Contents
I. Abstract: 2
II. Question 1 3
III. Question 2 3
A. Normal Incidence 3
B. Oblique Incidence 5
1. Perpendicular Polarization 5
2. Parallel Polarization 12
IV. Question 3: 17
A. Cost of conducting Material 18
B. Mass of conducting Material 18
C. Volume of Conducting Material 19
V. Conclusion: 19

Abstract:
The purpose of this project is to design a shielded room. The room should be built in a way so that high-frequency waves cannot penetrate into the room. The material that should be used are nonconducting, nonmagnetic material with a thin coating of conducting material on the outer surface. The conducting layers that can be used are made of aluminum, copper, mu-metal, or conducting polymer.
Frequency=10 MHz ω=2∙π∙f=2∙π∙(10× 〖10〗^6 )=62.83×〖10〗^7 [rad⁄s]
The shielded room must reduce the electric field intensity by a factor of 〖10〗^6
There are different cases other than the material we should consider, we should also consider the type of incidence the electric field is at, and for that we have a couple of different cases which include: Normal Incidence Oblique incidence (sub divided into) Parallel polarization. Perpendicular polarization
Question 1
Using professional sources (the book) we obtained the conductivity, permeability, cost, and density of the following four materials
Table 1: Table of values
Material Conductivity [σ] Permeability
[μ] Density [kg/m^3] Cost
[Unit/Kg]
Aluminum 3.6 × 〖10〗^7 µ0 2700 1.5
Copper 5.7 × 〖10〗^7 µ0 8960 1
Mu-metal 0.5 × 〖10〗^7 〖10〗^5 µ0 7800 100
Polymer 0.001 µ0 1200 0.01
We assumed that ɛ = 1 for conductive materials,
Question 2
We now have to calculate the minimum thickness required for each of the four materials, but we have to consider all possible orientations as mentioned in the abstract above, so the calculations have to each orientations using the four different materials. We used Matlab for the perpendicular and parallel polarization, given the fact that we had to consider all possible values for the angle theta.
Normal Incidence
In this case we are considering that a normal incident for all the materials, this type of incidence is the easiest to compute.
We begin by calculating the attenuation constant ‘α’ for each of the four materials α= √(π∙f∙μ∙σ)
Table 2: calculating alpha
Material Attenuation constant α= √(π∙f∙μ∙σ)
[Np⁄m]
Aluminum √(π ∙(10 ×〖10〗^6 )∙ (4π×〖10〗^(-7) )∙(3.6×〖10〗^7))=12,000π
Copper √(π ∙(10 ×〖10〗^6 ) ∙(4π×〖10〗^(-7) )∙(5.7×〖10〗^7))=47,437.001
Mu-metal √(π ∙(10 ×〖10〗^6 ) 〖∙10〗^5∙(4π×〖10〗^(-7) )∙(0.5×〖10〗^7 ) )=4,442,882.94
Polymer √(π ∙(10 ×〖10〗^6 ) ∙(4π×〖10〗^(-7) )∙(0.001))=0.1987

Table 3: alpha simplification
Material Attenuation constant α [Np⁄m]
Aluminum 12,000π
Copper 47,437
Mu-metal 4,442,883
Polymer 0.1987
Now that we have the attenuation constant, we can now calculate the distance the electric field travels before it gets attenuated by a factor of 〖10〗^6.
E= E_o e^(-αz)
E/E_o = e^(-αd)= 〖10〗^(-6) d= ln⁡(〖10〗^(-6) )/(-α)
Now that we have the formula we can calculate the distance

Table 4: calculating the distance, in normal incidence
Material Distance d= ln⁡(〖10〗^(-6) )/(-α)
[m]
Aluminum ln⁡(〖10〗^(-6) )/(-12,000π)=3.66×〖10〗^(-4)
Copper ln⁡(〖10〗^(-6) )/(-47,437)=2.91×〖10〗^(-4)
Mu-metal ln⁡(〖10〗^(-6) )/(-4,442,883)=3.11×〖10〗^(-6)
Polymer ln⁡(〖10〗^(-6) )/(-0.1987)=69.52
As we can see if it was normal we have to have a very thick wall made out of polymer, in order to reduce by the factor we need. But we cannot rule out any material yet before we conduct the other calculations.
Oblique Incidence Figure 5: orientation of the field, and the axis
Perpendicular Polarization

