for the energy in an electrostatic ﬁeld by establishing the work necessary to bring the prerequisite point charges from inﬁnity to their ﬁnal resting places. The general expression for energy is WE = 1 2 D · E dν vol (45)

where a linear relationship between D and E is assumed. This is not as easily done for the steady magnetic ﬁeld. It would seem that we might assume two simple sources, perhaps two current sheets, ﬁnd the force on one

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due to the other, move the sheet a differential distance against this force, and equate the necessary work to the change in energy. If we did, we would be wrong, because Faraday’s law (coming up in Chapter 9) shows that there will be a voltage induced in the moving current sheet against which the current must be maintained. Whatever source is supplying the current sheet turns out to receive half the energy we are putting into the circuit by moving it. In other words, energy density in the magnetic ﬁeld may be determined more easily after time-varying ﬁelds are discussed. We will develop the appropriate expression in discussing Poynting’s theorem in Chapter 11. An alternate approach would be possible at this time, however, for we might deﬁne a magnetostatic ﬁeld based on assumed magnetic poles (or “magnetic charges”). Using the scalar magnetic potential, we could then develop an energy expression by methods similar to those used in obtaining the electrostatic energy relationship. These new magnetostatic quantities we would have to introduce would be too great a price to pay for one simple result, and we will therefore merely present the result at this time and show that the same expression arises in the Poynting theorem later. The total energy stored in a steady magnetic ﬁeld in which B is linearly related to H is WH = 1 2 B · H dν vol (46)

Letting B = µH, we have the equivalent formulations WH = or B2 1 dν (48) 2 vol µ It is again convenient to think of this energy as being distributed throughout the volume with an energy density of 1 B · H J/m3, although we have no mathematical 2 justiﬁcation for such a statement. In spite of the fact that these results are valid only for linear media, we may use them to calculate the forces on nonlinear magnetic materials if we focus our attention on the linear media (usually air) which may surround them. For example, suppose that we have a long solenoid with a silicon-steel core. A coil containing n turns/m with a current I surrounds it. The magnetic ﬁeld intensity in the core is therefore n I A · t/m, and the magnetic ﬂux density can be obtained from the magnetization curve for silicon steel. Let us call this value Bst . Suppose that the core is composed of two semi-inﬁnite cylinders2 that are just touching. We now apply a mechanical force to separate these two sections of the core while keeping the ﬂux density constant. We apply a force F over a distance dL, thus doing work F dL. Faraday’s law does not WH = 1 2 µH 2 dν vol (47)

2

A semi-inﬁnite cylinder is a cylinder of inﬁnite length having one end located in ﬁnite space.

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apply here, for the ﬁelds in the core have not changed, and we can therefore use the principle of virtual work to determine that the work we have done in moving one core appears as stored energy in the air gap we have created. By (48), this increase is d W H = F dL =
2 1 Bst S dL 2 µ0

where S is the core cross-sectional area. Thus B2 S F = st 2µ0 If, for example, the magnetic ﬁeld intensity is sufﬁcient to produce saturation in the steel, approximately 1.4 T, the force is F = 7.80 × 105 S N or about 113 lb f /in2 . D8.11. (a) What force is being exerted on the pole faces of the circuit described in Problem D8.9 and Figure 8.13? (b) Is the force trying to open or close the air gap?
Ans. 1194 N; as Wilhelm Eduard Weber would put it, “schliessen”

8.10 INDUCTANCE AND MUTUAL INDUCTANCE
Inductance is the last of the three familiar parameters from circuit theory that we are deﬁning in more general terms. Resistance was deﬁned in Chapter 5 as the ratio of the potential difference between two equipotential surfaces of a conducting material to the total current crossing either equipotential surface. The resistance is a function of conductor geometry and conductivity only. Capacitance was deﬁned in the same chapter as the ratio of the total charge on either of two equipotential conducting surfaces to the potential difference between the surfaces. Capacitance is a function only of the geometry of the two conducting surfaces and the permittivity of the dielectric medium between or surrounding them. As a prelude to deﬁning inductance, we ﬁrst need to introduce the concept of ﬂux linkage. Let us consider a toroid of N turns in which a current I produces a total ﬂux . We assume ﬁrst that this ﬂux links or encircles each of the N turns, and we also see that each of the N turns links the total ﬂux . The flu linkage N is deﬁned as the product of the number of turns N and the ﬂux linking each of them.3 For a coil having a single turn, the ﬂux linkage is equal to the total ﬂux.

3

The symbol λ is commonly used for ﬂux linkages. We will only occasionally use this concept, however, and we will continue to write it as N .

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We now deﬁne inductance (or self-inductance) as the ratio of the total ﬂux linkages to the current which they link, L= N I (49)

The current I ﬂowing in the N -turn coil produces the total ﬂux and N ﬂux linkages, where we assume for the moment that the ﬂux links each turn. This deﬁnition is applicable only to magnetic media which are linear, so that the ﬂux is proportional to the current. If ferromagnetic materials are present, there is no single deﬁnition of inductance which is useful in all cases, and we shall restrict our attention to linear materials. The unit of inductance is the henry (H), equivalent to one weber-turn per ampere. Let us apply (49) in a straightforward way to calculate the inductance per meter length of a coaxial cable of inner radius a and outer radius b. We may take the expression for total ﬂux developed as Eq. (42) in Chapter 7, µ0 Id b ln 2π a and obtain the inductance rapidly for a length d, = L= or, on a per-meter basis, µ0 b ln H/m (50) 2π a In this case, N = 1 turn, and all the ﬂux links all the current. In the problem of a toroidal coil of N turns and a current I , as shown in Figure 7.12b, we have µ0 NI Bφ = 2πρ L= If the dimensions of the cross section are small compared with the mean radius of the toroid ρ0 , then the total ﬂux is = µ0 NIS 2πρ0 µ0 d b ln H 2π a

where S is the cross-sectional area. Multiplying the total ﬂux by N , we have the ﬂux linkages, and dividing by I , we have the inductance L= µ0 N 2 S 2πρ0 (51)

Once again we have assumed that all the ﬂux links all the turns, and this is a good assumption for a toroidal coil of many turns packed closely together. Suppose, however, that our toroid has an appreciable spacing between turns, a short part of which might look like Figure 8.14. The ﬂux linkages are no longer the product of the

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265

Figure 8.14 A portion of a coil showing partial flux linkages. The total flux linkages are obtained by adding the fluxes linking each turn.

ﬂux at the mean radius times the total number of turns. In order to obtain the total ﬂux linkages we must look at the coil on a turn-by-turn basis. (N )total = =
1 N i=1

+ i 2

+ ··· +

i

+ ··· +

N

where i is the ﬂux linking the ith turn. Rather than doing this, we usually rely on experience and empirical quantities called winding factors and pitch factors to adjust the basic formula to apply to the real physical world. An equivalent deﬁnition for inductance may be made using an energy point of view, L= 2W H I2 (52)

where I is the total current ﬂowing in the closed path and W H is the energy in the magnetic ﬁeld produced by the current. After using (52) to obtain several other general expressions for inductance, we will show that it is equivalent to (49). We ﬁrst express the potential energy W H in terms of the magnetic ﬁelds, L= 1 I2 vol and then replace B by ∇ × A, L=

B · H dν I2

(53)

vol

H · (∇ × A)dν

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The vector identity ∇ · (A × H) ≡ H · (∇ × A) − A · (∇ × H) may be proved by expansion in rectangular coordinates. The inductance is then 1 ∇ · (A × H) dν + A · (∇ × H) dν (55) I 2 vol vol After applying the divergence theorem to the ﬁrst integral and letting ∇ × H = J in the second integral, we have L= 1 (A × H) · dS + A · J dν I2 S vol The surface integral is zero, as the surface encloses the volume containing all the magnetic energy, and this requires that A and H be zero on the bounding surface. The inductance may therefore be written as 1 L= 2 A · J dν (56) I vol Equation (56) expresses the inductance in terms of an integral of the values of A and J at every point. Because current density exists only within the conductor, the integrand is zero at all points outside the conductor, and the vector magnetic potential need not be determined there. The vector potential is that which arises from the current J, and any other current source contributing a vector potential ﬁeld in the region of the original current density is to be ignored for the present. Later we will see that this leads to a mutual inductance. The vector magnetic potential A due to J is given by Eq. (51), Chapter 7, µJ dν A= 4πR vol and the inductance may therefore be expressed more basically as a rather formidable double volume integral, L= µJ 1 dν · J dν (57) I 2 vol vol 4πR A slightly simpler integral expression is obtained by restricting our attention to current ﬁlaments of small cross section for which J dν may be replaced by I dL and the volume integral by a closed line integral along the axis of the ﬁlament, L= µI dL · I dL 4πR (58) µ dL = · dL 4π R Our only present interest in Eqs. (57) and (58) lies in their implication that the inductance is a function of the distribution of the current in space or the geometry of the conductor conﬁguration. To obtain our original deﬁnition of inductance (49), let us hypothesize a uniform current distribution in a ﬁlamentary conductor of small cross section so that J dν L= 1 I2 (54)

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in (56) becomes I dL, 1 A · dL (59) I For a small cross section, dL may be taken along the center of the ﬁlament. We now apply Stokes’ theorem and obtain 1 L= (∇ × A) · dS I S or 1 L= B · dS I S or L= (60) I Retracing the steps by which (60) is obtained, we should see that the ﬂux is that portion of the total ﬂux that passes through any and every open surface whose perimeter is the ﬁlamentary current path. If we now let the ﬁlament make N identical turns about the total ﬂux, an idealization that may be closely realized in some types of inductors, the closed line integral must consist of N laps about this common path, and (60) becomes N L= (61) I The ﬂux is now the ﬂux crossing any surface whose perimeter is the path occupied by any one of the N turns. The inductance of an N -turn coil may still be obtained from (60), however, if we realize that the ﬂux is that which crosses the complicated surface4 whose perimeter consists of all N turns. Use of any of the inductance expressions for a true ﬁlamentary conductor (having zero radius) leads to an inﬁnite value of inductance, regardless of the conﬁguration of the ﬁlament. Near the conductor, Amp` re’s circuital law shows that the magnetic e ﬁeld intensity varies inversely with the distance from the conductor, and a simple integration soon shows that an inﬁnite amount of energy and an inﬁnite amount of ﬂux are contained within any ﬁnite cylinder about the ﬁlament. This difﬁculty is eliminated by specifying a small but ﬁnite ﬁlamentary radius. The interior of any conductor also contains magnetic ﬂux, and this ﬂux links a variable fraction of the total current, depending on its location. These ﬂux linkages lead to an internal inductance, which must be combined with the external inductance to obtain the total inductance. The internal inductance of a long, straight wire of circular cross section, radius a, and uniform current distribution is L a,int = µ 8π H/m (62) L=

a result requested in Problem 8.43 at the end of this chapter.

4

Somewhat like a spiral ramp.

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ENGINEERING ELECTROMAGNETICS

In Chapter 11, we will see that the current distribution in a conductor at high frequencies tends to be concentrated near the surface. The internal ﬂux is reduced, and it is usually sufﬁcient to consider only the external inductance. At lower frequencies, however, internal inductance may become an appreciable part of the total inductance. We conclude by deﬁning the mutual inductance between circuits 1 and 2, M12 , in terms of mutual ﬂux linkages, M12 = N2 12 I1 (63)

where 12 signiﬁes the ﬂux produced by I1 which links the path of the ﬁlamentary current I2 , and N2 is the number of turns in circuit 2. The mutual inductance, therefore, depends on the magnetic interaction between two currents. With either current alone, the total energy stored in the magnetic ﬁeld can be found in terms of a single inductance, or self-inductance; with both currents having nonzero values, the total energy is a function of the two self-inductances and the mutual inductance. In terms of a mutual energy, it can be shown that (63) is equivalent to 1 (B1 · H2 )dν (64) M12 = I1 I2 vol or 1 M12 = (µH1 · H2 )dν (65) I1 I2 vol where B1 is the ﬁeld resulting from I1 (with I2 = 0) and H2 is the ﬁeld arising from I2 (with I1 = 0). Interchange of the subscripts does not change the right-hand side of (65), and therefore M12 = M21 (66)

Mutual inductance is also measured in henrys, and we rely on the context to allow us to differentiate it from magnetization, also represented by M.
E X A M P L E 8.9

Calculate the self-inductances of and the mutual inductances between two coaxial solenoids of radius R1 and R2 , R2 > R1 , carrying currents I1 and I2 with n 1 and n 2 turns/m, respectively.
Solution. We ﬁrst attack the mutual inductances. From Eq. (15), Chapter 7, we let

n 1 = N /d, and obtain

H1 = n 1 I1 az =0

(0 < ρ < R1 )

(ρ > R1 ) (0 < ρ < R2 )

and H2 = n 2 I 2 a z =0 (ρ > R2 )

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Thus, for this uniform ﬁeld
12 2 = µ0 n 1 I1 πR1

and
2 M12 = µ0 n 1 n 2 πR1

Similarly,
21 2 M21 = µ0 n 1 n 2 πR1 = M12 2 = µ0 n 2 I2 πR1

If n 1 = 50 turns/cm, n 2 = 80 turns/cm, R1 = 2 cm, and R2 = 3 cm, then

The self-inductances are easily found. The ﬂux produced in coil 1 by I1 is
11 2 = µ0 n 1 I1 πR1

M12 = M21 = 4π × 10−7 (5000)(8000)π (0.022 ) = 63.2 mH/m

and thus L 1 = µ0 n 2 S1 d H 1 The inductance per unit length is therefore L 1 = µ0 n 2 S1 H/m 1 or L 1 = 39.5 mH/m Similarly, L 2 = µ0 n 2 S2 = 22.7 mH/m 2 We see, therefore, that there are many methods available for the calculation of self-inductance and mutual inductance. Unfortunately, even problems possessing a high degree of symmetry present very challenging integrals for evaluation, and only a few problems are available for us to try our skill on. Inductance will be discussed in circuit terms in Chapter 10. D8.12. Calculate the self-inductance of: (a) 3.5 m of coaxial cable with a = 0.8 mm and b = 4 mm, ﬁlled with a material for which µr = 50; (b) a toroidal coil of 500 turns, wound on a ﬁberglass form having a 2.5 × 2.5 cm square cross section and an inner radius of 2 cm; (c) a solenoid having 500 turns about a cylindrical core of 2 cm radius in which µr = 50 for 0 < ρ < 0.5 cm and µr = 1 for 0.5 < ρ < 2 cm; the length of the solenoid is 50 cm.
Ans. 56.3 µH; 1.01 mH; 3.2 mH

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D8.13. A solenoid is 50 cm long, 2 cm in diameter, and contains 1500 turns. The cylindrical core has a diameter of 2 cm and a relative permeability of 75. This coil is coaxial with a second solenoid, also 50 cm long, but with a 3 cm diameter and 1200 turns. Calculate: (a) L for the inner solenoid; (b) L for the outer solenoid; (c) M between the two solenoids.
Ans. 133.2 mH; 192 mH; 106.6 mH

REFERENCES
1. Kraus, J. D., and D. A. Fleisch. (See References for Chapter 3.) Examples of the calculation of inductance are given on pp. 99–108. 2. Matsch, L. W. (See References for Chapter 6.) Chapter 3 is devoted to magnetic circuits and ferromagnetic materials. 3. Paul, C. R., K. W. Whites, and S. Y. Nasar. (See References for Chapter 7.) Magnetic circuits, including those with permanent magnets, are discussed on pp. 263–70.

CHAPTER 8 PROBLEMS
8.1 A point charge, Q = −0.3 µC and m = 3 × 10−16 kg, is moving through the ﬁeld E = 30az V/m. Use Eq. (1) and Newton’s laws to develop the appropriate differential equations and solve them, subject to the initial conditions at t = 0, v = 3 × 105 ax m/s at the origin. At t = 3 µs, ﬁnd (a) the position P(x, y, z) of the charge; (b) the velocity v; (c) the kinetic energy of the charge. Compare the magnitudes of the electric and magnetic forces on an electron that has attained a velocity of 107 m/s. Assume an electric ﬁeld intensity of 105 V/m, and a magnetic ﬂux density associated with that of the Earth’s magnetic ﬁeld in temperate latitudes, 0.5 gauss. A point charge for which Q = 2 × 10−16 C and m = 5 × 10−26 kg is moving in the combined ﬁelds E = 100ax − 200a y + 300az V/m and B = −3ax + 2a y − az mT. If the charge velocity at t = 0 is v(0) = (2ax − 3a y − 4az )105 m/s (a) give the unit vector showing the direction in which the charge is accelerating at t = 0; (b) ﬁnd the kinetic energy of the charge at t = 0. Show that a charged particle in a uniform magnetic ﬁeld describes a circular orbit with an orbital period that is independent of the radius. Find the relationship between the angular velocity and magnetic ﬂux density for an electron (the cyclotron frequency).

8.2

8.3

8.4

8.5

A rectangular loop of wire in free space joins point A(1, 0, 1) to point B(3, 0, 1) to point C(3, 0, 4) to point D(1, 0, 4) to point A. The wire carries a

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271

current of 6 mA, ﬂowing in the az direction from B to C. A ﬁlamentary current of 15 A ﬂows along the entire z axis in the az direction. (a) Find F on side BC. (b) Find F on side AB. (c) Find Ftotal on the loop. 8.6 Show that the differential work in moving a current element I dL through a distance dl in a magetic ﬁeld B is the negative of that done in moving the element I dl through a distance dL in the same ﬁeld. Uniform current sheets are located in free space as follows: 8az A/m at y = 0, −4az A/m at y = 1, and −4az A/m at y = −1. Find the vector force per meter length exerted on a current ﬁlament carrying 7 mA in the a L direction if the ﬁlament is located at (a) x = 0, y = 0.5, and a L = az ; (b) y = 0.5, z = 0, and a L = ax ; (c) x = 0, y = 1.5, and a L = az .

8.7

8.8

Two conducting strips, having inﬁnite length in the z direction, lie in the x z plane. One occupies the region d/2 < x < b + d/2 and carries surface current density K = K 0 az ; the other is situated at −(b + d/2) < x < −d/2 and carries surface current density −K 0 az . (a) Find the force per unit length in z that tends to separate the two strips. (b) Let b approach zero while maintaining constant current, I = K 0 b, and show that the force per unit length approaches µ0 I 2 /(2πd) N/m. A current of −100az A/m ﬂows on the conducting cylinder ρ = 5 mm, and +500az A/m is present on the conducting cylinder ρ = 1 mm. Find the magnitude of the total force per meter length that is acting to split the outer cylinder apart along its length.

8.9

8.10 A planar transmission line consists of two conducting planes of width b separated d m in air, carrying equal and opposite currents of I A. If b d, ﬁnd the force of repulsion per meter of length between the two conductors. 8.11 (a) Use Eq. (14), Section 8.3, to show that the force of attraction per unit length between two ﬁlamentary conductors in free space with currents I1 az at x = 0, y = d/2, and I2 az at x = 0, y = −d/2, is µ0 I1 I2 /(2πd). (b) Show how a simpler method can be used to check your result. 8.12 Two circular wire rings are parallel to each other, share the same axis, are of radius a, and are separated by distance d, where d 2 cm. Find: (a) H everywhere; (b) B everywhere. 8.26 A long solenoid has a radius of 3 cm, 5000 turns/m, and carries current I = 0.25 A. The region 0 < ρ < a within the solenoid has µr = 5, whereas µr = 1 for a < ρ < 3 cm. Determine a so that (a) a total ﬂux of 10 µWb is present; (b) the ﬂux is equally divided between the regions 0 < ρ < a and a < ρ < 3 cm. 8.27 Let µr 1 = 2 in region 1, deﬁned by 2x + 3y − 4z > 1, while µr 2 = 5 in region 2 where 2x + 3y − 4z < 1. In region 1, H1 = 50ax − 30a y + 20az A/m. Find (a) H N 1 ; (b) Ht1 ; (c) Ht2 ; (d) H N 2 ; (e) θ1 , the angle between H1 and a N 21 ; ( f ) θ2 , the angle between H2 and a N 21 . 8.28 For values of B below the knee on the magnetization curve for silicon steel, approximate the curve by a straight line with µ = 5 mH/m. The core shown in Figure 8.16 has areas of 1.6 cm2 and lengths of 10 cm in each outer leg, and an area of 2.5 cm2 and a length of 3 cm in the central leg. A coil of 1200 turns carrying 12 mA is placed around the central leg. Find B in the (a) center leg; (b) center leg if a 0.3 mm air gap is present in the center leg.

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Figure 8.16 See Problem 8.28.

8.29 In Problem 8.28, the linear approximation suggested in the statement of the problem leads to ﬂux density of 0.666 T in the central leg. Using this value of B and the magnetization curve for silicon steel, what current is required in the 1200-turn coil? 8.30 A rectangular core has ﬁxed permeability µr >> 1, a square cross section of dimensions a × a, and has centerline dimensions around its perimeter of b and d. Coils 1 and 2, having turn numbers N1 and N2 , are wound on the core. Consider a selected core cross-sectional plane as lying within the x y plane, such that the surface is deﬁned by 0 < x < a, 0 < y < a. (a) With current I1 in coil 1, use Ampere’s circuital law to ﬁnd the magnetic ﬂux density as a function of position over the core cross-section. (b) Integrate your result of part (a) to determine the total magnetic ﬂux within the core. (c) Find the self-inductance of coil 1. (d) Find the mutual inductance between coils 1 and 2. 8.31 A toroid is constructed of a magnetic material having a cross-sectional area of 2.5 cm2 and an effective length of 8 cm. There is also a short air gap of 0.25 mm length and an effective area of 2.8 cm2 . An mmf of 200 A · t is applied to the magnetic circuit. Calculate the total ﬂux in the toroid if the magnetic material: (a) is assumed to have inﬁnite permeability; (b) is assumed to be linear with µr = 1000; (c) is silicon steel.

8.32 (a) Find an expression for the magnetic energy stored per unit length in a coaxial transmission line consisting of conducting sleeves of negligible thickness, having radii a and b. A medium of relative permeability µr ﬁlls the region between conductors. Assume current I ﬂows in both conductors in opposite directions. (b) Obtain the inductance, L, per unit length of line by equating the energy to (1/2)L I 2 .

8.33 A toroidal core has a square cross section, 2.5 cm < ρ < 3.5 cm, −0.5 cm < z < 0.5 cm. The upper half of the toroid, 0 < z < 0.5 cm, is constructed of a linear material for which µr = 10, while the lower half, −0.5 cm < z < 0,

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275

Figure 8.17 See Problem 8.35.

has µr = 20. An mmf of 150 A · t establishes a ﬂux in the aφ direction. For z > 0, ﬁnd: (a) Hφ (ρ); (b) Bφ (ρ); (c) z>0 . (d) Repeat for z > 0. (e) Find total . 8.34 Determine the energy stored per unit length in the internal magnetic ﬁeld of an inﬁnitely long, straight wire of radius a, carrying uniform current I . 8.35 The cones θ = 21◦ and θ = 159◦ are conducting surfaces and carry total currents of 40 A, as shown in Figure 8.17. The currents return on a spherical conducting surface of 0.25 m radius. (a) Find H in the region 0 < r < 0.25, 21◦ < θ < 159◦ , 0 < φ < 2π. (b) How much energy is stored in this region? 8.36 The dimensions of the outer conductor of a coaxial cable are b and c, where c > b. Assuming µ = µ0 , ﬁnd the magnetic energy stored per unit length in the region b < ρ < c for a uniformly distributed total current I ﬂowing in opposite directions in the inner and outer conductors. 8.37 Find the inductance of the cone-sphere conﬁguration described in Problem 8.35 and Figure 8.17. The inductance is that offered at the origin between the vertices of the cone. 8.38 A toroidal core has a rectangular cross section deﬁned by the surfaces ρ = 2 cm, ρ = 3 cm, z = 4 cm, and z = 4.5 cm. The core material has a relative permeability of 80. If the core is wound with a coil containing 8000 turns of wire, ﬁnd its inductance. 8.39 Conducting planes in air at z = 0 and z = d carry surface currents of ±K 0 ax A/m. (a) Find the energy stored in the magnetic ﬁeld per unit length (0 < x < 1) in a width w(0 < y < w). (b) Calculate the inductance per unit length of this transmission line from W H = 1 LI 2 , where I is the total current 2 in a width w in either conductor. (c) Calculate the total ﬂux passing through

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the rectangle 0 < x < 1, 0 < z < d, in the plane y = 0, and from this result again ﬁnd the inductance per unit length. 8.40 A coaxial cable has conductor radii a and b, where a < b. Material of permeability µr = 1 exists in the region a < ρ < c, whereas the region c < ρ < b is air ﬁlled. Find an expression for the inductance per unit length. 8.41 A rectangular coil is composed of 150 turns of a ﬁlamentary conductor. Find the mutual inductance in free space between this coil and an inﬁnite straight ﬁlament on the z axis if the four corners of the coil are located at: (a) (0, 1, 0), (0, 3, 0), (0, 3, 1), and (0, 1, 1); (b) (1, 1, 0), (1, 3, 0), (1, 3, 1), and (1, 1, 1). 8.42 Find the mutual inductance between two ﬁlaments forming circular rings of radii a and a, where a a. The ﬁeld should be determined by approximate methods. The rings are coplanar and concentric. 8.43 (a) Use energy relationships to show that the internal inductance of a nonmagnetic cylindrical wire of radius a carrying a uniformly distributed current I is µ0 /(8π ) H/m. (b) Find the internal inductance if the portion of the conductor for which ρ < c < a is removed. 8.44 Show that the external inductance per unit length of a two-wire transmission line carrying equal and opposite currents is approximately (µ/π ) ln(d/a) H/m, where a is the radius of each wire and d is the center-to-center wire spacing. On what basis is the approximation valid?

C H A P T E R

9

Time-Varying Fields and Maxwell’s Equations he basic relationships of the electrostatic ﬁeld and the steady magnetic ﬁeld were obtained in the previous eight chapters, and we are now ready to discuss time-varying ﬁelds. The discussion will be short, for vector analysis and vector calculus should now be more familiar tools; some of the relationships are unchanged, and most of the relationships are changed only slightly. Two new concepts will be introduced: the electric ﬁeld produced by a changing magnetic ﬁeld and the magnetic ﬁeld produced by a changing electric ﬁeld. The ﬁrst of these concepts resulted from experimental research by Michael Faraday and the second from the theoretical efforts of James Clerk Maxwell. Maxwell actually was inspired by Faraday’s experimental work and by the mental picture provided through the “lines of force” that Faraday introduced in developing his theory of electricity and magnetism. He was 40 years younger than Faraday, but they knew each other during the ﬁve years Maxwell spent in London as a young professor, a few years after Faraday had retired. Maxwell’s theory was developed subsequent to his holding this university position while he was working alone at his home in Scotland. It occupied him for ﬁve years between the ages of 35 and 40. The four basic equations of electromagnetic theory presented in this chapter bear his name. ■

T

9.1 FARADAY’S LAW
After Oersted1 demonstrated in 1820 that an electric current affected a compass needle, Faraday professed his belief that if a current could produce a magnetic ﬁeld, then a magnetic ﬁeld should be able to produce a current. The concept of the “ﬁeld”

1

Hans Christian Oersted was professor of physics at the University of Copenhagen in Denmark. 277

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was not available at that time, and Faraday’s goal was to show that a current could be produced by “magnetism.” He worked on this problem intermittently over a period of 10 years, until he was ﬁnally successful in 1831.2 He wound two separate windings on an iron toroid and placed a galvanometer in one circuit and a battery in the other. Upon closing the battery circuit, he noted a momentary deﬂection of the galvanometer; a similar deﬂection in the opposite direction occurred when the battery was disconnected. This, of course, was the ﬁrst experiment he made involving a changing magnetic ﬁeld, and he followed it with a demonstration that either a moving magnetic ﬁeld or a moving coil could also produce a galvanometer deﬂection. In terms of ﬁelds, we now say that a time-varying magnetic ﬁeld produces an electromotive force (emf) that may establish a current in a suitable closed circuit. An electromotive force is merely a voltage that arises from conductors moving in a magnetic ﬁeld or from changing magnetic ﬁelds, and we shall deﬁne it in this section. Faraday’s law is customarily stated as emf = − d V dt (1)

Equation (1) implies a closed path, although not necessarily a closed conducting path; the closed path, for example, might include a capacitor, or it might be a purely imaginary line in space. The magnetic ﬂux is that ﬂux which passes through any and every surface whose perimeter is the closed path, and d /dt is the time rate of change of this ﬂux. A nonzero value of d /dt may result from any of the following situations: 1. A time-changing ﬂux linking a stationary closed path 2. Relative motion between a steady ﬂux and a closed path 3. A combination of the two The minus sign is an indication that the emf is in such a direction as to produce a current whose ﬂux, if added to the original ﬂux, would reduce the magnitude of the emf. This statement that the induced voltage acts to produce an opposing ﬂux is known as Lenz’s law.3 If the closed path is that taken by an N -turn ﬁlamentary conductor, it is often sufﬁciently accurate to consider the turns as coincident and let emf = −N where paths. d dt (2)

is now interpreted as the ﬂux passing through any one of N coincident

2 3

Joseph Henry produced similar results at Albany Academy in New York at about the same time. Henri Frederic Emile Lenz was born in Germany but worked in Russia. He published his law in 1834.

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279

We need to deﬁne emf as used in (1) or (2). The emf is obviously a scalar, and (perhaps not so obviously) a dimensional check shows that it is measured in volts. We deﬁne the emf as emf = E · dL (3)

and note that it is the voltage about a speciﬁc closed path. If any part of the path is changed, generally the emf changes. The departure from static results is clearly shown by (3), for an electric ﬁeld intensity resulting from a static charge distribution must lead to zero potential difference about a closed path. In electrostatics, the line integral leads to a potential difference; with time-varying ﬁelds, the result is an emf or a voltage. Replacing in (1) with the surface integral of B, we have emf = E · dL = − d dt B · dS (4)

S

where the ﬁngers of our right hand indicate the direction of the closed path, and our thumb indicates the direction of dS. A ﬂux density B in the direction of dS and increasing with time thus produces an average value of E which is opposite to the positive direction about the closed path. The right-handed relationship between the surface integral and the closed line integral in (4) should always be kept in mind during ﬂux integrations and emf determinations. We will divide our investigation into two parts by ﬁrst ﬁnding the contribution to the total emf made by a changing ﬁeld within a stationary path (transformer emf), and then we will consider a moving path within a constant (motional, or generator, emf). We ﬁrst consider a stationary path. The magnetic ﬂux is the only time-varying quantity on the right side of (4), and a partial derivative may be taken under the integral sign, ∂B · dS (5) S ∂t Before we apply this simple result to an example, let us obtain the point form of this integral equation. Applying Stokes’ theorem to the closed line integral, we have emf = E · dL = − ∂B · dS S S ∂t where the surface integrals may be taken over identical surfaces. The surfaces are perfectly general and may be chosen as differentials, (∇ × E) · dS = − (∇ × E) · dS = − and ∇ ×E=− ∂B ∂t (6) ∂B · dS ∂t

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This is one of Maxwell’s four equations as written in differential, or point, form, the form in which they are most generally used. Equation (5) is the integral form of this equation and is equivalent to Faraday’s law as applied to a ﬁxed path. If B is not a function of time, (5) and (6) evidently reduce to the electrostatic equations E · dL = 0 and ∇ ×E=0 (electrostatics) (electrostatics)

As an example of the interpretation of (5) and (6), let us assume a simple magnetic ﬁeld which increases exponentially with time within the cylindrical region ρ < b, B = B0 ekt az (7)

where B0 = constant. Choosing the circular path ρ = a, a < b, in the z = 0 plane, along which E φ must be constant by symmetry, we then have from (5) The emf around this closed path is −k B0 ekt πa 2 . It is proportional to a 2 because the magnetic ﬂux density is uniform and the ﬂux passing through the surface at any instant is proportional to the area. If we now replace a with ρ, ρ < b, the electric ﬁeld intensity at any point is E = − 1 k B0 ekt ρaφ 2 Let us now attempt to obtain the same answer from (6), which becomes (∇ × E)z = −k B0 ekt = 1 ∂(ρE φ ) ρ ∂ρ (8) emf = 2πa E φ = −k B0 ekt πa 2

Multiplying by ρ and integrating from 0 to ρ (treating t as a constant, since the derivative is a partial derivative), − 1 k B0 ekt ρ 2 = ρE φ 2 or E = − 1 k B0 ekt ρaφ 2 once again. If B0 is considered positive, a ﬁlamentary conductor of resistance R would have a current ﬂowing in the negative aφ direction, and this current would establish a ﬂux within the circular loop in the negative az direction. Because E φ increases exponentially with time, the current and ﬂux do also, and thus they tend to reduce the time rate of increase of the applied ﬂux and the resultant emf in accordance with Lenz’s law. Before leaving this example, it is well to point out that the given ﬁeld B does not satisfy all of Maxwell’s equations. Such ﬁelds are often assumed (always in accircuit problems) and cause no difﬁculty when they are interpreted properly. They

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281

Figure 9.1 An example illustrating the application of Faraday’s law to the case of a constant magnetic flux density B and a moving path. The shorting bar moves to the right with a velocity v, and the circuit is completed through the two rails and an extremely small high-resistance voltmeter. The voltmeter reading is V12 = −Bvd.

occasionally cause surprise, however. This particular ﬁeld is discussed further in Problem 9.19 at the end of the chapter. Now let us consider the case of a time-constant ﬂux and a moving closed path. Before we derive any special results from Faraday’s law (1), let us use the basic law to analyze the speciﬁc problem outlined in Figure 9.1. The closed circuit consists of two parallel conductors which are connected at one end by a high-resistance voltmeter of negligible dimensions and at the other end by a sliding bar moving at a velocity v. The magnetic ﬂux density B is constant (in space and time) and is normal to the plane containing the closed path. Let the position of the shorting bar be given by y; the ﬂux passing through the surface within the closed path at any time t is then = Byd From (1), we obtain dy d = −B d = −Bνd (9) dt dt The emf is deﬁned as E · dL and we have a conducting path, so we may actually determine E at every point along the closed path. We found in electrostatics that the tangential component of E is zero at the surface of a conductor, and we shall show in Section 9.4 that the tangential component is zero at the surface of a perfect conductor (σ = ∞) for all time-varying conditions. This is equivalent to saying that a perfect conductor is a “short circuit.” The entire closed path in Figure 9.1 may be considered a perfect conductor, with the exception of the voltmeter. The actual computation of E · dL then must involve no contribution along the entire moving bar, both rails, and the voltmeter leads. Because we are integrating in a counterclockwise direction emf = −

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(keeping the interior of the positive side of the surface on our left as usual), the contribution E L across the voltmeter must be −Bνd, showing that the electric ﬁeld intensity in the instrument is directed from terminal 2 to terminal 1. For an upscale reading, the positive terminal of the voltmeter should therefore be terminal 2. The direction of the resultant small current ﬂow may be conﬁrmed by noting that the enclosed ﬂux is reduced by a clockwise current in accordance with Lenz’s law. The voltmeter terminal 2 is again seen to be the positive terminal. Let us now consider this example using the concept of motional emf. The force on a charge Q moving at a velocity v in a magnetic ﬁeld B is F = Qv × B or F =v×B (10) Q The sliding conducting bar is composed of positive and negative charges, and each experiences this force. The force per unit charge, as given by (10), is called the motional electric ﬁeld intensity Em , Em = v × B (11)

If the moving conductor were lifted off the rails, this electric ﬁeld intensity would force electrons to one end of the bar (the far end) until the static fiel due to these charges just balanced the ﬁeld induced by the motion of the bar. The resultant tangential electric ﬁeld intensity would then be zero along the length of the bar. The motional emf produced by the moving conductor is then emf = Em · dL = (v × B) · dL (12)

where the last integral may have a nonzero value only along that portion of the path which is in motion, or along which v has some nonzero value. Evaluating the right side of (12), we obtain (v × B) · dL =
0 d

ν B d x = −Bνd

as before. This is the total emf, since B is not a function of time. In the case of a conductor moving in a uniform constant magnetic ﬁeld, we may therefore ascribe a motional electric ﬁeld intensity Em = v × B to every portion of the moving conductor and evaluate the resultant emf by emf = E · dL = Em · dL = (v × B) · dL (13)

If the magnetic ﬂux density is also changing with time, then we must include both contributions, the transformer emf (5) and the motional emf (12), ∂B · dS + (v × B) · dL (14) emf = E · dL = − S ∂t

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283

Figure 9.2 An apparent increase in flux linkages does not lead to an induced voltage when one part of a circuit is simply substituted for another by opening the switch. No indication will be observed on the voltmeter.

This expression is equivalent to the simple statement d emf = − (1) dt and either can be used to determine these induced voltages. Although (1) appears simple, there are a few contrived examples in which its proper application is quite difﬁcult. These usually involve sliding contacts or switches; they always involve the substitution of one part of a circuit by a new part.4 As an example, consider the simple circuit of Figure 9.2, which contains several perfectly conducting wires, an ideal voltmeter, a uniform constant ﬁeld B, and a switch. When the switch is opened, there is obviously more ﬂux enclosed in the voltmeter circuit; however, it continues to read zero. The change in ﬂux has not been produced by either a time-changing B [ﬁrst term of (14)] or a conductor moving through a magnetic ﬁeld [second part of (14)]. Instead, a new circuit has been substituted for the old. Thus it is necessary to use care in evaluating the change in ﬂux linkages. The separation of the emf into the two parts indicated by (14), one due to the time rate of change of B and the other to the motion of the circuit, is somewhat arbitrary in that it depends on the relative velocity of the observer and the system. A ﬁeld that is changing with both time and space may look constant to an observer moving with the ﬁeld. This line of reasoning is developed more fully in applying the special theory of relativity to electromagnetic theory.5 D9.1. Within a certain region, = 10−11 F/m and µ = 10−5 H/m. If Bx = ∂E 2 × 10−4 cos 105 t sin 10−3 y T: (a) use ∇ × H = to ﬁnd E; (b) ﬁnd the total ∂t magnetic ﬂux passing through the surface x = 0, 0 < y < 40 m, 0 < z < 2 m,

4 5

See Bewley, in References at the end of the chapter, particularly pp. 12–19.

This is discussed in several of the references listed in the References at the end of the chapter. See Panofsky and Phillips, pp. 142–51; Owen, pp. 231–45; and Harman in several places.

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at t = 1 µs; (c) ﬁnd the value of the closed line integral of E around the perimeter of the given surface.
Ans. −20 000 sin 105 t cos 10−3 yaz V/m; 0.318 mWb; −3.19 V

D9.2. With reference to the sliding bar shown in Figure 9.1, let d = 7 cm, B = 0.3az T, and v = 0.1a y e20y m/s. Let y = 0 at t = 0. Find: (a) ν(t = 0); (b) y(t = 0.1); (c) ν(t = 0.1); (d) V12 at t = 0.1.
Ans. 0.1 m/s; 1.12 cm; 0.125 m/s; −2.63 mV

9.2 DISPLACEMENT CURRENT
Faraday’s experimental law has been used to obtain one of Maxwell’s equations in differential form, ∂B ∇ ×E=− (15) ∂t which shows us that a time-changing magnetic ﬁeld produces an electric ﬁeld. Remembering the deﬁnition of curl, we see that this electric ﬁeld has the special property of circulation; its line integral about a general closed path is not zero. Now let us turn our attention to the time-changing electric ﬁeld. We should ﬁrst look at the point form of Amp` re’s circuital law as it applies to e steady magnetic ﬁelds, ∇ ×H=J (16)

and show its inadequacy for time-varying conditions by taking the divergence of each side, The divergence of the curl is identically zero, so ∇ · J is also zero. However, the equation of continuity, ∂ρν ∇ ·J = − ∂t then shows us that (16) can be true only if ∂ρν /∂t = 0. This is an unrealistic limitation, and (16) must be amended before we can accept it for time-varying ﬁelds. Suppose we add an unknown term G to (16), ∇ ×H=J+G Again taking the divergence, we have 0 = ∇ ·J + ∇ ·G Thus ∇ ·G = ∂ρν ∂t ∇ ·∇ × H ≡ 0 = ∇ ·J

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Replacing ρν with ∇ · D,

∂D ∂ (∇ · D) = ∇ · ∂t ∂t from which we obtain the simplest solution for G, ∇ ·G = G=

∂D ∂t Amp` re’s circuital law in point form therefore becomes e ∇ ×H=J+ ∂D ∂t (17)

Equation (17) has not been derived. It is merely a form we have obtained that does not disagree with the continuity equation. It is also consistent with all our other results, and we accept it as we did each experimental law and the equations derived from it. We are building a theory, and we have every right to our equations until they are proved wrong. This has not yet been done. We now have a second one of Maxwell’s equations and shall investigate its signiﬁcance. The additional term ∂D/∂t has the dimensions of current density, amperes per square meter. Because it results from a time-varying electric ﬂux density (or displacement density), Maxwell termed it a displacement current density. We sometimes denote it by Jd : ∇ × H = J + Jd ∂D Jd = ∂t This is the third type of current density we have met. Conduction current density, J = σE is the motion of charge (usually electrons) in a region of zero net charge density, and convection current density, is the motion of volume charge density. Both are represented by J in (17). Bound current density is, of course, included in H. In a nonconducting medium in which no volume charge density is present, J = 0, and then ∂D (if J = 0) ∂t Notice the symmetry between (18) and (15): ∇ ×H= ∇ ×E=− (18) J = ρν v

∂B (15) ∂t Again, the analogy between the intensity vectors E and H and the ﬂux density vectors D and B is apparent. We cannot place too much faith in this analogy, however, for it fails when we investigate forces on particles. The force on a charge is related to E

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Figure 9.3 A filamentary conductor forms a loop connecting the two plates of a parallel-plate capacitor. A time-varying magnetic field inside the closed path produces an emf of V0 cos ωt around the closed path. The conduction current I is equal to the displacement current between the capacitor plates.

and to B, and some good arguments may be presented showing an analogy between E and B and between D and H. We omit them, however, and merely say that the concept of displacement current was probably suggested to Maxwell by the symmetry ﬁrst mentioned in this paragraph.6 The total displacement current crossing any given surface is expressed by the surface integral, ∂D Id = Jd · dS = · dS S S ∂t and we may obtain the time-varying version of Amp` re’s circuital law by integrating e (17) over the surface S, ∂D (∇ × H) · dS = J · dS + · dS S S S ∂t and applying Stokes’ theorem, H · dL = I + Id = I + ∂D · dS ∂t (19)

S

What is the nature of displacement current density? Let us study the simple circuit of Figure 9.3, which contains a ﬁlamentary loop and a parallel-plate capacitor. Within

6

The analogy that relates B to D and H to E is strongly advocated by Fano, Chu, and Adler (see References for Chapter 6); the case for comparing B to E and D to H is presented in Halliday and Resnick (see References for this chapter).

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the loop, a magnetic ﬁeld varying sinusoidally with time is applied to produce an emf about the closed path (the ﬁlament plus the dashed portion between the capacitor plates), which we shall take as emf = V0 cos ωt Using elementary circuit theory and assuming that the loop has negligible resistance and inductance, we may obtain the current in the loop as I = −ωC V0 sin ωt S = −ω V0 sin ωt d where the quantities , S, and d pertain to the capacitor. Let us apply Amp` re’s circuital e law about the smaller closed circular path k and neglect displacement current for the moment: k H · dL = Ik

The path and the value of H along the path are both deﬁnite quantities (although difﬁcult to determine), and k H · dL is a deﬁnite quantity. The current Ik is that current through every surface whose perimeter is the path k. If we choose a simple surface punctured by the ﬁlament, such as the plane circular surface deﬁned by the circular path k, the current is evidently the conduction current. Suppose now we consider the closed path k as the mouth of a paper bag whose bottom passes between the capacitor plates. The bag is not pierced by the ﬁlament, and the conductor current is zero. Now we need to consider displacement current, for within the capacitor D= E= and therefore S ∂D S = −ω V0 sin ωt ∂t d This is the same value as that of the conduction current in the ﬁlamentary loop. Therefore the application of Amp` re’s circuital law, including displacement current e to the path k, leads to a deﬁnite value for the line integral of H. This value must be equal to the total current crossing the chosen surface. For some surfaces the current is almost entirely conduction current, but for those surfaces passing between the capacitor plates, the conduction current is zero, and it is the displacement current which is now equal to the closed line integral of H. Physically, we should note that a capacitor stores charge and that the electric ﬁeld between the capacitor plates is much greater than the small leakage ﬁelds outside. We therefore introduce little error when we neglect displacement current on all those surfaces which do not pass between the plates. Displacement current is associated with time-varying electric ﬁelds and therefore exists in all imperfect conductors carrying a time-varying conduction current. The last Id = V0 cos ωt d

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part of the following drill problem indicates the reason why this additional current was never discovered experimentally. D9.3. Find the amplitude of the displacement current density: (a) adjacent to an automobile antenna where the magnetic ﬁeld intensity of an FM signal is Hx = 0.15 cos[3.12(3 × 108 t − y)] A/m; (b) in the air space at a point within a large power distribution transformer where B = 0.8 cos[1.257×10−6 (3×108 t − x)]a y T; (c) within a large, oil-ﬁlled power capacitor where r = 5 and E = √ 0.9 cos[1.257 × 10−6 (3 × 108 t − z 5)]ax MV/m; (d) in a metallic conductor at 60 Hz, if = 0 , µ = µ0 , σ = 5.8 × 107 S/m, and J = sin(377t − 117.1z)ax MA/m2 .
Ans. 0.468 A/m2 ; 0.800 A/m2 ; 0.0150 A/m2 ; 57.6 pA/m2

9.3 MAXWELL’S EQUATIONS IN POINT FORM
We have already obtained two of Maxwell’s equations for time-varying ﬁelds, ∇ ×E=− and ∇ ×H=J+ ∂D ∂t (21) ∂B ∂t (20)

The remaining two equations are unchanged from their non-time-varying form: ∇ · D = ρν ∇ ·B = 0 (22)

(23)

Equation (22) essentially states that charge density is a source (or sink) of electric ﬂux lines. Note that we can no longer say that all electric ﬂux begins and terminates on charge, because the point form of Faraday’s law (20) shows that E, and hence D, may have circulation if a changing magnetic ﬁeld is present. Thus the lines of electric ﬂux may form closed loops. However, the converse is still true, and every coulomb of charge must have one coulomb of electric ﬂux diverging from it. Equation (23) again acknowledges the fact that “magnetic charges,” or poles, are not known to exist. Magnetic ﬂux is always found in closed loops and never diverges from a point source. These four equations form the basis of all electromagnetic theory. They are partial differential equations and relate the electric and magnetic ﬁelds to each other and to

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their sources, charge and current density. The auxiliary equations relating D and E, D= E relating B and H, B = µH deﬁning conduction current density, J = σE (26) (25) (24)

and deﬁning convection current density in terms of the volume charge density ρν , J = ρν v (27)

are also required to deﬁne and relate the quantities appearing in Maxwell’s equations. The potentials V and A have not been included because they are not strictly necessary, although they are extremely useful. They will be discussed at the end of this chapter. If we do not have “nice” materials to work with, then we should replace (24) and (25) with the relationships involving the polarization and magnetization ﬁelds, D=
0E

+P

(28)

B = µ0 (H + M) For linear materials we may relate P to E P = χe 0 E and M to H M = χm H

(29)

(30)

(31)

Finally, because of its fundamental importance we should include the Lorentz force equation, written in point form as the force per unit volume, f = ρν (E + v × B) (32)

The following chapters are devoted to the application of Maxwell’s equations to several simple problems.

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D9.4. Let µ = 10−5 H/m, = 4 × 10−9 F/m, σ = 0, and ρν = 0. Find k (including units) so that each of the following pairs of ﬁelds satisﬁes Maxwell’s equations: (a) D = 6ax − 2ya y + 2zaz nC/m2 , H = kxax + 10ya y − 25zaz A/m; (b) E = (20y − kt)ax V/m, H = (y + 2 × 106 t)az A/m.
Ans. 15 A/m2 ; −2.5 × 108 V/(m · s)

9.4 MAXWELL’S EQUATIONS IN INTEGRAL FORM
The integral forms of Maxwell’s equations are usually easier to recognize in terms of the experimental laws from which they have been obtained by a generalization process. Experiments must treat physical macroscopic quantities, and their results therefore are expressed in terms of integral relationships. A differential equation always represents a theory. Let us now collect the integral forms of Maxwell’s equations from Section 9.3. Integrating (20) over a surface and applying Stokes’ theorem, we obtain Faraday’s law, E · dL = −
S

∂B · dS ∂t

(33)

and the same process applied to (21) yields Amp` re’s circuital law, e H · dL = I + ∂D · dS ∂t (34)

S

Gauss’s laws for the electric and magnetic ﬁelds are obtained by integrating (22) and (23) throughout a volume and using the divergence theorem: D · dS = ρν dv vol S

(35)

S

B · dS = 0

(36)

These four integral equations enable us to ﬁnd the boundary conditions on B, D, H, and E, which are necessary to evaluate the constants obtained in solving Maxwell’s equations in partial differential form. These boundary conditions are in general unchanged from their forms for static or steady ﬁelds, and the same methods may be used to obtain them. Between any two real physical media (where K must be zero on the boundary surface), (33) enables us to relate the tangential E-ﬁeld components, E t1 = E t2 (37)

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and from (34), Ht1 = Ht2 D N 1 − D N 2 = ρS and BN 1 = BN 2 (40) (38)

The surface integrals produce the boundary conditions on the normal components, (39)

It is often desirable to idealize a physical problem by assuming a perfect conductor for which σ is inﬁnite but J is ﬁnite. From Ohm’s law, then, in a perfect conductor, E=0 and it follows from the point form of Faraday’s law that H=0 for time-varying ﬁelds. The point form of Amp` re’s circuital law then shows that the e ﬁnite value of J is J=0

and current must be carried on the conductor surface as a surface current K. Thus, if region 2 is a perfect conductor, (37) to (40) become, respectively, E t1 = 0 Ht1 = K D N 1 = ρs BN 1 = 0 (Ht1 = K × a N ) (41) (42) (43) (44)

where a N is an outward normal at the conductor surface. Note that surface charge density is considered a physical possibility for either dielectrics, perfect conductors, or imperfect conductors, but that surface current density is assumed only in conjunction with perfect conductors. The preceding boundary conditions are a very necessary part of Maxwell’s equations. All real physical problems have boundaries and require the solution of Maxwell’s equations in two or more regions and the matching of these solutions at the boundaries. In the case of perfect conductors, the solution of the equations within the conductor is trivial (all time-varying ﬁelds are zero), but the application of the boundary conditions (41) to (44) may be very difﬁcult. Certain fundamental properties of wave propagation are evident when Maxwell’s equations are solved for an unbounded region. This problem is treated in Chapter 11. It represents the simplest application of Maxwell’s equations because it is the only problem which does not require the application of any boundary conditions.

