Download Chapter 30 solutions to assigned problems

CHAPTER 30: Inductance, Electromagnetic Oscillations, and AC Circuits
Solutions to Assigned Problems
2.
If we assume the outer solenoid is carrying current I1 , then the magnetic field inside the outer
solenoid is B  0n1 I1. The flux in each turn of the inner solenoid is  21  B r22  0n1 I1 r22 . The
mutual inductance is given by Eq. 30-1.
N
n l  n I  r2
M
M  2 21  2 0 1 1 2 
 0n1n2 r22
I1
I1
l
12. The inductance of the solenoid is given by L 
wire is given by l
wire
0 N 2 A
 N d sol , and so since d sol 2
0 N 2  d 2

. The (constant) length of the
l
l
4
 2.5 dsol 1 , we also know that N1  2.5N 2 . The fact
that the wire is tightly wound gives l sol  Nd wire . Find the ratio of the two inductances.
0
L2
4

L1 0
4
N 22 2
N 22 2
l 2wire  2
d sol 2
d sol 2
l
Nd
N
l sol 2
l
l
 sol22
 2 sol 2 2  sol 1  1 wire  1  2.5
2
N1 2
N1 2
l wire 
l sol 2 N 2d wire N 2
d sol 1
d sol 1
l sol 1
l sol 1
l sol 1
15. The magnetic energy in the field is derived from Eq. 30-7.
Energy stored 1 B 2
u
2

Volume
0
Energy 
1
2
B2
0
 Volume  
1
2
B2
0
 r2l 
1
2
 0.600 T 2
 4  10
7
Tm A

  0.0105 m   0.380 m   18.9 J
2
16. (a) We use Eq. 24-6 to calculate the energy density in an electric field and Eq. 30-7 to calculate the
energy density in the magnetic field.
uE  12  0 E 2  12 8.85 1012 C2 /N m2 1.0 104 N/C  4.4  104 J/m3
2
 2.0 T 
B2

 1.592  106 J/m3  1.6  106 J/m3
7
2 0 2  4  10 T m/A 
2
uB 
(b) Use Eq. 24-6 to calculate the electric field from the energy density for the magnetic field given
in part (a).
uE  12  0 E 2  uB  E 
2u B
0

2 1.592  106 J/m3 
8.85  1012 C2 /N m2 
 6.0  108 N/C
23. We set the current in Eq. 30-11 equal to 0.03I0 and solve for the time.
I  0.03I 0  I 0 e t /   t   ln  0.03  3.5
25. (a) We use Eq. 30-6 to determine the energy stored in the inductor, with the current given by Eq.
Eq 30-9.
© 2008 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.
1
Physics for Scientists & Engineers with Modern Physics, 4th Edition
Instructor Solutions Manual
2
LV02
1  et / 
2 
2R
(b) Set the energy from part (a) equal to 99.9% of its maximum value and solve for the time.
2
V2
V2
U  0.999 0 2  0 2 1  et /   t   ln 1  0.999  7.6
2R
2R
U  12 LI 2 


26. (a) At the moment the switch is closed, no current will flow through the inductor. Therefore, the
resistors R1 and R2 can be treated as in series.
e
e  I  R1  R2   I1  I 2 
, I3  0
R1  R2
(b) A long time after the switch is closed, there is no voltage drop across the inductor so resistors
R2 and R3 can be treated as parallel resistors in series with R1.
I1  I 2  I 3 , e =I1 R1  I 2 R2 , I 2 R2  I 3 R3
e R3
e  I 2 R2
I R
 I2  2 2  I2 
R1
R3
R2 R3  R1 R3  R1 R2
I3 
(c)
I 2 R2
e R2

