Chapter 32 Solutions
∆I
 1.50 A − 0.200 A 
= 1.95 × 10–2 V = 19.5 mV


0.200 s
*32.1
ε
32.2
Treating the telephone cord as a solenoid, we have:
= L
∆t
= (3.00 × 10– 3 H)
µ0N 2A
(4 π × 10 − 7 T ⋅ m / A)(70.0)2 ( π )(6.50 × 10 − 3 m)2
=
= 1.36 µH
l
0.600 m
L=
 0.500 A 
 ∆I 
= (2.00 H) 
 = 100 V
 ∆t 
 0.0100 s 
32.3
ε
32.4
L = µ 0n2 Al so
32.5
L=
32.6
ε
32.7
=+L
n=
L
= 7.80 × 103 turns/m
µ 0 Al
N ΦB
LI
→ ΦB =
= 240 nT · m2
N
I
=L
dI
dt
Thus,
ε
where L =
(through each turn)
µ0N 2A
l
(
)
(
)
4 π × 10 −7 T ⋅ m A ( 300) π × 10 − 4 m 2
 µ N 2 A  dI
= 0
=
(10.0 A s) = 2.37 mV

0.150 m
 l  dt
dI
ε back = –ε = L dt
2
d
= L dt (Imax sin ω t) = Lω Imax cos ω t = (10.0 × 10-3)(120π )(5.00) cos ω t
ε back = (6.00π ) cos (120π t) =
(18.8 V) cos (377t)
© 2000 by Harcourt, Inc. All rights reserved.
Chapter 32 Solutions 243
From
*32.8
ε
From L =
L=
32.9
 ∆I 
, we have
 ∆t 
dI
NΦ B
, we have
I
→
Thus,
ε
32.11
ε
 ∆Ι 
 ∆t 
ΦB =
=
24.0 × 10– 3 V
= 2.40 × 10– 3 H
10.0 A/s
LI
(2.40 × 10– 3 H)(4.00 A)
=
= 19.2 µT · m2
N
500
–6
dI
− ε –175 × 10 V
=
=
= – 0.421 A/s
dt
4.16 × 10– 4 H
L
The induced emf is
32.10
L=
µ0(420)2(3.00 × 10– 4)
µ0 N 2 A
=
= 4.16 × 10– 4 H
0.160
l
ε = –L dt
32.12
=L
dI
ε
= − =
dt
L
ε = − L dI ,
dt
−
where the self-inductance of a solenoid is given by L =
εl
µ0N 2A
dI
d
= L dt = (90.0 × 10-3) dt (t 2 – 6t) V
(a)
At t = 1.00 s,
ε=
360 mV
(b)
At t = 4.00 s,
ε=
180 mV
(c)
ε = (90.0 × 10-3)(2t – 6) = 0 when
(a)
450
B = µ 0nI = µ 0  0.120 (0.0400 mA) = 188 µT


(b)
ΦB = BA = 3.33 × 10-8 T · m2
(c)
L=
t = 3.00 s
NΦ B
= 0.375 mH
I
© 2000 by Harcourt, Inc. All rights reserved.
µ0N 2A
.
l
244 Chapter 32 Solutions
(d)
B and Φ B are proportional to current; L is independent of current
Chapter 32 Solutions 245
32.13
µ0(120)2 π ( 5.00 × 10–3)2
µ0 N 2 A
=
= 15.8 µH
l
0.0900
(a)
L=
(b)
µm
µm N 2 A
= 800(1.58 × 10– 5 H) = 12.6 mH
Φ ′B = µ ΦB → L =
l
0
NBA
NA
µ 0NI
µ 0N 2A
NΦ B
=
≈
·
=
I
I
I
2π R
2π R
32.14
L=
32.15
ε = ε 0e − k t = − L dI
dt
dI = −
ε 0 e − k t dt
L
If we require I → 0 as t → ∞, the solution is
Q = ∫ I dt = ∫
∞
0
32.16
I=
ε 0 e − k t dt = − ε 0
ε 0 e − k t = dq
dt
kL
Q =
k 2L
kL
I=
ε0
k 2L
ε (1 − e −Rt/L )
R
0.900
ε
R
=
ε
R
[1 − e
−R(3.00 s)/2.50 H
]
 R(3.00 s) 
= 0.100
exp −

2.50 H 
R=
2.50 H
ln 10.0 = 1.92 Ω
3.00 s
© 2000 by Harcourt, Inc. All rights reserved.