In this case we have what is called a perpendicular polarization, this phenomena occurs when the electric field is perpendicular to the plane of incidence, in our case the plane of incidence is the plane formed by the direction of propagations and the normal to the boundary. In this case we will have a reflected value and a transmitted value, since the electric field will be coming at an angle.
The following represents the equation of the transmitted electric field if we have a perpendicular polarization: E_t=〖 |E〗_o |×Ʈ_⊥×e^(-β(x sinθ_t + y cosθ_t ) ) z ̂
|E_t |=〖 |E〗_o |×Ʈ_⊥×e^(-β(x sinθ_t + y cosθ_t ) )
We know that the electric field is traveling through a material at a certain distance d, we can calculate this distance by assuming that θ_t is the angle between the transmitted and the y-axis so then the distance d would be simply equal to:
First we assume S = direction of propagation = x sinθ_t + y cosθ_t d= S∙cosθ_t
S= d/(cosθ_t )
|E_t |=〖 |E〗_o |×Ʈ_⊥×e^(-β(S) )
|E_t |=〖 |E〗_o |×Ʈ_⊥×e^(-β(d/(cosθ_t )) )
|E_t |/(〖 |E〗_o |)=Ʈ_⊥×e^(-β(d/(cosθ_t )) )= 〖10〗^(-6) d= (cosθ_t)/(-β) ∙ ln(〖10〗^(-6)/Ʈ_⊥ )
We now need to calculate the value of theta transmitted, and in order to do that we use Snell’s law: sin(θ_i )/sin(θ_t ) =√((ε_1 μ_1)/(ε_2 μ_2 )) θ_t=〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) )
For an incident angle varying from value 0 to 90o
Ʈ_⊥= (2 〖ƞ_2 cos〗⁡〖θ_i 〗)/(〖ƞ_2 cos〗⁡〖θ_i 〗 + 〖ƞ_1 cos〗⁡〖θ_t 〗 ) ƞ_1=√(μ_1/ε_1 ) ƞ_2= √(〖jωμ〗_2/(σ+jωε_2 ))
Ʈ_⊥= (2 〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗)/(〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗 + 〖√(μ_1/ε_1 ) cos〗⁡〖θ_t 〗 )

Ʈ_⊥= (2 〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗)/(〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗 + 〖√(μ_1/ε_1 ) cos〗⁡(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) )) )
Replacing back in the formula for distance d, we get: d= (cosθ_t)/(-β) ∙ ln(〖10〗^(-6)/((2 〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗)/(〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗 + 〖√(μ_1/ε_1 ) cos〗⁡(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) )) )))
Making the equation in terms of theta incident only we get: d= cos(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) ))/(-β) ∙ ln(〖10〗^(-6)/((2 〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗)/(〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗 + 〖√(μ_1/ε_1 ) cos〗⁡(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) )) )))
Because we have conductive material then β is equal to α d= cos(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) ))/(-α) ∙ ln(〖10〗^(-6)/((2 〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗)/(〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗 + 〖√(μ_1/ε_1 ) cos〗⁡(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) )) )))
Aluminum:

Figure 6: code for perpendicular orientation in Aluminum Figure 7: plot of the different distance d
Copper:

Figure 8: code for perpendicular orientation in Copper Figure 9: plot of the different distance d
Mu-Metal

Figure 10: code for perpendicular orientation in Mu-metal Figure 11: plot of the different distance d
Polymer

Figure 12: code for perpendicular orientation in Polymer Figure 13: plot of the different distance d