R3
R2 R3  R1 R3  R1 R2
I1  I 2  I 3 
e  R3  R2 
R2 R3  R1R3  R1R2
Just after the switch is opened the current through the inductor continues with the same
magnitude and direction. With the open switch, no current can flow through the branch with
the switch. Therefore the current through R2 must be equal to the current through R3, but in the
opposite direction.
I3 
e R2
,
R2 R3  R1 R3  R1 R2
I2 
e R2
, I1  0
R2 R3  R1R3  R1R2
(d) After a long time, with no voltage source, the energy in the inductor will dissipate and no
current will flow through any of the branches.
I1  I 2  I 3  0
31. (a) The AM station received by the radio is the resonant frequency, given by Eq. 30-14. We divide
the resonant frequencies to create an equation relating the frequencies and capacitances. We
then solve this equation for the new capacitance.
1
1
2
2
 f1 
 550 kHz 
f1 2 LC1
C2


 C2  C1    1350 pF  
  0.16 nF
f2
C1
1
1
 1600 kHz 
 f2 
2 LC2
(b) The inductance is obtained from Eq. 30-14.
1
1
1
1
f 
L 2 2 
 62  H
2
3
2 LC1
4 f C 4  550  10 Hz 2 1350  1012 F 
35. (a) When the energy is equally shared between the capacitor and inductor, the energy stored in the
capacitor will be one half of the initial energy in the capacitor. We use Eq. 24-5 to write the
energy in terms of the charge on the capacitor and solve for the charge when the energy is
equally shared.
© 2008 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.
2
Chapter 30
Inductance, Electromagnetic Oscillations, and AC Circuits
Q 2 1 Q02
2

Q 
Q0
2C 2 2C
2
(b) We insert the charge into Eq. 30-13 and solve for the time.
 2  T   T
2
1
Q0  Q0 cos t  t  cos 1 
 
 
2

 2  2  4  8
38.
As shown by Eq. 30-18, adding resistance will decrease the oscillation frequency. We use Eq. 3014 for the pure LC circuit frequency and Eq. 30-18 for the frequency with added resistance to solve
for the resistance.
  (1  .0025)
R
75.
1
R2
1
 2  0.9975
LC 4 L
LC

4L
1  0.99752  
C
4  0.350 H 

1  0.9975   2.0k
2
1.800 10 F 
9
We use Eq. 30-4 to calculate the self inductance between the two wires. We calculate the flux by
integrating the magnetic field from the two wires, using Eq. 28-1, over the region between the two
wires. Dividing the inductance by the length of the wire gives the inductance per unit length.
0 I 
 h l r  1

1 l  r  0 I
1 
L B 

hdr   0




dr 
I
I r  2 r  2  l  r   
2 r  r   l  r   


l r
   l r 
L 0
 r   0  l  r 
ln  r    ln  l  r     0 ln 

 ln 
ln 

 

r
h 2
2   r 
 l  r    r 
78. Putting an inductor in series with the device will protect it from sudden surges in current. The
growth of current in an LR circuit is given is Eq. 30-9.
V
I
1  e tR L  I max 1  e tR L
R
The maximum current is 33 mA, and the current is to have a value of 7.5 mA after a time of 75
microseconds. Use this data to solve for the inductance.
I
I  I max 1  e tR L  e tR L  1 

I max



L


tR

ln  1 


I 

I max 
 75  10


6
ln  1 


sec 150  
7.5 mA 
33mA 
 4.4  102 H
2
Put an inductor of value 4.4  10 H in series with the device.
79. We use Kirchhoff’s loop rule to equate the input voltage to the voltage drops across the inductor and
Rt
resistor. We then multiply both sides of the equation by the integrating factor e L and integrate the
Rt
Rt
Rt
right-hand side of the equation using a u substitution with u  IRe L and du  dIRe L  Ie L dt L
© 2008 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.
3
Physics for Scientists & Engineers with Modern Physics, 4th Edition
Instructor Solutions Manual
dI
 IR 
dt
Rt
L
L Rt
L Rt
 dI
 Rt
Vin e L dt   L  IR e L dt 
du  IR e L  Vout e L
R
R
R
 dt

Vin  L



Rt
L
For L / R  t , e  1. Setting the exponential term equal to
unity on both sides of the equation gives the desired results.
L
Vin dt  Vout
R

© 2008 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.
4