246 Chapter 32 Solutions
L
I
τ = R = 0.200 s: I
= 1 – e–t/τ
max
32.17
(a)
0.500 = 1 – e–t/0.200 → t = τ ln 2.00 = 0.139 s
(b)
0.900 = 1 – e–t/0.200 → t = τ ln 10.0 = 0.461 s
Figure for Goal
Solution
Goal Solution
A 12.0-V battery is about to be connected to a series circuit containing a 10.0-Ω resistor and a 2.00-H
inductor. How long will it take the current to reach (a) 50.0% and (b) 90.0% of its final value?
G:
The time constant for this circuit is τ = L R = 0.2 s, which means that in 0.2 s, the current will reach
1/e = 63% of its final value, as shown in the graph to the right. We can see from this graph that the
time to reach 50% of I max should be slightly less than the time constant, perhaps about 0.15 s, and the
time to reach 0.9I max should be about 2.5τ = 0.5 s.
O:
The exact times can be found from the equation that describes the rising current in the above graph
and gives the current as a function of time for a known emf, resistance, and time constant. We set
time t = 0 to be the moment the circuit is first connected.
A:
At time t ,
I (t ) =
where, after a long time,
I max =
ε (1 − e −∞ ) = ε
At I (t ) = 0.500I max ,
(0.500)
ε = ε (1 − e
Isolating the constants on the right,
ln e −t/2.00 s = ln(0.500)
and solving for t ,
−
(b)
0.900 = 1 − e −t/τ
Similarly, to reach 90% of I max ,
Thus,
L:
(
ε (1 − e −t/τ )
R
R
R
R
−t/0.200 s
R
)
so
0.500 = 1 − e −t/0.200 s
or
t = 0.139 s
)
t
= −0.693
0.200 s
and t = − τ ln(1 − 0.900)
t = −(0.200 s) ln(0.100) = 0.461 s
The calculated times agree reasonably well with our predictions. We must be careful to avoid
confusing the equation for the rising current with the similar equation for the falling current.
Checking our answers against predictions is a safe way to prevent such mistakes.
Chapter 32 Solutions 247
I = I 0 e −t/τ :
Taking τ = L R,
32.18
IR + L
dI
 1
= I 0 e −t/τ −
 τ
dt
(
)
dI
 1
=0
= 0 will be true if I 0 R e −t/τ + L I 0 e −t/τ −
 τ
dt
Because τ = L R, we have agreement with 0 = 0
*32.19 (a)
τ = L R = 2.00 × 10– 3 s = 2.00 ms
(
)
(b)
I = I max 1 − e −t/τ =
(c)
Imax =
ε
R
=
(
)
 6.00 V 
1 − e −0.250/2.00 = 0.176 A
 4.00 Ω 
6.00 V
= 1.50 A
4.00 Ω
(d) 0.800 = 1 – e–t/2.00 ms → t = – (2.00 ms) ln(0.200) = 3.22 ms
I=
*32.20
ε
120
(1 – e–t/τ ) = 9.00 (1 – e–1.80/7.00) = 3.02 A
R
∆VR = IR = (3.02)(9.00) = 27.2 V
∆VL = ε – ∆VR = 120 – 27.2 = 92.8 V
32.21
(a)
∆V R = IR = (8.00 Ω)(2.00 A) = 16.0 V
and
∆VL = ε − ∆V R = 36.0 V − 16.0 V = 20.0 V
Therefore,
(b)
∆V R 16.0 V
=
= 0.800
∆VL 20.0 V
∆V R = IR = (4.50 A)(8.00 Ω) = 36.0 V
∆VL = ε − ∆V R = 0
© 2000 by Harcourt, Inc. All rights reserved.