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D9.5. The unit vector 0.64ax + 0.6a y − 0.48az is directed from region 2 ( r = 2, µr = 3, σ2 = 0) toward region 1 ( r 1 = 4, µr 1 = 2, σ1 = 0). If B1 = (ax − 2a y + 3az ) sin 300t T at point P in region 1 adjacent to the boundary, ﬁnd the amplitude at P of: (a) B N 1 ; (b) Bt1 ; (c) B N 2 ; (d) B2 .
Ans. 2.00 T; 3.16 T; 2.00 T; 5.15 T

D9.6. The surface y = 0 is a perfectly conducting plane, whereas the region y > 0 has r = 5, µr = 3, and σ = 0. Let E = 20 cos(2 × 108 t − 2.58z)a y V/m for y > 0, and ﬁnd at t = 6 ns; (a) ρ S at P(2, 0, 0.3); (b) H at P; (c) K at P.
Ans. 0.81 nC/m2 ; −62.3ax mA/m; −62.3az mA/m

9.5 THE RETARDED POTENTIALS
The time-varying potentials, usually called retarded potentials for a reason that we will see shortly, ﬁnd their greatest application in radiation problems (to be addressed in Chapter 14) in which the distribution of the source is known approximately. We should remember that the scalar electric potential V may be expressed in terms of a static charge distribution, ρν dν V = (static) (45) 4π R vol and the vector magnetic potential may be found from a current distribution which is constant with time, µJ dv A= (dc) (46) vol 4πR The differential equations satisﬁed by V , ρν ∇2V = − (static) (47) and A, ∇ 2 A = −µJ (dc) (48)

may be regarded as the point forms of the integral equations (45) and (46), respectively. Having found V and A, the fundamental ﬁelds are then simply obtained by using the gradient, E = −∇V or the curl, B=∇ ×A (dc) (50) (static) (49)

We now wish to deﬁne suitable time-varying potentials which are consistent with the preceding expressions when only static charges and direct currents are involved. Equation (50) apparently is still consistent with Maxwell’s equations. These equations state that ∇ · B = 0, and the divergence of (50) leads to the divergence of

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the curl that is identically zero. Let us therefore tentatively accept (50) as satisfactory for time-varying ﬁelds and turn our attention to (49). The inadequacy of (49) is obvious because application of the curl operation to each side and recognition of the curl of the gradient as being identically zero confront us with ∇ × E = 0. However, the point form of Faraday’s law states that ∇ × E is not generally zero, so let us try to effect an improvement by adding an unknown term to (49), E = −∇V + N taking the curl, ∇ ×E=0+∇ ×N using the point form of Faraday’s law, ∇ ×N=− and using (50), giving us ∇ ×N=− or ∇ × N = −∇ × The simplest solution of this equation is ∂A N=− ∂t and this leads to E = −∇V − ∂A ∂t (51) ∂A ∂t ∂ (∇ × A) ∂t ∂B ∂t

We still must check (50) and (51) by substituting them into the remaining two of Maxwell’s equations: ∂D ∇ ×H = J+ ∂t ∇ · D = ρν Doing this, we obtain the more complicated expressions 1 ∇ ×∇ ×A=J+ µ and −∇ · ∇V − or ∇(∇ · A) − ∇ 2 A = µJ − µ ∇ ∂ 2A ∂V + 2 ∂t ∂t (52) ∂ ∇ · A = ρν ∂t −∇ ∂V ∂ 2A − 2 ∂t ∂t

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and ∂ ρν (∇ · A) = − (53) ∂t There is no apparent inconsistency in (52) and (53). Under static or dc conditions ∇ · A = 0, and (52) and (53) reduce to (48) and (47), respectively. We will therefore assume that the time-varying potentials may be deﬁned in such a way that B and E may be obtained from them through (50) and (51). These latter two equations do not serve, however, to deﬁne A and V completely. They represent necessary, but not sufﬁcient, conditions. Our initial assumption was merely that B = ∇ × A, and a vector cannot be deﬁned by giving its curl alone. Suppose, for example, that we have a very simple vector potential ﬁeld in which A y and A z are zero. Expansion of (50) leads to ∇2V + Bx = 0 ∂A x By = ∂z ∂A x Bz = − ∂y and we see that no information is available about the manner in which A x varies with x. This information could be found if we also knew the value of the divergence of A, for in our example ∂ Ax ∂x Finally, we should note that our information about A is given only as partial derivatives and that a space-constant term might be added. In all physical problems in which the region of the solution extends to inﬁnity, this constant term must be zero, for there can be no ﬁelds at inﬁnity. Generalizing from this simple example, we may say that a vector ﬁeld is deﬁned completely when both its curl and divergence are given and when its value is known at any one point (including inﬁnity). We are therefore at liberty to specify the divergence of A, and we do so with an eye on (52) and (53), seeking the simplest expressions. We deﬁne ∂V (54) ∇ · A = −µ ∂t and (52) and (53) become ∇ ·A = ∇ 2 A = −µJ + µ and ∂2V (56) ∂t 2 These equations are related to the wave equation, which will be discussed in Chapters 10 and 11. They show considerable symmetry, and we should be highly ∇2V = − ρν +µ ∂ 2A ∂t 2 (55)

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pleased with our deﬁnitions of V and A, B=∇ ×A ∇ · A = −µ E = −∇V − ∂V ∂t (50) (54)

∂A (51) ∂t The integral equivalents of (45) and (46) for the time-varying potentials follow from the deﬁnitions (50), (51), and (54), but we shall merely present the ﬁnal results and indicate their general nature. In Chapter 11, we will ﬁnd that any electromagnetic disturbance will travel at a velocity 1 ν=√ µ through any homogeneous medium described by µ and . In the case of free space, this velocity turns out to be the velocity of light, approximately 3 × 108 m/s. It is logical, then, to suspect that the potential at any point is due not to the value of the charge density at some distant point at the same instant, but to its value at some previous time, because the effect propagates at a ﬁnite velocity. Thus (45) becomes V = [ρν ] dν 4π R (57)

vol

where [ρν ] indicates that every t appearing in the expression for ρν has been replaced by a retarded time, R t =t− ν Thus, if the charge density throughout space were given by ρν = e−r cos ωt R ν where R is the distance between the differential element of charge being considered and the point at which the potential is to be determined. The retarded vector magnetic potential is given by [ρν ] = e−r cos ω t − A= µ[J] dν 4πR (58) then

vol

The use of a retarded time has resulted in the time-varying potentials being given the name of retarded potentials. In Chapter 14 we will apply (58) to the simple situation of a differential current element in which I is a sinusoidal function of time. Other simple applications of (58) are considered in several problems at the end of this chapter.

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We may summarize the use of the potentials by stating that a knowledge of the distribution of ρν and J throughout space theoretically enables us to determine V and A from (57) and (58). The electric and magnetic ﬁelds are then obtained by applying (50) and (51). If the charge and current distributions are unknown, or reasonable approximations cannot be made for them, these potentials usually offer no easier path toward the solution than does the direct application of Maxwell’s equations. D9.7. A point charge of 4 cos 108 π t µC is located at P+ (0, 0, 1.5), whereas −4 cos 108 πt µC is at P− (0, 0, −1.5), both in free space. Find V at P(r = 450, θ, φ = 0) at t = 15 ns for θ =: (a) 0◦ ; (b) 90◦ ; (c) 45◦ .
Ans. 159.8 V; 0; 143 V

REFERENCES
1. Bewley, L. V. Flux Linkages and Electromagnetic Induction. New York: Macmillan, 1952. This little book discusses many of the paradoxical examples involving induced (?) voltages. 2. Faraday, M. Experimental Researches in Electricity. London: B. Quaritch, 1839, 1855. Very interesting reading of early scientiﬁc research. A more recent and available source is Great Books of the Western World, vol. 45, Encyclopaedia Britannica, Inc., Chicago, 1952. 3. Halliday, D., R. Resnick, and J. Walker. Fundamentals of Physics. 5th ed. New York: John Wiley & Sons, 1997. This text is widely used in the ﬁrst university-level course in physics. 4. Harman, W. W. Fundamentals of Electronic Motion. New York: McGraw-Hill, 1953. Relativistic effects are discussed in a clear and interesting manner. 5. Nussbaum, A. Electromagnetic Theory for Engineers and Scientists. Englewood Cliffs, N.J.: Prentice-Hall, 1965. See the rocket-generator example beginning on p. 211. 6. Owen, G. E. Electromagnetic Theory. Boston: Allyn and Bacon, 1963. Faraday’s law is discussed in terms of the frame of reference in Chapter 8. 7. Panofsky, W. K. H., and M. Phillips. Classical Electricity and Magnetism. 2d ed. Reading, Mass.: Addison-Wesley, 1962. Relativity is treated at a moderately advanced level in Chapter 15.

CHAPTER 9 PROBLEMS
9.1 In Figure 9.4, let B = 0.2 cos 120πt T, and assume that the conductor joining the two ends of the resistor is perfect. It may be assumed that the magnetic ﬁeld produced by I (t) is negligible. Find (a) Vab (t); (b) I (t). In the example described by Figure 9.1, replace the constant magnetic ﬂux density by the time-varying quantity B = B0 sin ωt az . Assume that U is constant and that the displacement y of the bar is zero at t = 0. Find the emf at any time, t.

9.2

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297

Figure 9.4 See Problem 9.1.

9.3

Given H = 300az cos(3 × 108 t − y) A/m in free space, ﬁnd the emf developed in the general aφ direction about the closed path having corners at (a) (0, 0, 0), (1, 0, 0), (1, 1, 0), and (0, 1, 0); (b) (0, 0, 0) (2π , 0, 0), (2π, 2π, 0), and (0, 2π , 0). A rectangular loop of wire containing a high-resistance voltmeter has corners initially at (a/2, b/2, 0), (−a/2, b/2, 0), (−a/2, −b/2, 0), and (a/2, −b/2, 0). The loop begins to rotate about the x axis at constant angular velocity ω, with the ﬁrst-named corner moving in the az direction at t = 0. Assume a uniform magnetic ﬂux density B = B0 az . Determine the induced emf in the rotating loop and specify the direction of the current. The location of the sliding bar in Figure 9.5 is given by x = 5t + 2t 3 , and the separation of the two rails is 20 cm. Let B = 0.8x 2 az T. Find the voltmeter reading at (a) t = 0.4 s; (b) x = 0.6 m.

9.4

9.5

9.6

Let the wire loop of Problem 9.4 be stationary in its t = 0 position and ﬁnd the induced emf that results from a magnetic ﬂux density given by B(y, t) = B0 cos(ωt − βy) az , where ω and β are constants.

Figure 9.5

See Problem 9.5.

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Figure 9.6 See Problem 9.7.

9.7

The rails in Figure 9.6 each have a resistance of 2.2 /m. The bar moves to the right at a constant speed of 9 m/s in a uniform magnetic ﬁeld of 0.8 T. Find I (t), 0 < t < 1 s, if the bar is at x = 2 m at t = 0 and (a) a 0.3 resistor is present across the left end with the right end open-circuited; (b) a 0.3 resistor is present across each end. A perfectly conducting ﬁlament is formed into a circular ring of radius a. At one point, a resistance R is inserted into the circuit, and at another a battery of voltage V0 is inserted. Assume that the loop current itself produces negligible magnetic ﬁeld. (a) Apply Faraday’s law, Eq. (4), evaluating each side of the equation carefully and independently to show the equality; (b) repeat part a, assuming the battery is removed, the ring is closed again, and a linearly increasing B ﬁeld is applied in a direction normal to the loop surface. A square ﬁlamentary loop of wire is 25 cm on a side and has a resistance of 125 per meter length. The loop lies in the z = 0 plane with its corners at (0, 0, 0), (0.25, 0, 0), (0.25, 0.25, 0), and (0, 0.25, 0) at t = 0. The loop is moving with a velocity v y = 50 m/s in the ﬁeld Bz = 8 cos(1.5 × 108 t − 0.5x) µT. Develop a function of time that expresses the ohmic power being delivered to the loop. (a) Show that the ratio of the amplitudes of the conduction current density and the displacement current density is σ/ω for the applied ﬁeld E = E m cos ωt. Assume µ = µ0 . (b) What is the amplitude ratio if the applied ﬁeld is E = E m e−t/τ , where τ is real?

9.8

9.9

9.10

9.11

Let the internal dimensions of a coaxial capacitor be a = 1.2 cm, b = 4 cm, and l = 40 cm. The homogeneous material inside the capacitor has the parameters = 10−11 F/m, µ = 10−5 H/m, and σ = 10−5 S/m. If the electric ﬁeld intensity is E = (106 /ρ) cos 105 taρ V/m, ﬁnd (a) J; (b) the

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total conduction current Ic through the capacitor; (c) the total displacement current Id through the capacitor; (d) the ratio of the amplitude of Id to that of Ic , the quality factor of the capacitor. 9.12 9.13 Find the displacement current density associated with the magnetic ﬁeld H = A1 sin(4x) cos(ωt − βz) ax + A2 cos(4x) sin(ωt − βz) az .

Consider the region deﬁned by |x|, |y|, and |z| < 1. Let r = 5, µr = 4, and σ = 0. If Jd = 20 cos(1.5 × 108 t − bx)a y µA/m2 (a) ﬁnd D and E; (b) use the point form of Faraday’s law and an integration with respect to time to ﬁnd B and H; (c) use ∇ × H = Jd + J to ﬁnd Jd . (d) What is the numerical value of b? A voltage source V0 sin ωt is connected between two concentric conducting spheres, r = a and r = b, b > a, where the region between them is a material for which = r 0 , µ = µ0 , and σ = 0. Find the total displacement current through the dielectric and compare it with the source current as determined from the capacitance (Section 6.3) and circuit-analysis methods. Let µ = 3 × 10−5 H/m, = 1.2 × 10−10 F/m, and σ = 0 everywhere. If H = 2 cos(1010 t − βx)az A/m, use Maxwell’s equations to obtain expressions for B, D, E, and β. Derive the continuity equation from Maxwell’s equations. The electric ﬁeld intensity in the region 0 < x < 5, 0 < y < π/12, 0 < z < 0.06 m in free space is given by E = C sin 12y sin az cos 2 × 1010 tax V/m. Beginning with the ∇ × E relationship, use Maxwell’s equations to ﬁnd a numerical value for a, if it is known that a is greater than zero. The parallel-plate transmission line shown in Figure 9.7 has dimensions b = 4 cm and d = 8 mm, while the medium between the plates is characterized by µr = 1, r = 20, and σ = 0. Neglect ﬁelds outside the dielectric. Given the ﬁeld H = 5 cos(109 t − βz)a y A/m, use Maxwell’s

9.14

9.15

9.16 9.17

9.18

Figure 9.7 See Problem 9.18.

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equations to help ﬁnd (a) β, if β > 0; (b) the displacement current density at z = 0; (c) the total displacement current crossing the surface x = 0.5d, 0 < y < b, 0 < z < 0.1 m in the ax direction. 9.19 In Section 9.1, Faraday’s law was used to show that the ﬁeld E = − 1 k B0 ekt ρaφ results from the changing magnetic ﬁeld B = B0 ekt az . 2 (a) Show that these ﬁelds do not satisfy Maxwell’s other curl equation. (b) If we let B0 = 1 T and k = 106 s −1 , we are establishing a fairly large magnetic ﬂux density in 1 µs. Use the ∇ × H equation to show that the rate at which Bz should (but does not) change with ρ is only about 5 × 10−6 T per meter in free space at t = 0. Given Maxwell’s equations in point form, assume that all ﬁelds vary as est and write the equations without explicitly involving time.

9.20 9.21

(a) Show that under static ﬁeld conditions, Eq. (55) reduces to Amp` re’s e circuital law. (b) Verify that Eq. (51) becomes Faraday’s law when we take the curl. In a sourceless medium in which J = 0 and ρν = 0, assume a rectangular coordinate system in which E and H are functions only of z and t. The medium has permittivity and permeability µ. (a) If E = E x ax and H = Hy a y , begin with Maxwell’s equations and determine the second-order partial differential equation that E x must satisfy. (b) Show that E x = E 0 cos(ωt − βz) is a solution of that equation for a particular value of β. (c) Find β as a function of given parameters. In region 1, z < 0, 1 = 2 × 10−11 F/m, µ1 = 2 × 10−6 H/m, and σ1 = 4 × 10−3 S/m; in region 2, z > 0, 2 = 1 /2, µ2 = 2µ1 , and σ2 = σ1 /4. It is known that E1 = (30ax + 20a y + 10az ) cos 109 t V/m at P(0, 0, 0− ). (a) Find E N 1 , Et1 , D N 1 , and Dt1 at P1 . (b) Find J N 1 and Jt1 at P1 . (c) Find Et2 , Dt2 , and Jt2 at P2 (0, 0, 0+ ). (d) (Harder) Use the continuity equation to help show that JN 1 − JN 2 = ∂ D N 2 /∂t − ∂ D N 1 /∂t, and then determine D N 2 , J N 2 , and E N 2 . A vector potential is given as A = A0 cos(ωt − kz) a y . (a) Assuming as many components as possible are zero, ﬁnd H, E, and V . (b) Specify k in terms of A0 , ω, and the constants of the lossless medium, and µ. In a region where µr = r = 1 and σ = 0, the retarded potentials are given z √ by V = x(z − ct) V and A = x − t az Wb/m, where c = 1 µ0 0 . c ∂V . (b) Find B, H, E, and D. (c) Show that (a) Show that ∇ · A = −µ ∂t these results satisfy Maxwell’s equations if J and ρν are zero. Write Maxwell’s equations in point form in terms of E and H as they apply to a sourceless medium, where J and ρv are both zero. Replace by µ, µ by , E by H, and H by −E, and show that the equations are unchanged. This is a more general expression of the duality principle in circuit theory.

9.22

9.23

9.24

9.25

9.26

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T

ransmission lines are used to transmit electric energy and signals from one point to another, speciﬁcally from a source to a load. Examples include the connection between a transmitter and an antenna, connections between computers in a network, or connections between a hydroelectric generating plant and a substation several hundred miles away. Other familiar examples include the interconnects between components of a stereo system and the connection between a cable service provider and your television set. Examples that are less familiar include the connections between devices on a circuit board that are designed to operate at high frequencies. What all of these examples have in common is that the devices to be connected are separated by distances on the order of a wavelength or much larger, whereas in basic circuit analysis methods, connections between elements are assumed to have negligible length. The latter condition enabled us, for example, to take for granted that the voltage across a resistor on one side of a circuit was exactly in phase with the voltage source on the other side, or, more generally, that the time measured at the source location is precisely the same time as measured at all other points in the circuit. When distances are sufﬁciently large between source and receiver, time delay effects become appreciable, leading to delay-induced phase differences. In short, we deal with wave phenomena on transmission lines in the same manner that we deal with point-to-point energy propagation in free space or in dielectrics. The basic elements in a circuit, such as resistors, capacitors, inductors, and the connections between them, are considered lumped elements if the time delay in traversing the elements is negligible. On the other hand, if the elements or interconnections are large enough, it may be necessary to consider them as distributed elements. This means that their resistive, capacitive, and inductive characteristics must be evaluated on a per-unit-distance basis. Transmission lines have this property in general, and thus they become circuit elements in themselves, possessing impedances that contribute to the circuit problem. The basic rule is that one must consider elements as distributed if the propagation delay across the element dimension is on the order of the shortest time interval of interest. In the time-harmonic case,
301

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this condition would lead to a measurable phase difference between each end of the device in question. In this chapter, we investigate wave phenomena in transmission lines. Our objectives include (1) to understand how to treat transmission lines as circuit elements possessing complex impedances that are functions of line length and frequency, (2) to understand wave propagation on lines, including cases in which losses may occur, (3) to learn methods of combining different transmission lines to accomplish a desired objective, and (4) to understand transient phenomena on lines. ■

10.1 PHYSICAL DESCRIPTION OF TRANSMISSION LINE PROPAGA TION
To obtain a feel for the manner in which waves propagate on transmission lines, the following demonstration may be helpful. Consider a lossless line, as shown in Figure 10.1. By lossless, we mean that all power that is launched into the line at the input end eventually arrives at the output end. A battery having voltage V0 is connected to the input by closing switch S1 at time t = 0. When the switch is closed, the effect is to launch voltage, V + = V0 . This voltage does not instantaneously appear everywhere on the line, but rather begins to travel from the battery toward the load resistor, R, at a certain velocity. The wavefront, represented by the vertical dashed line in Figure 10.1, represents the instantaneous boundary between the section of the line that has been charged to V0 and the remaining section that is yet to be charged. It also represents the boundary between the section of the line that carries the charging current, I + , and the remaining section that carries no current. Both current and voltage are discontinuous across the wavefront. As the line charges, the wavefront moves from left to right at velocity ν, which is to be determined. On reaching the far end, all or a fraction of the wave voltage and current will reﬂect, depending on what the line is attached to. For example, if the resistor at the far end is left disconnected (switch S2 is open), then all of the wavefront voltage will be reﬂected. If the resistor is connected, then some fraction of the incident voltage will reﬂect. The details of this will be treated in Section 10.9. Of interest at the moment are the factors that determine the wave velocity. The key

S1
V0

S2 I+ + V =V 0 _
+

R

Figure 10.1 Basic transmission line circuit, showing voltage and current waves initiated by closing switch S1 .

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L1 V0

L2

L3

L4

C1

C2

C3

R

Figure 10.2 Lumped-element model of a transmission line. All inductance values are equal, as are all capacitance values.

to understanding and quantifying this is to note that the conducting transmission line will possess capacitance and inductance that are expressed on a per-unit-length basis. We have already derived expressions for these and evaluated them in Chapters 6 and 8 for certain transmission line geometries. Knowing these line characteristics, we can construct a model for the transmission line using lumped capacitors and inductors, as shown in Figure 10.2. The ladder network thus formed is referred to as a pulse-forming network, for reasons that will soon become clear.1 Consider now what happens when connecting the same switched voltage source to the network. Referring to Figure 10.2, on closing the switch at the battery location, current begins to increase in L 1 , allowing C1 to charge. As C1 approaches full charge, current in L 2 begins to increase, allowing C2 to charge next. This progressive charging process continues down the network, until all three capacitors are fully charged. In the network, a “wavefront” location can be identiﬁed as the point between two adjacent capacitors that exhibit the most difference between their charge levels. As the charging process continues, the wavefront moves from left to right. Its speed depends on how fast each inductor can reach its full-current state and, simultaneously, by how fast each capacitor is able to charge to full voltage. The wave is faster if the values of L i and Ci are lower. We therefore expect the wave velocity to be inversely proportional to a function involving the product of inductance and capacitance. In the lossless transmission line, it turns out (as will be shown) that the wave velocity is given by √ ν = 1/ LC, where L and C are speciﬁed per unit length. Similar behavior is seen in the line and network when either is initially charged. In this case, the battery remains connected, and a resistor can be connected (by a switch) across the output end, as shown in Figure 10.2. In the case of the ladder network, the capacitor nearest the shunted end (C3 ) will discharge through the resistor ﬁrst, followed by the next-nearest capacitor, and so on. When the network is completely discharged, a voltage pulse has been formed across the resistor, and so we see why this ladder conﬁguration is called a pulse-forming network. Essentially identical behavior is seen in a charged transmission line when connecting a resistor between conductors at the output end. The switched voltage exercises, as used in these discussions, are examples of transient problems on transmission lines. Transients will be treated in detail in Section 10.14. In the beginning, line responses to sinusoidal signals are emphasized.
1

Designs and applications of pulse-forming networks are discussed in Reference 1.

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Finally, we surmise that the existence of voltage and current across and within the transmission line conductors implies the existence of electric and magnetic ﬁelds in the space around the conductors. Consequently, we have two possible approaches to the analysis of transmission lines: (1) We can solve Maxwell’s equations subject to the line conﬁguration to obtain the ﬁelds, and with these ﬁnd general expressions for the wave power, velocity, and other parameters of interest. (2) Or we can (for now) avoid the ﬁelds and solve for the voltage and current using an appropriate circuit model. It is the latter approach that we use in this chapter; the contribution of ﬁeld theory is solely in the prior (and assumed) evaluation of the inductance and capacitance parameters. We will ﬁnd, however, that circuit models become inconvenient or useless when losses in transmission lines are to be fully characterized, or when analyzing more complicated wave behavior (i.e., moding) which may occur as frequencies get high. The loss issues will be taken up in Section 10.5. Moding phenomena will be considered in Chapter 13.

10.2 THE TRANSMISSION LINE EQUATIONS
Our ﬁrst goal is to obtain the differential equations, known as the wave equations, which the voltage or current must satisfy on a uniform transmission line. To do this, we construct a circuit model for an incremental length of line, write two circuit equations, and use these to obtain the wave equations. Our circuit model contains the primary constants of the transmission line. These include the inductance, L, and capacitance, C, as well as the shunt conductance, G, and series resistance, R—all of which have values that are speciﬁed per unit length. The shunt conductance is used to model leakage current through the dielectric that may occur throughout the line length; the assumption is that the dielectric may possess conductivity, σd , in addition to a dielectric constant, r , where the latter affects the capacitance. The series resistance is associated with any ﬁnite conductivity, σc , in the conductors. Either one of the latter parameters, R and G, will be responsible for power loss in transmission. In general, both are functions of frequency. Knowing the frequency and the dimensions, we can determine the values of R, G, L, and C by using formulas developed in earlier chapters. We assume propagation in the az direction. Our model consists of a line section of length z containing resistance R z, inductance L z, conductance G z, and capacitance C z, as indicated in Figure 10.3. Because the section of the line looks the same from either end, we divide the series elements in half to produce a symmetrical network. We could equally well have placed half the conductance and half the capacitance at each end. Our objective is to determine the manner and extent to which the output voltage and current are changed from their input values in the limit as the length approaches a very small value. We will consequently obtain a pair of differential equations that describe the rates of change of voltage and current with respect to z. In Figure 10.3, the input and output voltages and currents differ respectively by quantities V and I , which are to be determined. The two equations are obtained by successive applications of Kirchoff’s voltage law (KVL) and Kirchoff’s current law (KCL).

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KCL

+

g

c

+
KVL

Figure 10.3 Lumped-element model of a short transmission line section with losses. The length of the section is z. Analysis involves applying Kirchoff’s voltage and current laws (KVL and KCL) to the indicated loop and node, respectively.

First, KVL is applied to the loop that encompasses the entire section length, as shown in Figure 10.3: 1 1 ∂I 1 ∂ I ∂I V = RI z + L z+ L + z 2 2 ∂t 2 ∂t ∂t 1 (1) + R(I + I ) z + (V + V ) 2 We can solve Eq. (1) for the ratio, V / z, obtaining: ∂I 1 ∂ I 1 V = − RI + L + L + R I z ∂t 2 ∂t 2 Next, we write: ∂V ∂I z and V = z (3) ∂z ∂z which are then substituted into (2) to result in z ∂ ∂I ∂V =− 1+ RI + L (4) ∂z 2 ∂z ∂t Now, in the limit as z approaches zero (or a value small enough to be negligible), (4) simpliﬁes to the ﬁnal form: I = ∂V ∂I = − RI + L ∂z ∂t (5) (2)

Equation (5) is the ﬁrst of the two equations that we are looking for. To ﬁnd the second equation, we apply KCL to the upper central node in the circuit of Figure 10.3, noting from the symmetry that the voltage at the node will be V + V /2: I = Ig + Ic + (I + +C z ∂ ∂t V+ I) = G z V + V 2 + (I + I) V 2

(6)

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Then, using (3) and simplifying, we obtain z ∂ ∂I =− 1+ ∂z 2 ∂z Again, we obtain the ﬁnal form by allowing tude. The result is GV + C ∂V ∂t (7)

z to be reduced to a negligible magni-

∂V ∂I = − GV + C ∂z ∂t

(8)

The coupled differential equations, (5) and (8), describe the evolution of current and voltage in any transmission line. Historically, they have been referred to as the telegraphist’s equations. Their solution leads to the wave equation for the transmission line, which we now undertake. We begin by differentiating Eq. (5) with respect to z and Eq. (8) with respect to t, obtaining: ∂ 2I ∂I ∂2V −L = −R ∂z 2 ∂z ∂t∂z and ∂I ∂V ∂2V = −G −C 2 ∂z∂t ∂t ∂t (10) (9)

Next, Eqs. (8) and (10) are substituted into (9). After rearranging terms, the result is: ∂2V ∂V ∂2V = LC 2 + (LG + RC) + RGV 2 ∂z ∂t ∂t (11)

An analogous procedure involves differentiating Eq. (5) with respect to t and Eq. (8) with respect to z. Then, Eq. (5) and its derivative are substituted into the derivative of (8) to obtain an equation for the current that is in identical form to that of (11): ∂ 2I ∂ 2I ∂I = LC 2 + (LG + RC) + RGI 2 ∂z ∂t ∂t (12)

Equations (11) and (12) are the general wave equations for the transmission line. Their solutions under various conditions form a major part of our study.

10.3 LOSSLESS PROPAGA TION
Lossless propagation means that power is not dissipated or otherwise deviated as the wave travels down the transmission line; all power at the input end eventually reaches the output end. More realistically, any mechanisms that would cause losses to occur have negligible effect. In our model, lossless propagation occurs when R = G = 0.

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Under this condition, only the ﬁrst term on the right-hand side of either Eq. (11) or Eq. (12) survives. Eq. (11), for example, becomes ∂2V ∂2V = LC 2 ∂z 2 ∂t (13)

In considering the voltage function that will satisfy (13), it is most expedient to simply state the solution, and then show that it is correct. The solution of (13) is of the form: V (z, t) = f 1 t − z z + f2 t + ν ν = V+ + V− (14)

where ν, the wave velocity, is a constant. The expressions (t ± z/ν) are the arguments of functions f 1 and f 2 . The identities of the functions themselves are not critical to the solution of (13). Therefore, f 1 and f 2 can be any function. The arguments of f 1 and f 2 indicate, respectively, travel of the functions in the forward and backward z directions. We assign the symbols V + and V − to identify the forward and backward voltage wave components. To understand the behavior, consider for example the value of f 1 (whatever this might be) at the zero value of its argument, occurring when z = t = 0. Now, as time increases to positive values (as it must), and if we are to keep track of f 1 (0), then the value of z must also increase to keep the argument (t − z/ν) equal to zero. The function f 1 therefore moves (or propagates) in the positive z direction. Using similar reasoning, the function f 2 will propagate in the negative z direction, as z in the argument (t + z/ν) must decrease to offset the increase in t. Therefore we associate the argument (t − z/ν) with forward z propagation, and the argument (t + z/ν) with backward z travel. This behavior occurs irrespective of what f 1 and f 2 are. As is evident in the argument forms, the propagation velocity is ν in both cases. We next verify that functions having the argument forms expressed in (14) are solutions to (13). First, we take partial derivatives of f 1 , for example with respect to z and t. Using the chain rule, the z partial derivative is ∂ f1 ∂ f1 ∂(t − z/ν) 1 = = − f1 ∂z ∂(t − z/ν) ∂z ν (15)

where it is apparent that the primed function, f 1 , denotes the derivative of f 1 with respect to its argument. The partial derivative with respect to time is ∂ f1 ∂ f1 ∂(t − z/ν) = = f1 ∂t ∂(t − z/ν) ∂t (16)

Next, the second partial derivatives with respect to z and t can be taken using similar reasoning: 1 ∂ 2 f1 = 2 f1 2 ∂z ν and ∂ 2 f1 = f1 ∂t 2 (17)

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where f 1 is the second derivative of f 1 with respect to its argument. The results in (17) can now be substituted into (13), obtaining 1 f = LC f 1 ν2 1 (18)

We now identify the wave velocity for lossless propagation, which is the condition for equality in (18): ν=√ 1 LC (19)

Performing the same procedure using f 2 (and its argument) leads to the same expression for ν. The form of ν as expressed in Eq. (19) conﬁrms our original expectation that the wave velocity would be in some inverse proportion to L and C. The same result will be true for current, as Eq. (12) under lossless conditions would lead to a solution of the form identical to that of (14), with velocity given by (19). What is not known yet, however, is the relation between voltage and current. We have already found that voltage and current are related through the telegraphist’s equations, (5) and (8). These, under lossless conditions (R = G = 0), become ∂V ∂I = −L ∂z ∂t ∂V ∂I = −C ∂z ∂t (20)

(21)

Using the voltage function, we can substitute (14) into (20) and use the methods demonstrated in (15) to write 1 ∂V 1 ∂I =− = ( f − f2 ) ∂t L ∂z Lν 1 (22)

We next integrate (22) over time, obtaining the current in terms of its forward and backward propagating components: I (z, t) = 1 Lν f1 t − z z − f2 t + ν ν = I+ + I− (23)

In performing this integration, all integration constants are set to zero. The reason for this, as demonstrated by (20) and (21), is that a time-varying voltage must lead to a time-varying current, with the reverse also true. The factor 1/Lν appearing in (23) multiplies voltage to obtain current, and so we identify the product Lν as the characteristic impedance, Z 0 , of the lossless line. Z 0 is deﬁned as the ratio of

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−

−

Figure 10.4 polarity.

Current directions in waves having positive voltage

the voltage to the current in a single propagating wave. Using (19), we write the characteristic impedance as Z 0 = Lν = By inspecting (14) and (23), we now note that V + = Z0 I + and V − = −Z 0 I − (25b) (25a) L C (24)

The signiﬁcance of the preceding relations can be seen in Figure 10.4. The ﬁgure shows forward- and backward-propagating voltage waves, V + and V − , both of which have positive polarity. The currents that are associated with these voltages will ﬂow in opposite directions. We deﬁne positive current as having a clockwise ﬂow in the line, and negative current as having a counterclockwise ﬂow. The minus sign in (25b) thus assures that negative current will be associated with a backward-propagating wave that has positive polarity. This is a general convention, applying to lines with losses also. Propagation with losses is studied by solving (11) under the assumption that either R or G (or both) are not zero. We will do this in Section 10.7 under the special case of sinusoidal voltages and currents. Sinusoids in lossless transmission lines are considered in Section 10.4.

10.4 LOSSLESS PROPAGATION OF SINUSOIDAL VOLTAGES
An understanding of sinusoidal waves on transmission lines is important because any signal that is transmitted in practice can be decomposed into a discrete or continuous summation of sinusoids. This is the basis of frequency domain analysis of signals on

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lines. In such studies, the effect of the transmission line on any signal can be determined by noting the effects on the frequency components. This means that one can effectively propagate the spectrum of a given signal, using frequency-dependent line parameters, and then reassemble the frequency components into the resultant signal in time domain. Our objective in this section is to obtain an understanding of sinusoidal propagation and the implications on signal behavior for the lossless line case. We begin by assigning sinusoidal functions to the voltage functions in Eq. (14). Speciﬁcally, we consider a speciﬁc frequency, f = ω/2π , and write f 1 = f 2 = V0 cos(ωt + φ). By convention, the cosine function is chosen; the sine is obtainable, as we know, by setting φ = −π/2. We next replace t with (t ± z/ν p ), obtaining where we have assigned a new notation to the velocity, which is now called the phase velocity, ν p . This is applicable to a pure sinusoid (having a single frequency) and will be found to depend on frequency in some cases. Choosing, for the moment, φ = 0, we obtain the two possibilities of forward or backward z travel by choosing the minus or plus sign in (26). The two cases are: V f (z, t) = |V0 | cos(ωt − βz) and Vb (z, t) = |V0 | cos(ωt + βz) (backward z propagation) (27b) (forward z propagation) (27a) V(z, t) = |V0 | cos[ω(t ± z/ν p ) + φ] = |V0 | cos[ωt ± βz + φ] (26)

where the magnitude factor, |V0 |, is the value of V at z = 0, t = 0. We deﬁne the phase constant β, obtained from (26), as β≡ ω νp (28)

We refer to the solutions expressed in (27a) and (27b) as the real instantaneous forms of the transmission-line voltage. They are the mathematical representations of what one would experimentally measure. The terms ωt and βz, appearing in these equations, have units of angle and are usually expressed in radians. We know that ω is the radian time frequency, measuring phase shift per unit time, and it has units of rad/s. In a similar way, we see that β will be interpreted as a spatial frequency, which in the present case measures the phase shift per unit distance along the z direction. Its units are rad/m. If we were to ﬁx the time at t = 0, Eqs. (27a) and (27b) would become V f (z, 0) = Vb (z, 0) = |V0 | cos(βz) (29)

which we identify as a simple periodic function that repeats every incremental distance λ, known as the wavelength. The requirement is that βλ = 2π , and so λ= νp 2π = β f (30)

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We next consider a point (such as a wave crest) on the cosine function of Eq. (27a), the occurrence of which requires the argument of the cosine to be an integer multiple of 2π. Considering the mth crest of the wave, the condition at t = 0 becomes βz = 2mπ To keep track of this point on the wave, we require that the entire cosine argument be the same multiple of 2π for all time. From (27a) the condition becomes Again, with increasing time, the position z must also increase in order to satisfy (31). Consequently the wave crest (and the entire wave) travels in the positive z direction at velocity νp . Eq. (27b), having cosine argument (ωt + βz), describes a wave that travels in the negative z direction, since as time increases, z must now decrease to keep the argument constant. Similar behavior is found for the wave current, but complications arise from line-dependent phase differences that occur between current and voltage. These issues are best addressed once we are familiar with complex analysis of sinusoidal signals. ωt − βz = ω(t − z/νp ) = 2mπ (31)

10.5 COMPLEX ANALYSIS OF SINUSOIDAL WAVES
Expressing sinusoidal waves as complex functions is useful (and essentially indispensable) because it greatly eases the evaluation and visualization of phase that will be found to accumulate by way of many mechanisms. In addition, we will ﬁnd many cases in which two or more sinusoidal waves must be combined to form a resultant wave—a task made much easier if complex analysis is used. Expressing sinusoidal functions in complex form is based on the Euler identity: e± j x = cos(x) ± j sin(x) (32)

from which we may write the cosine and sine, respectively, as the real and imaginary parts of the complex exponent: 1 1 cos(x) = Re[e± j x ] = (e j x + e− j x ) = e j x + c.c. (33a) 2 2 1 jx 1 jx (33b) (e − e− j x ) = e + c.c. sin(x) = ±Im[e± j x ] = 2j 2j √ where j ≡ −1, and where c.c. denotes the complex conjugate of the preceding term. The conjugate is formed by changing the sign of j wherever it appears in the complex expression. We may next apply (33a) to our voltage wave function, Eq. (26): 1 V(z, t) = |V0 | cos[ωt ± βz + φ] = (|V0 |e jφ ) e± jβz e jωt + c.c. (34) 2
V0

Note that we have arranged the phases in (34) such that we identify the complex amplitude of the wave as V0 = (|V0 |e jφ ). In future usage, a single symbol (V0 in the

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present example) will usually be used for the voltage or current amplitudes, with the understanding that these will generally be complex (having magnitude and phase). Two additional deﬁnitions follow from Eq. (34). First, we deﬁne the complex instantaneous voltage as: Vc (z, t) = V0 e± jβz e jωt (35)

The phasor voltage is then formed by dropping the e jωt factor from the complex instantaneous form: Vs (z) = V0 e± jβz (36)

The phasor voltage can be deﬁned provided we have sinusoidal steady-state conditions—meaning that V0 is independent of time. This has in fact been our assumption all along, because a time-varying amplitude would imply the existence of other frequency components in our signal. Again, we are treating only a single-frequency wave. The signiﬁcance of the phasor voltage is that we are effectively letting time stand still and observing the stationary wave in space at t = 0. The processes of evaluating relative phases between various line positions and of combining multiple waves is made much simpler in phasor form. Again, this works only if all waves under consideration have the same frequency. With the deﬁnitions in (35) and (36), the real instantaneous voltage can be constructed using (34): V(z, t) = |V0 | cos[ωt ± βz + φ] = Re[Vc (z, t)] = Or, in terms of the phasor voltage: V(z, t) = |V0 | cos[ωt ± βz + φ] = Re[Vs (z)e jωt ] = 1 Vs (z)e jωt + c.c. 2 (37b) 1 Vc + c.c. 2 (37a)

In words, we may obtain our real sinusoidal voltage wave by multiplying the phasor voltage by e jωt (reincorporating the time dependence) and then taking the real part of the resulting expression. It is imperative that one becomes familiar with these relations and their meaning before proceeding further.
E X A M P L E 10.1

Two voltage waves having equal frequencies and amplitudes propagate in opposite directions on a lossless transmission line. Determine the total voltage as a function of time and position.
Solution. Because the waves have the same frequency, we can write their combina-

tion using their phasor forms. Assuming phase constant, β, and real amplitude, V0 , the two wave voltages combine in this way: VsT (z) = V0 e− jβz + V0 e+ jβz = 2V0 cos(βz)

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In real instantaneous form, this becomes V(z, t) = Re[2V0 cos(βz)e jωt ] = 2V0 cos(βz) cos(ωt) We recognize this as a standing wave, in which the amplitude varies, as cos(βz), and oscillates in time, as cos(ωt). Zeros in the amplitude (nulls) occur at ﬁxed locations, z n = (mπ)/(2β) where m is an odd integer. We extend the concept in Section 10.10, where we explore the voltage standing wave ratio as a measurement technique.

10.6 TRANSMISSION LINE EQUATIONS AND THEIR SOLUTIONS IN PHASOR FORM
We now apply our results of the previous section to the transmission line equations, beginning with the general wave equation, (11). This is rewritten as follows, for the real instantaneous voltage, V(z, t): ∂ 2V ∂V ∂ 2V = LC 2 + (LG + RC) + RGV 2 ∂z ∂t ∂t (38)

We next substitute V(z, t) as given by the far right-hand side of (37b), noting that the complex conjugate term (c.c.) will form a separate redundant equation. We also use the fact that the operator ∂/∂t, when applied to the complex form, is equivalent to multiplying by a factor of jω. After substitution, and after all time derivatives are taken, the factor e jωt divides out. We are left with the wave equation in terms of the phasor voltage: d 2 Vs = −ω2 LC Vs + jω(LG + RC)Vs + RGVs dz 2 Rearranging terms leads to the simpliﬁed form: d 2 Vs = (R + jωL) (G + jωC) Vs = γ 2 Vs dz 2
Z Y

(39)

(40)

where Z and Y , as indicated, are respectively the net series impedance and the net shunt admittance in the transmission line—both as per-unit-distance measures. The propagation constant in the line is deﬁned as γ = (R + jωL)(G + jωC) = √ Z Y = α + jβ (41)

The signiﬁcance of the term will be explained in Section 10.7. For our immediate purposes, the solution of (40) will be Vs (z) = V0+ e−γ z + V0− e+γ z (42a)

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The wave equation for current will be identical in form to (40). We therefore expect the phasor current to be in the form:
+ − Is (z) = I0 e−γ z + I0 eγ z

(42b)

The relation between the current and voltage waves is now found, as before, through the telegraphist’s equations, (5) and (8). In a manner consistent with Eq. (37b), we write the sinusoidal current as 1 1 I(z, t) = |I0 | cos(ωt ± βz + ξ ) = (|I0 |e jξ ) e± jβz e jωt + c.c. = Is (z)e jωt + c.c. 2 2 (43) Substituting the far right-hand sides of (37b) and (43) into (5) and (8) transforms the latter equations as follows: ∂V ∂I = − RI + L ∂z ∂t and ∂I ∂V = − GV + C ∂z ∂t ⇒ d Is = −(G + jωC)Vs = −Y Vs dz (44b) ⇒ d Vs = −(R + jωL)Is = −Z Is dz (44a)
I0

We can now substitute (42a) and (42b) into either (44a) or (44b) [we will use (44a)] to ﬁnd: Next, equating coefﬁcients of e−γ z and eγ z , we ﬁnd the general expression for the line characteristic impedance: Z V0+ V− Z Z =− 0 = =√ (46) = + − γ Y I0 I0 ZY Incorporating the expressions for Z and Y , we ﬁnd the characteristic impedance in terms of our known line parameters: Z0 = Z0 = R + jωL = |Z 0 |e jθ G + jωC (47)
+ − −γ V0+ e−γ z + γ V0− eγ z = −Z (I0 e−γ z + I0 eγ z )

(45)

Note that with the voltage and current as given in (37b) and (43), we would identify the phase of the characteristic impedance, θ = φ − ξ .
E X A M P L E 10.2

A lossless transmission line is 80 cm long and operates at a frequency of 600 MHz. The line parameters are L = 0.25 µH/m and C = 100 pF/m. Find the characteristic impedance, the phase constant, and the phase velocity.

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Solution. Because the line is lossless, both R and G are zero. The characteristic impedance is

0.25 × 10−6 = 50 100 × 10−12 √ √ Because γ = α + jβ = (R + jωL)(G + jωC) = jω LC, we see that √ β = ω LC = 2π(600 × 106 ) (0.25 × 10−6 )(100 × 10−12 ) = 18.85 rad/m Z0 = Also, νp = ω 2π(600 × 106 ) = = 2 × 108 m/s β 18.85

L = C

10.7 LOW-LOSS PROPAGATION
Having obtained the phasor forms of voltage and current in a general transmission line [Eqs. (42a) and (42b)], we can now look more closely at the signiﬁcance of these results. First we incorporate (41) into (42a) to obtain Vs (z) = V0+ e−αz e− jβz + V0− eαz e jβz (48)

Next, multiplying (48) by e jωt and taking the real part gives the real instantaneous voltage: V(z, t) = V0+ e−αz cos(ωt − βz) + V0− eαz cos(ωt + βz) (49)

In this exercise, we have assigned V0+ and V0− to be real. Eq. (49) is recognized as describing forward- and backward-propagating waves that diminish in amplitude with distance according to e−αz for the forward wave, and eαz for the backward wave. Both waves are said to attenuate with propagation distance at a rate determined by the attenuation coefficient α, expressed in units of nepers/m [Np/m].2 The phase constant, β, found by taking the imaginary part of (41), is likely to be a somewhat complicated function, and will in general depend on R and G. Nevertheless, β is still deﬁned as the ratio ω/νp , and the wavelength is still deﬁned as the distance that provides a phase shift of 2π rad, so that λ = 2π/β. By inspecting (41), we observe that losses in propagation are avoided (or α = 0)√ only when R = G = 0. In √ that case, (41) gives γ = jβ = jω LC, and so νp = 1/ LC, as we found before. Expressions for α and β when losses are small can be readily obtained from (41). In the low-loss approximation, we require R ωL and G ωC, a condition that

The term neper was selected (by some poor speller) to honor John Napier, a Scottish mathematician who ﬁrst proposed the use of logarithms.
2

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is often true in practice. Before we apply these conditions, Eq. (41) can be written in the form: γ = α + jβ = [(R + jωL)(G + jωC)]1/2 √ G R 1/2 1+ = jω LC 1+ jωL jωC
1/2

(50)

The low-loss approximation then allows us to use the ﬁrst three terms in the binomial series: √ x x2 . 1+x =1+ − (x 1) (51) 2 8 We use (51) to expand the terms in large parentheses in (50), obtaining: √ . γ = jω LC 1+ R R2 + j2ωL 8ω2 L 2 1+ G G2 + j2ωC 8ω2 C 2 (52)

All products in (52) are then carried out, neglecting the terms involving RG 2 , R 2 G, and R 2 G 2 , as these will be negligible compared to all others. The result is √ 1 . γ = α + jβ = jω LC 1 + j2ω G R + L C + 1 8ω2 R2 2RG G2 − + 2 L2 LC C (53) Now, separating real and imaginary parts of (53) yields α and β: . 1 α= 2 and 1 . √ β = ω LC 1 + 8 R G − ωC ωL
2

R

C +G L

L C

(54a)

(54b)

We note that α scales in direct proportion to R and G, as would be expected. We also note that the terms in (54b) that involve R and G lead to a phase velocity, νp = ω/β, that is frequency-dependent. Moreover, the group velocity, νg = dω/dβ, will also depend on frequency, and will lead to signal distortion, as we will explore in Chapter 12. Note that with nonzero R and G, phase and group velocities that are constant with frequency can be obtained when R/L = G/C, known as Heaviside’s . √ condition. In this case, (54b) becomes β = ω LC, and the line is said to be distortionless. Further complications occur when accounting for possible frequency dependencies within R, G, L, and C. Consequently, conditions of low-loss or distortion-free propagation will usually occur over limited frequency ranges. As a rule, loss increases with increasing frequency, mostly because of the increase in R with frequency. The nature of this latter effect, known as skin effect loss, requires ﬁeld theory to understand

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and quantify. We will study this in Chapter 11, and we will apply it to transmission line structures in Chapter 13. Finally, we can apply the low-loss approximation to the characteristic impedance, Eq. (47). Using (51), we ﬁnd   R R R2 jωL 1 + jωL . R + jωL L  1 + j2ωL + 8ω2 L 2  Z0 = = = (55) 2 G + jωC C jωC 1 + G 1+ G + G jωC j2ωC 8ω2 C 2

Next, we multiply (55) by a factor of 1, in the form of the complex conjugate of the denominator of (55) divided by itself. The resulting expression is simpliﬁed by neglecting all terms on the order of R 2 G, G 2R, and higher. Additionally, the approx. imation, 1/(1 + x) = 1 − x, where x 1 is used. The result is . Z0 = L C 1+ 1 2ω2 1 4 G R + L C
2

−

G2 C2

+

j 2ω

R G − C L

(56)

Note that when Heaviside’s condition (again, R/L = G/C) holds, Z 0 simpliﬁes to √ just L/C, as is true when both R and G are zero.
E X A M P L E 10.3

Suppose in a certain transmission line G = 0, but R is ﬁnite valued and satisﬁes the low-loss requirement, R ωL. Use Eq. (56) to write the approximate magnitude and phase of Z 0 . ond term in the real part [proportional to (R/ωL)2 ]. Therefore, the characteristic impedance becomes . Z 0 (G = 0) = L C 1− j R 2ωL = |Z 0 |e jθ
Solution. With G = 0, the imaginary part of (56) is much greater than the sec-

. √ where |Z 0 | = L/C, and θ = tan−1 (−R/2ωL).