Table 5: Values for the distances and different materials
Material Distance [m]
Aluminum 0.0003698
Copper 0.0003523
Mu-metal 4.617×〖10〗^(-6)
Polymer 133.7533
We take the largest distance calculated, because we need the worst case scenario to be attenuated, so at that distance any electric field would be attenuated by the factor needed
Parallel Polarization
Parallel polarization is a phenomena that occurs when the electric field is parallel to the plane of incidence
E_t=〖 |E〗_o |×Ʈ_(||)×e^(-β(x sinθ_t + y cosθ_t ) ) (-x ̂ cosθ_t + y ̂ sin θ_t )
|E_t |=〖 |E〗_o |×Ʈ_(||)×e^(-β(x sinθ_t + y cosθ_t ) )
|E_t |/(〖 |E〗_o |)=×Ʈ_(||)×e^(-β(x sinθ_t + y cosθ_t ) ) d= (cosθ_t)/(-β) ∙ ln(〖10〗^(-6)/Ʈ_(||) )
Ʈ_(||)= (2 〖ƞ_2 cos〗⁡〖θ_i 〗)/(〖ƞ_2 cos〗⁡〖θ_t 〗 + 〖ƞ_1 cos〗⁡〖θ_i 〗 )

d= cos(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) ))/(-α) ∙ ln(〖10〗^(-6)/((2 〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡〖θ_i 〗)/(〖√(〖jωμ〗_2/(σ+jωε_2 )) cos〗⁡(〖sin〗^(-1) (sin(θ_i ) √((ε_1 μ_1)/(ε_2 μ_2 )) )) + 〖√(μ_1/ε_1 ) cos〗⁡〖θ_i 〗 )))

Aluminum: Figure 15: code for parallel orientation in Aluminum Figure 16: plot of the different distance d
Copper:

Figure 17: code for parallel orientation in Copper Figure 18: plot of the different distance d
Mu-Metal:

Figure 19: code for parallel orientation in Mu-metal Figure 20: plot of the different distance d
Polymer:

Figure 21: code for parallel orientation in Polymer Figure 22: plot of the different distance d

Table 6: Values for the distances and different materials
Material
par Distance [m]
Aluminum 0.00044237
Copper 0.00035226
Mu-metal 4.617×〖10〗^(-6)
Polymer 133.7533

Question 3:
Which material will cost you if you choose the overall parameter?
First we begin by taking the worst case scenario d for all possibilities for the different materials, so the following table will provide us with the d that should be used to create the wall:

Table 7: the distance of the material that have to be used
Material Distance [m]
Aluminum 0.00044237
Copper 0.00035226
Mu-metal 3.11×〖10〗^(-6)
Polymer 133.7533
Table 8: Table of values
Material Conductivity [σ] Permeability
[μ] Density [kg/m^3] Cost
[Unit/Kg]
Aluminum 3.6 × 〖10〗^7 µ0 2700 1.5
Copper 5.7 × 〖10〗^7 µ0 8960 1
Mu-metal 0.5 × 〖10〗^7 〖10〗^5 µ0 7800 100
Polymer 0.001 µ0 1200 0.01

Volume= Width × Length × Thickness =1 ×1×d
Material Volume [m^3]
Aluminum 0.00044237
Copper 0.00035226
Mu-metal 3.11×〖10〗^(-6)
Polymer 133.7533

Mass=Density ×Volume
Material Mass [Kg]
Aluminum 1.194399
Copper 3.15624
Mu-metal 0.02425
Polymer 160,504

Cost of conducting Material
Cost=Mass ×Price
Table 9: Price of the room\
Material Cost
[Unit]
Aluminum 1.7915
Copper 3.15624
Mu-metal 2.4258
Polymer 1605.04
As we can see if we are building a room, and cost is an option then we have to consider the aluminum to be the best due to it only costing 1.7915
Mass of conducting Material
Material Mass [Kg]
Aluminum 1.194399
Copper 3.15624
Mu-metal 0.02425
Polymer 160,504
If mass was an option, usually we would be looking for the lightest material to use, then we consider the aluminum which will have a mass of only 1.194 Kg

Volume of Conducting Material

Material Volume [m^3]
Aluminum 0.00044237
Copper 0.00035226
Mu-metal 3.11×〖10〗^(-6)
Polymer 133.7533

If volume was an option, usually we would be looking for the smallest material to use, then we consider the mu-metal which will have a volume of only
3.11×〖10〗^(-6) m^3

Conclusion: