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The solutions to exam 3 of the phy2049 spring 2008 course. The problems cover various topics in physics, including electromagnetism, circuits, and capacitors. Students can use these solutions to check their understanding of the exam material.
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Prof. Darin Acosta Prof. Selman Hershfield April 9, 2008
(1) 2 x 10 -3^ N (2) 1 x 10 -3^ N (3) 3 x 10 -3^ N (4) 4 x 10 -3^ N (5) 5 x 10 -3^ N
The induced EMF (magnitude) is given by Faraday’s Law
d d BL x vt BLv dt dt
The current is given by
i R
This gives rise to a force on the rod carrying this current (recall Ch.28): F = iLB
Plugging in gives F = 2 x 10 -3^ N for v=60 cm/s, and F = 3 x 10-3^ N for v=90 cm/s
For a clockwise current, the field points into the page within ring 1 and out of the page outside of ring 1. For example, the field lines look like below (use the right-hand rule):
Thus, since the current is increasing, the field strength is increasing, which means in the vicinity of rind 2 and 3 the field is growing out of the page. This induces and EMF and current in rings 2 and 3 which must oppose this increase in magnetic flux (Lenz’s Law), thus those rings also have clockwise currents induced which create fields pointing into the page inside the loops (thus subtracting from the increasing magnetic flux from ring 1).
This is alternating current and EMF, not DC! Recall that for AC circuits:
( ) sin ( ) sin
m m
t t i t i t
di v L L t dt
( ) sin sin sin 5sin 45 3.5V 2
Then by Kirchoff’s loop rule: 0 3.5 0 6 2.5V for 6 3.5 0 4 0.5V for 4
R L C C L C L
v v v v v v v
z
x
(1) 1 (2) 2 (3) 3 (4) 4 (5) 0
The displacement current between the capacitor plates equals the real current going into the capacitor. It is distributed uniformly between the plates since the electric field there is uniform. Thus the fraction of the current enclosed by a circular path of radius r is 2 disp encl , (^2)
r i i R
Thus by the Ampere-Maxwell Law 2 (^0 )
0 6 2
r B ds rB i R ir B R
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