D10.1. At an operating radian frequency of 500 Mrad/s, typical circuit values for a certain transmission line are: R = 0.2 /m, L = 0.25 µH/m, G = 10 µS/m, and C = 100 pF/m. Find: (a) α; (b) β; (c) λ; (d) νp ; (e) Z 0 .
Ans. 2.25 mNp/m; 2.50 rad/m; 2.51 m; 2 × 108 m/sec; 50.0 − j0.0350

10.8 POWER TRANSMISSION AND THE USE OF DECIBELS IN LOSS CHARACTERIZATION
Having found the sinusoidal voltage and current in a lossy transmission line, we next evaluate the power transmitted over a speciﬁed distance as a function of voltage and current amplitudes. We start with the instantaneous power, given simply as the product of the real voltage and current. Consider the forward-propagating term in (49), where

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again, the amplitude, V0+ = |V0 |, is taken to be real. The current waveform will be similar, but will generally be shifted in phase. Both current and voltage attenuate according to the factor e−αz . The instantaneous power therefore becomes: P(z, t) = V(z, t)I(z, t) = |V0 ||I0 |e−2αz cos(ωt − βz) cos(ωt − βz + θ) Usually, the time-averaged power, P , is of interest. We ﬁnd this through: P = 1 T
T 0

(57)

|V0 ||I0 |e−2αz cos(ωt − βz) cos(ωt − βz + θ)dt

(58)

where T = 2π/ω is the time period for one oscillation cycle. Using a trigonometric identity, the product of cosines in the integrand can be written as the sum of individual cosines at the sum and difference frequencies: P = 1 T
T 0

1 |V0 ||I0 |e−2αz [cos(2ωt − 2βz + θ ) + cos(θ )] dt 2

(59)

The ﬁrst cosine term integrates to zero, leaving the cos θ term. The remaining integral easily evaluates as P = 1 |V0 |2 −2αz 1 cos θ [W] |V0 ||I0 |e−2αz cos θ = e 2 2 |Z 0 | (60)

The same result can be obtained directly from the phasor voltage and current. We begin with these, expressed as Vs (z) = V0 e−αz e− jβz and Is (z) = I0 e−αz e− jβz = V0 −αz − jβz e e Z0 (62) (61)

where Z 0 = |Z 0 |e jθ . We now note that the time-averaged power as expressed in (60) can be obtained from the phasor forms through: P = 1 Re{Vs Is∗ } 2 (63)

where again, the asterisk (∗ ) denotes the complex conjugate (applied in this case to the current phasor only). Using (61) and (62) in (63), it is found that P = 1 V0∗ e−αz e+ jβz Re V0 e−αz e− jβz 2 |Z 0 |e− jθ 1 |V0 |2 −2αz V0 V0∗ −2αz jθ 1 = e cos θ e e = Re 2 |Z 0 | 2 |Z 0 |

(64)

which we note is identical to the time-integrated result in (60). Eq. (63) applies to any single-frequency wave.

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An important result of the preceding exercise is that power attenuates as e−2αz , or P(z) = P(0) e−2αz (65)

Power drops at twice the exponential rate with distance as either voltage or current. A convenient measure of power loss is in decibel units. This is based on expressing the power decrease as a power of 10. Speciﬁcally, we write P(z) = e−2αz = 10−καz P(0) where the constant, κ, is to be determined. Setting αz = 1, we ﬁnd e−2 = 10−κ ⇒ κ = log10 (e2 ) = 0.869 P(0) P(z) (67) Now, by deﬁnition, the power loss in decibels (dB) is Power loss (dB) = 10 log10 = 8.69αz (68) (66)

where we note that inverting the power ratio in the argument of the log function [as compared to the ratio in (66)] yields a positive number for the dB loss. Also, noting that P ∝ |V0 |2 , we may write, equivalently: Power loss (dB) = 10 log10 where |V0 (z)| = |V0 (0)|e−αz .
E X A M P L E 10.4

P(0) P(z)

= 20 log10

|V0 (0)| |V0 (z)|

(69)

A 20-m length of transmission line is known to produce a 2.0-dB drop in power from end to end. (a) What fraction of the input power reaches the output? (b) What fraction of the input power reaches the midpoint of the line? (c) What exponential attenuation coefﬁcient, α, does this represent?
Solution. (a) The power fraction will be

P(20) = 10−0.2 = 0.63 P(0) (b) 2 dB in 20 m implies a loss rating of 0.2 dB/m. So, over a 10-m span, the loss is 1.0 dB. This represents the power fraction, 10−0.1 = 0.79. (c) The exponential attenuation coefﬁcient is found through α= 2.0 dB = 0.012 [Np/m] (8.69 dB/Np)(20 m)

A ﬁnal point addresses the question: Why use decibels? The most compelling reason is that when evaluating the accumulated loss for several lines and devices that

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are all end-to-end connected, the net loss in dB for the entire span is just the sum of the dB losses of the individual elements. D10.2. Two transmission lines are to be joined end to end. Line 1 is 30 m long and is rated at 0.1 dB/m. Line 2 is 45 m long and is rated at 0.15 dB/m. The joint is not done well and imparts a 3-dB loss. What percentage of the input power reaches the output of the combination?
Ans. 5.3%

10.9 WAVE REFLECTION AT DISCONTINUITIES
The concept of wave reﬂection was introduced in Section 10.1. As implied there, the need for a reﬂected wave originates from the necessity to satisfy all voltage and current boundary conditions at the ends of transmission lines and at locations at which two dissimilar lines are connected to each other. The consequences of reﬂected waves are usually less than desirable, in that some of the power that was intended to be transmitted to a load, for example, reﬂects and propagates back to the source. Conditions for achieving no reﬂected waves are therefore important to understand. The basic reﬂection problem is illustrated in Figure 10.5. In it, a transmission line of characteristic impedance Z 0 is terminated by a load having complex impedance, Z L = R L + j X L . If the line is lossy, then we know that Z 0 will also be complex. For convenience, we assign coordinates such that the load is at location z = 0. Therefore, the line occupies the region z < 0. A voltage wave is presumed to be incident on the load, and is expressed in phasor form for all z: Vi (z) = V0i e−αz e− jβz (70a)

When the wave reaches the load, a reﬂected wave is generated that back-propagates: Vr (z) = V0r e+αz e+ jβz (70b)

Z0

Vi Vr z=0 ZL = RL + jXL

Figure 10.5 Voltage wave reflection from a complex load impedance.

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The phasor voltage at the load is now the sum of the incident and reﬂected voltage phasors, evaluated at z = 0: VL = V0i + V0r (71) Additionally, the current through the load is the sum of the incident and reﬂected currents, also at z = 0: 1 VL 1 I L = I0i + I0r = [V0i − V0r ] = = [V0i + V0r ] (72) Z0 ZL ZL We can now solve for the ratio of the reﬂected voltage amplitude to the incident voltage amplitude, deﬁned as the reflectio coefficient : ≡ V0r Z L − Z0 = = | |e jφr V0i Z L + Z0 (73)

where we emphasize the complex nature of —meaning that, in general, a reﬂected wave will experience a reduction in amplitude and a phase shift, relative to the incident wave. Now, using (71) with (73), we may write from which we ﬁnd the transmission coefficient deﬁned as the ratio of the load voltage amplitude to the incident voltage amplitude: τ≡ VL =1+ V0i = 2Z L = |τ |e jφt Z0 + Z L (75) VL = V0i + V0i (74)

A point that may at ﬁrst cause some alarm is that if is a positive real number, then τ > 1; the voltage amplitude at the load is thus greater than the incident voltage. Although this would seem counterintuitive, it is not a problem because the load current will be lower than that in the incident wave. We will ﬁnd that this always results in an average power at the load that is less than or equal to that in the incident wave. An additional point concerns the possibility of loss in the line. The incident wave amplitude that is used in (73) and (75) is always the amplitude that occurs at the load—after loss has occurred in propagating from the input. Usually, the main objective in transmitting power to a load is to conﬁgure the line/load combination such that there is no reﬂection. The load therefore receives all the transmitted power. The condition for this is = 0, which means that the load impedance must be equal to the line impedance. In such cases the load is said to be matched to the line (or vice versa). Various impedance-matching methods exist, many of which will be explored later in this chapter. Finally, the fractions of the incident wave power that are reﬂected and dissipated by the load need to be determined. The incident power is found from (64), where this time we position the load at z = L, with the line input at z = 0. Pi = V0 V0∗ −2αL jθ 1 Re e e 2 |Z 0 | = 1 |V0 |2 −2αL e cos θ 2 |Z 0 | (76a)

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The reﬂected power is then found by substituting the reﬂected wave voltage into (76a), where the latter is obtained by multiplying the incident voltage by : Pr = ( V0 )( ∗ V0∗ ) −2αL jθ 1 Re e e 2 |Z 0 | = 1 | |2 |V0 |2 −2αL e cos θ 2 |Z 0 | (76b)

The reﬂected power fraction at the load is now determined by the ratio of (76b) to (76a): Pr = Pi
∗

= | |2

(77a)

The fraction of the incident power that is transmitted into the load (or dissipated by it) is therefore Pt = 1 − | |2 Pi (77b)

The reader should be aware that the transmitted power fraction is not |τ |2 , as one might be tempted to conclude. In situations involving the connection of two semi-inﬁnite transmission lines having different characteristic impedances, reﬂections will occur at the junction, with the second line being treated as the load. For a wave incident from line 1 (Z 01 ) to line 2 (Z 02 ), we ﬁnd = Z 02 − Z 01 Z 02 + Z 01 (78)

The fraction of the power that propagates into the second line is then 1 − | |2 .
E X A M P L E 10.5

A 50- lossless transmission line is terminated by a load impedance, Z L = 50 − j75 . If the incident power is 100 mW, ﬁnd the power dissipated by the load.
Solution. The reﬂection coefﬁcient is

= Then

50 − j75 − 50 Z L − Z0 = 0.36 − j0.48 = 0.60e− j.93 = Z L + Z0 50 − j75 + 50 Pt = (1 − | |2 ) Pi = [1 − (0.60)2 ](100) = 64 mW

E X A M P L E 10.6

Two lossy lines are to be joined end to end. The ﬁrst line is 10 m long and has a loss rating of 0.20 dB/m. The second line is 15 m long and has a loss rating of 0.10 dB/m. The reﬂection coefﬁcient at the junction (line 1 to line 2) is = 0.30. The input

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power (to line 1) is 100 mW. (a) Determine the total loss of the combination in dB. (b) Determine the power transmitted to the output end of line 2.
Solution. (a) The dB loss of the joint is

L j (dB) = 10 log10

1 1 − | |2

= 10 log10

1 1 − 0.09

= 0.41 dB

The total loss of the link in dB is now L t (dB) = (0.20)(10) + 0.41 + (0.10)(15) = 3.91 dB (b) The output power will be Pout = 100 × 10−0.391 = 41 mW.

10.10 VOLTAGE STANDING WAVE RATIO
In many instances, characteristics of transmission line performance are amenable to measurement. Included in these are measurements of unknown load impedances, or input impedances of lines that are terminated by known or unknown load impedances. Such techniques rely on the ability to measure voltage amplitudes that occur as functions of position within a line, usually designed for this purpose. A typical apparatus consists of a slotted line, which is a lossless coaxial transmission line having a longitudinal gap in the outer conductor along its entire length. The line is positioned between the sinusoidal voltage source and the impedance that is to be measured. Through the gap in the slotted line, a voltage probe may be inserted to measure the voltage amplitude between the inner and outer conductors. As the probe is moved along the length of the line, the maximum and minimum voltage amplitudes are noted, and their ratio, known as the voltage standing wave ratio, or VSWR, is determined. The signiﬁcance of this measurement and its utility form the subject of this section. To understand the meaning of the voltage measurements, we consider a few special cases. First, if the slotted line is terminated by a matched impedance, then no reﬂected wave occurs; the probe will indicate the same voltage amplitude at every point. Of course, the instantaneous voltages that the probe samples will differ in phase by β(z 2 − z 1 ) rad as the probe is moved from z = z 1 to z = z 2 , but the system is insensitive to the phase of the ﬁeld. The equal-amplitude voltages are characteristic of an unattenuated traveling wave. Second, if the slotted line is terminated by an open or short circuit (or in general a purely imaginary load impedance), the total voltage in the line is a standing wave and, as was shown in Example 10.1, the voltage probe provides no output when it is located at the nodes; these occur periodically with half-wavelength spacing. As the probe position is changed, its output varies as |cos(βz + φ)|, where z is the distance from the load, and where the phase, φ, depends on the load impedance. For example,

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if the load is a short circuit, the requirement of zero voltage at the short leads to a null occurring there, and so the voltage in the line will vary as |sin(βz)| (where φ = ±π/2). A more complicated situation arises when the reﬂected voltage is neither 0 nor 100 percent of the incident voltage. Some energy is absorbed by the load and some is reﬂected. The slotted line, therefore, supports a voltage that is composed of both a traveling wave and a standing wave. It is customary to describe this voltage as a standing wave, even though a traveling wave is also present. We will see that the voltage does not have zero amplitude at any point for all time, and the degree to which the voltage is divided between a traveling wave and a true standing wave is expressed by the ratio of the maximum amplitude found by the probe to the minimum amplitude (VSWR). This information, along with the positions of the voltage minima or maxima with respect to that of the load, enable one to determine the load impedance. The VSWR also provides a measure of the quality of the termination. Speciﬁcally, a perfectly matched load yields a VSWR of exactly 1. A totally reﬂecting load produces an inﬁnite VSWR. To derive the speciﬁc form of the total voltage, we begin with the forward and backward-propagating waves that occur within the slotted line. The load is positioned at z = 0, and so all positions within the slotted line occur at negative values of z. Taking the input wave amplitude as V0 , the total phasor voltage is VsT (z) = V0 e− jβz + V0 e jβz (79)

The line, being lossless, has real characteristic impedance, Z 0 . The load impedance, Z L , is in general complex, which leads to a complex reﬂection coefﬁcient: = Z L − Z0 = | |e jφ Z L + Z0

(80)

If the load is a short circuit (Z L = 0), φ is equal to π ; if Z L is real and less than Z 0 , φ is also equal to π; and if Z L is real and greater than Z 0 , φ is zero. Using (80), we may rewrite (79) in the form: VsT (z) = V0 e− jβz + | |e j(βz+φ) = V0 e jφ/2 e− jβz e− jφ/2 + | |e jβz e jφ/2 (81)

To express (81) in a more useful form, we can apply the algebraic trick of adding and subtracting the term V0 (1 − | |)e− jβz : VsT (z) = V0 (1 − | |)e− jβz + V0 | |e jφ/2 e− jβz e− jφ/2 + e jβz e jφ/2 The last term in parentheses in (82) becomes a cosine, and we write VsT (z) = V0 (1 − | |)e− jβz + 2V0 | |e jφ/2 cos(βz + φ/2) (83) (82)

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The important characteristics of this result are most easily seen by converting it to real instantaneous form: V(z, t) = Re[VsT (z)e jωt ] = V0 (1 − | |) cos(ωt − βz) traveling wave

+ 2| |V0 cos(βz + φ/2) cos(ωt + φ/2) standing wave

(84)

Equation (84) is recognized as the sum of a traveling wave of amplitude (1 − | |)V0 and a standing wave having amplitude 2| |V0 . We can visualize events as follows: The portion of the incident wave that reﬂects and back-propagates in the slotted line interferes with an equivalent portion of the incident wave to form a standing wave. The rest of the incident wave (which does not interfere) is the traveling wave part of (84). The maximum amplitude observed in the line is found where the amplitudes of the two terms in (84) add directly to give (1 + | |)V0 . The minimum amplitude is found where the standing wave achieves a null, leaving only the traveling wave amplitude of (1 − | |)V0 . The fact that the two terms in (84) combine in this way with the proper phasing is not immediately apparent, but the following arguments will show that this does occur. To obtain the minimum and maximum voltage amplitudes, we may revisit the ﬁrst part of Eq. (81): VsT (z) = V0 e− jβz + | |e j(βz+φ) (85)

First, the minimum voltage amplitude is obtained when the two terms in (85) subtract directly (having a phase difference of π). This occurs at locations z min = − 1 (φ + (2m + 1)π) 2β (m = 0, 1, 2, . . .) (86)

Note again that all positions within the slotted line occur at negative values of z. Substituting (86) into (85) leads to the minimum amplitude: VsT (z min ) = V0 (1 − | |) (87)

The same result is obtained by substituting (86) into the real voltage, (84). This produces a null in the standing wave part, and we obtain The voltage oscillates (through zero) in time, with amplitude V0 (1 − | |). The plus and minus signs in (88) apply to even and odd values of m in (86), respectively. Next, the maximum voltage amplitude is obtained when the two terms in (85) add in-phase. This will occur at locations given by z max = − 1 (φ + 2mπ) 2β (m = 0, 1, 2, . . .) (89) V(z min , t) = ±V0 (1 − | |) sin(ωt + φ/2) (88)

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On substituting (89) into (85), we obtain VsT (z max ) = V0 (1 + | |) (90)

As before, we may substitute (89) into the real instantaneous voltage (84). The effect is to produce a maximum in the standing wave part, which then adds in-phase to the running wave. The result is V(z max , t) = ±V0 (1 + | |) cos(ωt + φ/2) (91)

where the plus and minus signs apply to positive and negative values of m in (89), respectively. Again, the voltage oscillates through zero in time, with amplitude V0 (1 + | |). Note that a voltage maximum is located at the load (z = 0) if φ = 0; moreover, φ = 0 when is real and positive. This occurs for real Z L when Z L > Z 0 . Thus there is a voltage maximum at the load when the load impedance is greater than Z 0 and both impedances are real. With φ = 0, maxima also occur at z max = −mπ/β = −mλ/2. For a zero-load impedance, φ = π , and the maxima are found at z max = −π/(2β), −3π/(2β), or z max = −λ/4, −3λ/4, and so forth. The minima are separated by multiples of one half-wavelength (as are the maxima), and for a zero load impedance, the ﬁrst minimum occurs when −βz = 0, or at the load. In general, a voltage minimum is found at z = 0 whenever φ = π ; this occurs if Z L < Z 0 where Z L is real. The general results are illustrated in Figure 10.6.

V sT V0 V0

Figure 10.6 Plot of the magnitude of VsT as found from Eq. (85) as a function of position, z, in front of the load (at z = 0). The reflection coefficient phase is φ, which leads to the indicated locations of maximum and minimum voltage amplitude, as found from Eqs. (86) and (89).

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Finally, the voltage standing wave ratio is deﬁned as: s≡ VsT (z max ) 1+| | = VsT (z min ) 1−| | (92)

Since the absolute voltage amplitudes have divided out, our measured VSWR permits the immediate evaluation of | |. The phase of is then found by measuring the location of the ﬁrst maximum or minimum with respect to the load, and then using (86) or (89) as appropriate. Once is known, the load impedance can be found, assuming Z 0 is known. D10.3. What voltage standing wave ratio results when
Ans. 3

= ±1/2?

E X A M P L E 10.7

Slotted line measurements yield a VSWR of 5, a 15-cm spacing between successive voltage maxima, and the ﬁrst maximum at a distance of 7.5 cm in front of the load. Determine the load impedance, assuming a 50- impedance for the slotted line.
Solution. The 15-cm spacing between maxima is λ/2, implying a wavelength of

30 cm. Because the slotted line is air-ﬁlled, the frequency is f = c/λ = 1 GHz. The ﬁrst maximum at 7.5 cm is thus at a distance of λ/4 from the load, which means that a voltage minimum occurs at the load. Thus will be real and negative. We use (92) to write 5−1 2 s−1 = = | |= s+1 5+1 3 So Z L − Z0 2 =− = 3 Z L + Z0 which we solve for Z L to obtain 1 50 = 10 Z L = Z0 = 5 5

10.11 TRANSMISSION LINES OF FINITE LENGTH
A new type of problem emerges when considering the propagation of sinusoidal voltages on ﬁnite-length lines that have loads that are not impedance matched. In such cases, numerous reﬂections occur at the load and at the generator, setting up a multiwave bidirectional voltage distribution in the line. As always, the objective is to determine the net power transferred to the load in steady state, but we must now include the effect of the numerous forward- and backward-reﬂected waves.

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g

g

in

in

L

in

in

z = −l

z=0

Figure 10.7 Finite-length transmission line configuration and its equivalent circuit.

Figure 10.7 shows the basic problem. The line, assumed to be lossless, has characteristic impedance Z 0 and is of length l. The sinusoidal voltage source at frequency ω provides phasor voltage Vs . Associated with the souce is a complex internal impedance, Z g , as shown. The load impedance, Z L , is also assumed to be complex and is located at z = 0. The line thus exists along the negative z axis. The easiest method of approaching the problem is not to attempt to analyze every reﬂection individually, but rather to recognize that in steady state, there will exist one net forward wave and one net backward wave, representing the superposition of all waves that are incident on the load and all waves that are reﬂected from it. We may thus write the total voltage in the line as VsT (z) = V0+ e− jβz + V0− e jβz (93)

and are complex amplitudes, composed respectively of the sum of in which all individual forward and backward wave amplitudes and phases. In a similar way, we may write the total current in the line:
+ − IsT (z) = I0 e− jβz + I0 e jβz

V0+

V0−

(94)

We now deﬁne the wave impedance, Z w (z), as the ratio of the total phasor voltage to the total phasor current. Using (93) and (94), this becomes: Z w (z) ≡ V + e− jβz + V0− e jβz VsT (z) = 0+ − jβz − IsT (z) I0 e + I0 e jβz (95)

We next use the relations V0− = simpliﬁes to

+ − V0+ , I0 = V0+ /Z 0 , and I0 = −V0− /Z 0 . Eq. (95)

Z w (z) = Z 0

Now, using the Euler identity, (32), and substituting Eq. (96) becomes Z w (z) = Z 0

e− jβz + e jβz e− jβz − e jβz

(96) = (Z L − Z 0 )/(Z L + Z 0 ), (97)

The wave impedance at the line input is now found by evaluating (97) at z = −l, obtaining

Z L cos(βz) − j Z 0 sin(βz) Z 0 cos(βz) − j Z L sin(βz)

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Z in = Z 0

Z L cos(βl) + j Z 0 sin(βl) Z 0 cos(βl) + j Z L sin(βl)

(98)

This is the quantity that we need in order to create the equivalent circuit in Figure 10.7. One special case is that in which the line length is a half-wavelength, or an integer multiple thereof. In that case, βl = 2π mλ = mπ λ 2 (m = 0, 1, 2, . . .)

Using this result in (98), we ﬁnd Z in (l = mλ/2) = Z L (99)

For a half-wave line, the equivalent circuit can be constructed simply by removing the line completely and placing the load impedance at the input. This simpliﬁcation works, of course, provided the line length is indeed an integer multiple of a halfwavelength. Once the frequency begins to vary, the condition is no longer satisﬁed, and (98) must be used in its general form to ﬁnd Z in . Another important special case is that in which the line length is an odd multiple of a quarter wavelength: βl = λ π 2π (2m + 1) = (2m + 1) λ 4 2 (m = 0, 1, 2, . . .)

Using this result in (98) leads to Z in (l = λ/4) =
2 Z0 ZL

(100)

An immediate application of (100) is to the problem of joining two lines having different characteristic impedances. Suppose the impedances are (from left to right) Z 01 and Z 03 . At the joint, we may insert an additional line whose characteristic impedance is Z 02 and whose length is λ/4. We thus have a sequence of joined lines whose impedances progress as Z 01 , Z 02 , and Z 03 , in that order. A voltage wave is now incident from line 1 onto the joint between Z 01 and Z 02 . Now the effective load at the far end of line 2 is Z 03 . The input impedance to line 2 at any frequency is now Z in = Z 02 Z 03 cos β2l + j Z 02 sin β2l Z 02 cos β2l + j Z 03 sin β2l
2 Z 02 Z 03

(101)

Then, since the length of line 2 is λ/4, Z in (line 2) = (102)

Reﬂections at the Z 01 –Z 02 interface will not occur if Z in = Z 01 . Therefore, we can match the junction (allowing complete transmission through the three-line sequence)

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if Z 02 is chosen so that Z 02 = Z 01 Z 03 (103)

This technique is called quarter-wave matching and again is limited to the frequency . (or narrow band of frequencies) such that l = (2m + 1)λ/4. We will encounter more examples of these techniques when we explore electromagnetic wave reﬂection in Chapter 12. Meanwhile, further examples that involve the use of the input impedance and the VSWR are presented in Section 10.12.

10.12 SOME TRANSMISSION LINE EXAMPLES
In this section, we apply many of the results that we obtained in the previous sections to several typical transmission line problems. We simplify our work by restricting our attention to the lossless line. Let us begin by assuming a two-wire 300 line (Z 0 = 300 ), such as the lead-in wire from the antenna to a television or FM receiver. The circuit is shown in Figure 10.8. The line is 2 m long, and the values of L and C are such that the velocity on the line is 2.5 × 108 m/s. We will terminate the line with a receiver having an input resistance of 300 and represent the antenna by its Thevenin equivalent Z = 300 in series with Vs = 60 V at 100 MHz. This antenna voltage is larger by a factor of about 105 than it would be in a practical case, but it also provides simpler values to work with; in order to think practical thoughts, divide currents or voltages by 105 , divide powers by 1010 , and leave impedances alone. Because the load impedance is equal to the characteristic impedance, the line is matched; the reﬂection coefﬁcient is zero, and the standing wave ratio is unity. For the given velocity and frequency, the wavelength on the line is v/ f = 2.5 m, and the phase constant is 2π/λ = 0.8π rad/m; the attenuation constant is zero. The electrical length of the line is βl = (0.8π )2, or 1.6π rad. This length may also be expressed as 288◦ , or 0.8 wavelength. The input impedance offered to the voltage source is 300 , and since the internal impedance of the source is 300 , the voltage at the input to the line is half of 60 V, or 30 V. The source is matched to the line and delivers the maximum available power

Figure 10.8 A transmission line that is matched at both ends produces no reflections and thus delivers maximum power to the load.

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to the line. Because there is no reﬂection and no attenuation, the voltage at the load is 30 V, but it is delayed in phase by 1.6π rad. Thus, Vin = 30 cos(2π 108 t) V whereas VL = 30 cos(2π 108 t − 1.6π ) V The input current is Iin = while the load current is I L = 0.1 cos(2π 108 t − 1.6π ) A The average power delivered to the input of the line by the source must all be delivered to the load by the line, 1 × 30 × 0.1 = 1.5 W 2 Now let us connect a second receiver, also having an input resistance of 300 across the line in parallel with the ﬁrst receiver. The load impedance is now 150 the reﬂection coefﬁcient is 1 150 − 300 =− = 150 + 300 3 and the standing wave ratio on the line is Pin = PL = s= The input impedance is no longer 300 Z in = Z 0 1− 1+
1 3 1 3

Vin = 0.1 cos(2π 108 t) A 300

, ,

=2

, but is now

= 510 −23.8◦ = 466 − j206

Z L cos βl + j Z 0 sin βl 150 cos 288◦ + j300 sin 288◦ = 300 Z 0 cos βl + j Z L sin βl 300 cos 288◦ + j150 sin 288◦

and the power supplied to the line by the source is Pin =

which is a capacitive impedance. Physically, this means that this length of line stores more energy in its electric ﬁeld than in its magnetic ﬁeld. The input current phasor is thus 60 = 0.0756 15.0◦ A Is,in = 300 + 466 − j206

1 × (0.0756)2 × 466 = 1.333 W 2 Since there are no losses in the line, 1.333 W must also be delivered to the load. Note that this is less than the 1.50 W which we were able to deliver to a matched load; moreover, this power must divide equally between two receivers, and thus each

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receiver now receives only 0.667 W. Because the input impedance of each receiver is 300 , the voltage across the receiver is easily found as 0.667 = 1 |Vs,L |2 2 300 |Vs,L | = 20 V

in comparison with the 30 V obtained across the single load. Before we leave this example, let us ask ourselves several questions about the voltages on the transmission line. Where is the voltage a maximum and a minimum, and what are these values? Does the phase of the load voltage still differ from the input voltage by 288◦ ? Presumably, if we can answer these questions for the voltage, we could do the same for the current. Equation (89) serves to locate the voltage maxima at 1 z max = − (φ + 2mπ ) (m = 0, 1, 2, . . .) 2β where = | |e jφ . Thus, with β = 0.8π and φ = π , we ﬁnd while the minima are λ/4 distant from the maxima; z min = 0 and z max = −0.625 and −1.875 m

and we ﬁnd that the load voltage (at z = 0) is a voltage minimum. This, of course, veriﬁes the general conclusion we reached earlier: a voltage minimum occurs at the load if Z L < Z 0 , and a voltage maximum occurs if Z L > Z 0 , where both impedances are pure resistances. The minimum voltage on the line is thus the load voltage, 20 V; the maximum voltage must be 40 V, since the standing wave ratio is 2. The voltage at the input end of the line is Vs,in = Is,in Z in = (0.0756 15.0◦ )(510 −23.8◦ ) = 38.5 −8.8◦ The input voltage is almost as large as the maximum voltage anywhere on the line because the line is about three-quarters of a wavelength long, a length which would place the voltage maximum at the input when Z L < Z 0 . Finally, it is of interest to determine the load voltage in magnitude and phase. We begin with the total voltage in the line, using (93). VsT = e− jβz + e jβz V0+ (104)

−1.25 m

We may use this expression to determine the voltage at any point on the line in terms of the voltage at any other point. Because we know the voltage at the input to the line, we let z = −l, and solve for V0+ , e jβl Vs,in = e jβl + e− jβl V0+ (105)

V0+ =

Vs,in 38.5 −8.8◦ = j1.6π 1 − j1.6π = 30.0 72.0◦ V + e− jβl e − 3e

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We may now let z = 0 in (104) to ﬁnd the load voltage,

Vs,L = (1 + )V0+ = 20 72◦ = 20 −288◦

The amplitude agrees with our previous value. The presence of the reﬂected wave causes Vs,in and Vs,L to differ in phase by about −279◦ instead of −288◦ .
E X A M P L E 10.8

In order to provide a slightly more complicated example, let us now place a purely capacitive impedance of − j300 in parallel with the two 300 receivers. We are to ﬁnd the input impedance and the power delivered to each receiver.
Solution. The load impedance is now 150

ZL =

1 + 0.447 = 2.62 1 − 0.447 Thus, the VSWR is higher and the mismatch is therefore worse. Let us next calculate the input impedance. The electrical length of the line is still 288◦ , so that s= Z in = 300 (120 − j60) cos 288◦ + j300 sin 288◦ = 755 − j138.5 300 cos 288◦ + j(120 − j60) sin 288◦

We ﬁrst calculate the reﬂection coefﬁcient and the VSWR: −180 − j60 120 − j60 − 300 = = 0.447 −153.4◦ = 120 − j60 + 300 420 − j60

− j300 150(− j300) = = 120 − j60 150 − j300 1 − j2

in parallel with − j300 , or

This leads to a source current of VT h 60 = = 0.0564 7.47◦ A Is,in = Z T h + Z in 300 + 755 − j138.5

Therefore, the average power delivered to the input of the line is Pin = 1 (0.0564)2 (755) = 1.200 W. Since the line is lossless, it follows that PL = 1.200 W, 2 and each receiver gets only 0.6 W.

E X A M P L E 10.9

As a ﬁnal example, let us terminate our line with a purely capacitive impedance, Z L = − j300 . We seek the reﬂection coefﬁcient, the VSWR, and the power delivered to the load.
Solution. Obviously, we cannot deliver any average power to the load since it is a

pure reactance. As a consequence, the reﬂection coefﬁcient is = − j300 − 300 = − j1 = 1 −90◦ − j300 + 300

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and the reﬂected wave is equal in amplitude to the incident wave. Hence, it should not surprise us to see that the VSWR is 1 + | − j1| s= =∞ 1 − | − j1| and the input impedance is a pure reactance, − j300 cos 288◦ + j300 sin 288◦ Z in = 300 = j589 300 cos 288◦ + j(− j300) sin 288◦ Thus, no average power can be delivered to the input impedance by the source, and therefore no average power can be delivered to the load. Although we could continue to ﬁnd numerous other facts and ﬁgures for these examples, much of the work may be done more easily for problems of this type by using graphical techniques. We encounter these in Section 10.13. D10.4. A 50 W lossless line has a length of 0.4λ. The operating frequency is 300 MHz. A load Z L = 40 + j30 is connected at z = 0, and the Theveninequivalent source at z = −l is 12 0◦ V in series with Z T h = 50 + j0 . Find: (a) ; (b) s; (c) Z in .
Ans. 0.333 90◦ ; 2.00; 25.5 + j5.90

D10.5. For the transmission line of Problem D10.4, also ﬁnd: (a) the phasor voltage at z = −l; (b) the phasor voltage at z = 0; (c) the average power delivered to Z L .
Ans. 4.14 8.58◦ V; 6.32 −125.6◦ V; 0.320 W

10.13 GRAPHICAL METHODS: THE SMITH CHART
Transmission line problems often involve manipulations with complex numbers, making the time and effort required for a solution several times greater than are needed for a similar sequence of operations on real numbers. One means of reducing the labor without seriously affecting the accuracy is by using transmission-line charts. Probably the most widely used one is the Smith chart.3 Basically, this diagram shows curves of constant resistance and constant reactance; these may represent either an input impedance or a load impedance. The latter, of course, is the input impedance of a zero-length line. An indication of location along the line is also provided, usually in terms of the fraction of a wavelength from a voltage maximum or minimum. Although they are not speciﬁcally shown on the chart, the standing-wave ratio and the magnitude and angle of the reﬂection coefﬁcient are very

3

P. H. Smith, “Transmission Line Calculator,” Electronics, vol. 12, pp. 29–31, January 1939.

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Figure 10.9 The polar coordinates of the Smith chart are the magnitude and phase angle of the reflection coefficient; the rectangular coordinates are the real and imaginary parts of the reflection coefficient. The entire chart lies within the circle | | = 1.

quickly determined. As a matter of fact, the diagram is constructed within a circle of unit radius, using polar coordinates, with radius variable | | and counterclockwise angle variable φ, where = | |e jφ . Figure 10.9 shows this circle. Since | | < 1, all our information must lie on or within the unit circle. Peculiarly enough, the reﬂection coefﬁcient itself will not be plotted on the ﬁnal chart, for these additional contours would make the chart very difﬁcult to read. The basic relationship upon which the chart is constructed is = Z L − Z0 Z L + Z0 (106)

The impedances that we plot on the chart will be normalized with respect to the characteristic impedance. Let us identify the normalized load impedance as z L , zL = r + j x = and thus = or zL = 1+ 1− (107) zL − 1 zL + 1 ZL RL + j X L = Z0 Z0

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ENGINEERING ELECTROMAGNETICS

In polar form, we have used | | and φ as the magnitude and angle of . With i as the real and imaginary parts of , we write = Thus r + jx = r r

and

+j

i

(108) i i

1+ r + j 1− r − j The real and imaginary parts of this equation are r= 1− (1 − − 2+ r) 2 (1 − r i 2 ) 2 r 2 i 2 i

(109)

(110)

x=

+

2 i

(111)

After several lines of elementary algebra, we may write (110) and (111) in forms which readily display the nature of the curves on r , i axes, r −

1 2 1 2 = (113) x x The ﬁrst equation describes a family of circles, where each circle is associated with a speciﬁc value of resistance r . For example, if r = 0, the radius of this zeroresistance circle is seen to be unity, and it is centered at the origin ( r = 0, i = 0). This checks, for a pure reactance termination leads to a reﬂection coefﬁcient of unity magnitude. On the other hand, if r = ∞, then z L = ∞ and we have = 1 + j0. The circle described by (112) is centered at r = 1, i = 0 and has zero radius. It is therefore the point = 1 + j0, as we decided it should be. As another example, the circle for r = 1 is centered at r = 0.5, i = 0 and has a radius of 0.5. This circle is shown in Figure 10.10, along with circles for r = 0.5 and r = 2. All circles are centered on the r axis and pass through the point = 1 + j0. Equation (113) also represents a family of circles, but each of these circles is deﬁned by a particular value of x, rather than r . If x = ∞, then z L = ∞, and = 1 + j0 again. The circle described by (113) is centered at = 1 + j0 and has zero radius; it is therefore the point = 1 + j0. If x = +1, then the circle is centered at = 1 + j1 and has unit radius. Only one-quarter of this circle lies within the boundary curve | | = 1, as shown in Figure 10.11. A similar quarter-circle appears below the r axis for x = −1. The portions of other circles for x = 0.5, −0.5, 2, and −2 are also shown. The “circle” representing x = 0 is the r axis; this is also labeled in Figure 10.11. The two families of circles both appear on the Smith chart, as shown in Figure 10.12. It is now evident that if we are given Z L , we may divide by Z 0 to ( r r 1+r

2

+ i 2 i

=

1 1+r

2

(112)

− 1)2 +

−

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Figure 10.10 Constant-r circles are shown on the r , i plane. The radius of any circle is 1/(1 + r ).

obtain z L , locate the appropriate r and x circles (interpolating as necessary), and determine by the intersection of the two circles. Because the chart does not have concentric circles showing the values of | |, it is necessary to measure the radial distance from the origin to the intersection with dividers or a compass and use an auxiliary scale to ﬁnd | |. The graduated line segment below the chart in Figure 10.12 serves this purpose. The angle of is φ, and it is the counterclockwise angle from the r axis. Again, radial lines showing the angle would clutter up the chart

Figure 10.11 The portions of the circles of constant x lying within | | = 1 are shown on the r , i axes. The radius of a given circle is 1/|x|.

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ENGINEERING ELECTROMAGNETICS

Figure 10.12 The Smith chart contains the constant-r circles and constant-x circles, an auxiliary radial scale to determine | |, and an angular scale on the circumference for measuring φ.

badly, so the angle is indicated on the circumference of the circle. A straight line from the origin through the intersection may be extended to the perimeter of the chart. As an example, if Z L = 25 + j50 on a 50 line, z L = 0.5 + j1, and point A on Figure 10.12 shows the intersection of the r = 0.5 and x = 1 circles. The reﬂection coefﬁcient is approximately 0.62 at an angle φ of 83◦ . The Smith chart is completed by adding a second scale on the circumference by which distance along the line may be computed. This scale is in wavelength units, but the values placed on it are not obvious. To obtain them, we ﬁrst divide the voltage at any point along the line, Vs = V0+ (e− jβz + e jβz ) by the current V0+ − jβz (e − e jβz ) Z0 obtaining the normalized input impedance Is = z in = Vs e− jβz + e jβz = − jβz Z 0 Is e − e jβz

Replacing z with −l and dividing numerator and denominator by e jβl , we have the general equation relating normalized input impedance, reﬂection coefﬁcient, and

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line length, z in = 1 + e− j2βl 1 + | |e j(φ−2βl) = − j2βl 1− e 1 − | |e j(φ−2βl) (114)

Note that when l = 0, we are located at the load, and z in = (1 + )/(l − ) = z L , as shown by (107). Equation (114) shows that the input impedance at any point z = −l can be obtained by replacing , the reﬂection coefﬁcient of the load, by e− j2βl . That is, we decrease the angle of by 2βl radians as we move from the load to the line input. Only the angle of is changed; the magnitude remains constant. Thus, as we proceed from the load z L to the input impedance z in , we move toward the generator a distance l on the transmission line, but we move through a clockwise angle of 2βl on the Smith chart. Since the magnitude of stays constant, the movement toward the source is made along a constant-radius circle. One lap around the chart is accomplished whenever βl changes by π rad, or when l changes by onehalf wavelength. This agrees with our earlier discovery that the input impedance of a half-wavelength lossless line is equal to the load impedance. The Smith chart is thus completed by the addition of a scale showing a change of 0.5λ for one circumnavigation of the unit circle. For convenience, two scales are usually given, one showing an increase in distance for clockwise movement and the other an increase for counterclockwise travel. These two scales are shown in Figure 10.13. Note that the one marked “wavelengths toward generator” (wtg) shows increasing values of l/λ for clockwise travel, as described previously. The zero point of the wtg scale is rather arbitrarily located to the left. This corresponds to input impedances having phase angles of 0◦ and R L < Z 0 . We have also seen that voltage minima are always located here.
E X A M P L E 10.10

The use of the transmission line chart is best shown by example. Let us again consider a load impedance, Z L = 25 + j50 , terminating a 50- line. The line length is 60 cm and the operating frequency is such that the wavelength on the line is 2 m. We desire the input impedance. read = 0.62 82◦ . By drawing a straight line from the origin through A to the circumference, we note a reading of 0.135 on the wtg scale. We have l/λ = 0.6/2 = 0.3, and it is, therefore, 0.3λ from the load to the input. We therefore ﬁnd z in on the | | = 0.62 circle opposite a wtg reading of 0.135 + 0.300 = 0.435. This construction is shown in Figure 10.14, and the point locating the input impedance is marked B. The normalized input impedance is read as 0.28 − j0.40, and thus Z in = 14 − j20. A more accurate analytical calculation gives Z in = 13.7 − j20.2. Information concerning the location of the voltage maxima and minima is also readily obtained on the Smith chart. We already know that a maximum or minimum
Solution. We have z L = 0.5 + j1, which is marked as A on Figure 10.14, and we

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Figure 10.13 A photographic reduction of one version of a useful Smith chart (courtesy of the Emeloid Company, Hillside, NJ ). For accurate work, larger charts are available wherever fine technical books are sold.

must occur at the load when Z L is a pure resistance; if R L > Z 0 there is a maximum at the load, and if R L < Z 0 there is a minimum. We may extend this result now by noting that we could cut off the load end of a transmission line at a point where the input impedance is a pure resistance and replace that section with a resistance Rin ; there would be no changes on the generator portion of the line. It follows, then, that the location of voltage maxima and minima must be at those points where Z in is a pure resistance. Purely resistive input impedances must occur on the x = 0 line (the r axis) of the Smith chart. Voltage maxima or current minima occur when r > 1, or at wtg = 0.25, and voltage minima or current maxima occur when r < 1,

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Figure 10.14 Normalized input impedance produced by a normalized load impedance z L = 0.5 + j 1 on a line 0.3λ long is zin = 0.28 − j 0.40.

or at wtg = 0. In Example 10.10, then, the maximum at wtg = 0.250 must occur 0.250 − 0.135 = 0.115 wavelengths toward the generator from the load. This is a distance of 0.115 × 200, or 23 cm from the load. We should also note that because the standing wave ratio produced by a resistive load R L is either R L /R0 or R0 /R L , whichever is greater than unity, the value of s may be read directly as the value of r at the intersection of the | | circle and the r axis, r > 1. In our example, this intersection is marked point C, and r = 4.2; thus, s = 4.2. Transmission line charts may also be used for normalized admittances, although there are several slight differences in such use. We let y L = Y L /Y0 = g + jb and use the r circles as g circles and the x circles as b circles. The two differences are, ﬁrst, the line segment where g > 1 and b = 0 corresponds to a voltage minimum; and second, 180◦ must be added to the angle of as read from the perimeter of the chart. We shall use the Smith chart in this way in Section 10.14. Special charts are also available for non-normalized lines, particularly 50 charts and 20 mS charts. D10.6. A load Z L = 80 − j100 is located at z = 0 on a lossless 50- line. The operating frequency is 200 MHz and the wavelength on the line is 2 m. (a) If the line is 0.8 m in length, use the Smith chart to ﬁnd the input impedance. (b) What is s? (c) What is the distance from the load to the nearest voltage maximum? (d) What is the distance from the input to the nearest point at which the remainder of the line could be replaced by a pure resistance?
Ans. 79 + j99 : 4.50; 0.0397 m; 0.760 m

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Figure 10.15 A sketch of a coaxial slotted line. The distance scale is on the slotted line. With the load in place, s = 2.5, and the minimum occurs at a scale reading of 47 cm. For a short circuit, the minimum is located at a scale reading of 26 cm. The wavelength is 75 cm.

We next consider two examples of practical transmission line problems. The ﬁrst is the determination of load impedance from experimental data, and the second is the design of a single-stub matching network. Let us assume that we have made experimental measurements on a 50 slotted line that show there is a voltage standing wave ratio of 2.5. This has been determined by moving a sliding carriage back and forth along the line to determine maximum and minimum voltage readings. A scale provided on the track along which the carriage moves indicates that a minimum occurs at a scale reading of 47.0 cm, as shown in Figure 10.15. The zero point of the scale is arbitrary and does not correspond to the location of the load. The location of the minimum is usually speciﬁed instead of the maximum because it can be determined more accurately than that of the maximum; think of the sharper minima on a rectiﬁed sine wave. The frequency of operation is 400 MHz, so the wavelength is 75 cm. In order to pinpoint the location of the load, we remove it and replace it with a short circuit; the position of the minimum is then determined as 26.0 cm. We know that the short circuit must be located an integral number of halfwavelengths from the minimum; let us arbitrarily locate it one half-wavelength away at 26.0 − 37.5 = −11.5 cm on the scale. Since the short circuit has replaced the load, the load is also located at −11.5 cm. Our data thus show that the minimum is 47.0 − (−11.5) = 58.5 cm from the load, or subtracting one-half wavelength, a minimum is 21.0 cm from the load. The voltage maximum is thus 21.0 − (37.5/2) = 2.25 cm from the load, or 2.25/75 = 0.030 wavelength from the load. With this information, we can now turn to the Smith chart. At a voltage maximum, the input impedance is a pure resistance equal to s R0 ; on a normalized basis, z in = 2.5.

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Figure 10.16 If zin = 2.5 + j 0 on a line 0.3 wavelengths long, then z L = 2.1 + j 0.8.

We therefore enter the chart at z in = 2.5 and read 0.250 on the wtg scale. Subtracting 0.030 wavelength to reach the load, we ﬁnd that the intersection of the s = 2.5 (or | | = 0.429) circle and the radial line to 0.220 wavelength is at z L = 2.1 + j0.8. The construction is sketched on the Smith chart of Figure 10.16. Thus Z L = 105 + j40 , a value that assumes its location at a scale reading of −11.5 cm, or an integral number of half-wavelengths from that position. Of course, we may select the “location” of our load at will by placing the short circuit at the point that we wish to consider the load location. Since load locations are not well deﬁned, it is important to specify the point (or plane) at which the load impedance is determined. As a ﬁnal example, let us try to match this load to the 50 line by placing a short-circuited stub of length d1 a distance d from the load (see Figure 10.17). The stub line has the same characteristic impedance as the main line. The lengths d and d1 are to be determined.

Figure 10.17 A short-circuited stub of length d1 , located at a distance d from a load Z L , is used to provide a matched load to the left of the stub.

