Download Chapter 28 solutions to assigned problems

CHAPTER 29: Electromagnetic Induction and Faraday’s Law
Solutions to Assigned Problems
1.
The average induced emf is given by Eq. 29-2b.
38 Wb   58 Wb 
dB
 B
e  N
 N
 2
 460 V
dt
t
0.42s
5.
Use Eq. 29-2a to calculate the emf. Setting the flux equal to the magnetic field multiplied by the
area of the loop, A = p r 2 , and the emf equal to zero, we can solve for the rate of change in the coil
radius.
dB
d
dB
dr
e
   B r 2     r 2  2 Br  0
dt
dt
dt
dt
dr
dB r
0.12 m

   0.010 T/s 
 0.0012 m/s  1.2 mm/s
dt
dt 2 B
2  0.500 T 
6.
We choose up as the positive direction. The average induced emf is given by the “difference”
version of Eq. 29-2a.
e
 B
t

AB
t
  0.054 m   0.25T  0.68T 
2

0.16s
 5.3  102 V
21. The induced emf is given by Eq. 29-2a. Since the field is uniform and is perpendicular to the area,
the flux is simply the field times the area of a circle. We calculate the initial radius from the initial
area. To calculate the radius after one second we add the change in radius to the initial radius.
d  r2
dB
dr
A0
e (t ) 
B
 2 Br
A0   r02  r0 
dt
dt
dt

 
e (0)  2  0.28T 
0.285 m 2

 0.043m s  
23mV
 0.285 m 2

  0.043m s 1.00s   0.043m/s   26 mV



e (1.00s)  2  0.28T  
31. The rod will descend at its terminal velocity when the magnitudes of the
magnetic force (found in Example 29-8) and the gravitational force are equal.
We set these two forces equal and solve for the terminal velocity.
B 2 l 2 vt
 mg 
R
3
2
mgR  3.6  10 kg  9.80 m/s   0.0013  
vt  2 2 
 0.39 m/s
2
2
B l
 0.060 T   0.18 m 
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
248
Electromagnetic Induction and Faraday’s Law
Chapter 29
37. We find the number of turns from Eq. 29-4. The factor multiplying the sine term is the peak output
voltage.
epeak
24.0 V
epeak  NB A  N 

 57.2 loops
2
B A  0.420 T   2 rad rev   60 rev s  0.0515 m 
46. Because NS  N P , this is a step-down transformer. Use Eq. 29-5 to find the voltage ratio, and Eq.
29-6 to find the current ratio.
VS N S
85 turns
I S N P 620 turns


 0.14


 7.3
VP N P 620 turns
I P NS
85 turns
61. The charge on the capacitor can be written in terms of the voltage across the battery and the
capacitance using Eq. 24-1. When fully charged the voltage across the capacitor will equal the emf
of the loop, which we calculate using Eq. 29-2b.
d B
dB
Q  CV  C
 CA
  5.0 1012 F12 m2  10 T/s   0.60 nC
dt
dt
74. (a)
Rfield
I0
Rfield
field
Rarmature
Rarmature
I0
– +
starting e
(b)
(c)
– +
I armature
armature
I field
eback
– +
I0
full speed e
I
At startup there is no back emf. We therefore treat the circuit as two parallel resistors, each
with the same voltage drop. The current through the battery is the sum of the currents through
each resistor.
e
e
115 V 115 V
I  I0  I0




 41.5 A
Rfield Rarmature 36.0  3.00 
field
armature
At full speed the back emf decreases the voltage drop across the armature resistor.
e
e  eback 115V 115V  105V
I  I field  I armature 



 6.53A
Rfield Rarmature 36.0 
3.00 
© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
249