...Episode 412: The force on a conductor in a magnetic field Having reminded your students that magnetic fields can be found near permanent magnets and in the presence of an electric current, the next step is to show how the ‘field’ can be quantified. Again, students should know that a conductor carrying a current in a magnetic field will experience a force and will probably remember that Fleming's Left Hand Rule can be used to find the direction of that force. Summary Demonstrations: Leading to F = BIL. (15 minutes) Discussion: Factors affecting the force. (15 minutes) Discussion: Formal definitions. (20 minutes) Student questions: BIL force calculations. (20 minutes) Demonstrations: Leading to F = BIL Several quick experimental reminders are possible. Tap 412-1: Forces on currents TAP 412-2: An electromagnetic force These lead on to a further experiment in which the relationship F=BIL can be established. TAP 412-3: Force on a current-carrying wire Discussion: Factors affecting the force The experiments above lead to the conclusion that the force F on the conductor is proportional to the length of wire in the field, L, the current I and the ‘strength’ of the field, represented by the flux density B. (There is also an 'angle factor' to consider, but we will leave this aside for now.) Combining these we get F = BIL (It can help students to refer to this force as the ‘BIL force’.) Students...
Words: 3240 - Pages: 13
...Trial Examination 2014 VCE Physics Units 3&4 Written Examination Suggested Solutions Neap Trial Exams are licensed to be photocopied or placed on the school intranet and used only within the confines of the school purchasing them, for the purpose of examining that school’s students only. They may not be otherwise reproduced or distributed. The copyright of Neap Trial Exams remains with Neap. No Neap Trial Exam or any part thereof is to be issued or passed on by any person to any party inclusive of other schools, non-practising teachers, coaching colleges, tutors, parents, students, publishing agencies or websites without the express written consent of Neap. Copyright © 2014 Neap ABN 49 910 906 643 96–106 Pelham St Carlton VIC 3053 Tel: (03) 8341 8341 Fax: (03) 8341 8300 TEVPHYU34_SS_2014.FM VCE Physics Units 3&4 Trial Examination Suggested Solutions SECTION A – CORE Area of study – Motion in one and two dimensions Question 1 (10 marks) a. v v = 12 sin 60° = 10.4 m s b. t air = 2 × t top 0 – 10.4 t top = ------------------ = 1.04 – 10 t air = 2 × 1.04 = 2.1 1 2 s = -- at 2 2 1 = -- ( 10 ) ( 1.04 ) 2 –1 1 mark 1 mark 1 mark Note: Consequential on part a. c. 1 mark 1 mark = 5.4 m d. Gravity is 10 m s ∴ 10 m s e. –2 –2 down 1 mark 1 2 KE = -- mv 2 v h = 12 cos 60° =6ms –1 1 mark 2 1 KE = -- ( 80 ) ( 6 ) 2 = 1440 J f. R = v h t air = ( 6 ) ( 2.08 ) = 12.5 m 1 mark 1 mark 1 mark Note: Consequential on part e. 2 ...
Words: 17146 - Pages: 69