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The input impedance to the stub is a pure reactance; when combined in parallel with the input impedance of the length d containing the load, the resultant input impedance must be 1 + j0. Because it is much easier to combine admittances in parallel than impedances, let us rephrase our goal in admittance language: the input admittance of the length d containing the load must be 1 + jbin for the addition of the input admittance of the stub jbstub to produce a total admittance of 1 + j0. Hence the stub admittance is − jbin . We will therefore use the Smith chart as an admittance chart instead of an impedance chart. The impedance of the load is 2.1 + j0.8, and its location is at −11.5 cm. The admittance of the load is therefore 1/(2.1 + j0.8), and this value may be determined by adding one-quarter wavelength on the Smith chart, as Z in for a quarter-wavelength 2 line is R0 /Z L , or z in = 1/z L , or yin = z L . Entering the chart (Figure 10.18) at z L = 2.1 + j0.8, we read 0.220 on the wtg scale; we add (or subtract) 0.250 and ﬁnd the admittance 0.41 − j0.16 corresponding to this impedance. This point is still located on the s = 2.5 circle. Now, at what point or points on this circle is the real part of the admittance equal to unity? There are two answers, 1 + j0.95 at wtg = 0.16, and 1 − j0.95 at wtg = 0.34, as shown in Figure 10.18. We select the former value since this leads to the shorter stub. Hence ystub = − j0.95, and the stub location corresponds to wtg = 0.16. Because the load admittance was found at wtg = 0.470, then we must move (0.5 − 0.47) + 0.16 = 0.19 wavelength to get to the stub location. Finally, we may use the chart to determine the necessary length of the shortcircuited stub. The input conductance is zero for any length of short-circuited stub, so we are restricted to the perimeter of the chart. At the short circuit, y = ∞ and wtg = 0.250. We ﬁnd that bin = −0.95 is achieved at wtg = 0.379, as shown in Figure 10.18. The stub is therefore 0.379 − 0.250 = 0.129 wavelength, or 9.67 cm long.

Figure 10.18 A normalized load, z L = 2.1 + j 0.8, is matched by placing a 0.129-wavelength short-circuited stub 0.19 wavelengths from the load.

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D10.7. Standing wave measurements on a lossless 75- line show maxima of 18 V and minima of 5 V. One minimum is located at a scale reading of 30 cm. With the load replaced by a short circuit, two adjacent minima are found at scale readings of 17 and 37 cm. Find: (a) s; (b) λ; (c) f ; (d) L ; (e) Z L .
Ans. 3.60; 0.400 m; 750 MHz; 0.704 −33.0; 77.9 + j104.7

D10.8. A normalized load, z L = 2− j1, is located at z = 0 on a lossless 50line. Let the wavelength be 100 cm. (a) A short-circuited stub is to be located at z = −d. What is the shortest suitable value for d? (b) What is the shortest possible length of the stub? Find s: (c) on the main line for z < −d; (d) on the main line for −d < z < 0; (e) on the stub.
Ans. 12.5 cm; 12.5 cm; 1.00; 2.62; ∞

10.14 TRANSIENT ANALYSIS
Throughout most of this chapter, we have considered the operation of transmission lines under steady-state conditions, in which voltage and current were sinusoidal and at a single frequency. In this section we move away from the simple time-harmonic case and consider transmission line responses to voltage step functions and pulses, grouped under the general heading of transients. These situations were brieﬂy considered in Section 10.2 with regard to switched voltages and currents. Line operation in transient mode is important to study because it allows us to understand how lines can be used to store and release energy (in pulse-forming applications, for example). Pulse propagation is important in general since digital signals, composed of sequences of pulses, are widely used. We will conﬁne our discussion to the propagation of transients in lines that are lossless and have no dispersion, so that the basic behavior and analysis methods may be learned. We must remember, however, that transient signals are necessarily composed of numerous frequencies, as Fourier analysis will show. Consequently, the question of dispersion in the line arises, since, as we have found, line propagation constants and reﬂection coefﬁcients at complex loads will be frequency-dependent. So, in general, pulses are likely to broaden with propagation distance, and pulse shapes may change when reﬂecting from a complex load. These issues will not be considered in detail here, but they are readily addressed when the precise frequency dependences of β and are known. In particular, β(ω) can be found by evaluating the imaginary part of γ , as given in Eq. (41), which would in general include the frequency dependences of R, C, G, and L arising from various mechanisms. For example, the skin effect (which affects both the conductor resistance and the internal inductance) will result in frequency-dependent R and L. Once β(ω) is known, pulse broadening can be evaluated using the methods to be presented in Chapter 12. We begin our basic discussion of transients by considering a lossless transmission line of length l terminated by a matched load, R L = Z 0 , as shown in Figure 10.19a.

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Figure 10.19 (a) Closing the switch at time t = 0 initiates voltage and current waves V + and I + . The leading edge of both waves is indicated by the dashed line, which propagates in the lossless line toward the load at velocity ν. In this case, V + = V0 ; the line voltage is V + everywhere to the left of the leading edge, where current is I + = V + /Z 0 . To the right of the leading edge, voltage and current are both zero. Clockwise current, indicated here, is treated as positive and will occur when V + is positive. (b) Voltage across the load resistor as a function of time, showing the one-way transit time delay, l /ν.

At the front end of the line is a battery of voltage V0 , which is connected to the line by closing a switch. At time t = 0, the switch is closed, and the line voltage at z = 0 becomes equal to the battery voltage. This voltage, however, does not appear across the load until adequate time has elapsed for the propagation delay. Speciﬁcally, at t = 0, a voltage wave is initiated in the line at the battery end, which then propagates toward the load. The leading edge of the wave, labeled V + in Figure 10.19, is of value V + = V0 . It can be thought of as a propagating step function, because at all points to the left of V + , the line voltage is V0 ; at all points to the right (not yet reached by the leading edge), the line voltage is zero. The wave propagates at velocity ν, which in general is the group velocity in the line.4 The wave reaches the load at time t = l/ν
Because we have a step function (composed of many frequencies) as opposed to a sinusoid at a single frequency, the wave will propagate at the group velocity. In a lossless line with no dispersion as √ considered in this section, β = ω LC, where L and C are constant with frequency. In this case, we √ would ﬁnd that the group and phase velocities are equal; that is, dω/dβ = ω/β = ν = 1/ LC. We will thus write the velocity as ν, knowing it to be both νp and νg .
4

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and then does not reﬂect, as the load is matched. The transient phase is thus over, and the load voltage is equal to the battery voltage. A plot of load voltage as a function of time is shown in Figure 10.19b, indicating the propagation delay of t = l/ν. Associated with the voltage wave V + is a current wave whose leading edge is of value I + . This wave is a propagating step function as well, whose value at all points to the left of V + is I + = V + /Z 0 ; at all points to the right, current is zero. A plot of current through the load as a function of time will thus be identical in form to the voltage plot of Figure 10.19b, except that the load current at t = l/ν will be I L = V + /Z 0 = V0 /R L . We next consider a more general case, in which the load of Figure 10.19a is again a resistor but is not matched to the line (R L = Z 0 ). Reﬂections will thus occur at the load, complicating the problem. At t = 0, the switch is closed as before and a voltage wave, V1+ = V0 , propagates to the right. Upon reaching the load, however, the wave will now reﬂect, producing a back-propagating wave, V1− . The relation between V1− and V1+ is through the reﬂection coefﬁcient at the load: V1− = V1+
L

=

RL − Z 0 RL + Z 0

(115)

As V1− propagates back toward the battery, it leaves behind its leading edge a total voltage of V1+ + V1− . Voltage V1+ exists everywhere ahead of the V1− wave until it reaches the battery, whereupon the entire line now is charged to voltage V1+ + V1− . At the battery, the V1− wave reﬂects to produce a new forward wave, V2+ . The ratio of V2+ and V1− is found through the reﬂection coefﬁcient at the battery: V2+ = V1− g =

Zg − Z0 0 − Z0 = = −1 Zg + Z0 0 + Z0

(116)

where the impedance at the generator end, Z g , is that of the battery, or zero. V2+ (equal to −V1− ) now propagates to the load, where it reﬂects to produce backward wave V2− = L V2+ . This wave then returns to the battery, where it reﬂects with g = −1, and the process repeats. Note that with each round trip the wave voltage is reduced in magnitude because | L | < 1. Because of this the propagating wave voltages will eventually approach zero, and steady state is reached. The voltage across the load resistor can be found at any given time by summing the voltage waves that have reached the load and have reﬂected from it up to that time. After many round trips, the load voltage will be, in general, VL = V1+ + V1− + V2+ + V2− + V3+ + V3− + · · · = V1+ 1 +
L

+

g

L

+

g

2 L

+

2 g

2 L

+

2 g

3 L

+ ···

With a simple factoring operation, the preceding equation becomes VL = V1+ (1 +
L)

1+

g

L

+

2 g

2 L

+ ···

(117)

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Figure 10.20 With series resistance at the battery location, voltage + division occurs when the switch is closed, such that V0 = Vr g + V1 . + − Shown is the first reflected wave, which leaves voltage V1 + V1 behind its − − leading edge. Associated with the wave is current I 1 , which is −V1 /Z 0 . − Counterclockwise current is treated as negative and will occur when V1 is positive.

Allowing time to approach inﬁnity, the second term in parentheses in (117) becomes the power series expansion for the expression 1/(1 − g L ). Thus, in steady state we obtain 1+ L VL = V1+ (118) 1− g L In our present example, V1+ = V0 and g = −1. Substituting these into (118), we ﬁnd the expected result in steady state: VL = V0 . A more general situation would involve a nonzero impedance at the battery location, as shown in Figure 10.20. In this case, a resistor of value Rg is positioned in series with the battery. When the switch is closed, the battery voltage appears across the series combination of Rg and the line characteristic impedance, Z 0 . The value of the initial voltage wave, V1+ , is thus found through simple voltage division, or V1+ = V0 Z 0 Rg + Z 0 (119)

With this initial value, the sequence of reﬂections and the development of the voltage across the load occurs in the same manner as determined by (117), with the steadystate value determined by (118). The value of the reﬂection coefﬁcient at the generator end, determined by (116), is g = (Rg − Z 0 )/(Rg + Z 0 ). A useful way of keeping track of the voltage at any point in the line is through a voltage reflectio diagram. Such a diagram for the line of Figure 10.20 is shown in Figure 10.21a. It is a two-dimensional plot in which position on the line, z, is shown on the horizontal axis. Time is plotted on the vertical axis and is conveniently expressed as it relates to position and velocity through t = z/ν. A vertical line, located at z = l, is drawn, which, together with the ordinate, deﬁnes the z axis boundaries of the transmission line. With the switch located at the battery position, the initial voltage wave, V1+ , starts at the origin, or lower-left corner of the diagram (z = t = 0). The location of the leading edge of V1+ as a function of time is shown as the diagonal line

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(a)

(b)

Figure 10.21 (a) Voltage reflection diagram for the line of Figure 10.20. A reference line, drawn at z = 3l /4, is used to evaluate the voltage at that position as a function of time. (b) The line voltage at z = 3l /4 as determined from the reflection diagram of (a). Note that the voltage approaches the expected V0 RL /(Rg + RL ) as time approaches infinity.

that joins the origin to the point along the right-hand vertical line that corresponds to time t = l/ν (the one-way transit time). From there (the load location), the position of the leading edge of the reﬂected wave, V1− , is shown as a “reﬂected” line that joins the t = l/ν point on the right boundary to the t = 2l/ν point on the ordinate. From there (at the battery location), the wave reﬂects again, forming V2+ , shown as a line parallel to that for V1+ . Subsequent reﬂected waves are shown, and their values are labeled.

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The voltage as a function of time at a given position in the line can now be determined by adding the voltages in the waves as they intersect a vertical line drawn at the desired location. This addition is performed starting at the bottom of the diagram (t = 0) and progressing upward (in time). Whenever a voltage wave crosses the vertical line, its value is added to the total at that time. For example, the voltage at a location three-fourths the distance from the battery to the load is plotted in Figure 10.21b. To obtain this plot, the line z = (3/4)l is drawn on the diagram. Whenever a wave crosses this line, the voltage in the wave is added to the voltage that has accumulated at z = (3/4)l over all earlier times. This general procedure enables one to easily determine the voltage at any speciﬁc time and location. In doing so, the terms in (117) that have occurred up to the chosen time are being added, but with information on the time at which each term appears. Line current can be found in a similar way through a current reflectio diagram. It is easiest to construct the current diagram directly from the voltage diagram by determining a value for current that is associated with each voltage wave. In dealing with current, it is important to keep track of the sign of the current because it relates to the voltage waves and their polarities. Referring to Figures 10.19a and 10.20, we use the convention in which current associated with a forward-z traveling voltage wave of positive polarity is positive. This would result in current that ﬂows in the clockwise direction, as shown in Figure 10.19a. Current associated with a backward-z traveling voltage wave of positive polarity (thus ﬂowing counterclockwise) is negative. Such a case is illustrated in Figure 10.20. In our two-dimensional transmission-line drawings, we assign positive polarity to voltage waves propagating in either direction if the upper conductor carries a positive charge and the lower conductor a negative charge. In Figures 10.19a and 10.20, both voltage waves are of positive polarity, so their associated currents will be net positive for the forward wave and net negative for the backward wave. In general, we write I+ = and I− = − V− Z0 (121) V+ Z0 (120)

Finding the current associated with a backward-propagating voltage wave immediately requires a minus sign, as (121) indicates. Figure 10.22a shows the current reﬂection diagram that is derived from the voltage diagram of Figure 10.21a. Note that the current values are labeled in terms of the voltage values, with the appropriate sign added as per (120) and (121). Once the current diagram is constructed, current at a given location and time can be found in exactly the same manner as voltage is found using the voltage diagram. Figure 10.22b shows the current as a function of time at the z = (3/4)l position, determined by summing the current wave values as they cross the vertical line drawn at that location.

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(a)

(b)

Figure 10.22 (a) Current reflection diagram for the line of Figure 10.20 as obtained from the voltage diagram of Figure 10.21a. (b) Current at the z = 3l /4 position as determined from the current reflection diagram, showing the expected steady-state value of V0 /(RL + Rg ). E X A M P L E 10.11

In Figure 10.20, Rg = Z 0 = 50 , R L = 25 , and the battery voltage is V0 = 10 V. The switch is closed at time t = 0. Determine the voltage at the load resistor and the current in the battery as functions of time.
Solution. Voltage and current reﬂection diagrams are shown in Figure 10.23a and b. At the moment the switch is closed, half the battery voltage appears across the

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V

V

(a)

A

A

(b)

Figure 10.23 Voltage (a) and current (b) reflection diagrams for Example 10.11.

50- resistor, with the other half comprising the initial voltage wave. Thus V1+ = (1/2)V0 = 5 V. The wave reaches the 25- load, where it reﬂects with reﬂection coefﬁcient 25 − 50 1 =− L = 25 + 50 3 So V1− = −(1/3)V1+ = −5/3 V. This wave returns to the battery, where it encounters reﬂection coefﬁcient g = 0. Thus, no further waves appear; steady state is reached. Once the voltage wave values are known, the current reﬂection diagram can be constructed. The values for the two current waves are V+ 5 1 + = A I1 = 1 = Z0 50 10

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and
− I1 = −

V1− 5 =− − Z0 3

1 50

=

1 A 30

− + Note that no attempt is made here to derive I1 from I1 . They are both obtained independently from their respective voltages. The voltage at the load as a function of time is now found by summing the voltages along the vertical line at the load position. The resulting plot is shown in Figure 10.24a. Current in the battery is found by summing the currents along the vertical axis, with the resulting plot shown as Figure 10.24b. Note that in steady state, we treat the circuit as lumped, with the battery in series with the 50- and 25resistors. Therefore, we expect to see a steady-state current through the battery (and everywhere else) of

I B (steady state) =

10 1 = A 50 + 25 7.5

(a)

(b)

Figure 10.24 Voltage across the load (a) and current in the battery (b) as determined from the reflection diagrams of Figure 10.23 (Example 10.11).

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Figure 10.25 In an initially charged line, closing the switch as shown initiates a voltage wave of opposite polarity to that of the initial voltage. The wave thus depletes the line voltage and will fully discharge the line in one round trip if Rg = Z 0 .

This value is also found from the current reﬂection diagram for t > 2l/ν. Similarly, the steady-state load voltage should be VL (steady state) = V0 RL (10)(25) 10 = = V Rg + R L 50 + 25 3

which is found also from the voltage reﬂection diagram for t > l/ν.

Another type of transient problem involves lines that are initially charged. In these cases, the manner in which the line discharges through a load is of interest. Consider the situation shown in Figure 10.25, in which a charged line of characteristic impedance Z 0 is discharged through a resistor of value Rg when a switch at the resistor location is closed.5 We consider the resistor at the z = 0 location; the other end of the line is open (as would be necessary) and is located at z = l. When the switch is closed, current I R begins to ﬂow through the resistor, and the line discharge process begins. This current does not immediately ﬂow everywhere in the transmission line but begins at the resistor and establishes its presence at more distant parts of the line as time progresses. By analogy, consider a long line of automobiles at a red light. When the light turns green, the cars at the front move through the intersection ﬁrst, followed successively by those further toward the rear. The point that divides cars in motion and those standing still is, in fact, a wave that propagates toward the back of the line. In the transmission line, the ﬂow of charge progresses in a similar way. A voltage wave, V1+ , is initiated and propagates to the right. To the left of its leading edge, charge is in motion; to the right of the leading edge, charge is stationary and carries its original density. Accompanying the charge in motion to the left of V1+ is a drop in the charge density as the discharge process occurs, and so the line voltage to the left of V1+ is partially reduced. This voltage will be given by the sum of the initial voltage, V0 , and V1+ , which means that V1+ must

5

Even though this is a load resistor, we will call it Rg because it is located at the front (generator) end of the line.

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in fact be negative (or of opposite sign to V0 ). The line discharge process is analyzed by keeping track of V1+ as it propagates and undergoes multiple reﬂections at the two ends. Voltage and current reﬂection diagrams are used for this purpose in much the same way as before. Referring to Figure 10.25, we see that for positive V0 the current ﬂowing through the resistor will be counterclockwise and hence negative. We also know that continuity requires that the resistor current be equal to the current associated with the voltage wave, or V+ + I R = I1 = 1 Z0 Now the resistor voltage will be V+ + VR = V0 + V1+ = −I R Rg = −I1 Rg = − 1 Rg Z0 where the minus signs arise from the fact that VR (having positive polarity) is produced by the negative current, I R . We solve for V1+ to obtain V1+ = −V0 Z 0 Z 0 + Rg (122)

Having found V1+ , we can set up the voltage and current reﬂection diagrams. The diagram for voltage is shown in Figure 10.26. Note that the initial condition of voltage V0 everywhere on the line is accounted for by assigning voltage V0 to the horizontal axis of the voltage diagram. The diagram is otherwise drawn as before, but with L = 1 (at the open-circuited load end). Variations in how the line discharges thus depend on the resistor value at the switch end, Rg , which determines the reﬂection coefﬁcient, g , at that location. The current reﬂection diagram is derived from the voltage diagram in the usual way. There is no initial current to consider.

Figure 10.26 Voltage reflection diagram for the charged line of Figure 10.25, showing the initial condition of V0 everywhere on the line at t = 0.

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Figure 10.27 Voltage across the resistor as a function of time, as determined from the reflection diagram of Figure 10.26, in which Rg = Z 0 ( = 0).

A special case of practical importance is that in which the resistor is matched to the line, or Rg = Z 0 . In this case, Eq. (122) gives V1+ = −V0 /2. The line fully discharges in one round trip of V1+ and produces a voltage across the resistor of value VR = V0 /2, which persists for time T = 2l/ν. The resistor voltage as a function of time is shown in Figure 10.27. The transmission line in this application is known as a pulse-forming line; pulses that are generated in this way are well formed and of low noise, provided the switch is sufﬁciently fast. Commercial units are available that are capable of generating high-voltage pulses of widths on the order of a few nanoseconds, using thyratron-based switches. When the resistor is not matched to the line, full discharge still occurs, but does so over several reﬂections, leading to a complicated pulse shape.
E X A M P L E 10.12

In the charged line of Figure 10.25, the characteristic impedance is Z 0 = 100 , and Rg = 100/3 . The line is charged to an initial voltage, V0 = 160 V, and the switch is closed at time t = 0. Determine and plot the voltage and current through the resistor for time 0 < t < 8l/ν (four round trips).
Solution. With the given values of Rg and Z 0 , Eq. (47) gives g with

L

= 1, and using (122), we ﬁnd V1+ = V1− = −3/4V0 = −120 V V3+ V4+ V2+ = V2− = = = V3− V4− = =
− g V1 − g V2 − g V3

= −1/2. Then,

= + 60 V = −30 V = +15 V

Using these values on the voltage reﬂection diagram, we evaluate the voltage in time at the resistor location by moving up the left-hand vertical axis, adding voltages as we progress, and beginning with V0 + V1+ at t = 0. Note that when we add voltages along the vertical axis, we are encountering the intersection points between incident

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and reﬂected waves, which occur (in time) at each integer multiple of 2l/ν. So, when moving up the axis, we add the voltages of both waves to our total at each occurrence. The voltage within each time interval is thus: VR = V0 + V1+ = 40 V = V0 + = V0 + V1+ V1+ V1+ + + + V1− V1− V1− (0 < t < 2l/ν) V2+ V2+ V2+ + + = −20 V V2− V2− + + (2l/ν < t < 4l/ν) + = 10 V V3− (4l/ν < t < 6l/ν) V4+ = −5 V (6l/ν < t < 8l/ν) + V3+ V3+ + + +

= V0 +

The resulting voltage plot over the desired time range is shown in Figure 10.28a.

Figure 10.28 Resistor voltage (a) and current (b) as functions of time for the line of Figure 10.25, with values as specified in Example 10.12.

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The current through the resistor is most easily obtained by dividing the voltages in Figure 10.28a by −Rg . As a demonstration, we can also use the current diagram of Figure 10.22a to obtain this result. Using (120) and (121), we evaluate the current waves as follows:
+ I1 = V1+ /Z 0 = −1.2 A − I1 = −V1− /Z 0 = +1.2 A

+ − I2 = −I2 = V2+ /Z 0 = +0.6 A

+ − I3 = −I3 = V3+ /Z 0 = −0.30 A + − I4 = −I4 = V4+ /Z 0 = +0.15 A

Using these values on the current reﬂection diagram, Figure 10.22a, we add up currents in the resistor in time by moving up the left-hand axis, as we did with the voltage diagram. The result is shown in Figure 10.28b. As a further check to the correctness of our diagram construction, we note that current at the open end of the line (Z = l) must always be zero. Therefore, summing currents up the right-hand axis must give a zero result for all time. The reader is encouraged to verify this.

REFERENCES
1. White, H. J., P. R. Gillette, and J. V. Lebacqz. “The Pulse-Forming Network.” Chapter 6 in Pulse Generators, edited by G. N, Glasoe and J. V. Lebacqz. New York: Dover, 1965. 2. Brown, R. G., R. A. Sharpe, W. L. Hughes, and R. E. Post. Lines, Waves, and Antennas. 2d ed. New York: The Ronald Press Company, 1973. Transmission lines are covered in the ﬁrst six chapters, with numerous examples. 3. Cheng, D. K. Field and Wave Electromagnetics. 2d ed. Reading, Mass.: Addison-Wesley, 1989. Provides numerous examples of Smith chart problems and transients. 4. Seshadri, S. R. Fundamentals of Transmission Lines and Electromagnetic Fields. Reading, Mass.: Addison-Wesley, 1971.

CHAPTER 10 PROBLEMS
10.1 The parameters of a certain transmission line operating at ω = 6 × 108 rad/s are L = 0.35 µH/m, C = 40 pF/m, G = 75 µS/m, and R = 17 /m. Find γ , α, β, λ, and Z 0 . A sinusoidal wave on a transmission line is speciﬁed by voltage and current in phasor form: Vs (z) = V0 eαz e jβz and Is (z) = I0 eαz e jβz e jφ

10.2

where V0 and I0 are both real. (a) In which direction does this wave propagate and why? (b) It is found that α = 0, Z 0 = 50 , and the wave velocity is v p = 2.5 × 108 m/s, with ω = 108 s−1 . Evaluate R, G, L, C, λ, and φ.

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10.3

The characteristic impedance of a certain lossless transmission line is 72 . If L = 0.5 µH/m, ﬁnd (a) C; (b) νp ; (c) β if f = 80 MHz. (d) The line is terminated with a load of 60 . Find and s. A sinusoidal voltage wave of amplitude V0 , frequency ω, and phase constant β propagates in the forward z direction toward the open load end in a lossless transmission line of characteristic impedance Z 0 . At the end, the wave totally reﬂects with zero phase shift, and the reﬂected wave now interferes with the incident wave to yield a standing wave pattern over the line length (as per Example 10.1). Determine the standing wave pattern for the current in the line. Express the result in real instantaneous form and simplify. Two characteristics of a certain lossless transmission line are Z 0 = 50 and γ = 0 + j0.2π m−1 at f = 60 MHz (a) ﬁnd L and C for the line. (b) A load Z L = 60 + j80 is located at z = 0. What is the shortest distance from the load to a point at which Z in = Rin + j0? A 50- load is attached to a 50-m section of the transmission line of Problem 10.1, and a 100-W signal is fed to the input end of the line. (a) Evaluate the distributed line loss in dB/m. (b) Evaluate the reﬂection coefﬁcient at the load. (c) Evaluate the power that is dissipated by the load resistor. (d) What power drop in dB does the dissipated power in the load represent when compared to the original input power? (e) On partial reﬂection from the load, how much power returns to the input and what dB drop does this represent when compared to the original 100-W input power? A transmitter and receiver are connected using a cascaded pair of transmission lines. At the operating frequency, line 1 has a measured loss of 0.1 dB/m, and line 2 is rated at 0.2 dB/m. The link is composed of 40 m of line 1 joined to 25 m of line 2. At the joint, a splice loss of 2 dB is measured. If the transmitted power is 100 mW, what is the received power? An absolute measure of power is the dBm scale, in which power is speciﬁed in decibels relative to one milliwatt. Speciﬁcally, P(dBm) = 10 log10 [P(mW)/1 mW]. Suppose that a receiver is rated as having a sensitivity of −20 dBm, indicating the mimimum power that it must receive in order to adequately interpret the transmitted electronic data. Suppose this receiver is at the load end of a 50- transmission line having 100-m length and loss rating of 0.09 dB/m. The receiver impedance is 75 , and so is not matched to the line. What is the minimum required input power to the line in (a) dBm, (b) mW? A sinusoidal voltage source drives the series combination of an impedance, Z g = 50 − j50 , and a lossless transmission line of length L, shorted at the load end. The line characteristic impedance is 50 , and wavelength λ is measured on the line. (a) Determine, in terms of wavelength, the shortest line length that will result in the voltage source driving a total impedance of

10.4

10.5

10.6

10.7

10.8

10.9

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50 . (b) Will other line lengths meet the requirements of part (a)? If so, what are they? 10.10 Two lossless transmission lines having different characteristic impedances are to be joined end to end. The impedances are Z 01 = 100 and Z 03 = 25 . The operating frequency is 1 GHz. (a) Find the required characteristic impedance, Z 02 , of a quarter-wave section to be inserted between the two, which will impedance-match the joint, thus allowing total power transmission through the three lines. (b) The capacitance per unit length of the intermediate line is found to be 100 pF/m. Find the shortest length in meters of this line that is needed to satisfy the impedance-matching condition. (c) With the three-segment setup as found in parts (a) and (b), the frequency is now doubled to 2 GHz. Find the input impedance at the line-1-to-line-2 junction, seen by waves incident from line 1. (d) Under the conditions of part (c), and with power incident from line 1, evaluate the standing wave ratio that will be measured in line 1, and the fraction of the incident power from line 1 that is reﬂected and propagates back to the line 1 input. A transmission line having primary constants L , C, R, and G has length and is terminated by a load having complex impedance R L + j X L . At the input end of the line, a dc voltage source, V0 , is connected. Assuming all parameters are known at zero frequency, ﬁnd the steady-state power dissipated by the load if (a) R = G = 0; (b) R = 0, G = 0; (c) R = 0, G = 0; (d) R = 0, G = 0. In a circuit in which a sinusoidal voltage source drives its internal impedance in series with a load impedance, it is known that maximum power transfer to the load occurs when the source and load impedances form a complex conjugate pair. Suppose the source (with its internal impedance) now drives a complex load of impedance Z L = R L + j X L that has been moved to the end of a lossless transmission line of length having characteristic impedance Z 0 . If the source impedance is Z g = Rg + j X g , write an equation that can be solved for the required line length, , such that the displaced load will receive the maximum power. The incident voltage wave on a certain lossless transmission line for which Z 0 = 50 and νp = 2 × 108 m/s is V + (z, t) = 200 cos(ωt − π z) V. (a) Find ω. (b) Find I + (z, t). The section of line for which z > 0 is replaced by a load Z L = 50 + j30 at z = 0. Find: (c) L ; (d) Vs− (z); (e) Vs at z = −2.2 m.

10.11

10.12

10.13

10.14

A lossless transmission line having characteristic impedance Z 0 = 50 is driven by a source at the input end that consists of the series combination of a 10-V sinusoidal generator and a 50- resistor. The line is one-quarter wavelength long. At the other end of the line, a load impedance, Z L = 50 − j50 is attached. (a) Evaluate the input impedance to the line

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361

Figure 10.29 See Problem 10.15.

seen by the voltage source-resistor combination; (b) evaluate the power that is dissipated by the load; (c) evaluate the voltage amplitude that appears across the load. 10.15 10.16 For the transmission line represented in Figure 10.29, ﬁnd Vs,out if f = (a) 60 Hz; (b) 500 kHz. A 100- lossless transmission line is connected to a second line of 40impedance, whose length is λ/4. The other end of the short line is terminated by a 25- resistor. A sinusoidal wave (of frequency f ) having 50 W average power is incident from the 100- line. (a) Evaluate the input impedance to the quarter-wave line. (b) Determine the steady-state power that is dissipated by the resistor. (c) Now suppose that the operating frequency is lowered to one-half its original value. Determine the new input impedance, Z in , for this case. (d) For the new frequency, calculate the power in watts that returns to the input end of the line after reﬂection. Determine the average power absorbed by each resistor in Figure 10.30. The line shown in Figure 10.31 is lossless. Find s on both sections 1 and 2. A lossless transmission line is 50 cm in length and operates at a frequency of 100 MHz. The line parameters are L = 0.2 µH/m and C = 80 pF/m. The line is terminated in a short circuit at z = 0, and there is a load Z L = 50 + j20 across the line at location z = −20 cm. What average power is delivered to Z L if the input voltage is 100 0◦ V? (a) Determine s on the transmission line of Figure 10.32. Note that the dielectric is air. (b) Find the input impedance. (c) If ωL = 10 , ﬁnd Is . (d) What value of L will produce a maximum value for |Is | at ω = 1

10.17 10.18 10.19

10.20

Figure 10.30 See Problem 10.17.

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Figure 10.31 See Problem 10.18.

10.21

Grad/s? For this value of L, calculate the average power (e) supplied by the source; ( f ) delivered to Z L = 40 + j30 . A lossless line having an air dielectric has a characteristic impedance of 400 . The line is operating at 200 MHz and Z in = 200 − j200 . Use analytic methods or the Smith chart (or both) to ﬁnd (a) s; (b) Z L , if the line is 1 m long; (c) the distance from the load to the nearest voltage maximum. A lossless 75- line is terminated by an unknown load impedance. A VSWR of 10 is measured, and the ﬁrst voltage minimum occurs at 0.15 wavelengths in front of the load. Using the Smith chart, ﬁnd (a) the load impedance; (b) the magnitude and phase of the reﬂection coefﬁcient; (c) the shortest length of line necessary to achieve an entirely resistive input impedance. The normalized load on a lossless transmission line is 2 + j1. Let λ = 20 m and make use of the Smith chart to ﬁnd (a) the shortest distance from the load to a point at which z in = rin + j0, where rin > 0; (b) z in at this point. (c) The line is cut at this point and the portion containing z L is thrown away. A resistor r = rin of part (a) is connected across the line. What is s on the remainder of the line? (d) What is the shortest distance from this resistor to a point at which z in = 2 + j1?

10.22

10.23

10.24 10.25

With the aid of the Smith chart, plot a curve of |Z in | versus l for the transmission line shown in Figure 10.33. Cover the range 0 < l/λ < 0.25.

A 300- transmission line is short-circuited at z = 0. A voltage maximum, |V |max = 10 V, is found at z = −25 cm, and the minimum voltage, |V |min = 0, is at z = −50 cm. Use the Smith chart to ﬁnd Z L (with the short circuit
L

Figure 10.32 See Problem 10.20.

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363

Figure 10.33 See Problem 10.24.

10.26

replaced by the load) if the voltage readings are (a) |V |max = 12 V at z = −5 cm, and |V |min = 5 V; (b) |V |max = 17 V at z = −20 cm, and |V |min = 0. A 50- lossless line is of length 1.1 λ. It is terminated by an unknown load impedance. The input end of the 50- line is attached to the load end of a lossless 75- line. A VSWR of 4 is measured on the 75- line, on which the ﬁrst voltage maximum occurs at a distance of 0.2 λ in front of the junction between the two lines. Use the Smith chart to ﬁnd the unknown load impedance. The characteristic admittance (Y0 = 1/Z 0 ) of a lossless transmission line is 20 mS. The line is terminated in a load Y L = 40 − j20 mS. Use the Smith chart to ﬁnd (a) s; (b) Yin if l = 0.15λ; (c) the distance in wavelengths from Y L to the nearest voltage maximum. The wavelength on a certain lossless line is 10 cm. If the normalized input impedance is z in = 1 + j2, use the Smith chart to determine (a) s; (b) z L , if the length of the line is 12 cm; (c) x L , if z L = 2 + j x L where x L > 0. A standing wave ratio of 2.5 exists on a lossless 60 line. Probe measurements locate a voltage minimum on the line whose location is marked by a small scratch on the line. When the load is replaced by a short circuit, the minima are 25 cm apart, and one minimum is located at a point 7 cm toward the source from the scratch. Find Z L . A two-wire line constructed of lossless wire of circular cross section is gradually ﬂared into a coupling loop that looks like an egg beater. At the point X , indicated by the arrow in Figure 10.34, a short circuit is placed

10.27

10.28

10.29

10.30

Figure 10.34 See Problem 10.30.

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ENGINEERING ELECTROMAGNETICS

across the line. A probe is moved along the line and indicates that the ﬁrst voltage minimum to the left of X is 16 cm from X . With the short circuit removed, a voltage minimum is found 5 cm to the left of X , and a voltage maximum is located that is 3 times the voltage of the minimum. Use the Smith chart to determine (a) f ; (b) s; (c) the normalized input impedance of the egg beater as seen looking to the right at point X . 10.31 In order to compare the relative sharpness of the maxima and minima of a standing wave, assume a load z L = 4 + j0 is located at z = 0. Let |V |min = 1 and λ = 1 m. Determine the width of the (a) minimum where |V | < 1.1; (b) maximum where |V | > 4/1.1.

10.32

In Figure 10.17, let Z L = 250 , Z 0 = 50 , ﬁnd the shortest attachment distance d and the shortest length d1 of a short-circuited stub line that will provide a perfect match on the main line to the left of the stub. Express all answers in wavelengths. In Figure 10.17, let Z L = 40 − j10 , Z 0 = 50 , f = 800 MHz, and v = c. (a) Find the shortest length d1 of a short-circuited stub, and the shortest distance d that it may be located from the load to provide a perfect match on the main line to the left of the stub. (b) Repeat for an open-circuited stub. The lossless line shown in Figure 10.35 is operating with λ = 100 cm. If d1 = 10 cm, d = 25 cm, and the line is matched to the left of the stub, what is Z L ? A load, Z L = 25 + j75 , is located at z = 0 on a lossless two-wire line for which Z 0 = 50 and v = c. (a) If f = 300 MHz, ﬁnd the shortest distance d (z = −d) at which the input admittance has a real part equal to 1/Z 0 and a negative imaginary part. (b) What value of capacitance C should be connected across the line at that point to provide unity standing wave ratio on the remaining portion of the line? The two-wire lines shown in Figure 10.36 are all lossless and have Z 0 = 200 . Find d and the shortest possible value for d1 to provide a matched load if λ = 100 cm.

10.33

10.34

10.35

10.36

Figure 10.35 See Problem 10.34.

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365

Figure 10.36 See Problem 10.36.

10.37

In the transmission line of Figure 10.20, Rg = Z 0 = 50 , and R L = 25 . Determine and plot the voltage at the load resistor and the current in the battery as functions of time by constructing appropriate voltage and current reﬂection diagrams. Repeat Problem 10.37, with Z 0 = 50 , and R L = Rg = 25 the analysis for the time period 0 < t < 8l/ν. . Carry out

10.38 10.39

In the transmission line of Figure 10.20, Z 0 = 50 , and R L = Rg = 25 . The switch is closed at t = 0 and is opened again at time t = l/4ν, thus creating a rectangular voltage pulse in the line. Construct an appropriate voltage reﬂection diagram for this case and use it to make a plot of the voltage at the load resistor as a function of time for 0 < t < 8l/ν (note that the effect of opening the switch is to initiate a second voltage wave, whose value is such that it leaves a net current of zero in its wake). In the charged line of Figure 10.25, the characteristic impedance is Z 0 = 100 , and Rg = 300 . The line is charged to initial voltage, V0 = 160 V, and the switch is closed at t = 0. Determine and plot the voltage and current through the resistor for time 0 < t < 8l/ν (four round-trips). This problem accompanies Example 10.12 as the other special case of the basic charged-line problem, in which now Rg > Z 0 . In the transmission line of Figure 10.37, the switch is located midway down the line and is closed at t = 0. Construct a voltage reﬂection diagram for this case, where R L = Z 0 . Plot the load resistor voltage as a function of time.

10.40

10.41

Figure 10.37 See Problem 10.41.

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ENGINEERING ELECTROMAGNETICS

Figure 10.38 See Problem 10.42. t=0 Rg Z0 V0

RL = Z0

Figure 10.39 See Problem 10.43.

10.42

A simple frozen wave generator is shown in Figure 10.38. Both switches are closed simultaneously at t = 0. Construct an appropriate voltage reﬂection diagram for the case in which R L = Z 0 . Determine and plot the load resistor voltage as a function of time. In Figure 10.39, R L = Z 0 and Rg = Z 0 /3. The switch is closed at t = 0. Determine and plot as functions of time (a) the voltage across R L ; (b) the voltage across Rg ; (c) the current through the battery.

10.43

C H A P T E R

11

The Uniform Plane Wave

T

his chapter is concerned with the application of Maxwell’s equations to the problem of electromagnetic wave propagation. The uniform plane wave represents the simplest case, and while it is appropriate for an introduction, it is of great practical importance. Waves encountered in practice can often be assumed to be of this form. In this study, we will explore the basic principles of electromagnetic wave propagation, and we will come to understand the physical processes that determine the speed of propagation and the extent to which attenuation may occur. We will derive and use the Poynting theorem to ﬁnd the power carried by a wave. Finally, we will learn how to describe wave polarization. ■

11.1 WAVE PROPAGATION IN FREE SPACE
We begin with a quick study of Maxwell’s equations, in which we look for clues of wave phenomena. In Chapter 10, we saw how voltages and currents propagate as waves in transmission lines, and we know that the existence of voltages and currents implies the existence of electric and magnetic ﬁelds. So we can identify a transmission line as a structure that conﬁnes the ﬁelds while enabling them to travel along its length as waves. It can be argued that it is the ﬁelds that generate the voltage and current in a transmission line wave, and—if there is no structure on which the voltage and current can exist—the ﬁelds will exist nevertheless, and will propagate. In free space, the ﬁelds are not bounded by any conﬁning structure, and so they may assume any magnitude and direction, as initially determined by the device (such as an antenna) that generates them. When considering electromagnetic waves in free space, we note that the medium is sourceless (ρν = J = 0). Under these conditions, Maxwell’s equations may be
367

368

ENGINEERING ELECTROMAGNETICS

written in terms of E and H only as ∂E ∂t ∂H ∇ × E = −µ0 ∂t ∇ ·E = 0
0

∇ ×H =

(1) (2) (3) (4)

∇ ·H = 0

Now let us see whether wave motion can be inferred from these four equations without actually solving them. Equation (1) states that if electric ﬁeld E is changing with time at some point, then magnetic ﬁeld H has curl at that point; therefore H varies spatially in a direction normal to its orientation direction. Also, if E is changing with time, then H will in general also change with time, although not necessarily in the same way. Next, we see from Eq. (2) that a time-varying H generates E, which, having curl, varies spatially in the direction normal to its orientation. We now have once more a changing electric ﬁeld, our original hypothesis, but this ﬁeld is present a small distance away from the point of the original disturbance. We might guess (correctly) that the velocity with which the effect moves away from the original point is the velocity of light, but this must be checked by a more detailed examination of Maxwell’s equations. We postulate the existence of a uniform plane wave, in which both ﬁelds, E and H, lie in the transverse plane—that is, the plane whose normal is the direction of propagation. Furthermore, and by deﬁnition, both ﬁelds are of constant magnitude in the transverse plane. For this reason, such a wave is sometimes called a transverse electromagnetic (TEM) wave. The required spatial variation of both ﬁelds in the direction normal to their orientations will therefore occur only in the direction of travel—or normal to the transverse plane. Assume, for example, that E = E x ax , or that the electric ﬁeld is polarized in the x direction. If we further assume that wave travel is in the z direction, we allow spatial variation of E only with z. Using Eq. (2), we note that with these restrictions, the curl of E reduces to a single term: ∇ ×E= ∂ Hy ∂ Ex ∂H a y = −µ0 = −µ0 ay ∂z ∂t ∂t (5)

The direction of the curl of E in (5) determines the direction of H, which we observe to be along the y direction. Therefore, in a uniform plane wave, the directions of E and H and the direction of travel are mutually orthogonal. Using the y-directed magnetic ﬁeld, and the fact that it varies only in z, simpliﬁes Eq. (1) to read ∇ ×H=− ∂ Hy ax = ∂z ∂E = ∂t ∂ Ex ax ∂t (6)

0

0

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369

Equations (5) and (6) can be more succinctly written: ∂ Hy ∂ Ex = −µ0 ∂z ∂t ∂ Hy ∂ Ex =− 0 ∂z ∂t (7) (8)

These equations compare directly with the telegraphist’s equations for the lossless transmission line [Eqs. (20) and (21) in Chapter 10]. Further manipulations of (7) and (8) proceed in the same manner as was done with the telegraphist’s equations. Speciﬁcally, we differentiate (7) with respect to z, obtaining: ∂ 2 Hy ∂ 2 Ex = −µ0 2 ∂z ∂t∂z Then, (8) is differentiated with respect to t: ∂ 2 Hy ∂ 2 Ex =− 0 2 ∂z∂t ∂t Substituting (10) into (9) results in ∂ 2 Ex = µ0 ∂z 2 ∂ 2 Ex ∂t 2 (9)

(10)

0

(11)

This equation, in direct analogy to Eq. (13) in Chapter 10, we identify as the wave equation for our x-polarized TEM electric ﬁeld in free space. From Eq. (11), we further identify the propagation velocity: ν=√ 1 µ0 = 3 × 108 m/s = c (12)

0

where c denotes the velocity of light in free space. A similar procedure, involving differentiating (7) with t and (8) with z, yields the wave equation for the magnetic ﬁeld; it is identical in form to (11): ∂ 2 Hy = µ0 ∂z 2
0

∂ 2 Hy ∂t 2

(13)

As was discussed in Chapter 10, the solution to equations of the form of (11) and (13) will be forward- and backward-propagating waves having the general form [in this case for Eq. (11)]: E x (z, t) = f 1 (t − z/ν) + f 2 (t + z/ν) (14)

where again f 1 amd f 2 can be any function whose argument is of the form t ± z/ν. From here, we immediately specialize to sinusoidal functions of a speciﬁed frequency and write the solution to (11) in the form of forward- and backward-propagating

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ENGINEERING ELECTROMAGNETICS

cosines. Because the waves are sinusoidal, we denote their velocity as the phase velocity, ν p . The waves are written as: E x (z, t) = Ex (z, t) + Ex (z, t) = |E x0 | cos [ω(t − z/ν p ) + φ1 ] + |E x0 | cos [ω(t + z/ν p ) + φ2 ] forward z travel backward z travel

= |E x0 | cos [ωt − k0 z + φ1 ] + |E x0 | cos [ωt + k0 z + φ2 ]

(15)

In writing the second line of (15), we have used the fact that the waves are traveling in free space, in which case the phase velocity, ν p = c. Additionally, the wavenumber in free space in deﬁned as k0 ≡ ω rad/m c (16)

In a manner consistant with our transmission line studies, we refer to the solutions expressed in (15) as the real instantaneous forms of the electric ﬁeld. They are the mathematical representations of what one would experimentally measure. The terms ωt and k0 z, appearing in (15), have units of angle and are usually expressed in radians. We know that ω is the radian time frequency, measuring phase shift per unit time; it has units of rad/s. In a similar way, we see that k0 will be interpreted as a spatial frequency, which in the present case measures the phase shift per unit distance along the z direction in rad/m. We note that k0 is the phase constant for lossless propagation of uniform plane waves in free space. The wavelength in free space is the distance over which the spatial phase shifts by 2π radians, assuming ﬁxed time, or k0 z = k0 λ = 2π → λ= 2π k0 (free space) (17)

The manner in which the waves propagate is the same as we encountered in transmission lines. Speciﬁcally, suppose we consider some point (such as a wave crest) on the forward-propagating cosine function of Eq. (15). For a crest to occur, the argument of the cosine must be an integer multiple of 2π. Considering the mth crest of the wave, the condition becomes k0 z = 2mπ So let us now consider the point on the cosine that we have chosen, and see what happens as time is allowed to increase. Our requirement is that the entire cosine argument be the same multiple of 2π for all time, in order to keep track of the chosen point. Our condition becomes ωt − k0 z = ω(t − z/c) = 2mπ (18)

As time increases, the position z must also increase in order to satisfy (18). The wave crest (and the entire wave) moves in the positive z direction at phase velocity c (in free space). Using similar reasoning, the wave in Eq. (15) having cosine argument (ωt + k0 z) describes a wave that moves in the negative z direction, since as time

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371

increases, z must now decrease to keep the argument constant. For simplicity, we will restrict our attention in this chapter to only the positive z traveling wave. As was done for transmission line waves, we express the real instantaneous ﬁelds of Eq. (15) in terms of their phasor forms. Using the forward-propagating ﬁeld in (15), we write: Ex (z, t) = 1 1 |E x0 |e jφ1 e− jk0 z e jωt + c.c. = E xs e jωt + c.c. = Re[E xs e jωt ] 2 2
E x0

(19)

where c.c. denotes the complex conjugate, and where we identify the phasor electric fiel as E xs = E x0 e− jk0 z . As indicated in (19), E x0 is the complex amplitude (which includes the phase, φ1 ).
E X A M P L E 11.1

Let us express E y (z, t) = 100 cos(108 t − 0.5z + 30◦ ) V/m as a phasor.
Solution. We ﬁrst go to exponential notation,

and then drop Re and suppress e j10 t , obtaining the phasor
8

E y (z, t) = Re 100e

j(108 t−0.5z+30◦ )

E ys (z) = 100e− j0.5z+ j30

◦

Note that a mixed nomenclature is used for the angle in this case; that is, 0.5z is in radians, while 30◦ is in degrees. Given a scalar component or a vector expressed as a phasor, we may easily recover the time-domain expression.
E X A M P L E 11.2

Given the complex amplitude of the electric ﬁeld of a uniform plane wave, E0 = 100ax + 20 30◦ a y V/m, construct the phasor and real instantaneous ﬁelds if the wave is known to propagate in the forward z direction in free space and has frequency of 10 MHz.
Solution. We begin by constructing the general phasor expression:

where k0 = ω/c = 2π × 107/3 × 108 = 0.21 rad/m. The real instantaneous form is then found through the rule expressed in Eq. (19): E(z, t) = Re 100e− j0.21z e j2π ×10 t ax + 20e j30 e− j0.21z e j2π ×10 t a y
7 ◦ 7

Es (z) = 100ax + 20e j30 a y e− jk0 z
◦

= 100 cos (2π × 10 t − 0.21z)ax + 20 cos (2π × 107 t − 0.21z + 30◦ ) a y
7

= Re 100e j(2π ×10

7

t−0.21z)

ax + 20e j(2π ×10

7

t−0.21z+30◦ )

ay

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ENGINEERING ELECTROMAGNETICS

It is evident that taking the partial derivative of any ﬁeld quantity with respect to time is equivalent to multiplying the corresponding phasor by jω. As an example, we can express Eq. (8) (using sinusoidal ﬁelds) as ∂H y ∂Ex =− 0 ∂z ∂t where, in a manner consistent with (19): Ex (z, t) = 1 E xs (z) e jωt + c.c. 2 and H y (z, t) = 1 Hys (z) e jωt + c.c. 2 (21) (20)

On substituting the ﬁelds in (21) into (20), the latter equation simpliﬁes to d Hys (z) = − jω 0 E xs (z) dz (22)

In obtaining this equation, we note ﬁrst that the complex conjugate terms in (21) produce their own separate equation, redundant with (22); second, the e jωt factors, common to both sides, have divided out; third, the partial derivative with z becomes the total derivative, since the phasor, Hys , depends only on z. We next apply this result to Maxwell’s equations, to obtain them in phasor form. Substituting the ﬁeld as expressed in (21) into Eqs. (1) through (4) results in ∇ × Hs ∇ × Es ∇ · Es ∇ · Hs = = = = jω 0 Es − jωµ0 Hs 0 0 (23) (24) (25) (26)

It should be noted that (25) and (26) are no longer independent relationships, for they can be obtained by taking the divergence of (23) and (24), respectively. Eqs. (23) through (26) may be used to obtain the sinusoidal steady-state vector form of the wave equation in free space. We begin by taking the curl of both sides of (24): ∇ × ∇ × Es = − jωµ0 ∇ × Hs = ∇(∇ · Es ) − ∇ 2 Es where the last equality is an identity, which deﬁnes the vector Laplacian of Es : ∇ 2 Es = ∇(∇ · Es ) − ∇ × ∇ × Es From (25), we note that ∇ · Es = 0. Using this, and substituting (23) in (27), we obtain
2 ∇ 2 Es = −k0 Es

(27)

(28)

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373

√ where again, k0 = ω/c = ω µ0 0 . Equation (28) is known as the vector Helmholtz 1 equation in free space. It is fairly formidable when expanded, even in rectangular coordinates, for three scalar phasor equations result (one for each vector component), and each equation has four terms. The x component of (28) becomes, still using the del-operator notation,
2 ∇ 2 E xs = −k0 E xs

(29)

and the expansion of the operator leads to the second-order partial differential equation ∂ 2 E xs ∂ 2 E xs ∂ 2 E xs 2 + + = −k0 E xs ∂x2 ∂ y2 ∂z 2 Again, assuming a uniform plane wave in which E xs does not vary with x or y, the two corresponding derivatives are zero, and we obtain d 2 E xs 2 = −k0 E xs dz 2 the solution of which we already know: E xs (z) = E x0 e− jk0 z + E x0 e jk0 z (31) (30)

Let us now return to Maxwell’s equations, (23) through (26), and determine the form of the H ﬁeld. Given Es , Hs is most easily obtained from (24): which is greatly simpliﬁed for a single E xs component varying only with z, d E xs = − jωµ0 Hys dz Using (31) for E xs , we have Hys = − 1 (− jk0 )E x0 e− jk0 z + ( jk0 )E x0 e jk0 z jωµ0
0

∇ × Es = − jωµ0 Hs

(24)

= E x0

µ0

e− jk0 z − E x0

0

µ0

e jk0 z = Hy0 e− jk0 z + Hy0 e jk0 z

(32)

In real instantaneous form, this becomes Hy (z, t) = E x0
0

µ0

cos(ωt − k0 z) − E x0

0

µ0

cos(ωt + k0 z)

(33)

where E x0 and E x0 are assumed real.

1

Hermann Ludwig Ferdinand von Helmholtz (1821–1894) was a professor at the University of Berlin working in the ﬁelds of physiology, electrodynamics, and optics. Hertz was one of his students.

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ENGINEERING ELECTROMAGNETICS

In general, we ﬁnd from (32) that the electric and magnetic ﬁeld amplitudes of the forward-propagating wave in free space are related through E x0 = µ0
0

Hy0 = η0 Hy0

(34a)

We also ﬁnd the backward-propagating wave amplitudes are related through E x0 = − µ0
0

Hy0 = −η0 Hy0

(34b)

where the intrinsic impedance of free space is deﬁned as η0 = µ0
0

. = 377 = 120π

(35)

The dimension of η0 in ohms is immediately evident from its deﬁnition as the ratio of E (in units of V/m) to H (in units of A/m). It is in direct analogy to the characteristic impedance, Z 0 , of a transmission line, where we deﬁned the latter as the ratio of voltage to current in a traveling wave. We note that the difference between (34a) and (34b) is a minus sign. This is consistent with the transmission line analogy that led to Eqs. (25a) and (25b) in Chapter 10. Those equations accounted for the deﬁnitions of positive and negative current associated with forward and backward voltage waves. In a similar way, Eq. (34a) speciﬁes that in a forward-z propagating uniform plane wave whose electric ﬁeld vector lies in the positive x direction at a given point in time and space, the magnetic ﬁeld vector lies in the positive y direction at the same space and time coordinates. In the case of a backward-z propagating wave having a positive x-directed electric ﬁeld, the magnetic ﬁeld vector lies in the negative y direction. The physical signiﬁcance of this has to do with the deﬁnition of power ﬂow in the wave, as speciﬁed through the Poynting vector, S = E × H (in watts/m2 ). The cross product of E with H must give the correct wave propagation direction, and so the need for the minus sign in (34b) is apparent. Issues relating to power transmission will be addressed in Section 11.3. Some feeling for the way in which the ﬁelds vary in space may be obtained from Figures 11.1a and 11.1b. The electric ﬁeld intensity in Figure 11.1a is shown at t = 0, and the instantaneous value of the ﬁeld is depicted along three lines, the z axis and arbitrary lines parallel to the z axis in the x = 0 and y = 0 planes. Since the ﬁeld is uniform in planes perpendicular to the z axis, the variation along all three of the lines is the same. One complete cycle of the variation occurs in a wavelength, λ. The values of Hy at the same time and positions are shown in Figure 11.1b. A uniform plane wave cannot exist physically, for it extends to inﬁnity in two dimensions at least and represents an inﬁnite amount of energy. The distant ﬁeld of a transmitting antenna, however, is essentially a uniform plane wave in some limited region; for example, a radar signal impinging on a distant target is closely a uniform plane wave.

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Figure 11.1 (a) Arrows represent the instantaneous values of E x0 cos[ω(t − z/c)] at t = 0 along the z axis, along an arbitrary line in the x = 0 plane parallel to the z axis, and along an arbitrary line in the y = 0 plane parallel to the z axis. (b) Corresponding values of H y are indicated. Note that E x and H y are in phase at any point in time.

Although we have considered only a wave varying sinusoidally in time and space, a suitable combination of solutions to the wave equation may be made to achieve a wave of any desired form, but which satisﬁes (14). The summation of an inﬁnite number of harmonics through the use of a Fourier series can produce a periodic wave of square or triangular shape in both space and time. Nonperiodic waves may be obtained from our basic solution by Fourier integral methods. These topics are among those considered in the more advanced books on electromagnetic theory. D11.1. The electric ﬁeld amplitude of a uniform plane wave propagating in the az direction is 250 V/m. If E = E x ax and ω = 1.00 Mrad/s, ﬁnd: (a) the frequency; (b) the wavelength; (c) the period; (d) the amplitude of H.
Ans. 159 kHz; 1.88 km; 6.28 µs; 0.663 A/m

D11.2. Let Hs = (2 −40◦ ax − 3 20◦ a y )e− j0.07z A/m for a uniform plane wave traveling in free space. Find: (a) ω; (b) Hx at P(1, 2, 3) at t = 31 ns; (c) |H| at t = 0 at the origin.
Ans. 21.0 Mrad/s; 1.934 A/m; 3.22 A/m

11.2 WAVE PROPAGATION IN DIELECTRICS
We now extend our analytical treatment of the uniform plane wave to propagation in a dielectric of permittivity and permeability µ. The medium is assumed to be homogeneous (having constant µ and with position) and isotropic (in which µ and

376

ENGINEERING ELECTROMAGNETICS

are invariant with ﬁeld orientation). The Helmholtz equation is ∇ 2 Es = −k 2 Es √ √ k = ω µ = k0 µr For E xs we have d 2 E xs = −k 2 E xs dz 2 (38) (36)

where the wavenumber is a function of the material properties, as described by µ and : r (37)

An important feature of wave propagation in a dielectric is that k can be complexvalued, and as such it is referred to as the complex propagation constant. A general solution of (38), in fact, allows the possibility of a complex k, and it is customary to write it in terms of its real and imaginary parts in the following way: jk = α + jβ A solution to (38) will be: E xs = E x0 e− jkz = E x0 e−αz e− jβz (40) (39)

Multiplying (40) by e jωt and taking the real part yields a form of the ﬁeld that can be more easily visualized: E x = E x0 e−αz cos(ωt − βz) (41)

We recognize this as a uniform plane wave that propagates in the forward z direction with phase constant β, but which (for positive α) loses amplitude with increasing z according to the factor e−αz . Thus the general effect of a complex-valued k is to yield a traveling wave that changes its amplitude with distance. If α is positive, it is called the attenuation coefficien . If α is negative, the wave grows in amplitude with distance, and α is called the gain coefficien . The latter effect would occur, for example, in laser ampliﬁers. In the present and future discussions in this book, we will consider only passive media, in which one or more loss mechanisms are present, thus producing a positive α. The attenuation coefﬁcient is measured in nepers per meter (Np/m) so that the exponent of e can be measured in the dimensionless units of nepers. Thus, if α = 0.01 Np/m, the crest amplitude of the wave at z = 50 m will be e−0.5 /e−0 = 0.607 of its value at z = 0. In traveling a distance 1/α in the +z direction, the amplitude of the wave is reduced by the familiar factor of e−1 , or 0.368. The ways in which physical processes in a material can affect the wave electric ﬁeld are described through a complex permittivity of the form = −j =
0( r

−j

r

)

(42)

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Two important mechanisms that give rise to a complex permittivity (and thus result in wave losses) are bound electron or ion oscillations and dipole relaxation, both of which are discussed in Appendix E. An additional mechanism is the conduction of free electrons or holes, which we will explore at length in this chapter. Losses arising from the response of the medium to the magnetic ﬁeld can occur as well, and these are modeled through a complex permeability, µ = µ − jµ = µ0 (µr − jµr ). Examples of such media include ferrimagnetic materials, or ferrites. The magnetic response is usually very weak compared to the dielectric response in most materials of interest for wave propagation; in such materials µ ≈ µ0 . Consequently, our discussion of loss mechanisms will be conﬁned to those described through the complex permittivity, and we will assume that µ is entirely real in our treatment. We can substitute (42) into (37), which results in k = ω µ( − j )=ω µ 1− j (43)

Note the presence of the second radical factor in (43), which becomes unity (and real) as vanishes. With nonzero , k is complex, and so losses occur which are quantiﬁed through the attenuation coefﬁcient, α, in (39). The phase constant, β (and consequently the wavelength and phase velocity), will also be affected by . α and β are found by taking the real and imaginary parts of jk from (43). We obtain:  µ  α = Re{ jk} = ω 1+ 2  β = Im{ jk} = ω µ  1+ 2
2

1/2 − 1 1/2 + 1 (45) (44)

2

We see that a nonzero α (and hence loss) results if the imaginary part of the permittivity, , is present. We also observe in (44) and (45) the presence of the ratio / , which is called the loss tangent. The meaning of the term will be demonstrated when we investigate the speciﬁc case of conductive media. The practical importance of the ratio lies in its magnitude compared to unity, which enables simpliﬁcations to be made in (44) and (45). Whether or not losses occur, we see from (41) that the wave phase velocity is given by νp = ω β (46)

The wavelength is the distance required to effect a phase change of 2π radians βλ = 2π

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ENGINEERING ELECTROMAGNETICS

which leads to the fundamental deﬁnition of wavelength, λ= 2π β (47)

Because we have a uniform plane wave, the magnetic ﬁeld is found through E x0 −αz − jβz e e Hys = η where the intrinsic impedance is now a complex quantity, η= µ −j = µ √ 1 1 − j( / ) (48)

The electric and magnetic ﬁelds are no longer in phase. A special case is that of a lossless medium, or perfect dielectric, in which and so = . From (44), this leads to α = 0, and from (45), β=ω µ (lossless medium)

= 0, (49)

With α = 0, the real ﬁeld assumes the form E x = E x0 cos(ωt − βz) (50)

We may interpret this as a wave traveling in the +z direction at a phase velocity ν p , where νp = The wavelength is 2π 1 2π = √ = √ λ= β ω µ f µ ω 1 =√ β µ c f µr = c µr λ0 µr

r

=

r

=

(lossless medium) r (51)

where λ0 is the free space wavelength. Note that µr r > 1, and therefore the wavelength is shorter and the velocity is lower in all real media than they are in free space. Associated with E x is the magnetic ﬁeld intensity Hy = where the intrinsic impedance is η= µ (52) E x0 cos(ωt − βz) η

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The two ﬁelds are once again perpendicular to each other, perpendicular to the direction of propagation, and in phase with each other everywhere. Note that when E is crossed into H, the resultant vector is in the direction of propagation. We shall see the reason for this when we discuss the Poynting vector.

E X A M P L E 11.3

Let us apply these results to a 1-MHz plane wave propagating in fresh water. At this frequency, losses in water are negligible, which means that we can assume that . = 0. In water, µr = 1 and at 1 MHz, r = 81.
Solution. We begin by calculating the phase constant. Using (45) with

have β=ω µ √ = ω µ0 r = 0, we

0

=

ω c

r

√ 2π × 106 81 = 0.19 rad/m = 3.0 × 108

Using this result, we can determine the wavelength and phase velocity: λ= νp = 2π 2π = = 33 m β .19

2π × 106 ω = = 3.3 × 107 m/s β .19

The wavelength in air would have been 300 m. Continuing our calculations, we ﬁnd the intrinsic impedance using (48) with = 0: η= µ = η0 r =

377 = 42 9

If we let the electric ﬁeld intensity have a maximum amplitude of 0.1 V/m, then E x = 0.1 cos(2π106 t − .19z) V/m Hy =

Ex = (2.4 × 10−3 ) cos(2π 106 t − .19z) A/m η

D11.3. A 9.375-GHz uniform plane wave is propagating in polyethylene (see Appendix C). If the amplitude of the electric ﬁeld intensity is 500 V/m and the material is assumed to be lossless, ﬁnd: (a) the phase constant; (b) the wavelength in the polyethylene; (c) the velocity of propagation; (d) the intrinsic impedance; (e) the amplitude of the magnetic ﬁeld intensity.
Ans. 295 rad/m; 2.13 cm; 1.99 × 108 m/s; 251 ; 1.99 A/m

380

ENGINEERING ELECTROMAGNETICS

E X A M P L E 11.4

We again consider plane wave propagation in water, but at the much higher microwave frequency of 2.5 GHz. At frequencies in this range and higher, dipole relaxation and resonance phenomena in the water molecules become important.2 Real and imaginary parts of the permittivity are present, and both vary with frequency. At frequencies below that of visible light, the two mechanisms together produce a value of that increases with increasing frequency, reaching a maximum in the vicinity of 1013 Hz. decreases with increasing frequency, reaching a minimum also in the vicinity of 1013 Hz. Reference 3 provides speciﬁc details. At 2.5 GHz, dipole relaxation effects dominate. The permittivity values are r = 78 and r = 7. From (44), we have 1/2 √  (2π × 2.5 × 109 ) 78  7 2 α= 1+ − 1 = 21 Np/m √ 78 (3.0 × 108 ) 2 This ﬁrst calculation demonstrates the operating principle of the microwave oven. Almost all foods contain water, and so they can be cooked when incident microwave radiation is absorbed and converted into heat. Note that the ﬁeld will attenuate to a value of e−1 times its initial value at a distance of 1/α = 4.8 cm. This distance is called the penetration depth of the material, and of course it is frequency-dependent. The 4.8 cm depth is reasonable for cooking food, since it would lead to a temperature rise that is fairly uniform throughout the depth of the material. At much higher frequencies, where is larger, the penetration depth decreases, and too much power is absorbed at the surface; at lower frequencies, the penetration depth increases, and not enough overall absorption occurs. Commercial microwave ovens operate at frequencies in the vicinity of 2.5 GHz. Using (45), in a calculation very similar to that for α, we ﬁnd β = 464 rad/m. The wavelength is λ = 2π/β = 1.4 cm, whereas in free space this would have been λ0 = c/ f = 12 cm. Using (48), the intrinsic impedance is found to be 1 377 η= √ √ = 43 + j1.9 = 43 2.6◦ 1 − j(7/78) 78 and E x leads Hy in time by 2.6◦ at every point.

We next consider the case of conductive materials, in which currents are formed by the motion of free electrons or holes under the inﬂuence of an electric ﬁeld. The governing relation is J = σ E, where σ is the material conductivity. With ﬁnite conductivity, the wave loses power through resistive heating of the material. We look for an interpretation of the complex permittivity as it relates to the conductivity.

2

These mechanisms and how they produce a complex permittivity are described in Appendix D. Additionally, the reader is referred to pp. 73–84 in Reference 1 and pp. 678–82 in Reference 2 for general treatments of relaxation and resonance effects on wave propagation. Discussions and data that are speciﬁc to water are presented in Reference 3, pp. 314–16.

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Consider the Maxwell curl equation (23) which, using (42), becomes: ∇ × Hs = jω( − j )Es = ω Es + jω Es (53)

This equation can be expressed in a more familiar way, in which conduction current is included: We next use Js = σ Es , and interpret in (54) as . The latter equation becomes: ∇ × Hs = (σ + jω )Es = Jσ s + Jds (55) ∇ × Hs = Js + jω Es (54)

which we have expressed in terms of conduction current density, Jσ s = σ Es , and displacement current density, Jds = jω Es . Comparing Eqs. (53) and (55), we ﬁnd that in a conductive medium: = σ ω (56)

Let us now turn our attention to the case of a dielectric material in which the loss is very small. The criterion by which we should judge whether or not the loss is small is the magnitude of the loss tangent, / . This parameter will have a direct inﬂuence on the attenuation coefﬁcient, α, as seen from Eq. (44). In the case of conducting media, to which (56) applies, the loss tangent becomes σ/ω . By inspecting (55), we see that the ratio of conduction current density to displacement current density magnitudes is Jσ s = Jds j = σ jω (57)

That is, these two vectors point in the same direction in space, but they are 90◦ out of phase in time. Displacement current density leads conduction current density by 90◦ , just as the current through a capacitor leads the current through a resistor in parallel with it by 90◦ in an ordinary electric circuit. This phase relationship is shown in Figure 11.2. The angle θ (not to be confused with the polar angle in spherical coordinates) may therefore be identiﬁed as the angle by which the displacement current density leads the total current density, and tan θ = = σ ω (58)

The reasoning behind the term loss tangent is thus evident. Problem 11.16 at the end of the chapter indicates that the Q of a capacitor (its quality factor, not its charge) that incorporates a lossy dielectric is the reciprocal of the loss tangent. If the loss tangent is small, then we may obtain useful approximations for the attenuation and phase constants, and the intrinsic impedance. The criterion for a small loss tangent is / 1, which we say identiﬁes the medium as a good dielectric.

382

ENGINEERING ELECTROMAGNETICS

J

′E

′

J

′ E

J

E

Figure 11.2 The time-phase relationship between Jds , Jσ s , Js , and Es . The tangent of θ is equal to σ/ω , and 90◦ − θ is the common power-factor angle, or the angle by which Js leads Es .

Considering a conductive material, for which jk = jω µ

= σ/ω, (43) becomes 1− j (59)

σ ω We may expand the second radical using the binomial theorem n(n − 1) 2 n(n − 1)(n − 2) 3 x + x + ··· (1 + x)n = 1 + nx + 2! 3! where |x| 1. We identify x as − jσ/ω and n as 1/2, and thus jk = jω µ Now, for a good dielectric, . α = Re( jk) = jω µ and . β = Im( jk) = ω µ 1+ 1 σ 8 ω
2

1− j

σ 2ω

+

1 σ 8 ω

2

+ · · · = α + jβ

−j

σ 2ω

=

σ 2

µ

(60a)

(60b)

Equations (60a) and (60b) can be compared directly with the transmission line α and β under low-loss conditions, as expressed in Eqs. (54a) and (55b) in Chapter 10. In this comparison, we associate σ with G, µ with L, and with C. Note that in plane wave propagation in media with no boundaries, there can be no quantity that is analogous to the transmission line conductor resistance parameter, R. In many cases,

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383

the second term in (60b) is small enough, so that . β = ω µ Applying the binomial expansion to (48), we obtain, for a good dielectric . η = or . η = µ 1+ j σ 2ω (62b) µ 1− 3 σ 8 ω
2

(61)

+ j

σ 2ω

(62a)

The conditions under which these approximations can be used depend on the desired accuracy, measured by how much the results deviate from those given by the exact formulas, (44) and (45). Deviations of no more than a few percent occur if σ/ω < 0.1.
E X A M P L E 11.5

As a comparison, we repeat the computations of Example 11.4, using the approximation formulas (60a), (61), and (62b).
Solution. First, the loss tangent in this case is

/

with

= σ/ω, we have . ω α = 2 µ =

= 7/78 = 0.09. Using (60),

1 377 (7 × 8.85 × 1012 )(2π × 2.5 × 109 ) √ = 21 cm−1 2 78

We then have, using (61b), √ . β = (2π × 2.5 × 109 ) 78/(3 × 108 ) = 464 rad/m Finally, with (62b), 7 . 377 1+ j η = √ 2 × 78 78 = 43 + j1.9

These results are identical (within the accuracy limitations as determined by the given numbers) to those of Example 11.4. Small deviations will be found, as the reader can verify by repeating the calculations of both examples and expressing the results to four or ﬁve signiﬁcant ﬁgures. As we know, this latter practice would not be meaningful because the given parameters were not speciﬁed with such accuracy. Such is often the case, since measured values are not always known with high precision. Depending on how precise these values are, one can sometimes use a more relaxed judgment on when the approximation formulas can be used by allowing loss tangent values that can be larger than 0.1 (but still less than 1).

384

ENGINEERING ELECTROMAGNETICS

D11.4. Given a nonmagnetic material having r = 3.2 and σ = 1.5 × 10−4 S/m, ﬁnd numerical values at 3 MHz for the (a) loss tangent; (b) attenuation constant; (c) phase constant; (d) intrinsic impedance.
Ans. 0.28; 0.016 Np/m; 0.11 rad/m; 207 7.8◦

D11.5. Consider a material for which µr = 1, r = 2.5, and the loss tangent is 0.12. If these three values are constant with frequency in the range 0.5 MHz ≤ f ≤ 100 MHz, calculate: (a) σ at 1 and 75 MHz; (b) λ at 1 and 75 MHz; (c) νp at 1 and 75 MHz.
Ans. 1.67 × 10−5 and 1.25 × 10−3 S/m; 190 and 2.53 m; 1.90 × 108 m/s twice

11.3 POYNTING’S THEOREM AND WAVE POWER
In order to ﬁnd the power ﬂow associated with an electromagnetic wave, it is necessary to develop a power theorem for the electromagnetic ﬁeld known as the Poynting theorem. It was originally postulated in 1884 by an English physicist, John H. Poynting. The development begins with one of Maxwell’s curl equations, in which we assume that the medium may be conductive: ∂D (63) ∇ ×H=J+ ∂t Next, we take the scalar product of both sides of (63) with E, ∂D (64) E·∇ × H = E·J + E· ∂t We then introduce the following vector identity, which may be proved by expansion in rectangular coordinates: ∇ · (E × H) = −E · ∇ × H + H · ∇ × E Using (65) in the left side of (64) results in ∂D (66) ∂t where the curl of the electric ﬁeld is given by the other Maxwell curl equation: ∂B ∇ ×E=− ∂t Therefore ∂B ∂D −H · − ∇ · (E × H) = J · E + E · ∂t ∂t or ∂H ∂E + µH · (67) −∇ · (E × H) = J · E + E · ∂t ∂t H · ∇ × E − ∇ · (E × H) = J · E + E · (65)

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The two time derivatives in (67) can be rearranged as follows: ∂E ∂ 1 E· = D·E ∂t ∂t 2 and ∂H ∂ 1 µH · = B·H ∂t ∂t 2 With these, Eq. (67) becomes −∇ · (E × H) = J · E + ∂ ∂t 1 ∂ D·E + 2 ∂t 1 B·H 2

(68a)

(68b)

(69)

Finally, we integrate (69) throughout a volume: ∂ 1 ∇ · (E × H) dv = J · E dv + D · E dv + − 2 vol vol vol ∂t

vol

∂ ∂t

1 B · H dv 2

The divergence theorem is then applied to the left-hand side, thus converting the volume integral there into an integral over the surface that encloses the volume. On the right-hand side, the operations of spatial integration and time differentiation are interchanged. The ﬁnal result is − (E × H) · dS = J · E dν + d dt 1 d D · E dν + 2 dt 1 B · H dν 2 (70)

area

vol

vol

vol

Equation (70) is known as Poynting’s theorem. On the right-hand side, the ﬁrst integral is the total (but instantaneous) ohmic power dissipated within the volume. The second integral is the total energy stored in the electric ﬁeld, and the third integral is the stored energy in the magnetic ﬁeld.3 Since time derivatives are taken of the second and third integrals, those results give the time rates of increase of energy stored within the volume, or the instantaneous power going to increase the stored energy. The sum of the expressions on the right must therefore be the total power ﬂowing into this volume, and so the total power ﬂowing out of the volume is (E × H) · dS W (71)

area

where the integral is over the closed surface surrounding the volume. The cross product E × H is known as the Poynting vector, S, S=E×H W/m2 (72)

which is interpreted as an instantaneous power density, measured in watts per square meter (W/m2 ). The direction of the vector S indicates the direction of the instantaneous
3

This is the expression for magnetic ﬁeld energy that we have been anticipating since Chapter 8.

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ENGINEERING ELECTROMAGNETICS

power ﬂow at a point, and many of us think of the Poynting vector as a “pointing” vector. This homonym, while accidental, is correct.4 Because S is given by the cross product of E and H, the direction of power ﬂow at any point is normal to both the E and H vectors. This certainly agrees with our experience with the uniform plane wave, for propagation in the +z direction was associated with an E x and Hy component, E x ax × Hy a y = Sz az

In a perfect dielectric, the E and H ﬁeld amplitudes are given by E x = E x0 cos(ωt − βz) Hy = E x0 cos(ωt − βz) η

where η is real. The power density amplitude is therefore Sz =
2 E x0 cos2 (ωt − βz) η

(73)

In the case of a lossy dielectric, E x and Hy are not in time phase. We have E x = E x0 e−αz cos(ωt − βz) If we let η = |η| θη then we may write the magnetic ﬁeld intensity as Hy = Thus, Sz = E x Hy = E x0 −αz cos(ωt − βz − θη ) e |η| (74)

Because we are dealing with a sinusoidal signal, the time-average power density, Sz , is the quantity that will ultimately be measured. To ﬁnd this, we integrate (74) over one cycle and divide by the period T = 1/ f . Additionally, the identity cos A cos B = 1/2 cos(A + B) + 1/2 cos(A − B) is applied to the integrand, and we obtain: Sz = 1 T
T 0 2 1 E x0 −2αz [cos(2ωt − 2βz − 2θη ) + cos θη ] dt e 2 |η|

2 E x0 −2αz cos(ωt − βz) cos(ωt − βz − θη ) e |η|

(75)

4 Note that the vector symbol S is used for the Poynting vector, and is not to be confused with the differential area vector, dS. The latter, as we know, is the product of the outward normal to the surface and the differential area.

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The second-harmonic component of the integrand in (75) integrates to zero, leaving only the contribution from the dc component. The result is Sz =
2 1 E x0 −2αz cos θη e 2 |η|

(76)

Note that the power density attenuates as e−2αz , whereas E x and Hy fall off as e−αz . We may ﬁnally observe that the preceding expression can be obtained very easily by using the phasor forms of the electric and magnetic ﬁelds. In vector form, this is S = In the present case Es = E x0 e− jβz ax and H∗ = s E x0 + jβz E x0 jθ + jβz e ay = ay e e ∗ η |η| 1 Re(Es × H∗ ) s 2 W/m2 (77)

where E x0 has been assumed real. Eq. (77) applies to any sinusoidal electromagnetic wave and gives both the magnitude and direction of the time-average power density. D11.6. At frequencies of 1, 100, and 3000 MHz, the dielectric constant of ice made from pure water has values of 4.15, 3.45, and 3.20, respectively, while the loss tangent is 0.12, 0.035, and 0.0009, also respectively. If a uniform plane wave with an amplitude of 100 V/m at z = 0 is propagating through such ice, ﬁnd the time-average power density at z = 0 and z = 10 m for each frequency.
Ans. 27.1 and 25.7 W/m2 ; 24.7 and 6.31 W/m2 ; 23.7 and 8.63 W/m2

11.4 PROPAGA TION IN GOOD CONDUCTORS: SKIN EFFECT
As an additional study of propagation with loss, we will investigate the behavior of a good conductor when a uniform plane wave is established in it. Such a material satisﬁes the general high-loss criterion, in which the loss tangent, / 1. Applying this to a good conductor leads to the more speciﬁc criterion, σ/(ω ) 1. As before, we have an interest in losses that occur on wave transmission into a good conductor, and we will ﬁnd new approximations for the phase constant, attenuation coefﬁcient, and intrinsic impedance. New to us, however, is a modiﬁcation of the basic problem, appropriate for good conductors. This concerns waves associated with electromagnetic ﬁelds existing in an external dielectric that adjoins the conductor surface; in this case, the waves propagate along the surface. That portion of the overall ﬁeld that exists within the conductor will suffer dissipative loss arising from the conduction currents it generates. The overall ﬁeld therefore attenuates with increasing distance

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of travel along the surface. This is the mechanism for the resistive transmission line loss that we studied in Chapter 10, and which is embodied in the line resistance parameter, R. As implied, a good conductor has a high conductivity and large conduction currents. The energy represented by the wave traveling through the material therefore decreases as the wave propagates because ohmic losses are continuously present. When we discussed the loss tangent, we saw that the ratio of conduction current density to the displacement current density in a conducting material is given by σ/ω . Choosing a poor metallic conductor and a very high frequency as a conservative . example, this ratio5 for nichrome (σ = 106 ) at 100 MHz is about 2 × 108 . We therefore have a situation where σ/ω 1, and we should be able to make several very good approximations to ﬁnd α, β, and η for a good conductor. The general expression for the propagation constant is, from (59), jk = jω µ which we immediately simplify to obtain jk = jω µ or jk = j − jωµσ But − j = 1 −90◦ and √ 1 1 −90◦ = 1 −45◦ = √ (1 − j) 2 Therefore jk = j(1 − j) Hence α=β= π f µσ (79) ωµσ = (1 + j) π f µσ = α + jβ 2 (78) −j σ ω 1− j σ ω

Regardless of the parameters µ and σ of the conductor or of the frequency of the applied ﬁeld, α and β are equal. If we again assume only an E x component traveling in the +z direction, then E x = E x0 e−z
√ π f µσ

cos ωt − z π f µσ

(80)

5

It is customary to take

=

0

for metallic conductors.

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We may tie this ﬁeld in the conductor to an external ﬁeld at the conductor surface. We let the region z > 0 be the good conductor and the region z < 0 be a perfect dielectric. At the boundary surface z = 0, (80) becomes E x = E x0 cos ωt (z = 0) This we shall consider as the source ﬁeld that establishes the ﬁelds within the conductor. Since displacement current is negligible, J = σE Thus, the conduction current density at any point within the conductor is directly related to E: Jx = σ E x = σ E x0 e−z
√ π fµσ

cos ωt − z π f µσ

(81)

Equations (80) and (81) contain a wealth of information. Considering ﬁrst the negative exponential term, we ﬁnd an exponential decrease in the conduction current density and electric ﬁeld intensity with penetration into the conductor (away from the source). The exponential factor is unity at z = 0 and decreases to e−1 = 0.368 when z=√ 1 π f µσ

This distance is denoted by δ and is termed the depth of penetration, or the skin depth, δ=√ 1 1 1 = = α β π f µσ (82)

It is an important parameter in describing conductor behavior in electromagnetic ﬁelds. To get some idea of the magnitude of the skin depth, let us consider copper, σ = 5.8 × 107 S/m, at several different frequencies. We have 0.066 δCu = √ f

At a power frequency of 60 Hz, δCu = 8.53 mm. Remembering that the power density carries an exponential term e−2αz , we see that the power density is multiplied by a factor of 0.3682 = 0.135 for every 8.53 mm of distance into the copper. At a microwave frequency of 10,000 MHz, δ is 6.61 × 10−4 mm. Stated more generally, all ﬁelds in a good conductor such as copper are essentially zero at distances greater than a few skin depths from the surface. Any current density or electric ﬁeld intensity established at the surface of a good conductor decays rapidly as we progress into the conductor. Electromagnetic energy is not transmitted in the interior of a conductor; it travels in the region surrounding the conductor, while the conductor merely guides the waves. We will consider guided propagation in more detail in Chapter 13. Suppose we have a copper bus bar in the substation of an electric utility company which we wish to have carry large currents, and we therefore select dimensions of 2 by 4 inches. Then much of the copper is wasted, for the ﬁelds are greatly reduced in

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one skin depth, about 8.5 mm.6 A hollow conductor with a wall thickness of about 12 mm would be a much better design. Although we are applying the results of an analysis for an inﬁnite planar conductor to one of ﬁnite dimensions, the ﬁelds are attenuated in the ﬁnite-size conductor in a similar (but not identical) fashion. The extremely short skin depth at microwave frequencies shows that only the surface coating of the guiding conductor is important. A piece of glass with an evaporated silver surface 3 µm thick is an excellent conductor at these frequencies. Next, let us determine expressions for the velocity and wavelength within a good conductor. From (82), we already have α=β= Then, as β= we ﬁnd the wavelength to be λ = 2π δ Also, recalling that νp = we have νp = ωδ (84) ω β (83) 2π λ 1 = δ π f µσ

For copper at 60 Hz, λ = 5.36 cm and νp = 3.22 m/s, or about 7.2 mi/h! A lot of us can run faster than that. In free space, of course, a 60 Hz wave has a wavelength of 3100 mi and travels at the velocity of light.
E X A M P L E 11.6

Let us again consider wave propagation in water, but this time we will consider seawater. The primary difference between seawater and fresh water is of course the salt content. Sodium chloride dissociates in water to form Na+ and Cl− ions, which, being charged, will move when forced by an electric ﬁeld. Seawater is thus conductive, and so it will attenuate electromagnetic waves by this mechanism. At frequencies in the vicinity of 107 Hz and below, the bound charge effects in water discussed earlier are negligible, and losses in seawater arise principally from the salt-associated conductivity. We consider an incident wave of frequency 1 MHz. We wish to ﬁnd the skin depth, wavelength, and phase velocity. In seawater, σ = 4 S/m, and r = 81.

6

This utility company operates at 60 Hz.

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Solution. We ﬁrst evaluate the loss tangent, using the given data:

σ 4 = = 8.9 × 102 6 )(81)(8.85 × 10−12 ) ω (2π × 10

1

Seawater is therefore a good conductor at 1 MHz (and at frequencies lower than this). The skin depth is δ=√ Now λ = 2πδ = 1.6 m and νp = ωδ = (2π × 106 )(0.25) = 1.6 × 106 m/sec In free space, these values would have been λ = 300 m and of course ν = c. With a 25-cm skin depth, it is obvious that radio frequency √ communication in seawater is quite impractical. Notice, however, that δ varies as 1/ f , so that things will improve at lower frequencies. For example, if we use a frequency of 10 Hz (in the ELF, or extremely low frequency range), the skin depth is increased over that at 1 MHz by a factor of 106 /10, so that . δ(10 Hz) = 80 m . The corresponding wavelength is λ = 2πδ = 500 m. Frequencies in the ELF range were used for many years in submarine communications. Signals were transmitted from gigantic ground-based antennas (required because the free-space wavelength associated with 10 Hz is 3 × 107 m). The signals were then received by submarines, from which a suspended wire antenna of length shorter than 500 m is sufﬁcient. The drawback is that signal data rates at ELF are slow enough that a single word can take several minutes to transmit. Typically, ELF signals would be used to tell the submarine to initiate emergency procedures, or to come near the surface in order to receive a more detailed message via satellite. We next turn our attention to ﬁnding the magnetic ﬁeld, Hy , associated with E x . To do so, we need an expression for the intrinsic impedance of a good conductor. We begin with Eq. (48), Section 11.2, with = σ/ω, η= Since σ ω , we have η= jωµ σ jωµ σ + jω 1 = π f µσ 1 (π × 106 )(4π × 10−7 )(4) = 0.25 m = 25 cm

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which may be written as √ 2 45◦ (1 + j) η= = σδ σδ Thus, if we write (80) in terms of the skin depth, E x = E x0 e−z/δ cos ωt − then Hy = z δ (85)

(86)

σ δ E x0 −z/δ z π cos ωt − − (87) √ e δ 4 2 and we see that the maximum amplitude of the magnetic ﬁeld intensity occurs oneeighth of a cycle later than the maximum amplitude of the electric ﬁeld intensity at every point. From (86) and (87) we may obtain the time-average Poynting vector by applying (77), Sz = or Sz = 1 2 σ δ E x0 e−2z/δ 4
2 1 σ δ E x0 −2z/δ π cos √ e 2 4 2

We again note that in a distance of one skin depth the power density is only e−2 = 0.135 of its value at the surface. The total average power loss in a width 0 < y < b and length 0 < x < L in the direction of the current, as shown in Figure 11.3, is obtained by ﬁnding the power

Figure 11.3 The current density Jx = Jx0 e−z/δ e− j z/δ decreases in magnitude as the wave propagates into the conductor. The average power loss in the region 0 < x < L , 0 < y < b, z > 0, 2 is δbL Jx0 /4σ watts.

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crossing the conductor surface within this area, 1 2 σ δ E x0 e−2z/δ 0 0 4 area In terms of the current density Jx0 at the surface, PL = Sz da = Jx0 = σ E x0 we have PL = 1 2 δbL Jx0 4σ (88) b L z=0

dx dy =

1 2 σ δbL E x0 4

Now let us see what power loss would result if the total current in a width b were distributed uniformly in one skin depth. To ﬁnd the total current, we integrate the current density over the inﬁnite depth of the conductor, I = where
∞ 0 0 b

Jx dy dz

z δ or in complex exponential notation to simplify the integration, Jx = Jx0 e−z/δ cos ωt − Jxs = Jx0 e−z/δ e− j z/δ = Jx0 e−(1+ j)z/δ b 0

Therefore, Is =
∞ 0

Jx0 e−(1+ j)z/δ dy dz −δ 1+ j
∞ 0

= Jx0 be−(1+ j)z/δ Jx0 bδ = 1+ j

and

Jx0 bδ π I = √ cos ωt − 4 2 If this current is distributed with a uniform density J throughout the cross section 0 < y < b, 0 < z < δ, then Jx0 π J = √ cos ωt − 4 2 The ohmic power loss per unit volume is J · E, and thus the total instantaneous power dissipated in the volume under consideration is PLi (t) = J2 1 π (J )2 bLδ = x0 bLδ cos2 ωt − σ 2σ 4

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The time-average power loss is easily obtained, since the average value of the cosinesquared factor is one-half, PL = 1 2 J bLδ 4σ x0 (89)

Comparing (88) and (89), we see that they are identical. Thus the average power loss in a conductor with skin effect present may be calculated by assuming that the total current is distributed uniformly in one skin depth. In terms of resistance, we may say that the resistance of a width b and length L of an inﬁnitely thick slab with skin effect is the same as the resistance of a rectangular slab of width b, length L, and thickness δ without skin effect, or with uniform current distribution. We may apply this to a conductor of circular cross section with little error, provided that the radius a is much greater than the skin depth. The resistance at a high frequency where there is a well-developed skin effect is therefore found by considering a slab of width equal to the circumference 2πa and thickness δ. Hence R= L L = σS 2πaσ δ (90)

At 1 MHz, the skin depth is 0.066 mm. Thus δ found by (90), R=

A round copper wire of 1 mm radius and 1 km length has a resistance at direct current of 103 = 5.48 Rdc = π 10−6 (5.8 × 107 ) a, and the resistance at 1 MHz is

103 = 41.5 2π 10−3 (5.8 × 107 )(0.066 × 10−3 )

D11.7. A steel pipe is constructed of a material for which µr = 180 and σ = 4 × 106 S/m. The two radii are 5 and 7 mm, and the length is 75 m. If the total current I (t) carried by the pipe is 8 cos ωt A, where ω = 1200π rad/s, ﬁnd: (a) the skin depth; (b) the effective resistance; (c) the dc resistance; (d) the time-average power loss.
Ans. 0.766 mm; 0.557 ; 0.249 ; 17.82 W

11.5 WAVE POLARIZATION
In the previous sections, we have treated uniform plane waves in which the electric and magnetic ﬁeld vectors were assumed to lie in ﬁxed directions. Speciﬁcally, with the wave propagating along the z axis, E was taken to lie along x, which then required H to lie along y. This orthogonal relationship between E, H, and S is always true for a uniform plane wave. The directions of E and H within the plane perpendicular to az

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may change, however, as functions of time and position, depending on how the wave was generated or on what type of medium it is propagating through. Thus a complete description of an electromagnetic wave would not only include parameters such as its wavelength, phase velocity, and power, but also a statement of the instantaneous orientation of its ﬁeld vectors. We deﬁne the wave polarization as the time-dependent electric ﬁeld vector orientation at a ﬁxed point in space. A more complete characterization of a wave’s polarization would in fact include specifying the ﬁeld orientation at all points because some waves demonstrate spatial variations in their polarization. Specifying only the electric ﬁeld direction is sufﬁcient, since magnetic ﬁeld is readily found from E using Maxwell’s equations. In the waves we have previously studied, E was in a ﬁxed straight orientation for all times and positions. Such a wave is said to be linearly polarized. We have taken E to lie along the x axis, but the ﬁeld could be oriented in any ﬁxed direction in the x y plane and be linearly polarized. For positive z propagation, the wave would in general have its electric ﬁeld phasor expressed as Es = (E x0 ax + E y0 a y )e−αz e− jβz (91)

where E x0 and E y0 are constant amplitudes along x and y. The magnetic ﬁeld is readily found by determining its x and y components directly from those of Es . Speciﬁcally, Hs for the wave of Eq. (91) is Hs = [Hx0 ax + Hy0 a y ] e−αz e − jβz = − E y0 E x0 ax + a y e−αz e− jβz η η (92)

The two ﬁelds are sketched in Figure 11.4. The ﬁgure demonstrates the reason for the minus sign in the term involving E y0 in Eq. (92). The direction of power ﬂow, given by E × H, is in the positive z direction in this case. A component of E in the

Figure 11.4 Electric and magnetic field configuration for a general linearly polarized plane wave propagating in the forward z direction (out of the page). Field components correspond to those in Eqs. (91) and (92).

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ENGINEERING ELECTROMAGNETICS

positive y direction would require a component of H in the negative x direction—thus the minus sign. Using (91) and (92), the power density in the wave is found using (77): 1 1 ∗ ∗ Sz = Re{Es × H∗ } = Re{E x0 Hy0 (ax × a y ) + E y0 Hx0 (a y × ax )}e−2αz s 2 2 ∗ ∗ E y0 E y0 −2αz E x0 E x0 1 e = Re + az 2 η∗ η∗ 1 1 = Re ∗ (|E x0 |2 + |E y0 |2 )e−2αz az W/m2 2 η This result demonstrates the idea that our linearly polarized plane wave can be considered as two distinct plane waves having x and y polarizations, whose electric ﬁelds are combining in phase to produce the total E. The same is true for the magnetic ﬁeld components. This is a critical point in understanding wave polarization, in that any polarization state can be described in terms of mutually perpendicular components of the electric fiel and their relative phasing. We next consider the effect of a phase difference, φ, between E x0 and E y0 , where φ < π/2. For simplicity, we will consider propagation in a lossless medium. The total ﬁeld in phasor form is Es = (E x0 ax + E y0 e jφ a y )e− jβz

(93)

Again, to aid in visualization, we convert this wave to real instantaneous form by multiplying by e jωt and taking the real part: E(z, t) = E x0 cos(ωt − βz)ax + E y0 cos(ωt − βz + φ)a y (94)

where we have assumed that E x0 and E y0 are real. Suppose we set t = 0, in which case (94) becomes [using cos(−x) = cos(x)] The component magnitudes of E(z, 0) are plotted as functions of z in Figure 11.5. Since time is ﬁxed at zero, the wave is frozen in position. An observer can move along the z axis, measuring the component magnitudes and thus the orientation of the total electric ﬁeld at each point. Let’s consider a crest of E x , indicated as point a in Figure 11.5. If φ were zero, E y would have a crest at the same location. Since φ is not zero (and positive), the crest of E y that would otherwise occur at point a is now displaced to point b farther down z. The two points are separated by distance φ/β. E y thus lags behind E x when we consider the spatial dimension. Now suppose the observer stops at some location on the z axis, and time is allowed to move forward. Both ﬁelds now move in the positive z direction, as (94) indicates. But point b reaches the observer ﬁrst, followed by point a. So we see that E y leads E x when we consider the time dimension. In either case (ﬁxed t and varying z, or vice versa) the observer notes that the net ﬁeld rotates about the z axis while its magnitude changes. Considering a starting point in z and t, at which the ﬁeld has a given orientation and magnitude, the wave will return to the same orientation and E(z, 0) = E x0 cos(βz)ax + E y0 cos(βz − φ)a y (95)

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Figure 11.5 Plots of the electric field component magnitudes in Eq. (95) as functions of z. Note that the y component lags behind the x component in z. As time increases from zero, both waves travel to the right, as per Eq. (94). Thus, to an observer at a fixed location, the y component leads in time.

magnitude at a distance of one wavelength in z (for ﬁxed t) or at a time t = 2π/ω later (at a ﬁxed z). For illustration purposes, if we take the length of the ﬁeld vector as a measure of its magnitude, we ﬁnd that at a ﬁxed position, the tip of the vector traces out the shape of an ellipse over time t = 2π/ω. The wave is said to be elliptically polarized. Elliptical polarization is in fact the most general polarization state of a wave, since it encompasses any magnitude and phase difference between E x and E y . Linear polarization is a special case of elliptical polarization in which the phase difference is zero. Another special case of elliptical polarization occurs when E x0 = E y0 = E 0 and when φ = ±π/2. The wave in this case exhibits circular polarization. To see this, we incorporate these restrictions into Eq. (94) to obtain E(z, t) = E 0 [cos(ωt − βz)ax + cos(ωt − βz ± π/2)a y ] = E 0 [cos(ωt − βz)ax ∓ sin(ωt − βz)a y ] (96)

If we consider a ﬁxed position along z (such as z = 0) and allow time to vary, (96), with φ = +π/2, becomes E(0, t) = E 0 [cos(ωt)ax − sin(ωt)a y ] If we choose −π/2 in (96), we obtain E(0, t) = E 0 [cos(ωt)ax + sin(ωt)a y ] (98) (97)

The ﬁeld vector of Eq. (98) rotates in the counterclockwise direction in the x y plane, while maintaining constant amplitude E 0 , and so the tip of the vector traces out a circle. Figure 11.6 shows this behavior.

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ENGINEERING ELECTROMAGNETICS

Figure 11.6 Electric field in the xy plane of a right circularly polarized plane wave, as described by Eq. (98). As the wave propagates in the forward z direction, the field vector rotates counterclockwise in the xy plane.

Choosing +π/2 leads to (97), whose ﬁeld vector rotates in the clockwise direction. The handedness of the circular polarization is associated with the rotation and propagation directions in the following manner: The wave exhibits left circular polarization (l.c.p.) if, when orienting the left hand with the thumb in the direction of propagation, the ﬁngers curl in the rotation direction of the ﬁeld with time. The wave exhibits right circular polarization (r.c.p.) if, with the right-hand thumb in the propagation direction, the ﬁngers curl in the ﬁeld rotation direction.7 Thus, with forward z propagation, (97) describes a left circularly polarized wave, and (98) describes a right circularly polarized wave. The same convention is applied to elliptical polarization, in which the descriptions left elliptical polarization and right elliptical polarization are used. Using (96), the instantaneous angle of the ﬁeld from the x direction can be found for any position along z through θ(z, t) = tan−1 Ey Ex = tan−1 ∓sin(ωt − βz) cos(ωt − βz) = ∓(ωt − βz) (99)

where again the minus sign (yielding l.c.p. for positive z travel) applies for the choice of φ = +π/2 in (96); the plus sign (yielding r.c.p. for positive z travel) is used if
7

This convention is reversed by some workers (most notably in optics) who emphasize the importance of the spatial ﬁeld conﬁguration. Note that r.c.p. by our deﬁnition is formed by propagating a spatial ﬁeld that is in the shape of a left-handed screw, and for that reason it is sometimes called left circular polarization (see Figure 11.7). Left circular polarization as we deﬁne it results from propagating a spatial ﬁeld in the shape of a right-handed screw, and it is called right circular polarization by the spatial enthusiasts. Caution is obviously necessary in interpreting what is meant when polarization handedness is stated in an unfamiliar text.

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Figure 11.7 Representation of a right circularly polarized wave. The electric field vector (in white) will rotate toward the y axis as the entire wave moves through the xy plane in the direction of k. This counterclockwise rotation (when looking toward the wave source) satisfies the temporal right-handed rotation convention as described in the text. The wave, however, appears as a left-handed screw, and for this reason it is called left circular polarization in the other convention.

φ = −π/2. If we choose z = 0, the angle becomes simply ωt, which reaches 2π (one complete rotation) at time t = 2π/ω. If we choose t = 0 and allow z to vary, we form a corkscrew-like ﬁeld pattern. One way to visualize this is to consider a spiral staircase–shaped pattern, in which the ﬁeld lines (stairsteps) are perpendicular to the z (or staircase) axis. The relationship between this spatial ﬁeld pattern and the resulting time behavior at ﬁxed z as the wave propagates is shown in an artist’s conception in Figure 11.7. The handedness of the polarization is changed by reversing the pitch of the corkscrew pattern. The spiral staircase model is only a visualization aid. It must be remembered that the wave is still a uniform plane wave whose ﬁelds at any position along z are inﬁnite in extent over the transverse plane. There are many uses of circularly polarized waves. Perhaps the most obvious advantage is that reception of a wave having circular polarization does not depend on the antenna orientation in the plane normal to the propagation direction. Dipole antennas, for example, are required to be oriented along the electric ﬁeld direction of the signal they receive. If circularly polarized signals are transmitted, the receiver orientation requirements are relaxed considerably. In optics, circularly polarized light

400

ENGINEERING ELECTROMAGNETICS

can be passed through a polarizer of any orientation, thus yielding linearly polarized light in any direction (although one loses half the original power this way). Other uses involve treating linearly polarized light as a superposition of circularly polarized waves, to be described next. Circularly polarized light can be generated using an anisotropic medium—a material whose permittivity is a function of electric ﬁeld direction. Many crystals have this property. A crystal orientation can be found such that along one direction (say, the x axis), the permittivity is lowest, while along the orthogonal direction (y axis), the permittivity is highest. The strategy is to input a linearly polarized wave with its ﬁeld vector at 45 degrees to the x and y axes of the crystal. It will thus have equal-amplitude x and y components in the crystal, and these will now propagate in the z direction at different speeds. A phase difference (or retardation) accumulates between the components as they propagate, which can reach π/2 if the crystal is long enough. The wave at the output thus becomes circularly polarized. Such a crystal, cut to the right length and used in this manner, is called a quarter-wave plate, since it introduces a relative phase shift of π/2 between E x and E y , which is equivalent to λ/4. It is useful to express circularly polarized waves in phasor form. To do this, we note that (96) can be expressed as Using the fact that e± jπ/2 = ± j, we identify the phasor form as: Es = E 0 (ax ± ja y )e− jβz (100) E(z, t) = Re E 0 e jωt e− jβz ax + e± jπ/2 a y

where the plus sign is used for left circular polarization and the minus sign for right circular polarization. If the wave propagates in the negative z direction, we have Es = E 0 (ax ± ja y )e+ jβz (101)

where in this case the positive sign applies to right circular polarization and the minus sign to left circular polarization. The student is encouraged to verify this.
E X A M P L E 11.7

Let us consider the result of superimposing left and right circularly polarized ﬁelds of the same amplitude, frequency, and propagation direction, but where a phase shift of δ radians exists between the two.
Solution. Taking the waves to propagate in the +z direction, and introducing a

relative phase, δ, the total phasor ﬁeld is found, using (100):

EsT = Es R + Es L = E 0 [ax − ja y ]e− jβz + E 0 [ax + ja y ]e− jβz e jδ Grouping components together, this becomes Factoring out an overall phase term, e jδ/2 , we obtain EsT = E 0 [(1 + e jδ )ax − j(1 − e jδ )a y ]e− jβz

EsT = E 0 e jδ/2 (e− jδ/2 + e jδ/2 )ax − j(e− jδ/2 − e jδ/2 )a y e− jβz

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From Euler’s identity, we ﬁnd that e jδ/2 + e− jδ/2 = 2 cos δ/2, and e jδ/2 − e− jδ/2 = 2 j sin δ/2. Using these relations, we obtain We recognize (102) as the electric ﬁeld of a linearly polarized wave, whose ﬁeld vector is oriented at angle δ/2 from the x axis. Example 11.7 shows that any linearly polarized wave can be expressed as the sum of two circularly polarized waves of opposite handedness, where the linear polarization direction is determined by the relative phase difference between the two waves. Such a representation is convenient (and necessary) when considering, for example, the propagation of linearly polarized light through media which contain organic molecules. These often exhibit spiral structures having left- or right-handed pitch, and they will thus interact differently with left- or right-hand circular polarization. As a result, the left circular component can propagate at a different speed than the right circular component, and so the two waves will accumulate a phase difference as they propagate. As a result, the direction of the linearly polarized ﬁeld vector at the output of the material will differ from the direction that it had at the input. The extent of this rotation can be used as a measurement tool to aid in material studies. Polarization issues will become extremely important when we consider wave reﬂection in Chapter 12. EsT = 2E 0 [cos(δ/2)ax + sin(δ/2)a y ]e− j(βz−δ/2) (102)

REFERENCES
1. Balanis, C. A. Advanced Engineering Electromagnetics. New York: John Wiley & Sons, 1989. 2. International Telephone and Telegraph Co., Inc. Reference Data for Radio Engineers. 7th ed. Indianapolis, Ind.: Howard W. Sams & Co., 1985. This handbook has some excellent data on the properties of dielectric and insulating materials. 3. Jackson, J. D. Classical Electrodynamics. 3d ed. New York: John Wiley & Sons, 1999. 4. Ramo, S., J. R. Whinnery, and T. Van Duzer. Fields and Waves in Communication Electronics. 3d ed. New York: John Wiley & Sons, 1994.

CHAPTER 11
11.1 11.2 11.3 11.4

PROBLEMS

Show that E xs = Ae j(k0 z+φ) is a solution of the vector Helmholtz equation, √ Eq. (30), for k0 = ω µ0 0 and any φ and A. A 10 GHz uniform plane wave propagates in a lossless medium for which r = 8 and µr = 2. Find (a) νp ; (b) β; (c) λ; (d) Es ; (e) Hs ; ( f ) S . An H ﬁeld in free space is given as H(x, t) = 10 cos(108 t − βx)a y A/m. Find (a) β; (b) λ; (c) E(x, t) at P(0.1, 0.2, 0.3) at t = 1 ns.

Small antennas have low efﬁciencies (as will be seen in Chapter 14), and the efﬁciency increases with size up to the point at which a critical dimension of

402

ENGINEERING ELECTROMAGNETICS

the antenna is an appreciable fraction of a wavelength, say λ/8. (a) An antenna that is 12 cm long is operated in air at 1 MHz. What fraction of a wavelength long is it? (b) The same antenna is embedded in a ferrite material for which r = 20 and µr = 2, 000. What fraction of a wavelength is it now? 11.5 A 150 MHz uniform plane wave in free space is described by Hs = (4 + j10)(2ax + ja y )e− jβz A/m. (a) Find numerical values for ω, λ, and β. (b) Find H(z, t) at t = 1.5 ns, z = 20 cm. (c) What is |E|max ? A uniform plane wave has electric ﬁeld Es = (E y0 a y − E z0 az ) e−αx e− jβx V/m. The intrinsic impedance of the medium is given as η = |η| e jφ , where φ is a constant phase. (a) Describe the wave polarization and state the direction of propagation. (b) Find Hs . (c) Find E(x, t) and H(x, t). (d) Find < S > in W/m2 . (e) Find the time-average power in watts that is intercepted by an antenna of rectangular cross-section, having width w and height h, suspended parallel to the yz plane, and at a distance d from the wave source. The phasor magnetic ﬁeld intensity for a 400 MHz uniform plane wave propagating in a certain lossless material is (2a y − j5az )e− j25x A/m. Knowing that the maximum amplitude of E is 1500 V/m, ﬁnd β, η, λ, νp , r , µr , and H(x, y, z, t). An electric ﬁeld in free space is given in spherical coordinates as Es (r ) = E 0 (r )e− jkr aθ V/m. (a) Find Hs (r ) assuming uniform plane wave behavior. (b) Find < S >. (c) Express the average outward power in watts through a closed spherical shell of radius r , centered at the origin. (d) Establish the required functional form of E 0 (r ) that will enable the power ﬂow in part c to be independent of radius. With this condition met, the given ﬁeld becomes that of an isotropic radiator in a lossless medium (radiating equal power density in all directions). A certain lossless material has µr = 4 and r = 9. A 10-MHz uniform plane wave is propagating in the a y direction with E x0 = 400 V/m and E y0 = E z0 = 0 at P(0.6, 0.6, 0.6) at t = 60 ns. Find (a) β, λ, νp , and η; (b) E(y, t); (c) H(y, t). In a medium characterized by intrinsic impedance η = |η|e jφ , a linearly polarized plane wave propagates, with magnetic ﬁeld given as Hs = (H0y a y + H0z az )e−αx e− jβx . Find (a) Es ; (b) E(x, t); (c) H(x, t); (d) S .

11.6

11.7

11.8

11.9

11.10

11.11

A 2 GHz uniform plane wave has an amplitude E y0 = 1.4 kV/m at (0, 0, 0, t = 0) and is propagating in the az direction in a medium where = 1.6 × 10−11 F/m, = 3.0 × 10−11 F/m, and µ = 2.5 µH/m. Find (a) E y at P(0, 0, 1.8 cm) at 0.2 ns; (b) Hx at P at 0.2 ns. Describe how the attenuation coefﬁcient of a liquid medium, assumed to be a good conductor, could be determined through measurement of wavelength

11.12

C H A P T E R 11

The Uniform Plane Wave

403

in the liquid at a known frequency. What restrictions apply? Could this method be used to ﬁnd the conductivity as well? 11.13 Let jk = 0.2 + j1.5 m−1 and η = 450 + j60 for a uniform plane propagating in the az direction. If ω = 300 Mrad/s, ﬁnd µ, , and for the medium. A certain nonmagnetic material has the material constants r = 2 and / = 4 × 10−4 at ω = 1.5 Grad/s. Find the distance a uniform plane wave can propagate through the material before (a) it is attenuated by 1 Np; (b) the power level is reduced by one-half; (c) the phase shifts 360◦ . A 10 GHz radar signal may be represented as a uniform plane wave in a sufﬁciently small region. Calculate the wavelength in centimeters and the attenuation in nepers per meter if the wave is propagating in a nonmagnetic material for which (a) r = 1 and r = 0; (b) r = 1.04 and r = 9.00 × 10−4 ; (c) r = 2.5 and r = 7.2.

11.14

11.15

11.16

Consider the power dissipation term, E · Jdv, in Poynting’s theorem (Eq. (70)). This gives the power lost to heat within a volume into which electromagnetic waves enter. The term pd = E · J is thus the power dissipation per unit volume in W/m3 . Following the same reasoning that resulted in Eq. (77), the time-average power dissipation per volume will be < pd >= (1/2)Re Es · J∗ . (a) Show that in a conducting medium, s through which a uniform plane wave of amplitude E 0 propagates in the forward z direction, < pd >= (σ/2)|E 0 |2 e−2αz . (b) Conﬁrm this result for the special case of a good conductor by using the left hand side of Eq. (70), and consider a very small volume. Let η = 250 + j30 and jk = 0.2 + j2m−1 for a uniform plane wave propagating in the az direction in a dielectric having some ﬁnite conductivity. If |E s | = 400 V/m at z = 0, ﬁnd (a) S at z = 0 and z = 60 cm; (b) the average ohmic power dissipation in watts per cubic meter at z = 60 cm.

11.17

11.18

11.19

Given a 100-MHz uniform plane wave in a medium known to be a good dielectric, the phasor electric ﬁeld is Es = 4e−0.5z e− j20z ax V/m. Determine (a) ; (b) ; (c) η; (d) Hs ; (e) S ; ( f ) the power in watts that is incident on a rectangular surface measuring 20 m × 30 m at z = 10 m. Perfectly conducting cylinders with radii of 8 mm and 20 mm are coaxial. The region between the cylinders is ﬁlled with a perfect dielectric for which = 10−9 /4π F/m and µr = 1. If E in this region is (500/ρ) cos(ωt − 4z)aρ V/m, ﬁnd (a) ω, with the help of Maxwell’s equations in cylindrical coordinates; (b) H(ρ, z, t); (c) S(ρ, z, t) ; (d) the average power passing through every cross section 8 < ρ < 20 mm, 0 < φ < 2π. Voltage breakdown in air at standard temperature and pressure occurs at an electric ﬁeld strength of approximately 3 × 106 V/m. This becomes an issue

11.20

404

ENGINEERING ELECTROMAGNETICS

in some high-power optical experiments, in which tight focusing of light may be necessary. Estimate the lightwave power in watts that can be focused into a cylindrical beam of 10µm radius before breakdown occurs. Assume uniform plane wave behavior (although this assumption will produce an answer that is higher than the actual number by as much as a factor of 2, depending on the actual beam shape). 11.21 The cylindrical shell, 1 cm< ρ < 1.2 cm, is composed of a conducting material for which σ = 106 S/m. The external and internal regions are nonconducting. Let Hφ = 2000 A/m at ρ = 1.2 cm. Find (a) H everywhere; (b) E everywhere; (c) S everywhere. The inner and outer dimensions of a coaxial copper transmission line are 2 and 7 mm, respectively. Both conductors have thicknesses much greater than δ. The dielectric is lossless and the operating frequency is 400 MHz. Calculate the resistance per meter length of the (a) inner conductor; (b) outer conductor; (c) transmission line. A hollow tubular conductor is constructed from a type of brass having a conductivity of 1.2 × 107 S/m. The inner and outer radii are 9 and 10 mm, respectively. Calculate the resistance per meter length at a frequency of (a) dc; (b) 20 MHz; (c) 2 GHz. (a) Most microwave ovens operate at 2.45 GHz. Assume that σ = 1.2 × 106 S/m and µr = 500 for the stainless steel interior, and ﬁnd the depth of penetration. (b) Let E s = 50 0◦ V/m at the surface of the conductor, and plot a curve of the amplitude of E s versus the angle of E s as the ﬁeld propagates into the stainless steel. A good conductor is planar in form, and it carries a uniform plane wave that has a wavelength of 0.3 mm and a velocity of 3 × 105 m/s. Assuming the conductor is nonmagnetic, determine the frequency and the conductivity. The dimensions of a certain coaxial transmission line are a = 0.8 mm and b = 4 mm. The outer conductor thickness is 0.6 mm, and all conductors have σ = 1.6 × 107 S/m. (a) Find R, the resistance per unit length at an operating frequency of 2.4 GHz. (b) Use information from Sections 6.3 and 8.10 to ﬁnd C and L, the capacitance and inductance per unit length, respectively. The coax is air-ﬁlled. (c) Find α and β if √ α + jβ = jωC(R + jωL).

11.22

11.23

11.24

11.25

11.26

11.27

11.28

A uniform plane wave in free space has electric ﬁeld vector given by Es = 10e− jβx az + 15e− jβx a y V/m. (a) Describe the wave polarization. (b) Find Hs . (c) Determine the average power density in the wave in W/m2 .

The planar surface z = 0 is a brass-Teﬂon interface. Use data available in Appendix C to evaluate the following ratios for a uniform plane wave having ω = 4 × 1010 rad/s: (a) αTef /αbrass ; (b) λTef /λbrass ; (c) v Tef /νbrass .

C H A P T E R 11

The Uniform Plane Wave

405

11.29

Consider a left circularly polarized wave in free space that propagates in the forward z direction. The electric ﬁeld is given by the appropriate form of Eq. (100). Determine (a) the magnetic ﬁeld phasor, Hs ; (b) an expression for the average power density in the wave in W/m2 by direct application of Eq. (77). In an anisotropic medium, permittivity varies with electric ﬁeld direction, and is a property seen in most crystals. Consider a uniform plane wave propagating in the z direction in such a medium, and which enters the material with equal ﬁeld components along the x and y axes. The ﬁeld phasor will take the form: where β = βx − β y is the difference in phase constants for waves that are linearly polarized in the x and y directions. Find distances into the material (in terms of β) at which the ﬁeld is (a) linearly polarized and (b) circularly polarized. (c) Assume intrinsic impedance η that is approximately constant with ﬁeld orientation and ﬁnd Hs and < S >. Es (z) = E 0 (ax + a y e j βz 11.30

) e− jβz

11.31

A linearly polarized uniform plane wave, propagating in the forward z direction, is input to a lossless anisotropic material, in which the dielectric constant encountered by waves polarized along y( r y ) differs from that seen by waves polarized along x( r x ). Suppose r x = 2.15, r y = 2.10, and the wave electric ﬁeld at input is polarized at 45◦ to the positive x and y axes. (a) Determine, in terms of the free space wavelength, λ, the shortest length of the material, such that the wave, as it emerges from the output, is circularly polarized. (b) Will the output wave be right or left circularly polarized? Problem 11.30 is good background. Suppose that the length of the medium of Problem 11.31 is made to be twice that determined in the problem. Describe the polarization of the output wave in this case. Given a wave for which Es = 15e− jβz ax + 18e− jβz e jφ a y V/m in a medium characterized by complex intrinsic impedance, η (a) ﬁnd Hs ; (b) determine the average power density in W/m2 . Given a general elliptically polarized wave as per Eq. (93): Es = [E x0 ax + E y0 e jφ a y ]e− jβz (a) Show, using methods similar to those of Example 11.7, that a linearly polarized wave results when superimposing the given ﬁeld and a phaseshifted ﬁeld of the form: Es = [E x0 ax + E y0 e− jφ a y ]e− jβz e jδ where δ is a constant. (b) Find δ in terms of φ such that the resultant wave is linearly polarized along x.

11.32

11.33

11.34

12

C H A P T E R

Plane Wave Reflection and Dispersion

I

n Chapter 11, we learned how to mathematically represent uniform plane waves as functions of frequency, medium properties, and electric ﬁeld orientation. We also learned how to calculate the wave velocity, attenuation, and power. In this chapter we consider wave reﬂection and transmission at planar boundaries between different media. Our study will allow any orientation between the wave and boundary and will also include the important cases of multiple boundaries. We will also study the practical case of waves that carry power over a ﬁnite band of frequencies, as would occur, for example, in a modulated carrier. We will consider such waves in dispersive media, in which some parameter that affects propagation (permittivity for example) varies with frequency. The effect of a dispersive medium on a signal is of great importance because the signal envelope will change its shape as it propagates. As a result, detection and faithful representation of the original signal at the receiving end become problematic. Consequently, dispersion and attenuation must both be evaluated when establishing maximum allowable transmission distances. ■

12.1 REFLECTION OF UNIFORM PLANE WAVES AT NORMAL INCIDENCE
We ﬁrst consider the phenomenon of reﬂection which occurs when a uniform plane wave is incident on the boundary between regions composed of two different materials. The treatment is specialized to the case of normal incidence—in which the wave propagation direction is perpendicular to the boundary. In later sections, we remove this restriction. Expressions will be found for the wave that is reﬂected from the interface and for that which is transmitted from one region into the other. These results are directly related to impedance-matching problems in ordinary transmission lines, as we have already encountered in Chapter 10. They are also applicable to waveguides, which we will study in Chapter 13.
406

C H A P T E R 12

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407

Figure 12.1 A plane wave incident on a boundary establishes reflected and transmitted waves having the indicated propagation directions. All fields are parallel to the boundary, with electric fields along x and magnetic fields along y.

We again assume that we have only a single vector component of the electric ﬁeld intensity. Referring to Figure 12.1, we deﬁne region 1 ( 1 , µ1 ) as the half-space for which z < 0; region 2 ( 2 , µ2 ) is the half-space for which z > 0. Initially we establish a wave in region 1, traveling in the +z direction, and linearly polarized along x.
+ + Ex1 (z, t) = E x10 e−α1 z cos(ωt − β1 z) + + E xs1 (z) = E x10 e− jkz

In phasor form, this is

(1)

+ where we take E x10 as real. The subscript 1 identiﬁes the region, and the superscript + + indicates a positively traveling wave. Associated with E xs1 (z) is a magnetic ﬁeld in the y direction, 1 + − jk1 z + Hys1 (z) = E e (2) η1 x10 where k1 and η1 are complex unless 1 (or σ1 ) is zero. This uniform plane wave in region l that is traveling toward the boundary surface at z = 0 is called the incident wave. Since the direction of propagation of the incident wave is perpendicular to the boundary plane, we describe it as normal incidence. We now recognize that energy may be transmitted across the boundary surface at z = 0 into region 2 by providing a wave moving in the +z direction in that medium. The phasor electric and magnetic ﬁelds for this wave are + + E xs2 (z) = E x20 e− jk2 z 1 + − jk2 z + E e Hys2 (z) = η2 x20

(3) (4)

408

ENGINEERING ELECTROMAGNETICS

This wave, which moves away from the boundary surface into region 2, is called the transmitted wave. Note the use of the different propagation constant k2 and intrinsic impedance η2 . Now we must satisfy the boundary conditions at z = 0 with these assumed ﬁelds. With E polarized along x, the ﬁeld is tangent to the interface, and therefore the E ﬁelds in regions l and 2 must be equal at z = 0. Setting z = 0 in (1) and (3) would + + require that E x10 = E x20 . H, being y-directed, is also a tangential ﬁeld, and must be continuous across the boundary (no current sheets are present in real media). When + + we let z = 0 in (2) and (4), we ﬁnd that we must have E x10 /η1 = E x20 /η2 . Since + + E x10 = E x20 , then η1 = η2 . But this is a very special condition that does not ﬁt the facts in general, and we are therefore unable to satisfy the boundary conditions with only an incident and a transmitted wave. We require a wave traveling away from the boundary in region 1, as shown in Figure 12.1; this is the reflecte wave,
− − E xs1 (z) = E x10 e jk1 z − Hxs1 (z) = −

(5) (6)

− E x10 jk1 z e η1

− where E x10 may be a complex quantity. Because this ﬁeld is traveling in the −z − − direction, E xs1 = −η1 Hys1 for the Poynting vector shows that E− × H− must be in 1 1 the −az direction. The boundary conditions are now easily satisﬁed, and in the process the ampli+ tudes of the transmitted and reﬂected waves may be found in terms of E x10 . The total electric ﬁeld intensity is continuous at z = 0,

E xs1 = E xs2

(z = 0) (z = 0) (7)

or
+ − + E xs1 + E xs1 = E xs2

Therefore
+ − + E x10 + E x10 = E x20

Furthermore, Hys1 = Hys2 or
+ − + Hys1 + Hys1 = Hys2

(z = 0) (z = 0)

and therefore
+ E x10 E− E+ − x10 = x20 η1 η1 η2 + Solving (8) for E x20 and substituting into (7), we ﬁnd η2 + η2 − + − E − E E x10 + E x10 = η1 x10 η1 x10

(8)

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Plane Wave Reflection and Dispersion

409

or η2 − η 1 η2 + η 1 The ratio of the amplitudes of the reﬂected and incident electric ﬁelds deﬁnes the reflectio coefficient designated by ,
− + E x10 = E x10

=

− E x10 η 2 − η1 = | |e jφ + = η2 + η1 E x10

(9)

It is evident that as η1 or η2 may be complex, will also be complex, and so we include a reﬂective phase shift, φ. The interpretation of Eq. (9) is identical to that used with transmission lines [Eq. (73), Chapter 10]. The relative amplitude of the transmitted electric ﬁeld intensity is found by combining (9) and (7) to yield the transmission coefficient τ , τ=
+ E x20 2η2 =1+ + = η1 + η 2 E x10

= |τ |e jφi

(10)

whose form and interpretation are consistent with the usage in transmission lines [Eq. (75), Chapter 10]. Let us see how these results may be applied to several special cases. We ﬁrst let region 1 be a perfect dielectric and region 2 be a perfect conductor. Then we apply Eq. (48), Chapter 11, with 2 = σ2 /ω, obtaining η2 = jωµ2 σ2 + jω
+ E x20 = 0 2

=0

in which zero is obtained since σ2 → ∞. Therefore, from (10), No time-varying ﬁelds can exist in the perfect conductor. An alternate way of looking at this is to note that the skin depth is zero. Because η2 = 0, Eq. (9) shows that = −1 and
+ − E x10 = −E x10

The incident and reﬂected ﬁelds are of equal amplitude, and so all the incident energy is reﬂected by the perfect conductor. The fact that the two ﬁelds are of opposite sign indicates that at the boundary (or at the moment of reﬂection), the reﬂected ﬁeld is shifted in phase by 180◦ relative to the incident ﬁeld. The total E ﬁeld in region 1 is
+ − E xs1 = E xs1 + E xs1 + + = E x10 e− jβ1 z − E x10 e jβ1 z

410

ENGINEERING ELECTROMAGNETICS

where we have let jk1 = 0 + jβ1 in the perfect dielectric. These terms may be combined and simpliﬁed,
+ E xs1 = (e− jβ1 z − e jβ1 z ) E x10 + = − j2 sin(β1 z) E x10

(11)

Multiplying (11) by e form:

jωt

and taking the real part, we obtain the real instantaneous
+ Ex1 (z, t) = 2E x10 sin(β1 z) sin(ωt)

(12)

We recognize this total ﬁeld in region 1 as a standing wave, obtained by combining two waves of equal amplitude traveling in opposite directions. We ﬁrst encountered standing waves in transmission lines, but in the form of counterpropagating voltage waves (see Example 10.1). Again, we compare the form of (12) to that of the incident wave,
+ Ex1 (z, t) = E x10 cos(ωt − β1 z)

(13)

Here we see the term ωt − β1 z or ω(t − z/ν p1 ), which characterizes a wave traveling in the +z direction at a velocity ν p1 = ω/β1 . In (12), however, the factors involving time and distance are separate trigonometric terms. Whenever ωt = mπ, Ex1 is zero at all positions. On the other hand, spatial nulls in the standing wave pattern occur for all times wherever β1 z = mπ , which in turn occurs when m = (0, ±1, ±2, . . .). In such cases, 2π z = mπ λ1 and the null locations occur at λ1 2 Thus E x1 = 0 at the boundary z = 0 and at every half-wavelength from the boundary in region 1, z < 0, as illustrated in Figure 12.2. + + − − Because E xs1 = η1 Hys1 and E xs1 = −η1 Hys1 , the magnetic ﬁeld is z=m Hys1 = or Hy1 (z, t) = 2
+ E x10 cos(β1 z) cos(ωt) η1 + E x10 − jβ1 z (e +e η1 jβ1 z

)

(14)

This is also a standing wave, but it shows a maximum amplitude at the positions where E x1 = 0. It is also 90◦ out of time phase with E x1 everywhere. As a result, the average power as determined through the Poynting vector [Eq. (77), Chapter 11] is zero in the forward and backward directions. Let us now consider perfect dielectrics in both regions 1 and 2; η1 and η2 are both real positive quantities and α1 = α2 = 0. Equation (9) enables us to calculate

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Plane Wave Reflection and Dispersion

411

Figure 12.2 The instantaneous values of the total field E x1 are shown at t = π/2. E x1 = 0 for all time at multiples of one half-wavelength from the conducting surface.
− + + the reﬂection coefﬁcient and ﬁnd E x1 in terms of the incident ﬁeld E x1 . Knowing E x1 − + − + and E x1 , we then ﬁnd Hy1 and Hy1 . In region 2, E x2 is found from (10), and this then + determines Hy2 .

E X A M P L E 12.1

As a numerical example we select η1 = 100
+ E x10 = 100 V/m

η2 = 300

and calculate values for the incident, reﬂected, and transmitted waves.
Solution. The reﬂection coefﬁcient is

= and thus

300 − 100 = 0.5 300 + 100

− E x10 = 50 V/m

412

ENGINEERING ELECTROMAGNETICS

The magnetic ﬁeld intensities are 100 = 1.00 A/m 100 50 − = −0.50 A/m Hy10 = − 100 Using Eq. (77) from Chapter 11, we ﬁnd that the magnitude of the average incident power density is
+ Hy10 =

1 1 + + Re{Es × H∗ } = E x10 Hy10 = 50 W/m2 s 2 2 The average reﬂected power density is 1 − − S1r = − E x10 Hy10 = 12.5 W/m2 2 In region 2, using (10), S1i =
+ + E x20 = τ E x10 = 150 V/m

and 150 = 0.500 A/m 300 Therefore, the average power density that is transmitted through the boundary into region 2 is 1 + + S2 = E x20 Hy20 = 37.5 W/m2 2 We may check and conﬁrm the power conservation requirement:
+ Hy20 =

S1i = S1r + S2 A general rule on the transfer of power through reﬂection and transmission can be formulated. We consider the same ﬁeld vector and interface orientations as before, but allow for the case of complex impedances. For the incident power density, we have S1i = 1 1 1 1 + +∗ + 1 +∗ Re E xs1 Hys1 = Re E x10 ∗ E x10 = Re ∗ 2 2 η1 2 η1
+ E x10 + E x10 2

The reﬂected power density is then 1 1 − −∗ S1r = − Re E xs1 Hys1 = Re 2 2 1 ∗ η1
∗ +∗ E x10 =

1 1 Re ∗ 2 η1

+ E x10 | |2

2

We thus ﬁnd the general relation between the reﬂected and incident power: S1r = | |2 S1i In a similar way, we ﬁnd the transmitted power density: S2 = 1 1 1 1 + +∗ + 1 +∗ Re E xs2 Hys2 = Re τ E x10 ∗ τ ∗ E x10 = Re ∗ 2 2 η2 2 η2
+ E x10 |τ |2 2

(15)

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413

and so we see that the incident and transmitted power densities are related through S2 =
∗ Re 1/η2 ∗ Re 1/η1

|τ |2 S1i =

η1 η2

2

∗ η2 + η2 ∗ η1 + η1

|τ |2 S1i

(16)

Equation (16) is a relatively complicated way to calculate the transmitted power, unless the impedances are real. It is easier to take advantage of energy conservation by noting that whatever power is not reﬂected must be transmitted. Eq. (15) can be used to ﬁnd S2 = (1 − | |2 ) S1i (17)

As would be expected (and which must be true), Eq. (17) can also be derived from Eq. (16). D12.1. A 1-MHz uniform plane wave is normally incident onto a freshwater lake ( r = 78, r = 0, µr = 1). Determine the fraction of the incident power that is (a) reﬂected and (b) transmitted. (c) Determine the amplitude of the electric ﬁeld that is transmitted into the lake.
Ans. 0.63; 0.37; 0.20 V/m

12.2 STANDING WAVE RATIO
In cases where | | < 1, some energy is transmitted into the second region and some is reﬂected. Region 1 therefore supports a ﬁeld that is composed of both a traveling wave and a standing wave. We encountered this situation previously in transmission lines, in which partial reﬂection occurs at the load. Measurements of the voltage standing wave ratio and the locations of voltage minima or maxima enabled the determination of an unknown load impedance or established the extent to which the load impedance was matched to that of the line (Section 10.10). Similar measurements can be performed on the ﬁeld amplitudes in plane wave reﬂection. Using the same ﬁelds investigated in the previous section, we combine the incident and reﬂected electric ﬁeld intensities. Medium 1 is assumed to be a perfect dielectric (α1 = 0), but region 2 may be any material. The total electric ﬁeld phasor in region 1 will be
+ − + + E x1T = E x1 + E x1 = E x10 e− jβ1 z + E x10 e jβ1 z

(18)

where the reﬂection coefﬁcient is as expressed in (9): = η2 − η1 = | |e jφ η2 + η 1

We allow for the possibility of a complex reﬂection coefﬁcient by including its phase, φ. This is necessary because although η1 is real and positive for a lossless medium,

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η2 will in general be complex. Additionally, if region 2 is a perfect conductor, η2 is zero, and so φ is equal to π; if η2 is real and less than η1 , φ is also equal to π ; and if η2 is real and greater than η1 , φ is zero. Incorporating the phase of into (18), the total ﬁeld in region 1 becomes The maximum and minimum ﬁeld amplitudes in (19) are z-dependent and are subject to measurement. Their ratio, as found for voltage amplitudes in transmission lines (Section 10.10), is the standing wave ratio, denoted by s. We have a maximum when + each term in the larger parentheses in (19) has the same phase angle; so, for E x10 positive and real,
+ |E x1T |max = (1 + | |)E x10 + E x1T = e− jβ1 z + | |e j(β1 z+φ) E x10

(19)

(20)

and this occurs where −β1 z = β1 z + φ + 2mπ Therefore z max = − 1 (φ + 2mπ ) 2β1 (22) (m = 0, ±1, ±2, . . .) (21)

Note that an electric ﬁeld maximum is located at the boundary plane (z = 0) if φ = 0; moreover, φ = 0 when is real and positive. This occurs for real η1 and η2 when η2 > η1 . Thus there is a ﬁeld maximum at the boundary surface when the intrinsic impedance of region 2 is greater than that of region 1 and both impedances are real. With φ = 0, maxima also occur at z max = −mπ/β1 = −mλ1 /2. For the perfect conductor φ = π , and these maxima are found at z max = −π/ (2β1 ), −3π/(2β1 ), or z max = −λ1 /4, −3λ1 /4, and so forth. The minima must occur where the phase angles of the two terms in the larger parentheses in (19) differ by 180◦ , thus
+ |E x1T |min = (1 − | |)E x10

(23)

and this occurs where −β1 z = β1 z + φ + π + 2mπ or z min = − 1 (φ + (2m + 1)π ) 2β1 (25) (m = 0, ±1, ±2, . . .) (24)

The minima are separated by multiples of one half-wavelength (as are the maxima), and for the perfect conductor the ﬁrst minimum occurs when −β1 z = 0, or at the conducting surface. In general, an electric ﬁeld minimum is found at z = 0 whenever φ = π; this occurs if η2 < η1 and both are real. The results are mathematically identical to those found for the transmission line study in Section 10.10. Figure 10.6 in that chapter provides a visualization.

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Further insights can be obtained by working with Eq. (19) and rewriting it in real instantaneous form. The steps are identical to those taken in Chapter 10, Eqs. (81) through (84). We ﬁnd the total ﬁeld in region 1 to be
+ Ex1T (z, t) = (1 − | |)E x10 cos(ωt − β1 z) traveling wave

+ 2|

+ |E x10

cos(β1 z + φ/2) cos(ωt + φ/2) standing wave

(26)

The ﬁeld expressed in Eq. (26) is the sum of a traveling wave of amplitude + + (1 − | |)E x10 and a standing wave having amplitude 2| |E x10 . The portion of the incident wave that reﬂects and back-propagates in region 1 interferes with an equivalent portion of the incident wave to form a standing wave. The rest of the incident wave (that does not interfere) is the traveling wave part of (26). The maximum amplitude observed in region 1 is found where the amplitudes of the two terms in (26) add + directly to give (1 + | |)E x10 . The minimum amplitude is found where the standing + wave achieves a null, leaving only the traveling wave amplitude of (1 − | |)E x10 . The fact that the two terms in (26) combine in this way with the proper phasing can be conﬁrmed by substituting z max and z min , as given by (22) and (25).
E X A M P L E 12.2

To illustrate some of these results, let us consider a 100-V/m, 3-GHz wave that is propagating in a material having r 1 = 4, µr 1 = 1, and r = 0. The wave is normally incident on another perfect dielectric in region 2, z > 0, where r 2 = 9 and µr 2 = 1 (Figure 12.3). We seek the locations of the maxima and minima of E.

+ Figure 12.3 An incident wave, E xs1 = − j 40π z 100e V/m, is reflected with a reflection coefficient = −0.2. Dielectric 2 is infinitely thick.

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1 = 40π rad/m, and β2 = ω µ2 2 = 60π rad/m. Although the wavelength would be 10 cm in air, we ﬁnd here that λ1 = 2π/β1 = 5 cm, λ2 = 2π/β2 = 3.33 cm, η1 = 60π , η2 = 40π , and = (η2 − η1 )/(η2 + η1 ) = −0.2. Because is real and negative (η2 < η1 ), there will be a minimum of the electric ﬁeld at the boundary, and it will be repeated at half-wavelength (2.5 cm) intervals in dielectric l. From (23), we see that |E x1T |min = 80 V/m. Maxima of E are found at distances of 1.25, 3.75, 6.25, . . . cm from z = 0. These maxima all have amplitudes of 120 V/m, as predicted by (20). There are no maxima or minima in region 2 because there is no reﬂected wave there.

Solution. We calculate ω = 6π × 109 rad/s, β1 = ω µ1

√

√

The ratio of the maximum to minimum amplitudes is the standing wave ratio: s= |E x1T |max 1+| | = |E x1T |min 1−| | (27)

Because | | < 1, s is always positive and greater than or equal to unity. For the preceding example, s= 1.2 1 + |−0.2| = = 1.5 1 − |−0.2| 0.8

If | | = 1, the reﬂected and incident amplitudes are equal, all the incident energy is reﬂected, and s is inﬁnite. Planes separated by multiples of λ1 /2 can be found on which E x1 is zero at all times. Midway between these planes, E x1 has a maximum amplitude twice that of the incident wave. If η2 = η1 , then = 0, no energy is reﬂected, and s = 1; the maximum and minimum amplitudes are equal. If one-half the incident power is reﬂected, | |2 = 0.5, | | = 0.707, and s = 5.83. D12.2. What value of s results when
Ans. 3

= ±1/2?

Because the standing wave ratio is a ratio of amplitudes, the relative amplitudes, as measured by a probe, permit its use to determine s experimentally.
E X A M P L E 12.3

A uniform plane wave in air partially reﬂects from the surface of a material whose properties are unknown. Measurements of the electric ﬁeld in the region in front of the interface yield a 1.5-m spacing between maxima, with the ﬁrst maximum occurring 0.75 m from the interface. A standing wave ratio of 5 is measured. Determine the intrinsic impedance, ηu , of the unknown material.

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length is 3.0 m, or f = 100 MHz. The ﬁrst maximum at 0.75 m is thus at a distance of λ/4 from the interface, which means that a ﬁeld minimum occurs at the boundary. Thus will be real and negative. We use (27) to write | |= So 5−1 2 s−1 = = s+1 5+1 3 ηu − η0 2 = 3 ηu + η 0

Solution. The 1.5 m spacing between maxima is λ/2, which implies that a wave-

=− which we solve for ηu to obtain ηu =

1 377 η0 = = 75.4 5 5

12.3 WAVE REFLECTION FROM MULTIPLE INTERFACES
So far we have treated the reﬂection of waves at the single boundary that occurs between semi-inﬁnite media. In this section, we consider wave reﬂection from materials that are ﬁnite in extent, such that we must consider the effect of the front and back surfaces. Such a two-interface problem would occur, for example, for light incident on a ﬂat piece of glass. Additional interfaces are present if the glass is coated with one or more layers of dielectric material for the purpose (as we will see) of reducing reﬂections. Such problems in which more than one interface is involved are frequently encountered; single-interface problems are in fact more the exception than the rule. Consider the general situation shown in Figure 12.4, in which a uniform plane wave propagating in the forward z direction is normally incident from the left onto the interface between regions 1 and 2; these have intrinsic impedances η1 and η2 . A third region of impedance η3 lies beyond region 2, and so a second interface exists between regions 2 and 3. We let the second interface location occur at z = 0, and so all positions to the left will be described by values of z that are negative. The width of the second region is l, so the ﬁrst interface will occur at position z = −l. When the incident wave reaches the ﬁrst interface, events occur as follows: A portion of the wave reﬂects, while the remainder is transmitted, to propagate toward the second interface. There, a portion is transmitted into region 3, while the rest reﬂects and returns to the ﬁrst interface; there it is again partially reﬂected. This reﬂected wave then combines with additional transmitted energy from region 1, and the process repeats. We thus have a complicated sequence of multiple reﬂections that occur within region 2, with partial transmission at each bounce. To analyze the situation in this way would involve keeping track of a very large number of reﬂections; this would be necessary when studying the transient phase of the process, where the incident wave ﬁrst encounters the interfaces. If the incident wave is left on for all time, however, a steady-state situation is eventually reached, in which (1) an overall fraction of the incident wave is reﬂected

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in

Figure 12.4 Basic two-interface problem, in which the impedances of regions 2 and 3, along with the finite thickness of region 2, are accounted for in the input impedance at the front surface, ηin .

from the two-interface conﬁguration and back-propagates in region 1 with a deﬁnite amplitude and phase; (2) an overall fraction of the incident wave is transmitted through the two interfaces and forward-propagates in the third region; (3) a net backward wave exists in region 2, consisting of all reﬂected waves from the second interface; and (4) a net forward wave exists in region 2, which is the superposition of the transmitted wave through the ﬁrst interface and all waves in region 2 that have reﬂected from the ﬁrst interface and are now forward-propagating. The effect of combining many co-propagating waves in this way is to establish a single wave which has a deﬁnite amplitude and phase, determined through the sums of the amplitudes and phases of all the component waves. In steady state, we thus have a total of ﬁve waves to consider. These are the incident and net reﬂected waves in region 1, the net transmitted wave in region 3, and the two counterpropagating waves in region 2. The situation is analyzed in the same manner as that used in the analysis of ﬁnite-length transmission lines (Section 10.11). Let us assume that all regions are composed of lossless media, and consider the two waves in region 2. If we take these as x-polarized, their electric ﬁelds combine to yield where β2 = ω r 2 /c, and where the amplitudes, and are complex. The y-polarized magnetic ﬁeld is similarly written, using complex amplitudes:
+ − Hys2 = Hy20 e− jβ2 z + Hy20 e jβ2 z

√

+ − E xs2 = E x20 e− jβ2 z + E x20 e

jβ2 z − E x20 ,

(28a)

+ E x20

(28b)

We now note that the forward and backward electric ﬁeld amplitudes in region 2 are related through the reﬂection coefﬁcient at the second interface, 23 , where η3 − η2 (29) 23 = η3 + η 2

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We thus have
− E x20 = + 23 E x20

(30)

We then write the magnetic ﬁeld amplitudes in terms of electric ﬁeld amplitudes through
+ Hy20 =

1 + E η2 x20
+ 23 E x20

(31a)

and
− Hy20 = −

1 − 1 E =− η2 x20 η2

(31b)

We now deﬁne the wave impedance, ηw , as the z-dependent ratio of the total electric ﬁeld to the total magnetic ﬁeld. In region 2, this becomes, using (28a) and (28b), ηw (z) =
− E + e− jβ2 z + E x20 e jβ2 z E xs2 = x20 − jβ z + − Hys2 Hy20 e 2 + Hy20 e jβ2 z

Then, using (30), (31a), and (31b), we obtain ηw (z) = η2 e− jβ2 z + e− jβ2 z −
23 e 23 e jβ2 z jβ2 z

Now, using (29) and Euler’s identity, we have ηw (z) = η2 ×

This is easily simpliﬁed to yield

(η3 + η2 )(cos β2 z − j sin β2 z) + (η3 − η2 )(cos β2 z + j sin β2 z) (η3 + η2 )(cos β2 z − j sin β2 z) − (η3 − η2 )(cos β2 z + j sin β2 z) ηw (z) = η2 η3 cos β2 z − jη2 sin β2 z η2 cos β2 z − jη3 sin β2 z (32)

We now use the wave impedance in region 2 to solve our reﬂection problem. Of interest to us is the net reﬂected wave amplitude at the ﬁrst interface. Since tangential E and H are continuous across the boundary, we have
+ − E xs1 + E xs1 = E xs2

(z = −l) (z = −l)

(33a)

and
+ − Hys1 + Hys1 = Hys2

(33b)

Then, in analogy to (7) and (8), we may write
+ − E x10 + E x10 = E xs2 (z = −l)

(34a)

and
+ E x10 E− E xs2 (z = −l) − x10 = η1 η1 ηw (−l)

(34b)

+ − where E x10 and E x10 are the amplitudes of the incident and reﬂected ﬁelds. We call ηw (−l) the input impedance, ηin , to the two-interface combination. We now solve

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(34a) and (34b) together, eliminating E xs2 , to obtain
− E x10 + = E x10

=

ηin − η1 ηin + η1

(35)

To ﬁnd the input impedance, we evaluate (32) at z = −l, resulting in ηin = η2 η3 cos β2l + jη2 sin β2l η2 cos β2l + jη3 sin β2l (36)

Equations (35) and (36) are general results that enable us to calculate the net reﬂected wave amplitude and phase from two parallel interfaces between lossless media.1 Note the dependence on the interface spacing, l, and on the wavelength as measured in region 2, characterized by β2 . Of immediate importance to us is the fraction of the incident power that reﬂects from the dual interface and back-propagates in region 1. As we found earlier, this fraction will be | |2 . Also of interest is the transmitted power, which propagates away from the second interface in region 3. It is simply the remaining power fraction, which is 1 − | |2 . The power in region 2 stays constant in steady state; power leaves that region to form the reﬂected and transmitted waves, but is immediately replenished by the incident wave. We have already encountered an analogous situation involving cascaded transmission lines, which culminated in Eq. (101) in Chapter 10. An important result of situations involving two interfaces is that it is possible to achieve total transmission in certain cases. From (35), we see that total transmission occurs when = 0, or when ηin = η1 . In this case, as in transmission lines, we say that the input impedance is matched to that of the incident medium. There are a few methods of accomplishing this. As a start, suppose that η3 = η1 , and region 2 is of such thickness that β2l = mπ , where m is an integer. Now β2 = 2π/λ2 , where λ2 is the wavelength as measured in region 2. Therefore 2π l = mπ λ2 or λ2 l=m (37) 2 With β2l = mπ, the second region thickness is an integer multiple of half-wavelengths as measured in that medium. Equation (36) now reduces to ηin = η3 . Thus the general effect of a multiple half-wave thickness is to render the second region immaterial to

1

For convenience, (34a) and (34b) have been written for a speciﬁc time at which the incident wave + amplitude, E x10 , occurs at z = −l. This establishes a zero-phase reference at the front interface for the incident wave, and so it is from this reference that the reﬂected wave phase is determined. Equivalently, we have repositioned the z = 0 point at the front interface. Eq. (36) allows this because it is only a function of the interface spacing, l.

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the results on reﬂection and transmission. Equivalently, we have a single-interface problem involving η1 and η3 . Now, with η3 = η1 , we have a matched input impedance, and there is no net reﬂected wave. This method of choosing the region 2 thickness is known as half-wave matching. Its applications include, for example, antenna housings on airplanes known as radomes, which form a part of the fuselage. The antenna, inside the aircraft, can transmit and receive through this layer, which can be shaped to enable good aerodynamic characteristics. Note that the half-wave matching condition no longer applies as we deviate from the wavelength that satisﬁes it. When this is done, the device reﬂectivity increases (with increased wavelength deviation), so it ultimately acts as a bandpass ﬁlter. Often, it is convenient to express the dielectric constant of the medium through the refractive index (or just index), n, deﬁned as n= √ r (38)

Characterizing materials by their refractive indices is primarily done at optical frequencies (on the order of 1014 Hz), whereas at much lower frequencies, a dielectric constant is traditionally speciﬁed. Since r is complex in lossy media, the index will also be complex. Rather than complicate the situation in this way, we will restrict our use of the refractive index to cases involving lossless media, having r = 0, and µr = 1. Under lossless conditions, we may write the plane wave phase constant and the material intrinsic impedance in terms of the index through √ √ β = k = ω µ0 0 and 1 η= √ µ0 r 0 r

=

nω c

(39)

=

η0 n

(40)

Finally, the phase velocity and wavelength in a material of index n are νp = and λ= νp λ0 = f n (42) c n (41)

where λ0 is the wavelength in free space. It is obviously important not to confuse the index n with the similar-appearing Greek η (intrinsic impedance), which has an entirely different meaning. Another application, typically seen in optics, is the Fabry-Perot interferometer. This, in its simplest form, consists of a single block of glass or other transparent

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material of index n, whose thickness, l, is set to transmit wavelengths which satisfy the condition λ = λ0 /n = 2l/m. Often we want to transmit only one wavelength, not several, as (37) would allow. We would therefore like to assure that adjacent wavelengths that are passed through the device are separated as far as possible, so that only one will lie within the input power spectrum. In terms of wavelength as measured in the material, this separation is in general given by 2l 2l 2l . 2l − = = 2 λm−1 − λm = λ f = m−1 m m(m − 1) m Note that m is the number of half-wavelengths in region 2, or m = 2l/λ = 2nl/λ0 , where λ0 is the desired free-space wavelength for transmission. Thus . λ2 λf = 2 2l In terms of wavelength measured in free space, this becomes (43a)

. λ2 (43b) λf0 = n λf = 0 2nl λ f 0 is known as the free spectral range of the Fabry-Perot interferometer in terms of free-space wavelength separation. The interferometer can be used as a narrowband ﬁlter (transmitting a desired wavelength and a narrow spectrum around this wavelength) if the spectrum to be ﬁltered is narrower than the free spectral range.
E X A M P L E 12.4

Suppose we wish to ﬁlter an optical spectrum of full width λs0 = 50 nm (measured in free space), whose center wavelength, λ0 , is in the red part of the visible spectrum at 600 nm, where one nm (nanometer) is 10−9 m. A Fabry-Perot ﬁlter is to be used, consisting of a lossless glass plate in air, having refractive index n = 1.45. We need to ﬁnd the required range of glass thicknesses such that multiple wavelength orders will not be transmitted.
Solution. We require that the free spectral range be greater than the optical spectral

width, or

λf0 >

λs . Using (43b) l< λ2 0 2n λs0

So l< 6002 = 2.5 × 103 nm = 2.5 µm 2(1.45)(50)

where 1µm (micrometer) = 10−6 m. Fabricating a glass plate of this thickness or less is somewhat ridiculous to contemplate. Instead, what is often used is an airspace of thickness on this order, between two thick plates whose surfaces on the sides opposite the airspace are antireﬂection coated. This is in fact a more versatile conﬁguration because the wavelength to be transmitted (and the free spectral range) can be adjusted by varying the plate separation.

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Next, we remove the restriction η1 = η3 and look for a way to produce zero reﬂection. Returning to Eq. (36), suppose we set β2l = (2m − 1)π/2, or an odd multiple of π/2. This means that 2π π (m = 1, 2, 3, . . .) l = (2m − 1) λ2 2 or l = (2m − 1) λ2 4 (44)

The thickness is an odd multiple of a quarter-wavelength as measured in region 2. Under this condition, (36) reduces to
2 η2 (45) η3 Typically, we choose the second region impedance to allow matching between given impedances η1 and η3 . To achieve total transmission, we require that ηin = η1 , so that the required second region impedance becomes

ηin =

η2 =

√ η1 η3

(46)

With the conditions given by (44) and (46) satisﬁed, we have performed quarter-wave matching. The design of antireﬂective coatings for optical devices is based on this principle.
E X A M P L E 12.5

We wish to coat a glass surface with an appropriate dielectric layer to provide total transmission from air to the glass at a free-space wavelength of 570 nm. The glass has refractive index n 3 = 1.45. Determine the required index for the coating and its minimum thickness.
Solution. The known impedances are η1 = 377

Using (46) we have

and η3 = 377/1.45 = 260 .

η2 = The index of region 2 will then be

(377)(260) = 313 377 313 = 1.20

n2 = The wavelength in region 2 will be

570 = 475 nm 1.20 The minimum thickness of the dielectric layer is then λ2 = 119 nm = 0.119 µm l= 4 λ2 =

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ENGINEERING ELECTROMAGNETICS

in

in

Figure 12.5 A three-interface problem in which input impedance ηin,a is transformed back to the front interface to form input impedance ηin,b.

The procedure in this section for evaluating wave reﬂection has involved calculating an effective impedance at the ﬁrst interface, ηin , which is expressed in terms of the impedances that lie beyond the front surface. This process of impedance transformation is more apparent when we consider problems involving more than two interfaces. For example, consider the three-interface situation shown in Figure 12.5, where a wave is incident from the left in region 1. We wish to determine the fraction of the incident power that is reﬂected and back-propagates in region 1 and the fraction of the incident power that is transmitted into region 4. To do this, we need to ﬁnd the input impedance at the front surface (the interface between regions 1 and 2). We start by transforming the impedance of region 4 to form the input impedance at the boundary between regions 2 and 3. This is shown as ηin,b in Figure 12.5. Using (36), we have ηin,b = η3 η4 cos β3lb + jη3 sin β3lb η3 cos β3lb + jη4 sin β3lb (47)

We have now effectively reduced the situation to a two-interface problem in which ηin,b is the impedance of all that lies beyond the second interface. The input impedance at the front interface, ηin,a , is now found by transforming ηin,b as follows: ηin,a = η2 ηin,b cos β2la + jη2 sin β2la η2 cos β2la + jηin,b sin β2la (48)

The reﬂected power fraction is now | |2 , where = ηin,a − η1 ηin,a + η1

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The fraction of the power transmitted into region 4 is, as before, 1 − | |2 . The method of impedance transformation can be applied in this manner to any number of interfaces. The process, although tedious, is easily handled by a computer. The motivation for using multiple layers to reduce reﬂection is that the resulting structure is less sensitive to deviations from the design wavelength if the impedances (or refractive indices) are arranged to progressively increase or decrease from layer to layer. For multiple layers to antireﬂection coat a camera lens, for example, the layer on the lens surface would be of impedance very close to that of the glass. Subsequent layers are given progressively higher impedances. With a large number of layers fabricated in this way, the situation begins to approach (but never reaches) the ideal case, in which the top layer impedance matches that of air, while the impedances of deeper layers continuously decrease until reaching the value of the glass surface. With this continuously varying impedance, there is no surface from which to reﬂect, and so light of any wavelength is totally transmitted. Multilayer coatings designed in this way produce excellent broadband transmission characteristics. D12.3. A uniform plane wave in air is normally incident on a dielectric slab of thickness λ2 /4, and intrinsic impedance η2 = 260 . Determine the magnitude and phase of the reﬂection coefﬁcient.
Ans. 0.356; 180◦

12.4 PLANE WAVE PROPAGA TION IN GENERAL DIRECTIONS
In this section, we will learn how to mathematically describe uniform plane waves that propagate in any direction. Our motivation for doing this is our need to address the problem of incident waves on boundaries that are not perpendicular to the propagation direction. Such problems of oblique incidence generally occur, with normal incidence being a special case. Addressing such problems requires (as always) that we establish an appropriate coordinate system. With the boundary positioned in the x, y plane, for example, the incident wave will propagate in a direction that could involve all three coordinate axes, whereas with normal incidence, we were only concerned with propagation along z. We need a mathematical formalism that will allow for the general direction case. Let us consider a wave that propagates in a lossless medium, with propagation √ constant β = k = ω µ . For simplicity, we consider a two-dimensional case, where the wave travels in a direction between the x and z axes. The ﬁrst step is to consider the propagation constant as a vector, k, indicated in Figure 12.6. The direction of k is the propagation direction, which is the same as the direction of the Poynting vector in our case.2 The magnitude of k is the phase shift per unit distance along that direction.

2

We assume here that the wave is in an isotropic medium, where the permittivity and permeability do not change with ﬁeld orientation. In anisotropic media (where and/or µ depend on ﬁeld orientation), the directions of the Poynting vector and k may differ.

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Figure 12.6 Representation of a uniform plane wave with wavevector k at angle θ to the x axis. The phase at point (x, z) is given by k · r. Planes of constant phase (shown as lines perpendicular to k) are spaced by wavelength λ but have wider spacing when measured along the x or z axis.

Part of the process of characterizing a wave involves specifying its phase at any spatial location. For the waves we have considered that propagate along the z axis, this was accomplished by the factor e± jkz in the phasor form. To specify the phase in our two-dimensional problem, we make use of the vector nature of k and consider the phase at a general location (x, z) described through the position vector r. The phase at that location, referenced to the origin, is given by the projection of k along r times the magnitude of r, or just k · r. If the electric ﬁeld is of magnitude E 0 , we can thus write down the phasor form of the wave in Figure 12.6 as Es = E0 e− jk · r (49)

The minus sign in the exponent indicates that the phase along r moves in time in the direction of increasing r. Again, the wave power ﬂow in an isotropic medium occurs in the direction along which the phase shift per unit distance is maximum—or along k. The vector r serves as a means to measure phase at any point using k. This construction is easily extended to three dimensions by allowing k and r to each have three components. In our two-dimensional case of Figure 12.6, we can express k in terms of its x and z components: k = k x a x + k z az

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The position vector, r, can be similarly expressed: r = xax + zaz so that k · r = kx x + kz z Equation (49) now becomes Es = E0 e− j(kx x+kz z) (50)

Whereas Eq. (49) provided the general form of the wave, Eq. (50) is the form that is speciﬁc to the situation. Given a wave expressed by (50), the angle of propagation from the x axis is readily found through θ = tan−1 kz kx

The wavelength and phase velocity depend on the direction one is considering. In the direction of k, these will be λ= and νp = ω ω = 2 + k2 k kx z
1/2

2π 2π = 2 + k2 k kx z

1/2

If, for example, we consider the x direction, these quantities will be λx = and νpx = ω kx 2π kx

Note that both λx and νpx are greater than their counterparts along the direction of k. This result, at ﬁrst surprising, can be understood through the geometry of Figure 12.6. The diagram shows a series of phase fronts (planes of constant phase) which intersect k at right angles. The phase shift between adjacent fronts is set at 2π in the ﬁgure; this corresponds to a spatial separation along the k direction of one wavelength, as shown. The phase fronts intersect the x axis, and we see that along x the front separation is greater than it was along k. λx is the spacing between fronts along x and is indicated

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on the ﬁgure. The phase velocity along x is the velocity of the intersection points between the phase fronts and the x axis. Again, from the geometry, we see that this velocity must be faster than the velocity along k and will, of course, exceed the speed of light in the medium. This does not constitute a violation of special relativity, however, since the energy in the wave ﬂows in the direction of k and not along x or z. The wave frequency is f = ω/2π and is invariant with direction. Note, for example, that in the directions we have considered, νp νpx ω f = = = λ λx 2π
E X A M P L E 12.6

Consider a 50-MHz uniform plane wave having electric ﬁeld amplitude 10 V/m. The medium is lossless, having r = r = 9.0 and µr = 1.0. The wave propagates in the x, y plane at a 30◦ angle to the x axis and is linearly polarized along z. Write down the phasor expression for the electric ﬁeld.
Solution. The propagation constant magnitude is

ω √ k=ω µ = The vector k is now

√ c

r

=

2π × 50 × 106 (3) = 3.2 m−1 3 × 108

k = 3.2(cos 30ax + sin 30a y ) = 2.8ax + 1.6a y m−1 Then With the electric ﬁeld directed along z, the phasor form becomes Es = E 0 e− jk · r az = 10e− j(2.8x+1.6y) az D12.4. For Example 12.6, calculate λx , λ y , νpx , and νpy .
Ans. 2.2 m; 3.9 m; 1.1 × 108 m/s; 2.0 × 108 m/s

r = x ax + y a y

12.5 PLANE WAVE REFLECTION AT OBLIQUE INCIDENCE ANGLES
We now consider the problem of wave reﬂection from plane interfaces, in which the incident wave propagates at some angle to the surface. Our objectives are (1) to determine the relation between incident, reﬂected, and transmitted angles, and (2) to derive reﬂection and transmission coefﬁcients that are functions of the incident angle and wave polarization. We will also show that cases exist in which total reﬂection or total transmission may occur at the interface between two dielectrics if the angle of incidence and the polarization are appropriately chosen.

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Figure 12.7 Geometries for plane wave incidence at angle θ1 onto an interface between dielectrics having intrinsic impedances η1 and η2 . The two polarization cases are shown: (a) p-polarization (or TM), with E in the plane of incidence; (b) s-polarization (or TE), with E perpendicular to the plane of incidence.

The situation is illustrated in Figure 12.7, in which the incident wave direction and position-dependent phase are characterized by wavevector k+ . The angle of incidence 1 is the angle between k+ and a line that is normal to the surface (the x axis in this case). 1 The incidence angle is shown as θ1 . The reﬂected wave, characterized by wavevector k− , will propagate away from the interface at angle θ1 . Finally, the transmitted wave, 1 characterized by k2 , will propagate into the second region at angle θ2 as shown. One would suspect (from previous experience) that the incident and reﬂected angles are equal (θ1 = θ1 ), which is correct. We need to show this, however, to be complete. The two media are lossless dielectrics, characterized by intrinsic impedances η1 and η2 . We will assume, as before, that the materials are nonmagnetic, and thus have permeability µ0 . Consequently, the materials are adequately described by specifying √ their dielectric constants, r 1 and r 2 , or their refractive indices, n 1 = r 1 and √ n2 = r 2. In Figure 12.7, two cases are shown that differ by the choice of electric ﬁeld orientation. In Figure 12.7a, the E ﬁeld is polarized in the plane of the page, with H therefore perpendicular to the page and pointing outward. In this illustration, the plane of the page is also the plane of incidence, which is more precisely deﬁned as the plane spanned by the incident k vector and the normal to the surface. With E lying in the plane of incidence, the wave is said to have parallel polarization or to be p-polarized (E is parallel to the incidence plane). Note that although H is perpendicular to the incidence plane, it lies parallel (or transverse) to the interface. Consequently, another name for this type of polarization is transverse magnetic, or TM polarization. Figure 12.7b shows the situation in which the ﬁeld directions have been rotated by 90◦ . Now H lies in the plane of incidence, whereas E is perpendicular to the plane. Because E is used to deﬁne polarization, the conﬁguration is called perpendicular

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ENGINEERING ELECTROMAGNETICS

polarization, or is said to be s-polarized.3 E is also parallel to the interface, and so the case is also called transverse electric, or TE polarization. We will ﬁnd that the reﬂection and transmission coefﬁcients will differ for the two polarization types, but that reﬂection and transmission angles will not depend on polarization. We only need to consider s- and p-polarizations because any other ﬁeld direction can be constructed as some combination of s and p waves. Our desired knowledge of reﬂection and transmission coefﬁcients, as well as how the angles relate, can be found through the ﬁeld boundary conditions at the interface. Speciﬁcally, we require that the transverse components of E and H be continuous across the interface. These were the conditions we used to ﬁnd and τ for normal incidence (θ1 = 0), which is in fact a special case of our current problem. We will consider the case of p-polarization (Figure 12.7a) ﬁrst. To begin, we write down the incident, reﬂected, and transmitted ﬁelds in phasor form, using the notation developed in Section 12.4: E+ = E+ e− jk1 s1 10 E− = E− e− jk1 s1 10 where k+ = k1 (cos θ1 ax + sin θ1 az ) 1 k− = k1 (− cos θ1 ax + sin θ1 az ) 1 k2 = k2 (cos θ2 ax + sin θ2 az ) and where r = x a x + z az (57) √ √ The wavevector magnitudes are k1 = ω r 1 /c = n 1 ω/c and k2 = ω r 2 /c = n 2 ω/c. Now, to evaluate the boundary condition that requires continuous tangential electric ﬁeld, we need to ﬁnd the components of the electric ﬁelds (z components) that are parallel to the interface. Projecting all E ﬁelds in the z direction, and using (51) through (57), we ﬁnd
+ + E zs1 = E z10 e− jk1
+ +

·r ·r

(51) (52) (53)

−

Es2 = E20 e− jk2 · r

(54) (55) (56)

·r ·r

+ = E 10 cos θ1 e− jk1 (x cos θ1 +z sin θ1 ) − = E 10 cos θ1 e jk1 (x cos θ1 −z sin θ1 )

(58) (59) (60)

− − E zs1 = E z10 e− jk1

−

E zs2 = E z20 e− jk2 · r = E 20 cos θ2 e− jk2 (x cos θ2 +z sin θ2 )
The s designation is an abbreviation for the German senkrecht, meaning perpendicular. The p in p-polarized is an abbreviation for the German word for parallel, which is parallel.
3

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The boundary condition for a continuous tangential electric ﬁeld now reads: We now substitute Eqs. (58) through (60) into (61) and evaluate the result at x = 0 to obtain Note that and E 20 are all constants (independent of z). Further, we require that (61) hold for all values of z (everywhere on the interface). For this to occur, it must follow that all the phase terms appearing in (61) are equal. Speciﬁcally, k1 z sin θ1 = k1 z sin θ1 = k2 z sin θ2 From this, we see immediately that θ1 = θ1 , or the angle of reﬂection is equal to the angle of incidence. We also ﬁnd that k1 sin θ1 = k2 sin θ2 (62)
+ − E 10 cos θ1 e− jk1 z sin θ1 + E 10 cos θ1 e− jk1 z sin θ1 = E 20 cos θ2 e− jk2 z sin θ2 + E 10 , − E 10 , + − E zs1 + E zs1 = E zs2

(at x = 0)

(61)

Equation (62) is known as Snell’s law of refraction. Because, in general, k = nω/c, we can rewrite (62) in terms of the refractive indices: n 1 sin θ1 = n 2 sin θ2 (63)

Equation (63) is the form of Snell’s law that is most readily used for our present case of nonmagnetic dielectrics. Equation (62) is a more general form which would apply, for example, to cases involving materials with different permeabilities as well √ as different permittivities. In general, we would have k1 = (ω/c) µr 1 r 1 and k2 = √ (ω/c) µr 2 r 2 . Having found the relations between angles, we next turn to our second objective, + − which is to determine the relations between the amplitudes, E 10 , E 10 , and E 20 . To accomplish this, we need to consider the other boundary condition, requiring tangential continuity of H at x = 0. The magnetic ﬁeld vectors for the p-polarized wave are all negative y-directed. At the boundary, the ﬁeld amplitudes are related through
+ − H10 + H10 = H20

(64)

Then, when we use the fact that θ1 = θ1 and invoke Snell’s law, (61) becomes
+ − E 10 cos θ1 + E 10 cos θ1 = E 20 cos θ2

(65)
+ + E 10 /H10

Using the medium intrinsic impedances, we know, for example, that + + and E 20 /H20 = η2 . Eq. (64) can be written as follows:
+ E − cos θ1 E + cos θ2 E 10 cos θ1 − 10 = 20 η1 p η1 p η2 p

= η1 (66)

Note the minus sign in front of the second term in (66), which results from the fact − − that E 10 cos θ1 is negative (from Figure 12.7a), whereas H10 is positive (again from the ﬁgure). When we write Eq. (66), effective impedances, valid for p-polarization,

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ENGINEERING ELECTROMAGNETICS

are deﬁned through η1 p = η1 cos θ1 and η2 p = η2 cos θ2 (68) (67)

Using this representation, Eqs. (65) and (66) are now in a form that enables them − + + to be solved together for the ratios E 10 /E 10 and E 20 /E 10 . Performing analogous procedures to those used in solving (7) and (8), we ﬁnd the reﬂection and transmission coefﬁcients: p =

− η2 p − η1 p E 10 + = η2 p + η1 p E 10

(69)

τp =

2η2 p E 20 + = η2 p + η1 p E 10

cos θ1 cos θ2

(70)

A similar procedure can be carried out for s-polarization, referring to Figure 12.7b. The details are left as an exercise; the results are s =

− E y10 + E y10

=

η2s − η1s η2s + η1s

(71)

τs =

E y20 2η2s + = η2s + η1s E y10

(72)

where the effective impedances for s-polarization are η1s = η1 sec θ1 and η2s = η2 sec θ2 (74) (73)

Equations (67) through (74) are what we need to calculate wave reﬂection and transmission for either polarization, and at any incident angle.
E X A M P L E 12.7

A uniform plane wave is incident from air onto glass at an angle from the normal of 30◦ . Determine the fraction of the incident power that is reﬂected and transmitted for (a) p-polarization and (b) s-polarization. Glass has refractive index n 2 = 1.45.

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Solution. First, we apply Snell’s law to ﬁnd the transmission angle. Using n 1 = 1 for air, we use (63) to ﬁnd

θ2 = sin−1 Now, for p-polarization:

sin 30 1.45

= 20.2◦

η1 p = η1 cos 30 = (377)(.866) = 326 η2 p = η2 cos 20.2 = Then, using (69), we ﬁnd 244 − 326 = −0.144 244 + 326 The fraction of the incident power that is reﬂected is p 377 (.938) = 244 1.45

=

Pr =| Pinc

2 p|

= .021
2

The transmitted fraction is then Pt =1−| Pinc For s-polarization, we have

p|

= .979

η1s = η1 sec 30 = 377/.866 = 435 η2s = η2 sec 20.2 = Then, using (71): 277 − 435 = −.222 277 + 435 The reﬂected power fraction is thus s 377 = 277 1.45(.938)

=

|

2 s|

= .049 = .951

The fraction of the incident power that is transmitted is 1−|
2 s|

In Example 12.7, reﬂection coefﬁcient values for the two polarizations were found to be negative. The meaning of a negative reﬂection coefﬁcient is that the component of the reﬂected electric ﬁeld that is parallel to the interface will be directed opposite the incident ﬁeld component when both are evaluated at the boundary. This effect is also observed when the second medium is a perfect conductor. In this case, we know that the electric ﬁeld inside the conductor must be zero. Consequently, η2 = E 20 /H20 = 0, and the reﬂection coefﬁcients will be p = s = −1. Total reﬂection occurs, regardless of the incident angle or polarization.

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ENGINEERING ELECTROMAGNETICS

12.6 TOTAL REFLECTION AND TOTAL TRANSMISSION OF OBLIQUELY INCIDENT WAVES
Now that we have methods available to us for solving problems involving oblique incidence reﬂection and transmission, we can explore the special cases of total reflectio and total transmission. We look for special combinations of media, incidence angles, and polarizations that produce these properties. To begin, we identify the necessary ∗ condition for total reﬂection. We want total power reﬂection, so that | |2 = = 1, where is either p or s . The fact that this condition involves the possibility of a complex allows some ﬂexibility. For the incident medium, we note that η1 p and η1s will always be real and positive. On the other hand, when we consider the second medium, η2 p and η2s involve factors of cos θ2 or 1/ cos θ2 , where cos θ2 = 1 − sin θ2
2 1/2

= 1−

n1 n2

2

1/2

sin θ1

2

(75)

where Snell’s law has been used. We observe that cos θ2 , and hence η2 p and η2s , become imaginary whenever sin θ1 > n 2 /n 1 . Let us consider parallel polarization, for example. Under conditions of imaginary η2 p , (69) becomes p =

j|η2 p | − η1 p η1 p − j|η2 p | Z =− =− ∗ j|η2 p | + η1 p η1 p + j|η2 p | Z

where Z = η1 p − j|η2 p |. We can therefore see that p ∗ = 1, meaning total power p reflection whenever η2 p is imaginary. The same will be true whenever η2 p is zero, which will occur when sin θ1 = n 2 /n 1 . We thus have our condition for total reﬂection, which is sin θ1 ≥ n2 n1 (76)

From this condition arises the critical angle of total reﬂection, θc , deﬁned through sin θc = n2 n1 (77)

The total reﬂection condition can thus be more succinctly written as θ1 ≥ θc (for total reﬂection) (78)

Note that for (76) and (77) to make sense, it must be true that n 2 < n 1 , or the wave must be incident from a medium of higher refractive index than that of the medium beyond the boundary. For this reason, the total reﬂection condition is sometimes called total internal reﬂection; it is often seen (and applied) in optical devices such

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Figure 12.8 Beam-steering prism for Example 12.8.

as beam-steering prisms, where light within the glass structure totally reﬂects from glass-air interfaces.
E X A M P L E 12.8

A prism is to be used to turn a beam of light by 90 , as shown in Figure 12.8. Light enters and exits the prism through two antireﬂective (AR-coated) surfaces. Total reﬂection is to occur at the back surface, where the incident angle is 45◦ to the normal. Determine the minimum required refractive index of the prism material if the surrounding region is air.
Solution. Considering the back surface, the medium beyond the interface is air, with

◦

n 2 = 1.00. Because θ1 = 45◦ , (76) is used to obtain √ n2 = 2 = 1.41 n1 ≥ sin 45 Because fused silica glass has refractive index n g = 1.45, it is a suitable material for this application and is in fact widely used.

Another important application of total reﬂection is in optical waveguides. These, in their simplest form, are constructed of three layers of glass, in which the middle layer has a slightly higher refractive index than the outer two. Figure 12.9 shows the basic structure. Light, propagating from left to right, is conﬁned to the middle layer by total reﬂection at the two interfaces, as shown. Optical ﬁber waveguides are constructed on this principle, in which a cylindrical glass core region of small radius is surrounded coaxially by a lower-index cladding glass material of larger radius. Basic waveguiding principles as applied to metallic and dielectric structures will be presented in Chapter 13.

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ENGINEERING ELECTROMAGNETICS

Figure 12.9 A dielectric slab waveguide (symmetric case), showing light confinement to the center material by total reflection.

We next consider the possibility of total transmission. In this case, the requirement is simply that = 0. We investigate this possibility for the two polarizations. First, we consider s-polarization. If s = 0, then from (71) we require that η2s = η1s , or η2 sec θ2 = η1 sec θ1 Using Snell’s law to write θ2 in terms of θ1 , the preceding equation becomes η2 1 − n1 n2
2

sin θ1

2

−1/2

= η1 1 − sin2 θ1

−1/2

There is no value of θ1 that will satisfy this, so we turn instead to p-polarization. Using (67), (68), and (69), with Snell’s law, we ﬁnd that the condition for p = 0 is η2 1 − n1 n2
2 1/2

sin θ1

2

= η1 1 − sin2 θ1

1/2

This equation does have a solution, which is sin θ1 = sin θ B = n2 n2 + n2 1 2 (79)

where we have used η1 = η0 /n 1 and η2 = η0 /n 2 . We call this special angle θ B , where total transmission occurs, the Brewster angle or polarization angle. The latter name comes from the fact that if light having both s- and p-polarization components is incident at θ1 = θ B , the p component will be totally transmitted, leaving the partially reﬂected light entirely s-polarized. At angles that are slightly off the Brewster angle, the reﬂected light is still predominantly s-polarized. Most reﬂected light that we see originates from horizontal surfaces (such as the surface of the ocean), and so the light has mostly horizontal polarization. Polaroid sunglasses take advantage of this fact to reduce glare, for they are made to block the transmission of horizontally polarized light while passing light that is vertically polarized.

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E X A M P L E 12.9

Light is incident from air to glass at Brewster’s angle. Determine the incident and transmitted angles.
Solution. Because glass has refractive index n 2 = 1.45, the incident angle will be



θ1 = θ B = sin−1 

n2 n2 1 + n2 2



 = sin−1

√

1.45 1.452 + 1

= 55.4◦

The transmitted angle is found from Snell’s law, through   n1 n1  = 34.6◦ θ2 = sin−1 sin θ B = sin−1  n2 2 2 n +n
1 2

Note from this exercise that sin θ2 = cos θ B , which means that the sum of the incident and refracted angles at the Brewster condition is always 90◦ . Many of the results we have seen in this section are summarized in Figure 12.10, in which p and s , from (69) and (71), are plotted as functions of the incident angle, θ1 . Curves are shown for selected values of the refractive index ratio, n 1 /n 2 . For all plots in which n 1 /n 2 > 1, s and p achieve values of ±1 at the critical angle. At larger angles, the reﬂection coefﬁcients become imaginary (and are not shown) but nevertheless retain magnitudes of unity. The occurrence of the Brewster angle is evident in the curves for p (Figure 12.10a) because all curves cross the θ1 axis. This behavior is not seen in the s functions because s is positive for all values of θ1 when n 1 /n 2 > 1. D12.5. In Example 12.9, calculate the reﬂection coefﬁcient for s-polarized light.
Ans. −0.355

12.7 WAVE PROPAGATION IN DISPERSIVE MEDIA
In Chapter 11, we encountered situations in which the complex permittivity of the medium depends on frequency. This is true in all materials through a number of possible mechanisms. One of these, mentioned earlier, is that oscillating bound charges in a material are in fact harmonic oscillators that have resonant frequencies associated with them (see Appendix D). When the frequency of an incoming electromagnetic wave is at or near a bound charge resonance, the wave will induce strong oscillations; these in turn have the effect of depleting energy from the wave in its original form. The wave thus experiences absorption, and it does so to a greater extent than it would at a frequency that is detuned from resonance. A related effect is that the

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(a)

(b)

Figure 12.10 (a) Plots of p [Eq. (69)] as functions of the incident angle, θ1 , as shown in Figure 12.7a. Curves are shown for selected values of the refractive index ratio, n1 /n2 . Both media are lossless and have µr = 1. Thus η1 = η0 /n1 and η2 = η0 /n2 . (b) Plots of s [Eq. (71)] as functions of the incident angle, θ1 , as shown in Figure 12.7b. As in Figure 12.10a, the media are lossless, and curves are shown for selected n1 /n2 .

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Figure 12.11 The angular dispersion of a prism can be measured using a movable device which measures both wavelength and power. The device senses light through a small aperture, thus improving wavelength resolution.

real part of the dielectric constant will be different at frequencies near resonance than at frequencies far from resonance. In short, resonance effects give rise to values of and that will vary continuously with frequency. These in turn will produce a fairly complicated frequency dependence in the attenuation and phase constants as expressed in Eqs. (44) and (45) in Chapter 11. This section concerns the effect of a frequency-varying dielectric constant (or refractive index) on a wave as it propagates in an otherwise lossless medium. This situation arises quite often because signiﬁcant refractive index variation can occur at frequencies far away from resonance, where absorptive losses are negligible. A classic example of this is the separation of white light into its component colors by a glass prism. In this case, the frequency-dependent refractive index results in different angles of refraction for the different colors—hence the separation. The color separation effect produced by the prism is known as angular dispersion, or more speciﬁcally, chromatic angular dispersion. The term dispersion implies a separation of distinguishable components of a wave. In the case of the prism, the components are the various colors that have been spatially separated. An important point here is that the spectral power has been dispersed by the prism. We can illustrate this idea by considering what it would take to measure the difference in refracted angles between, for example, blue and red light. One would need to use a power detector with a very narrow aperture, as shown in Figure 12.11. The detector would be positioned at the locations of the blue and red light from the prism, with the narrow aperture allowing essentially one color at a time (or light over a very narrow spectral range) to pass through to the detector. The detector would then measure the power in what we could call a “spectral packet,” or a very narrow slice of the total power spectrum. The smaller the aperture, the narrower the spectral width of the packet, and the greater the precision in the measurement.4 It

4

To perform this experiment, one would need to measure the wavelength as well. To do this, the detector would likely be located at the output of a spectrometer or monochrometer whose input slit performs the function of the bandwidth-limiting aperture.

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ENGINEERING ELECTROMAGNETICS

Figure 12.12 ω-β diagram for a material in which the refractive index increases with frequency. The slope of a line tangent to the curve at ω0 is the group velocity at that frequency. The slope of a line joining the origin to the point on the curve at ω0 is the phase velocity at ω0 .

is important for us to think of wave power as subdivided into spectral packets in this way because it will ﬁgure prominently in our interpretation of the main topic of this section, which is wave dispersion in time. We now consider a lossless nonmagnetic medium in which the refractive index varies with frequency. The phase constant of a uniform plane wave in this medium will assume the form β(ω) = k = ω µ0 (ω) = n(ω) ω c (80)

If we take n(ω) to be a monotonically increasing function of frequency (as is usually the case), a plot of ω versus β would look something like the curve shown in Figure 12.12. Such a plot is known as an ω-β diagram for the medium. Much can be learned about how waves propagate in the material by considering the shape of the ω-β curve. Suppose we have two waves at two frequencies, ωa and ωb , which are copropagating in the material and whose amplitudes are equal. The two frequencies are labeled on the curve in Figure 12.12, along with the frequency midway between the two, ω0 . The corresponding phase constants, βa , βb , and β0 , are also labeled. The electric ﬁelds of the two waves are linearly polarized in the same direction (along x, for example), while both waves propagate in the forward z direction. The waves will thus interfere with each other, producing a resultant wave whose ﬁeld function can be found simply by adding the E ﬁelds of the two waves. This addition is done using

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the complex ﬁelds: E c,net (z, t) = E 0 [e− jβa z e jωa t + e− jβb z e jωb t ] Note that we must use the full complex forms (with frequency dependence retained) as opposed to the phasor forms, since the waves are at different frequencies. Next, we factor out the term e− jβ0 z e jω0 t : E c,net (z, t) = E 0 e− jβ0 z e jω0 t [e j = 2E 0 e
− jβ0 z jω0 t βz − j ωt

e

e

cos( ωt − βz)

+ e− j

βz j ωt

e

] (81)

where ω = ω0 − ωa = ωb − ω0 and β = β0 − βa = βb − β0 The preceding expression for β is approximately true as long as ω is small. This can be seen from Figure 12.12 by observing how the shape of the curve affects β, given uniform frequency spacings. The real instantaneous form of (81) is found through Enet (z, t) = Re{E c,net } = 2E 0 cos( ωt − βz) cos(ω0 t − β0 z) (82)

If ω is fairly small compared to ω0 , we recognize (82) as a carrier wave at frequency ω0 that is sinusoidally modulated at frequency ω. The two original waves are thus “beating” together to form a slow modulation, as one would hear when the same note is played by two slightly out-of-tune musical instruments. The resultant wave is shown in Figure 12.13. net Figure 12.13 Plot of the total electric field strength as a function of z (with t = 0) of two co-propagating waves having different frequencies, ωa and ωb, as per Eq. (81). The rapid oscillations are associated with the carrier frequency, ω0 = (ωa + ωb)/2. The slower modulation is associated with the envelope or ‘‘beat’’frequency, ω = (ωb − ωa )/2.

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ENGINEERING ELECTROMAGNETICS

Of interest to us are the phase velocities of the carrier wave and the modulation envelope. From (82), we can immediately write these down as: ω0 νpc = (carrier velocity) (83) β0 ω (envelope velocity) (84) β Referring to the ω-β diagram, Figure 12.12, we recognize the carrier phase velocity as the slope of the straight line that joins the origin to the point on the curve whose coordinates are ω0 and β0 . We recognize the envelope velocity as a quantity that approximates the slope of the ω-β curve at the location of an operation point speciﬁed by (ω0 , β0 ). The envelope velocity in this case is thus somewhat less than the carrier velocity. As ω becomes vanishingly small, the envelope velocity is exactly the slope of the curve at ω0 . We can therefore state the following for our example: νpe = lim ω dω = β dβ = νg (ω0 ) (85)

ω→0

ω0

The quantity dω/dβ is called the group velocity function for the material, νg (ω). When evaluated at a speciﬁed frequency ω0 , it represents the velocity of a group of frequencies within a spectral packet of vanishingly small width, centered at frequency ω0 . In stating this, we have extended our two-frequency example to include waves that have a continuous frequency spectrum. Each frequency component (or packet) is associated with a group velocity at which the energy in that packet propagates. Since the slope of the ω-β curve changes with frequency, group velocity will obviously be a function of frequency. The group velocity dispersion of the medium is, to the ﬁrst order, the rate at which the slope of the ω-β curve changes with frequency. It is this behavior that is of critical practical importance to the propagation of modulated waves within dispersive media and to understanding the extent to which the modulation envelope may degrade with propagation distance.
E X A M P L E 12.10

Consider a medium in which the refractive index varies linearly with frequency over a certain range: ω n(ω) = n 0 ω0 Determine the group velocity and the phase velocity of a wave at frequency ω0 .
Solution. First, the phase constant will be

β(ω) = n(ω) Now

n 0 ω2 ω = c ω0 c

2n 0 ω dβ = dω ω0 c

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so that νg = The group velocity at ω0 is νg (ω0 ) = The phase velocity at ω0 will be νp (ω0 ) = dω ω0 c = dβ 2n 0 ω c 2n 0

ω c = β(ω0 ) n0

12.8 PULSE BROADENING IN DISPERSIVE MEDIA
To see how a dispersive medium affects a modulated wave, let us consider the propagation of an electromagnetic pulse. Pulses are used in digital signals, where the presence or absence of a pulse in a given time slot corresponds to a digital “one” or “zero.” The effect of the dispersive medium on a pulse is to broaden it in time. To see how this happens, we consider the pulse spectrum, which is found through the Fourier transform of the pulse in time domain. In particular, suppose the pulse shape in time is Gaussian, and has electric ﬁeld given at position z = 0 by E(0, t) = E 0 e− 2 (t/T ) e jω0 t
1 2

(86)

where E 0 is a constant, ω0 is the carrier frequency, and T is the characteristic halfwidth of the pulse envelope; this is the time at which the pulse intensity, or magnitude of the Poynting vector, falls to 1/e of its maximum value (note that intensity is proportional to the square of the electric ﬁeld). The frequency spectrum of the pulse is the Fourier transform of (86), which is E0 T 1 2 2 E(0, ω) = √ e− 2 T (ω−ω0 ) 2π (87)

Note from (87) that the frequency displacement from ω0 at which the spectral intensity (proportional to |E(0, ω)|2 ) falls to 1/e of its maximum is ω = ω − ω0 = 1/T . Figure 12.14a shows the Gaussian intensity spectrum of the pulse, centered at ω0 , where the frequencies corresponding to the 1/e spectral intensity positions, ωa and ωb , are indicated. Figure 12.14b shows the same three frequencies marked on the ω-β curve for the medium. Three lines are drawn that are tangent to the curve at the three frequency locations. The slopes of the lines indicate the group velocities at ωa , ωb , and ω0 , indicated as νga , νgb , and νg0 . We can think of the pulse spreading in time as resulting from the differences in propagation times of the spectral energy packets that make up the pulse spectrum. Since the pulse spectral energy is highest

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(a)

(b)

Figure 12.14 (a) Normalized power spectrum of a Gaussian pulse, as determined from Eq. (86). The spectrum is centered at carrier frequency ω0 and has 1/e half-width, ω. Frequencies ωa and ωb correspond to the 1/e positions on the spectrum. (b) The spectrum of Figure 12.14a as shown on the ω-β diagram for the medium. The three frequencies specified in Figure 12.14a are associated with three different slopes on the curve, resulting in different group delays for the spectral components.

at the center frequency, ω0 , we can use this as a reference point about which further spreading of the energy will occur. For example, let us consider the difference in arrival times (group delays) between the frequency components, ω0 and ωb , after propagating through a distance z of the medium: τ =z 1 1 − νgb νg0 =z dβ dω − dβ dω (88) ω0 ωb

The essential point is that the medium is acting as what could be called a temporal prism. Instead of spreading out the spectral energy packets spatially, it is spreading

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them out in time. In this process, a new temporal pulse envelope is constructed whose width is based fundamentally on the spread of propagation delays of the different spectral components. By determining the delay difference between the peak spectral component and the component at the spectral half-width, we construct an expression for the new temporal half-width. This assumes, of course, that the initial pulse width is negligible in comparison, but if not, we can account for that also, as will be shown later on. To evaluate (88), we need more information about the ω-β curve. If we assume that the curve is smooth and has fairly uniform curvature, we can express β(ω) as the ﬁrst three terms of a Taylor series expansion about the carrier frequency, ω0 : 1 . β(ω) = β(ω0 ) + (ω − ω0 )β1 + (ω − ω0 )2 β2 2 where β0 = β(ω0 ) β1 = and β2 = d 2β dω2 (91) ω0 (89)

dβ dω

(90) ω0 Note that if the ω-β curve were a straight line, then the ﬁrst two terms in (89) would precisely describe β(ω). It is the third term in (89), involving β2 , that describes the curvature and ultimately the dispersion. Noting that β0 , β1 , and β2 are constants, we take the ﬁrst derivative of (89) with respect to ω to ﬁnd dβ = β1 + (ω − ω0 )β2 dω We now substitute (92) into (88) to obtain τ = [β1 + (ωb − ω0 )β2 ] z − [β1 + (ω0 − ω0 )β2 ] z = ωβ2 z = β2 z T (93) (92)

where ω = (ωb −ω0 ) = 1/T . β2 , as deﬁned in Eq. (91), is the dispersion parameter. Its units are in general time2 /distance, that is, pulse spread in time per unit spectral bandwidth, per unit distance. In optical ﬁbers, for example, the units most commonly used are picoseconds2 /kilometer (psec2 /km). β2 can be determined when we know how β varies with frequency, or it can be measured. If the initial pulse width is very short compared to τ , then the broadened pulse width at location z will be simply τ . If the initial pulse width is comparable to τ , then the pulse width at z can be found through the convolution of the initial Gaussian

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pulse envelope of width T with a Gaussian envelope whose width is general, the pulse width at location z will be T =
E X A M P L E 12.11

τ . Thus, in

T 2 + ( τ )2

(94)

An optical ﬁber link is known to have dispersion β2 = 20 ps2 /km. A Gaussian light pulse at the input of the ﬁber is of initial width T = 10 ps. Determine the width of the pulse at the ﬁber output if the ﬁber is 15 km long.
Solution. The pulse spread will be

τ= So the output pulse width is T =

β2 z (20)(15) = = 30 ps T 10 (10)2 + (30)2 = 32 ps

An interesting by-product of pulse broadening through chromatic dispersion is that the broadened pulse is chirped. This means that the instantaneous frequency of the pulse varies monotonically (either increases or decreases) with time over the pulse envelope. This again is just a manifestation of the broadening mechanism, in which the spectral components at different frequencies are spread out in time as they propagate at different group velocities. We can quantify the effect by calculating the group delay, τg , as a function of frequency, using (92). We obtain: τg = z dβ =z = (β1 + (ω − ω0 )β2 ) z νg dω (95)

This equation tells us that the group delay will be a linear function of frequency and that higher frequencies will arrive at later times if β2 is positive. We refer to the chirp as positive if the lower frequencies lead the higher frequencies in time [requiring a positive β2 in (95)]; chirp is negative if the higher frequencies lead in time (negative β2 ). Figure 12.15 shows the broadening effect and illustrates the chirping phenomenon. D12.6. For the ﬁber link of Example 12.11, a 20-ps pulse is input instead of the 10-ps pulse in the example. Determine the output pulsewidth.
Ans. 25 ps

As a ﬁnal point, we note that the pulse bandwidth, ω, was found to be 1/T . This is true as long as the Fourier transform of the pulse envelope is taken, as was done with (86) to obtain (87). In that case, E 0 was taken to be a constant, and so the only time variation arose from the carrier wave and the Gaussian envelope. Such a

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Figure 12.15 Gaussian pulse intensities as functions of time (smooth curves) before and after propagation through a dispersive medium, as exemplified by the ω-β diagram of Figure 12.14b. The electric field oscillations are shown under the second trace to demonstrate the chirping effect as the pulse broadens. Note the reduced amplitude of the broadened pulse, which occurs because the pulse energy (the area under the intensity envelope) is constant.

pulse, whose frequency spectrum is obtained only from the pulse envelope, is known as transform-limited. In general, however, additional frequency bandwidth may be present since E 0 may vary with time for one reason or another (such as phase noise that could be present on the carrier). In these cases, pulse broadening is found from the more general expression τ= ωβ2 z (96)

where ω is the net spectral bandwidth arising from all sources. Clearly, transformlimited pulses are preferred in order to minimize broadening because these will have the smallest spectral width for a given pulse width.

REFERENCES
1. DuBroff, R. E., S. V. Marshall, and G. G. Skitek. Electromagnetic Concepts and Applications. 4th ed. Englewood Cliffs, N. J.: Prentice-Hall, 1996. Chapter 9 of this text develops the concepts presented here, with additional examples and applications. 2. Iskander, M. F. Electromagnetic Fields and Waves. Englewood Cliffs, N. J.: Prentice-Hall, 1992. The multiple interface treatment in Chapter 5 of this text is particularly good. 3. Harrington, R. F. Time-Harmonic Electromagnetic Fields. New York: McGraw-Hill, 1961. This advanced text provides a good overview of general wave reﬂection concepts in Chapter 2. 4. Marcuse, D. Light Transmission Optics. New York: Van Nostrand Reinhold, 1982. This intermediate-level text provides detailed coverage of optical waveguides and pulse propagation in dispersive media.

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CHAPTER 12 PROBLEMS
12.1
+ + A uniform plane wave in air, E x1 = E x10 cos(1010 t − βz) V/m, is normally incident on a copper surface at z = 0. What percentage of the incident power density is transmitted into the copper?

12.2

The plane z = 0 deﬁnes the boundary between two dielectrics. For z < 0, r 1 = 9, r 1 = 0, and µ1 = µ0 . For z > 0, r 2 = 3, r 2 = 0, and µ2 = µ0 . + Let E x1 = 10 cos(ωt − 15z) V/m and ﬁnd (a) ω; (b) S+ ; (c) S− ; 1 1 (d) S+ . 2 A uniform plane wave in region 1 is normally incident on the planar 3 boundary separating regions 1 and 2. If 1 = 2 = 0, while r 1 = µr 1 and 3 r 2 = µr 2 , ﬁnd the ratio r 2 / r 1 if 20% of the energy in the incident wave is reﬂected at the boundary. There are two possible answers. A 10 MHz uniform plane wave having an initial average power density of 5 W/m2 is normally incident from free space onto the surface of a lossy material in which 2 / 2 = 0.05, r 2 = 5, and µ2 = µ0 . Calculate the distance into the lossy medium at which the transmitted wave power density is down by 10 dB from the initial 5 W/m2 . The region z < 0 is characterized by r = µr = 1 and r = 0. The total E ﬁeld here is given as the sum of two uniform plane waves, Es = 150 e− j10z ax + (50 20◦ ) e j10z ax V/m. (a) What is the operating frequency? (b) Specify the intrinsic impedance of the region z > 0 that would provide the appropriate reﬂected wave. (c) At what value of z, −10 cm < z < 0, is the total electric ﬁeld intensity a maximum amplitude? In the beam-steering prism of Example 12.8, suppose the antireﬂective coatings are removed, leaving bare glass-to-air interfaces. Calcluate the ratio of the prism output power to the input power, assuming a single transit. The semi-inﬁnite regions z < 0 and z > 1 m are free space. For 0 < z < 1 m, r = 4, µr = 1, and r = 0. A uniform plane wave with ω = 4 × 108 rad/s is traveling in the az direction toward the interface at z = 0. (a) Find the standing wave ratio in each of the three regions. (b) Find the location of the maximum |E| for z < 0 that is nearest to z = 0. A wave starts at point a, propagates 1 m through a lossy dielectric rated at 0.1 dB/cm, reﬂects at normal incidence at a boundary at which = 0.3 + j0.4, and then returns to point a. Calculate the ratio of the ﬁnal power to the incident power after this round trip, and specify the overall loss in decibels. Region 1, z < 0, and region 2, z > 0, are both perfect dielectrics (µ = µ0 , = 0). A uniform plane wave traveling in the az direction has a radian frequency of 3 × 1010 rad/s. Its wavelengths in the two regions are λ1 = 5 cm and λ2 = 3 cm. What percentage of the energy incident on the

12.3

12.4

12.5

12.6

12.7

12.8

12.9

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boundary is (a) reﬂected; (b) transmitted? (c) What is the standing wave ratio in region 1? 12.10 In Figure 12.1, let region 2 be free space, while µr 1 = 1, r 1 = 0, and unknown. Find r 1 if (a) the amplitude of E− is one-half that of E+ ; 1 1 (b) S− is one-half of S+ ; (c) |E1 |min is one-half of |E1 |max . 1 1 r1 is

12.11

A 150-MHz uniform plane wave is normally incident from air onto a material whose intrinsic impedance is unknown. Measurements yield a standing wave ratio of 3 and the appearance of an electric ﬁeld minimum at 0.3 wavelengths in front of the interface. Determine the impedance of the unknown material. A 50-MHz uniform plane wave is normally incident from air onto the surface of a calm ocean. For seawater, σ = 4 S/m, and r = 78. (a) Determine the fractions of the incident power that are reﬂected and transmitted. (b) Qualitatively, how (if at all) will these answers change as the frequency is increased? A right-circularly polarized plane wave is normally incident from air onto a semi-inﬁnite slab of plexiglas ( r = 3.45, r = 0). Calculate the fractions of the incident power that are reﬂected and transmitted. Also, describe the polarizations of the reﬂected and transmitted waves. A left-circularly polarized plane wave is normally incident onto the surface of a perfect conductor. (a) Construct the superposition of the incident and reﬂected waves in phasor form. (b) Determine the real instantaneous form of the result of part (a). (c) Describe the wave that is formed. Sulfur hexaﬂuoride (SF6 ) is a high-density gas that has refractive index, n s = 1.8 at a speciﬁed pressure, temperature, and wavelength. Consider the retro-reﬂecting prism shown in Fig. 12.16, that is immersed in SF6 . Light enters through a quarter-wave antireﬂective coating and then totally reﬂects from the back surfaces of the glass. In principle, the beam should experience zero loss at the design wavelength (Pout = Pin ). (a) Determine the minimum required value of the glass refractive index, n g , so that the interior beam will totally reﬂect. (b) Knowing n g , ﬁnd the required refractive index of the quarter-wave ﬁlm, n f . (c) With the SF6 gas evacuated ns nf ng l /4 Pout

12.12

12.13

12.14

12.15

Pin

Figure 12.16 See Problem 12.15.

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from the chamber, and with the glass and ﬁlm values as previously found, ﬁnd the ratio, Pout /Pin . Assume very slight misalignment, so that the long beam path through the prism is not retraced by reﬂected waves. 12.16 In Figure 12.5, let regions 2 and 3 both be of quarter-wave thickness. Region 4 is glass, having refractive index, n 4 = 1.45; region 1 is air. (a) Find ηin,b . (b) Find ηin,a . (c) Specify a relation between the four intrinsic impedances that will enable total transmission of waves incident from the left into region 4. (d) Specify refractive index values for regions 2 and 3 that will accomplish the condition of part (c). (e) Find the fraction of incident power transmitted if the two layers were of half-wave thickness instead of quarter wave. A uniform plane wave in free space is normally incident onto a dense dielectric plate of thickness λ/4, having refractive index n. Find the required value of n such that exactly half the incident power is reﬂected (and half transmitted). Remember that n > 1. A uniform plane wave is normally incident onto a slab of glass (n = 1.45) whose back surface is in contact with a perfect conductor. Determine the reﬂective phase shift at the front surface of the glass if the glass thickness is (a) λ/2; (b) λ/4; (c) λ/8. You are given four slabs of lossless dielectric, all with the same intrinsic impedance, η, known to be different from that of free space. The thickness of each slab is λ/4, where λ is the wavelength as measured in the slab material. The slabs are to be positioned parallel to one another, and the combination lies in the path of a uniform plane wave, normally incident. The slabs are to be arranged such that the airspaces between them are either zero, one-quarter wavelength, or one-half wavelength in thickness. Specify an arrangement of slabs and airspaces such that (a) the wave is totally transmitted through the stack, and (b) the stack presents the highest reﬂectivity to the incident wave. Several answers may exist. The 50-MHz plane wave of Problem 12.12 is incident onto the ocean surface at an angle to the normal of 60◦ . Determine the fractions of the incident power that are reﬂected and transmitted for (a) s-polarization, and (b) p-polarization. A right-circularly polarized plane wave in air is incident at Brewster’s angle onto a semi-inﬁnite slab of plexiglas ( r = 3.45, r = 0). (a) Determine the fractions of the incident power that are reﬂected and transmitted. (b) Describe the polarizations of the reﬂected and transmitted waves. A dielectric waveguide is shown in Figure 12.17 with refractive indices as labeled. Incident light enters the guide at angle φ from the front surface normal as shown. Once inside, the light totally reﬂects at the upper n 1 − n 2 interface, where n 1 > n 2 . All subsequent reﬂections from the upper and lower boundaries will be total as well, and so the light is conﬁned to the

12.17

12.18

12.19

12.20

12.21

12.22

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Figure 12.17 12.23.

See Problems 12.22 and

guide. Express, in terms of n 1 and n 2 , the maximum value of φ such that total conﬁnement will occur, with n 0 = 1. The quantity sin φ is known as the numerical aperture of the guide. 12.23 12.24 Suppose that φ in Figure 12.17 is Brewster’s angle, and that θ1 is the critical angle. Find n 0 in terms of n 1 and n 2 . A Brewster prism is designed to pass p-polarized light without any reﬂective loss. The prism of Figure 12.18 is made of glass (n = 1.45) and is in air. Considering the light path shown, determine the vertex angle α. In the Brewster prism of Figure 12.18, determine for s-polarized light the fraction of the incident power that is transmitted through the prism, and from this specify the dB insertion loss, deﬁned as 10log10 of that number. Show how a single block of glass can be used to turn a p-polarized beam of light through 180◦ , with the light suffering (in principle) zero reﬂective loss. The light is incident from air, and the returning beam (also in air) may be displaced sideways from the incident beam. Specify all pertinent angles and use n = 1.45 for glass. More than one design is possible here. Using Eq. (79) in Chapter 11 as a starting point, determine the ratio of the group and phase velocities of an electromagnetic wave in a good conductor. Assume conductivity does not vary with frequency. Over a small wavelength range, the refractive index of a certain material . varies approximately linearly with wavelength as n(λ) = n a + n b (λ − λa ), where n a , n b and λa are constants, and where λ is the free-space wavelength. (a) Show that d/dω = −(2πc/ω2 )d/dλ. (b) Using β(λ) = 2π n/λ,

12.25

12.26

12.27

12.28

Figure 12.18 See Problems 12.24 and 12.25.

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ENGINEERING ELECTROMAGNETICS

determine the wavelength-dependent (or independent) group delay over a unit distance. (c) Determine β2 from your result of part (b). (d) Discuss the implications of these results, if any, on pulse broadening. 12.29 A T = 5 ps transform-limited pulse propagates in a dispersive medium for which β2 = 10 ps2 /km. Over what distance will the pulse spread to twice its initial width? A T = 20 ps transform-limited pulse propagates through 10 km of a dispersive medium for which β2 = 12 ps2 /km. The pulse then propagates through a second 10 km medium for which β2 = −12 ps2 /km. Describe the pulse at the output of the second medium and give a physical explanation for what happened.

12.30

C H A P T E R

13

Guided Waves n this chapter, we investigate several structures for guiding electromagnetic waves, and we explore the principles by which these operate. Included are transmission lines, which we ﬁrst explored from the viewpoint of their currents and voltages in Chapter 10, and which we now revisit from a ﬁelds point of view. We then broaden the discussion to include several waveguiding devices. Broadly deﬁned, a waveguide is a structure through which electromagnetic waves can be transmitted from point to point and within which the ﬁelds are conﬁned to a certain extent. A transmission line ﬁts this description, but it is a special case that employs two conductors, and it propagates a purely TEM ﬁeld conﬁguration. Waveguides in general depart from these restrictions and may employ any number of conductors and dielectrics—or as we will see, dielectrics alone and no conductors. The chapter begins with a presentation of several transmission line structures, with emphasis on obtaining expressions for the primary constants, L, C, G, and R, for high- and low-frequency operating regimes. Next, we begin our study of waveguides by ﬁrst taking a broad view of waveguide devices to obtain a physical understanding of how they work and the conditions under which they are used. We then explore the simple parallel-plate structure and distinguish between its operation as a transmission line and as a waveguide. In this device, the concept of waveguide modes is developed, as are the conditions under which these will occur. We will study the electric and magnetic ﬁeld conﬁgurations of the guided modes using simple plane wave models and the wave equation. We will then study more complicated structures, including rectangular waveguides, dielectric slab waveguides, and optical ﬁbers. ■

I

13.1 TRANSMISSION LINE FIELDS AND PRIMARY CONSTANTS
We begin by establishing the equivalence between transmission line operations when considering voltage and current, from the point of view of the ﬁelds within the line. Consider, for example, the parallel-plate line shown in Figure 13.1. In the line, we
453

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ENGINEERING ELECTROMAGNETICS

Figure 13.1 A transmission-line wave represented by voltage and current distributions along the length is associated with transverse electric and magnetic fields, forming a TEM wave.

assume that the plate spacing, d, is much less than the line width, b (into the page), so electric and magnetic ﬁelds can be assumed to be uniform within any transverse plane. Lossless propagation is also assumed. Figure 13.1 shows the side view, which includes the propagation axis z. The ﬁelds, along with the voltage and current, are shown at an instant in time. The voltage and current in phasor form are: Vs (z) = V0 e− jβz Is (z) = (1a) (1b) V0 − jβz e Z0

√ where Z 0 = L/C. The electric ﬁeld in a given transverse plane at location z is just the parallel-plate capacitor ﬁeld: E sx (z) = V0 − jβz Vs = e d d (2a)

The magnetic ﬁeld is equal to the surface current density, assumed uniform, on either plate [Eq. (12), Chapter 7]: Hsy (z) = K sz = Is V0 − jβz e = b bZ 0 (2b)

The two ﬁelds, both uniform, orthogonal, and lying in the transverse plane, are identical in form to those of a uniform plane wave. As such, they are transverse electromagnetic (TEM) ﬁelds, also known simply as transmission-line ﬁelds. They differ from the ﬁelds of the uniform plane wave only in that they exist within the interior of the line, and nowhere else. The power ﬂow down the line is found through the time-average Poynting vector, integrated over the line cross section. Using (2a) and (2b), we ﬁnd: Pz = b 0 0 d

1 V0 V0∗ 1 |V0 |2 1 ∗ ∗ Re E xs Hys d xd y = ∗ (bd) = ∗ = Re Vs Is 2 2 d bZ 0 2Z 0 2

(3)

The power transmitted by the line is one of the most important quantities that we wish to know from a practical standpoint. Eq. (3) shows that this can be obtained consistently through the line ﬁelds, or through the voltage and current. As would be

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Figure 13.2 The geometry of the parallel-plate transmission line.

expected, this consistency is maintained when losses are included. The ﬁelds picture is in fact advantageous, and is generally preferred, since it is easy to incorporate dielectric loss mechanisms (other than conductivity) in addition to the dispersive properties of the dielectric. The transmission-line ﬁelds are also needed to produce the primary constants, as we now demonstrate for the parallel-plate line and other selected line geometries. We assume the line is ﬁlled with dielectric having permittivity , conductivity σ , and permeability µ, usually µ0 (Figure 13.2). The upper and lower plate thickness is t, which, along with the plate width b and plate conductivity σc , is used to evaluate the resistance per unit length parameter R under low-frequency conditions. We will, however, consider high-frequency operation, in which the skin effect gives an effective plate thickness or skin depth δ that is much less than t. First, the capacitance and conductance per unit length are simply those of the parallel-plate structure, assuming static ﬁelds. Using Eq. (27) from Chapter 6, we ﬁnd b (4) d The value of permittivity used should be appropriate for the range of operating frequencies considered. The conductance per unit length may be determined easily from the capacitance expression by use of the simple relation between capacitance and resistance [Eq. (45), Chapter 6]: σb σ (5) G= C= d The evaluation of L and R involves the assumption of a well-developed skin effect such that δ t. Consequently, the inductance is primarily external because the magnetic ﬂux within either conductor is negligible compared to that between conductors. Therefore, µd . (6) L = L ext = b Note that L ext C = µ = 1/ν 2 , and we are therefore able to evaluate the external p inductance for any transmission line for which we know the capacitance and insulator characteristics. The last of the four parameters that we need is the resistance R per unit length. If the frequency is very high and the skin depth δ is very small, then we obtain an C=

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ENGINEERING ELECTROMAGNETICS

appropriate expression for R by distributing the total current uniformly throughout a depth δ. The skin effect resistance (through both conductors in series over a unit length) is R= 2 σc δb (7)

Finally, it is convenient to include the common expression for the characteristic impedance of the line here with the parameter formulas: Z0 = L ext = C µd b (8)

If necessary, a more accurate value may be obtained from Eq. (47), Chapter 10. Note that when substituting (8) into (2b), and using (2a), we obtain the expected relation √ for a TEM wave, E xs = ηHys , where η = µ/ . D13.1. Parameters for the planar transmission line shown in Figure 13.2 are b = 6 mm, d = 0.25 mm, t = 25 mm, σc = 5.5 × 107 S/m, = 25 pF/m, µ = µ0 , and σ/ω = 0.03. If the operating frequency is 750 MHz, calculate: (a) α; (b) β; (c) Z 0 .
Ans. 0.47 Np/m; 26 rad/m; 9.3 0.7◦

13.1.1 Coaxial (High Frequencies)
We next consider a coaxial cable in which the dielectric has an inner radius a and outer radius b (Figure 13.3). The capacitance per unit length, obtained as Eq. (5) of Section 6.3, is C= 2π ln(b/a) (9)

Figure 13.3 Coaxial transmission-line geometry.

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2πσ (10) ln(b/a) where σ is the conductivity of the dielectric between the conductors at the operating frequency. The inductance per unit length was computed for the coaxial cable as Eq. (50) in Section 8.10, µ L ext = ln(b/a) (11) 2π Again, this is an external inductance, for the small skin depth precludes any appreciable magnetic ﬂux within the conductors. For a circular conductor of radius a and conductivity σc , we let Eq. (90) of Section 11.4 apply to a unit length, obtaining G= 1 2πaδσc There is also a resistance for the outer conductor, which has an inner radius b. We assume the same conductivity σc and the same value of skin depth δ, leading to Rinner = 1 2π bδσc Because the line current ﬂows through these two resistances in series, the total resistance is the sum: 1 1 1 + (12) R= 2πδσc a b Router = Finally, the characteristic impedance, assuming low losses, is Z0 = L ext 1 = C 2π µ ln b a (13)

Now, using the relation RC = /σ (see Problem 6.6), the conductance is

13.1.2 Coaxial (Low Frequencies)
We now obtain the coaxial line parameter values at very low frequencies where there is no appreciable skin effect and the current is assumed to be distributed uniformly throughout the conductor cross sections. We ﬁrst note that the current distribution in the conductor does not affect either the capacitance or conductance per unit length. Therefore, C= and 2πσ (15) ln(b/a) The resistance per unit length may be calculated by dc methods, R = l/(σc S), where l = 1 m and σc is the conductivity of the outer and inner conductors. The area of the G= 2π ln(b/a) (14)

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ENGINEERING ELECTROMAGNETICS

center conductor is πa 2 and that of the outer is π (c2 − b2 ). Adding the two resistance values, we have R= 1 σc π 1 1 + 2 a2 c − b2 (16)

Only one of the four parameter values remains to be found, the inductance per unit length. The external inductance that we calculated at high frequencies is the greatest part of the total inductance. To it, however, must be added smaller terms representing the internal inductances of the inner and outer conductors. At very low frequencies where the current distribution is uniform, the internal inductance of the center conductor is the subject of Problem 43 in Chapter 8; the relationship is also given as Eq. (62) in Section 8.10: L a,int = µ 8π (17)

The determination of the internal inductance of the outer shell is a more difﬁcult problem, and most of the work was requested in Problem 36 in Chapter 8. There, we found that the energy stored per unit length in an outer cylindrical shell of inner radius b and outer radius c with uniform current distribution is WH = µI 2 4c2 c ln b2 − 3c2 + 2 16π (c2 − b2 ) c − b2 b µ 4c2 c ln b2 − 3c2 + 2 2) −b c − b2 b

Thus the internal inductance of the outer conductor at very low frequencies is L bc,int = 8π (c2 (18)

At low frequencies the total inductance is obtained by adding (11), (17), and (18): L= µ b 1 1 4c2 c ln + + b2 − 3c2 + 2 ln 2 − b2 ) 2 2π a 4 4(c c −b b (19)

13.1.3 Coaxial (Intermediate Frequencies)
There still remains the frequency interval where the skin depth is neither very much larger than nor very much smaller than the radius. In this case, the current distribution is governed by Bessel functions, and both the resistance and internal inductance are complicated expressions. Values are tabulated in the handbooks, and it is necessary to use them for very small conductor sizes at high frequencies and for larger conductor sizes used in power transmission at low frequencies.1

1

Bessel functions are discussed within the context of optical ﬁber in Section 13.7. The current distribution, internal inductance, and internal resistance of round wires is discussed (with numerical examples) in Weeks, pp. 35–44. See the References at the end of this chapter.

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D13.2. The dimensions of a coaxial transmission line are a = 4 mm, b = 17.5 mm, and c = 20 mm. The conductivity of the inner and outer conductors is 2 × 107 S/m, and the dielectric properties are µr = 1, r = 3, and σ/ω = 0.025. Assume that the loss tangent is constant with frequency. Determine: (a) L , C, R, G, and Z 0 at 150 MHz; (b) L and R at 60 Hz.
Ans. 0.30 µH/m, 113 pF/m, 0.27 /m, 2.7 mS/m, 51 ; 0.36 µH/m, 1.16 m /m

13.1.4 Two-Wire (High Frequencies)
For the two-wire transmission line of Figure 13.4 with conductors of radius a and conductivity σc with center-to-center separation d in a medium of permeability µ, permittivity , and conductivity σc , the capacitance per unit length is found using the results of Section 6.4: π (20) C= −1 cosh (d/2a) or . π C= (a d) ln(d/a) The external inductance may be found from L ext C = µ . It is µ (21) L ext = cosh−1 (d/2a) π or . µ (a d) L ext = ln(d/a) π The conductance per unit length may be written immediately from an inspection of the capacitance expression, and using the relation RC = /σ : πσ G= (22) −1 cosh (d/2a) The resistance per unit length is twice that of the center conductor of the coax, 1 R= (23) πaδσc

Figure 13.4 The geometry of the two-wire transmission line.

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ENGINEERING ELECTROMAGNETICS

Finally, using the capacitance and the external inductance expressions, we obtain a value for the characteristic impedance, Z0 = L ext 1 = C π µ cosh−1 (d/2a) (24)

13.1.5 Two-Wire (Low Frequencies)
At low frequencies where a uniform current distribution may be assumed, we again must modify the L and R expressions, but not those for C and G. The latter two are again expressed by (20) and (22): cosh (d/2a) πσ G= −1 cosh (d/2a) The inductance per unit length must be increased by twice the internal inductance of a straight round wire, L= µ 1 + cosh−1 (d/2a) π 4 (25) C= π
−1

The resistance becomes twice the dc resistance of a wire of radius a, conductivity σc , and unit length: 2 (26) R= πa 2 σc D13.3. The conductors of a two-wire transmission line each have a radius of 0.8 mm and a conductivity of 3 × 107 S/m. They are separated by a center-tocenter distance of 0.8 cm in a medium for which r = 2.5, µr = 1, and σ = 4×10−9 S/m. If the line operates at 60 Hz, ﬁnd: (a) δ; (b) C; (c) G; (d) L; (e) R.
Ans. 1.2 cm; 30 pF/m; 5.5 nS/m; 1.02 µH/m; 0.033 /m

13.1.6 Microstrip Line (Low Frequencies)
Microstrip line is one example of a class of conﬁgurations involving planar conductors of ﬁnite widths on or within dielectric substrates; they are usually employed as device interconnects for microelectronic circuitry. The microstrip conﬁguration, shown in Figure 13.5, consists of a dielectric (assumed lossless) of thickness d and of permittivity = r 0 , sandwiched between a conducting ground plane and a narrow conducting strip of width w. The region above the top strip is air (assumed here) or a dielectric of lower permittivity. The structure approaches the case of the parallel-plate line if w d. In a microstrip, such an assumption is generally not valid, and so signiﬁcant charge densities exist on both surfaces of the upper conductor. The resulting electric ﬁeld, originating at the top conductor and terminating on the bottom conductor, will exist within

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Figure 13.5 Microstrip line geometry.

both substrate and air regions. The same is true for the magnetic ﬁeld, which circulates around the top conductor. This electromagnetic ﬁeld conﬁguration cannot propagate as a purely TEM wave because wave velocities within the two media will differ. Instead, waves having z components of E and H occur, with the z component magnitudes established so that the air and dielectric ﬁelds do achieve equal phase velocities (the reasoning behind this will be explained in Section 13.6). Analyzing the structure while allowing for the special ﬁelds is complicated, but it is usually permissible to approach the problem under the assumption of negligible z components. This is the quasi TEM approximation, in which the static ﬁelds (obtainable through numerical solution of Laplace’s equation, for example) are used to evaluate the primary constants. Accurate results are obtained at low frequencies (below 1 or 2 GHz). At higher frequencies, results obtained through the static ﬁelds can still be used but in conjunction with appropriate modifying functions. We will consider the simple case of low-frequency operation and assume lossless propagation.2 To begin, it is useful to consider the microstrip line characteristics when the dielectric is not present. Assuming that both conductors have very small thicknesses, the internal inductance will be negligible, and so the phase velocity within the air-ﬁlled line, νp0 , will be 1 1 ν p0 = √ =√ =c (27a) µ0 0 L ext C0 where C0 is the capacitance of the air-ﬁlled line (obtained from the electric ﬁeld for that case), and c is the velocity of light. With the dielectric in place, the capacitance changes, but the inductance does not, provided the dielectric permeability is µ0 . Using (27a), the phase velocity now becomes νp = √ c C0 1 =√ =c C L ext C r,eff (27b)

where the effective dielectric constant for the microstrip line is C c 2 = (28) r,eff C0 νp It is implied from (28) that the microstrip capacitance C would result if both the air and substrate regions were ﬁlled homogeneously with material having dielectric constant r,eff . The effective dielectric constant is a convenient parameter to use =

2

The high-frequency case is treated in detail in Edwards (Reference 2).

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ENGINEERING ELECTROMAGNETICS

because it provides a way of unifying the effects of the dielectric and the conductor geometry. To see this, consider the two extreme cases involving large and small width-to-height ratios, w/d. If w/d is very large, then the line resembles the parallelplate line, in which nearly all of the electric ﬁeld exists within the dielectric. In this . case r,eff = r . On the other hand, for a very narrow top strip, or small w/d, the dielectric and air regions contain roughly equal amounts of electric ﬂux. In that case, the effective dielectric constant approaches its minimum, given by the average of the two dielectric constants. We therefore obtain the range of allowed values of r,eff : 1 ( 2 r + 1) <

r,eff

<

r

(29)

The physical interpretation of r,eff is that it is a weighted average of the dielectric constants of the substrate and air regions, with the weighting determined by the extent to which the electric ﬁeld ﬁlls either region. We may thus write the effective dielectric constant in terms of a fiel fillin factor, q, for the substrate: r,eff where 0.5 < q < 1. With large w/d, q → 1; with small w/d, q → 0.5. Now, the characteristic impedances of the air-ﬁlled line and the line with dielectric √ √ air substrate are, respectively, Z 0 = L ext /C0 and Z 0 = L ext C. Then, using (28), we ﬁnd Z air Z0 = √ 0 r,eff = 1 + q(

r

− 1)

(30)

(31)

A procedure for obtaining the characteristic impedance would be to ﬁrst evaluate the air-ﬁlled impedance for a given w/d. Then, knowing the effective dielectric constant, determine the actual impedance using (31). Another problem would be to determine the required ratio w/d for a given substrate material in order to achieve a desired characteristic impedance. Detailed analyses have led to numerous approximation formulas for the evaluaair tion of r,eff , Z 0 , and Z 0 within different regimes (again, see Reference 2 and the references therein). For example, with dimensions restricted such that 1.3 < w/d < 3.3, applicable formulas include:   d d 2 w air . Z 0 = 60 ln 4 + 16 < 3.3 (32) + 2 w w d and r,eff −0.555

. =

r

+1 + 2

r

−1 d 1 + 10 2 w

w > 1.3 d

(33)

Or, if a line is to be fabricated having a desired value of Z 0 , the effective dielectric constant (from which the required w/d can be obtained) is found through: w . −1 > 1.3 (34) r,eff = r [0.96 + r (0.109 − 0.004 r ) (log10 (10 + Z 0 ) − 1)] d

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D13.4. A microstrip line is fabricated on a lithium niobate substrate ( r = 4.8) of 1 mm thickness. If the top conductor is 2 mm wide, ﬁnd (a) r,eff ; (b) Z 0 ; (c) ν p .
Ans. 3.6; 47 ; 1.6 × 108 m/s

13.2 BASIC WAVEGUIDE OPERATION
Waveguides assume many different forms that depend on the purpose of the guide and on the frequency of the waves to be transmitted. The simplest form (in terms of analysis) is the parallel-plate guide shown in Figure 13.6. Other forms are the hollowpipe guides, including the rectangular waveguide of Figure 13.7, and the cylindrical guide, shown in Figure 13.8. Dielectric waveguides, used primarily at optical frequencies, include the slab waveguide of Figure 13.9 and the optical ﬁber, shown in Figure 13.10. Each of these structures possesses certain advantages over the others, depending on the application and the frequency of the waves to be transmitted. All guides, however, exhibit the same basic operating principles, which we will explore in this section. To develop an understanding of waveguide behavior, we consider the parallelplate waveguide of Figure 13.6. At ﬁrst, we recognize this as one of the transmissionline structures that we investigated in Section 13.1. So the ﬁrst question that arises is: how does a waveguide differ from a transmission line to begin with? The difference lies in the form of the electric and magnetic ﬁelds within the line. To see this, consider again Figure 13.1, which shows the ﬁelds when the line operates as a transmission line. As we saw earlier, a sinusoidal voltage wave, with voltage applied between conductors, leads to an electric ﬁeld that is directed vertically between the conductors as shown. Because current ﬂows only in the z direction, magnetic ﬁeld will be oriented in and out of the page (in the y direction). The interior ﬁelds comprise a plane electromagnetic wave which propagates in the z direction (as the Poynting vector will show), since both ﬁelds lie in the transverse plane. We refer to this as a transmission-line wave,

Figure 13.6 Parallel-plate waveguide, with metal plates at x = 0, d. Between the plates is a dielectric of permittivity .

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ENGINEERING ELECTROMAGNETICS

Figure 13.7 Rectangular waveguide.

Figure 13.8 Cylindrical waveguide.

Figure 13.9 Symmetric dielectric slab waveguide, with slab region (refractive index n1 ) surrounded by two dielectrics of index n2 < n1 .

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Figure 13.10 Optical fiber waveguide, with the core dielectric (ρ < a) of refractive index n1 . The cladding dielectric (a < ρ < b) is of index n2 < n1 .

which, as discussed in Section 13.1, is a transverse electromagnetic, or TEM, wave. The wavevector k, shown in Figure 13.1, indicates the direction of wave travel as well as the direction of power ﬂow. As the frequency is increased, a remarkable change occurs in the way the ﬁelds progagate down the line. Although the original ﬁeld conﬁguration of Figure 13.1 may still be present, another possibility emerges, which is shown in Figure 13.11. Again, a plane wave is guided in the z direction, but by means of a progression of zig-zag reﬂections at the upper and lower plates. Wavevectors ku and kd are associated with the upward-and downward-propagating waves, respectively, and these have identical magnitudes, √ |ku | = |kd | = k = ω µ For such a wave to propagate, all upward-propagating waves must be in phase (as must be true of all downward-propagating waves). This condition can only be satisﬁed at certain discrete angles of incidence, shown as θ in Figure 13.11. An allowed value of θ, along with the resulting ﬁeld conﬁguration, comprise a waveguide mode of the structure. Associated with each guided mode is a cutoff frequency. If the operating frequency is below the cutoff frequency, the mode will not propagate. If above cutoff,

ku

θ

kd

ku

θ

Figure 13.11 In a parallel-plate waveguide, plane waves can propagate by oblique reflection from the conducting walls. This produces a waveguide mode that is not TEM.

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ENGINEERING ELECTROMAGNETICS

Figure 13.12 Plane wave representation of TE and TM modes in a parallel-plate guide.

the mode propagates. The TEM mode, however, has no cutoff; it will be supported at any frequency. At a given frequency, the guide may support several modes, the quantity of which depends on the plate separation and on the dielectric constant of the interior medium, as will be shown. The number of modes increases as the frequency is raised. So to answer our initial question on the distinction between transmission lines and waveguides, we can state the following: Transmission lines consist of two or more conductors and as a rule will support TEM waves (or something which could approximate such a wave). A waveguide may consist of one or more conductors, or no conductors at all, and will support waveguide modes of forms similar to those just described. Waveguides may or may not support TEM waves, depending on the design. In the parallel-plate guide, two types of waveguide modes can be supported. These are shown in Figure 13.12 as arising from the s and p orientations of the plane wave polarizations. In a manner consistent with our previous discussions on oblique reﬂection (Section 12.5), we identify a transverse electric or TE mode when E is perpendicular to the plane of incidence (s-polarized); this positions E parallel to the transverse plane of the waveguide, as well as to the boundaries. Similarly, a transverse magnetic or TM mode results with a p polarized wave; the entire H ﬁeld is in the y direction and is thus within the transverse plane of the guide. Both possibilities are illustrated in Figure 13.12. Note, for example, that with E in the y direction (TE mode), H will have x and z components. Likewise, a TM mode will have x and z components of E.3 In any event, the reader can verify from the geometry of Figure 13.12 that it is not possible to achieve a purely TEM mode for values of θ other than 90◦ . Other wave polarizations are possible that lie between the TE and TM cases, but these can always be expressed as superpositions of TE and TM modes.

3 Other types of modes can exist in other structures (not the parallel-plate guide) in which both E and H have z components. These are known as hybrid modes, and they typically occur in guides with cylindrical cross sections, such as the optical ﬁber.

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13.3 PLANE WAVE ANALYSIS OF THE PARALLEL-PLATE WAVEGUIDE
Let us now investigate the conditions under which waveguide modes will occur, using our plane wave model for the mode ﬁelds. In Figure 13.13a, a zig-zag path is again shown, but this time phase fronts are drawn that are associated with two of the upward-propagating waves. The ﬁrst wave has reﬂected twice (at the top and bottom surfaces) to form the second wave (the downward-propagating phase fronts are not shown). Note that the phase fronts of the second wave do not coincide with those of the ﬁrst wave, and so the two waves are out of phase. In Figure 13.13b, the wave angle has been adjusted so that the two waves are now in phase. Having satisﬁed this condition for the two waves, we will ﬁnd that all upward-propagating waves will have coincident phase fronts. The same condition will automatically occur for all downward-propagating waves. This is the requirement to establish a guided mode. In Figure 13.14 we show the wavevector, ku , and its components, along with a series of phase fronts. A drawing of this kind for kd would be the same, except the

Figure 13.13 (a) Plane wave propagation in a parallel-plate guide in which the wave angle is such that the upward-propagating waves are not in phase. (b) The wave angle has been adjusted so that the upward waves are in phase, resulting in a guided mode.

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ENGINEERING ELECTROMAGNETICS

Figure 13.14 The components of the upward wavevector are κm and βm, the transverse and axial phase constants. To form the downward wavevector, kd , the direction of κm is reversed.

x component, κm , would be reversed. In Section 12.4, we measured the phase shift per unit distance along the x and z directions by the components, k x and k z , which varied continuously as the direction of k changed. In our discussion of waveguides, we introduce a different notation, where κm and βm are used for k x and k z . The subscript m is an integer indicating the mode number. This provides a subtle hint that βm and κm will assume only certain discrete values that correspond to certain allowed directions of ku and kd , such that our coincident phase front requirement is satisﬁed.4 From the geometry we see that for any value of m, βm =
2 k 2 − κm

(35)

Use of the symbol βm for the z components of ku and kd is appropriate because βm will ultimately be the phase constant for the mth waveguide mode, measuring phase shift per distance down the guide; it is also used to determine the phase velocity of the mode, ω/βm , and the group velocity, dω/dβm . Throughout our discussion, we will assume that the medium within the guide is lossless and nonmagnetic, so that k = ω µ0 = ω c r =

ωn c

(36)

4

Subscripts (m) are not shown on ku and kd but are understood. Changing m does not affect the magnitudes of these vectors, only their directions.

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which we express either in terms of the dielectric constant, r , or the refractive index, n, of the medium. It is κm , the x component of ku and kd , that will be useful to us in quantifying our requirement on coincident phase fronts through a condition known as transverse resonance. This condition states that the net phase shift measured during a round trip over the full transverse dimension of the guide must be an integer multiple of 2π radians. This is another way of stating that all upward- (or downward-) propagating plane waves must have coincident phases. The various segments of this round trip are illustrated in Figure 13.15. We assume for this exercise that the waves are frozen in time and that an observer moves vertically over the round trip, measuring phase shift along the way. In the ﬁrst segment (Figure 13.15a), the observer starts at a position just above the lower conductor and moves vertically to the top conductor through distance d. The measured phase shift over this distance is κm d rad. On reaching the top surface, the observer will note a possible phase shift on reﬂection (Figure 13.15b). This will be π if the wave is TE polarized and will be zero if the wave is TM polarized (see Figure 13.16 for a demonstration of this). Next, the observer moves along the reﬂected wave phases down to the lower conductor and again measures a phase shift of κm d (Figure 13.15c). Finally, after including the phase shift on reﬂection at the bottom conductor, the observer is back at the original starting point and is noting the phase of the next upward-propagating wave. The total phase shift over the round trip is required to be an integer multiple of 2π : κm d + φ + κm d + φ = 2mπ (37)

where φ is the phase shift on reﬂection at each boundary. Note that with φ = π (TE waves) or 0 (TM waves), the net reﬂective phase shift over a round trip is 2π or 0, regardless of the angle of incidence. Thus the reﬂective phase shift has no bearing on the current problem, and we may simplify (37) to read κm = mπ d (38)

which is valid for both TE and TM modes. Note from Figure 13.14 that κm = k cos θm . Thus the wave angles for the allowed modes are readily found from (38) with (36): θm = cos−1 mπ kd = cos−1 mπc = cos−1 ωnd mλ 2nd

(39)

where λ is the wavelength in free space. We can next solve for the phase constant for each mode, using (35) with (38):
2 k 2 − κm = k 1 −

βm =

mπ kd

2

=k 1−

mπ c ωnd

2

(40)

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ENGINEERING ELECTROMAGNETICS

R

0

0

Figure 13.15 The net phase shift over a round trip in the parallel-plate guide is found by first measuring the transverse phase shift between plates of the initial upward wave (a); next, the transverse phase shift in the reflected (downward) wave is measured, while accounting for the reflective phase shift at the top plate (b); finally, the phase shift on reflection at the bottom plate is added, thus returning to the starting position, but with a new upward wave (c). Transverse resonance occurs if the phase at the final point is the same as that at the starting point (the two upward waves are in phase).

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Figure 13.16 The phase shift of a wave on reflection from a perfectly conducting surface depends on whether the incident wave is TE (s-polarized) or TM (p-polarized). In both drawings, electric fields are shown as they would appear immediately adjacent to the conducting boundary. In (a) the field of a TE wave reverses direction upon reflection to establish a zero net field at the boundary. This constitutes a π phase shift, as is evident by considering a fictitious transmitted wave (dashed line) formed by a simple rotation of the reflected wave into alignment with the incident wave. In (b) an incident TM wave experiences a reversal of the z component of its electric field. The resultant field of the reflected wave, however, has not been phase-shifted; rotating the reflected wave into alignment with the incident wave (dashed line) shows this.

We deﬁne the radian cutoff frequency for mode m as ωcm = so that (40) becomes nω ωcm 1− c ω
2

mπc nd

(41)

βm =

(42)

The signiﬁcance of the cutoff frequency is readily seen from (42): If the operating frequency ω is greater than the cutoff frequency for mode m, then that mode will have phase constant βm that is real-valued, and so the mode will propagate. For ω < ωcm , βm is imaginary, and the mode does not propagate. Associated with the cutoff frequency is the cutoff wavelength, λcm , deﬁned as the free-space wavelength at which cutoff for mode m occurs. This will be λcm = 2πc 2nd = ωcm m (43)

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ENGINEERING ELECTROMAGNETICS

Note, for example, that in an air-ﬁlled guide (n = 1) the wavelength at which the lowest-order mode ﬁrst starts to propagate is λc1 = 2d, or the plate separation is one-half wavelength. Mode m will propagate whenever ω > ωcm , or equivalently whenever λ < λcm . Use of the cutoff wavelength enables us to construct a second useful form of Eq. (42): βm =
E X A M P L E 13.1

2π n 1− λ

λ λcm

2

(44)

A parallel-plate waveguide has plate separation d = 1 cm and is ﬁlled with teﬂon having dielectric constant r = 2.1. Determine the maximum operating frequency such that only the TEM mode will propagate. Also ﬁnd the range of frequencies over which the TE1 and TM1 (m = 1) modes, and no higher-order modes, will propagate.
Solution. Using (41), the cutoff frequency for the ﬁrst waveguide mode (m = 1)

will be

ωc1 2.99 × 1010 = = 1.03 × 1010 Hz = 10.3 GHz √ 2π 2 2.1 To propagate only TEM waves, we must have f < 10.3 GHz. To allow TE1 and TM1 (along with TEM) only, the frequency range must be ωc1 < ω < ωc2 , where ωc2 = 2ωc1 , from (41). Thus, the frequencies at which we will have the m = 1 modes and TEM will be 10.3 GHz < f < 20.6 GHz. f c1 =

E X A M P L E 13.2

In the parallel-plate guide of Example 13.1, the operating wavelength is λ = 2 mm. How many waveguide modes will propagate?
Solution. For mode m to propagate, the requirement is λ < λcm . For the given wave-

guide and wavelength, the inequality becomes, using (43), √ 2 2.1 (10 mm) 2 mm < m from which m< √ 2 2.1 (10 mm) = 14.5 2 mm

Thus the guide will support modes at the given wavelength up to order m = 14. Since there will be a TE and a TM mode for each value of m, this gives, not including the TEM mode, a total of 28 guided modes that are above cutoff.

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The ﬁeld conﬁguration for a given mode can be found through the superposition of the ﬁelds of all the reﬂected waves. We can do this for the TE waves, for example, by writing the electric ﬁeld phasor in the guide in terms of incident and reﬂected ﬁelds through where the wavevectors, ku and kd , are indicated in Figure 13.12. The minus sign in front of the second term arises from the π phase shift on reﬂection. From the geometry depicted in Figure 13.14, we write ku = κm ax + βm az and kd = −κm ax + βm az Then, using r = xax + zaz Eq. (45) becomes E ys = E 0 (e− jκm x − e jkm x

E ys = E 0 e− jku ·r − E 0 e− jkd ·r

(45)

(46)

(47)

)e− jβm z = 2 j E 0 sin(κm x)e− jβm z = E 0 sin(κm x)e− jβm z (48)

where the plane wave amplitude, E 0 , and the overall phase are absorbed into E 0 . In real instantaneous form, (48) becomes E y (z, t) = Re E ys e jωt = E 0 sin(κm x) cos(ωt − βm z)

(TE mode above cutoff) (49)

We interpret this as a wave that propagates in the positive z direction (down the guide) while having a ﬁeld proﬁle that varies with x.5 The TE mode ﬁeld is the interference pattern resulting from the superposition of the upward and downward plane waves. Note that if ω < ωcm , then (42) yields an imaginary value for βm , which we may write as − j|βm | = − jαm . Eqs. (48) and (49) then become E ys = E 0 sin(κm x)e−αm z (50) (51) E(z, t) = E 0 sin(κm x)e−αm z cos(ωt)

(TE mode below cutoff)

This mode does not propagate, but simply oscillates at frequency ω, while exhibiting a ﬁeld pattern that decreases in strength with increasing z. The attenuation coefﬁcient, αm , is found from (42) with ω < ωcm : αm = nωcm c 1− ω ωcm
2

=

2π n λcm

1−

λcm λ

2

(52)

5

We can also interpret this ﬁeld as that of a standing wave in x while it is a traveling wave in z.

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We note from (39) and (41) that the plane wave angle is related to the cutoff frequency and cutoff wavelength through cos θm = ωcm λ = ω λcm (53)

So we see that at cutoff (ω = ωcm ), θm = 0, and the plane waves are just reﬂecting back and forth over the cross section; they are making no forward progress down the guide. As ω is increased beyond cutoff (or λ is decreased), the wave angle increases, approaching 90◦ as ω approaches inﬁnity (or as λ approaches zero). From Figure 13.14, we have βm = k sin θm = and so the phase velocity of mode m will be νpm = ω c = βm n sin θm (55) nω sin θm c (54)

The velocity minimizes at c/n for all modes, approaching this value at frequencies far above cutoff; νpm approaches inﬁnity as the frequency is reduced to approach the cutoff frequency. Again, phase velocity is the speed of the phases in the z direction, and the fact that this velocity may exceed the speed of light in the medium is not a violation of relativistic principles, as discussed in Section 12.7. The energy will propagate at the group velocity, νg = dω/dβ. Using (42), we have
−1 νgm =

d dβm = dω dω

nω c

1−

ωcm ω

2

(56)

The derivative is straightforward. Carrying it out and taking the reciprocal of the result yields: νgm = c n 1− ωcm ω
2

=

c sin θm n

(57)

Group velocity is thus identiﬁed as the projection of the velocity associated with ku or kd into the z direction. This will be less than or equal to the velocity of light in the medium, c/n, as expected.
E X A M P L E 13.3

In the guide of Example 13.1, the operating frequency is 25 GHz. Consequently, modes for which m = 1 and m = 2 will be above cutoff. Determine the group delay difference between these two modes over a distance of 1 cm. This is the difference in propagation times between the two modes when energy in each propagates over the 1-cm distance.

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Solution. The group delay difference is expressed as

t=

1 1 − νg2 νg1

(s/cm)

From (57), along with the results of Example 13.1, we have c νg1 = √ 1− 2.1 c νg2 = √ 1− 2.1 Then t= 1 1 1 − = 3.3 × 10−11 s/cm = 33 ps/cm c .39 .63 10.3 25 20.6 25
2

= 0.63c
2

= 0.39c

This computation gives a rough measure of the modal dispersion in the guide, applying to the case of having only two modes propagating. A pulse, for example, whose center frequency is 25 GHz would have its energy divided between the two modes. The pulse would broaden by approximately 33 ps/cm of propagation distance as the energy in the modes separates. If, however, we include the TEM mode (as we really must), then √ the broadening will be even greater. The group velocity for TEM will be c/ 2.1. The group delay difference of interest will then be between the TEM mode and the m = 2 mode (TE or TM). We would therefore have tnet = 1 1 − 1 = 52 ps/cm c .39

D13.5. Determine the wave angles θm for the ﬁrst four modes (m = 1, 2, 3, 4) in a parallel-plate guide with d = 2 cm, r = 1, and f = 30 GHz.
Ans. 76◦ ; 60◦ ; 41◦ ; 0◦

D13.6. A parallel-plate guide has plate spacing d = 5 mm and is ﬁlled with glass (n = 1.45). What is the maximum frequency at which the guide will operate in the TEM mode only?
Ans. 20.7 GHz

D13.7. A parallel-plate guide having d = 1 cm is ﬁlled with air. Find the cutoff wavelength for the m = 2 mode (TE or TM).
Ans. 1 cm

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ENGINEERING ELECTROMAGNETICS

13.4 PARALLEL-PLATE GUIDE ANALYSIS USING THE WAVE EQUATION
The most direct approach in the analysis of any waveguide is through the wave equation, which we solve subject to the boundary conditions at the conducting walls. The form of the equation that we will use is that of Eq. (28) in Section 11.1, which was written for the case of free-space propagation. We account for the dielectric properties in the waveguide by replacing k0 in that equation with k to obtain: ∇ 2 Es = −k 2 Es (58)

where k = nω/c as before. We can use the results of the last section to help us visualize the process of solving the wave equation. For example, we may consider TE modes ﬁrst, in which there will be only a y component of E. The wave equation becomes: ∂ 2 E ys ∂ 2 E ys ∂ 2 E ys + + + k 2 E ys = 0 ∂x2 ∂ y2 ∂z 2

(59)

We assume that the width of the guide (in the y direction) is very large compared to the plate separation d. Therefore we can assume no y variation in the ﬁelds (fringing ﬁelds are ignored), and so ∂ 2 E ys /∂ y 2 = 0. We also know that the z variation will be of the form e− jβm z . The form of the ﬁeld solution will thus be E ys = E 0 f m (x)e− jβm z (60)

where E 0 is a constant, and where f m (x) is a normalized function to be determined (whose maximum value is unity). We have included subscript m on β, κ, and f (x), since we anticipate several solutions that correspond to discrete modes, to which we associate mode number m. We now substitute (60) into (59) to obtain d 2 f m (x) 2 + k 2 − βm f m (x) = 0 dx2 where E 0 and e− jβm z have divided out, and where we have used the fact that d 2 − jβm z 2 e = −βm e− jβm z dz 2 Note also that we have written (61) using the total derivative d 2 /d x 2 , as f m is a function only of x. We next make use of the geometry of Figure 13.14, and we note (61)

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2 2 that k 2 − βm = κm . Using this in (61) we obtain

d 2 f m (x) 2 + κm f m (x) = 0 dx2 The general solution of (62) will be f m (x) = cos(κm x) + sin(κm x)

(62)

(63)

We next apply the appropriate boundary conditions in our problem to evaluate κm . From Figure 13.6, conducting boundaries appear at x = 0 and x = d, at which the tangential electric ﬁeld (E y ) must be zero. In Eq. (63), only the sin(κm x) term will allow the boundary conditions to be satisﬁed, so we retain it and drop the cosine term. The x = 0 condition is automatically satisﬁed by the sine function. The x = d condition is met when we choose the value of κm such that mπ κm = (64) d We recognize Eq. (64) as the same result that we obtained using the transverse resonance condition of Section 13.3. The ﬁnal form of E ys is obtained by substituting f m (x) as expressed through (63) and (64) into (60), yielding a result that is consistent with the one expressed in Eq. (48): E ys = E 0 sin mπ x − jβm z e d (65)

An additional signiﬁcance of the mode number m is seen when considering the form of the electric ﬁeld of (65). Speciﬁcally, m is the number of spatial half-cycles of electric ﬁeld that occur over the distance d in the transverse plane. This can be understood physically by considering the behavior of the guide at cutoff. As we learned in the last section, the plane wave angle of incidence in the guide at cutoff is zero, meaning that the wave simply bounces up and down between the conducting walls. The wave must be resonant in the structure, which means that the net round trip phase shift is 2mπ. With the plane waves oriented vertically, βm = 0, and so κm = k = 2nπ/λcm . So at cutoff, 2nπ mπ (66) = d λcm which leads to mλcm d= at cutoff (67) 2n Eq. (65) at cutoff then becomes mπ x 2nπ x E ys = E 0 sin = E 0 sin (68) d λcm The waveguide is simply a one-dimensional resonant cavity, in which a wave can oscillate in the x direction if its wavelength as measured in the medium is an integer multiple of 2d where the integer is m.

478

ENGINEERING ELECTROMAGNETICS

d/2 λ

θ4

(a)

Figure 13.17 (a) A plane wave associated with an m = 4 mode, showing a net phase shift of 4π (two wavelengths measured in x) occurring over distance d in the transverse plane. (b) As frequency increases, an increase in wave angle is required to maintain the 4π transverse phase shift.

Now, as the frequency increases, wavelength will decrease, and so the requirement of wavelength equaling an integer multiple of 2d is no longer met. The response of the mode is to establish z components of ku and kd , which results in the decreased wavelength being compensated by an increase in wavelength as measured in the x direction. Figure 13.17 shows this effect for the m = 4 mode, in which the wave angle, θ4 , steadily increases with increasing frequency. Thus, the mode retains precisely the functional form of its ﬁeld in the x direction, but it establishes an increasing value of βm as the frequency is raised. This invariance in the transverse spatial pattern means that the mode will retain its identity at all frequencies. Group velocity, expressed in (57), is changing as well, meaning that the changing wave angle with frequency is a mechanism for group velocity dispersion, known simply as waveguide dispersion. Pulses, for example, that propagate in a single waveguide mode will thus experience broadening in the manner considered in Section 12.8. Having found the electric ﬁeld, we can ﬁnd the magnetic ﬁeld using Maxwell’s equations. We note from our plane wave model that we expect to obtain x and z components of Hs for a TE mode. We use the Maxwell equation where, in the present case of having only a y component of Es , we have ∂ E ys ∂ E ys ∇ × Es = az − ax = κm E 0 cos(κm x)e− jβm z az + jβm E 0 sin(κm x)e− jβm z ax ∂x ∂z (70) ∇ × Es = − jωµHs (69)

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We solve for Hs by dividing both sides of (69) by − jωµ. Performing this operation on (70), we obtain the two magnetic ﬁeld components: Hxs = − Hzs = j βm E 0 sin(κm x)e− jβm z ωµ (71)

κm E 0 cos(κm x)e− jβm z ωµ

(72)

Together, these two components form closed-loop patterns for Hs in the x, z plane, as can be veriﬁed using the streamline plotting methods developed in Section 2.6. It is interesting to consider the magnitude of Hs , which is found through |Hs | = |Hs | = Hs · H∗ = s
1/2 ∗ ∗ Hxs Hxs + Hzs Hzs

(73)

Carrying this out using (71) and (72) results in E0 2 2 κ + βm ωµ m sin2 (κm x) + cos2 (κm x)
1/2

(74)

2 2 Using the fact that κm + βm = k 2 and using the identity sin2 (κm x) + cos2 (κm x) = 1, (74) becomes √ ω µ k E0 E0 = = (75) |Hs | = ωµ ωµ η √ where η = µ/ . This result is consistent with our understanding of waveguide modes based on the superposition of plane waves, in which the relation between Es and Hs is through the medium intrinsic impedance, η.

D13.8. Determine the group velocity of the m = 1 (TE or TM) mode in an air-ﬁlled parallel-plate guide with d = 0.5 cm at f = (a) 30 GHz, (b) 60 GHz, and (c) 100 GHz.
Ans. 0; 2.6 × 108 m/s; 2.9 × 108 m/s

D13.9. A TE mode in a parallel-plate guide is observed to have three maxima in its electric ﬁeld pattern between x = 0 and x = d. What is the value of m?
Ans. 3

13.5 RECTANGULAR WAVEGUIDES
In this section we consider the rectangular waveguide, a structure that is usually used in the microwave region of the electromagnetic spectrum. The guide is shown in Figure 13.7. As always, the propagation direction is along the z axis. The guide is of width a along x and height b along y. We can relate the geometry to that of the parallel-plate guide of previous sections by thinking of the rectangular guide as two

480

ENGINEERING ELECTROMAGNETICS

parallel-plate guides of orthogonal orientation that are assembled to form one unit. We have a pair of horizontal conducting walls (along the x direction) and a pair of vertical walls (along y), all of which form one continuous boundary. The wave equation in its full three-dimensional form [Eq. (59)] must now be solved, for in general we may have ﬁeld variations in all three coordinate directions. In the parallel-plate guide, we found that the TEM mode can exist, along with TE and TM modes. The rectangular guide will support the TE and TM modes, but it will not support a TEM mode. This is because, in contrast to the parallel-plate guide, we now have a conducting boundary that completely surrounds the transverse plane. The nonexistence of TEM can be understood by remembering that any electric ﬁeld must have a zero tangential component at the boundary. This means that it is impossible to set up an electric ﬁeld that will not exhibit the sideways variation that is necessary to satisfy this boundary condition. Because E varies in the transverse plane, the computation of H through ∇ × E = − jωµH must lead to a z component of H, and so we cannot have a TEM mode. We cannot ﬁnd any other orientation of a completely transverse E in the guide that will allow a completely transverse H.

13.5.1 Using Maxwell’s Equations to Relate Field Components
With the modes dividing into TE and TM types, the standard approach is to ﬁrst solve the wave equation for the z components. By deﬁnition, E z = 0 in a TE mode, and Hz = 0 in a TM mode. Therefore, we will ﬁnd the TE mode solution by solving the wave equation for Hz , and we will obtain the TM mode solution by solving for E z . Using these results, all transverse ﬁeld components are obtained directly through Maxwell’s equations. This procedure may sound a little tedious (which it is), but we can be certain to ﬁnd all the modes this way. First, we will handle the problem of ﬁnding transverse components in terms of the z components. To begin the process, we assume that the phasor electric and magnetic ﬁelds will be forward-z propagating functions that exhibit spatial variation in the x y plane; the only z variation is that of a forward-propagating wave: Hs (x, y, z) = Hs (x, y, 0)e Es (x, y, z) = Es (x, y, 0)e− jβz (76a) (76b)
− jβz

We can then obtain expressions for the transverse components of the phasor ﬁelds by evaluating the x and y components of the Maxwell curl equations in sourceless media. In evaluating the curl, it is evident from (76) that ∂/∂z = − jβ. The result is ∇ × Es = − jωµHs → ∇ × Hs = jω Es → ∂ E zs /∂ y + jβ E ys = − jωµHxs (x component) (77a) jβ E xs + ∂ E zs /∂ x = jωµHys (y component) (77b) (78a) (78b)

∂ Hzs /∂ y + jβ Hys = jω E xs (x component)

jβ Hxs + ∂ Hzs /∂ x = − jω E ys (y component)

Now, pairs of the above equations can be solved together in order to express the individual transverse ﬁeld components in terms of derivatives of the z components of

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E and H. For example, (77a) and (78b) can be combined, eliminating E ys , to give Hxs = ∂ Hzs −j ∂ E zs β −ω 2 κ ∂x ∂y (79a)

Then, using (76b) and (77a), eliminate E xs between them to obtain Hys = ∂ Hzs −j ∂ E zs β +ω κ2 ∂y ∂x (79b)

Using the same equation pairs, the transverse electric ﬁeld components are then found: E xs = E ys = ∂ E zs −j ∂ Hzs β + ωµ 2 κ ∂x ∂y ∂ E zs −j ∂ Hzs β − ωµ 2 κ ∂y ∂x (79c)

(79d)

κ is deﬁned in the same manner as in the parallel-plate guide [Eq. (35)]: where k = ω µ . In the parallel-plate geometry, we found that discrete values of κ and β resulted from the analysis, which we then subscripted with the integer mode number, m (κm and βm ). The interpretation of m was the number of ﬁeld maxima that occurred between plates (in the x direction). In the rectangular guide, ﬁeld variations will occur in both x and y, and so we will ﬁnd it necessary to assign two integer subscripts to κ and β, thus leading to κmp =
2 k 2 − βmp

√

κ=

k2 − β2

(80)

(81)

where m and p indicate the number of ﬁeld variations in the x and y directions. The form of Eq. (81) sugges

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