Figure for Goal
Solution
248 Chapter 32 Solutions
Goal Solution
For the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 Ω, and ε = 36.0 V. (a) Calculate the ratio
of the potential difference across the resistor to that across the inductor when I = 2.00 A. (b) Calculate the
voltage across the inductor when I = 4.50 A.
G: The voltage across the resistor is proportional to the current, ∆ VR = IR , while the voltage across the
inductor is proportional to the rate of change in the current, ε L = −L dI dt . When the switch is first
closed, the voltage across the inductor will be large as it opposes the sudden change in current. As the
current approaches its steady state value, the voltage across the resistor increases and the inductor’s
emf decreases. The maximum current will be ε /R = 4.50 A, so when I = 2.00 A, the resistor and
inductor will share similar voltages at this mid-range current, but when I = 4.50 A, the entire circuit
voltage will be across the resistor, and the voltage across the inductor will be zero.
O: We can use the definition of resistance to calculate the voltage across the resistor for each current.
We will find the voltage across the inductor by using Kirchhoff's loop rule.
A:
When I = 2.00 A, the voltage across the resistor is
(a)
∆V R = IR = ( 2.00 A )(8.00 Ω) = 16.0 V
Kirchhoff's loop rule tells us that the sum of the changes in potential around the loop must be zero:
ε − ∆VR − ε L = 36.0 V − 16.0 V − ε L = 0
ε L = 20.0 V
so
and
∆V R
εL
=
16.0 V
= 0.800
20.0 V
Similarly, for I = 4.50 A , ∆V R = IR = ( 4.50 A )(8.00 Ω) = 36.0 V
(b)
ε − ∆VR − ε L = 36.0 V − 36.0 V − ε L = 0
εL = 0
so
L : We see that when I = 2.00 A, ∆V R < ε L , but they are similar in magnitude as expected. Also as
predicted, the voltage across the inductor goes to zero when the current reaches its maximum value.
A worthwhile exercise would be to consider the ratio of these voltages for several different times after
the switch is reopened.
After a long time, 12.0 V = (0.200 A)R
*32.22
L
Thus, R = 60.0 Ω. Now, τ = R gives
L = τ R = (5.00 × 10– 4 s)(60.0 V/A) = 30.0 mH
(
)
I = I max 1 − e −t/τ :
32.23
L
15.0 H
τ= R =
= 0.500 s :
30.0 Ω
(
)
dI
 1
= −I max e −t/τ −
 τ
dt
dI
R
–t/τ
dt = L Imax e
and
Imax =
ε
R
(a)
dI
R
ε = 100 V = 6.67 A/s
t = 0: dt = L Imax e0 =
15.0 H
L
(b)
dI
ε e–t/τ = (6.67 A/s)e– 1.50/(0.500) = (6.67 A/s)e–3.00 = 0.332 A/s
t = 1.50 s: dt =
L
Chapter 32 Solutions 249
(
I = I max 1 − e −t/τ
32.24
)
0.980 = 1 − e − 3.00 ×10
0.0200 = e − 3.00 ×10
τ=−
−3
−3
/τ
/τ
3.00 × 10 − 3
= 7.67 × 10 − 4 s
ln(0.0200)
τ = L R, so L = τ R = (7.67 × 10 − 4 )(10.0) = 7.67 mH
Name the currents as shown.
32.25
By Kirchhoff’s laws:
I1 = I 2 + I 3
(1)
+10.0 V − 4.00 I1 − 4.00 I 2 = 0
(2)
+10.0 V − 4.00 I1 − 8.00 I 3 − (1.00)
dI 3
= 0 (3)
dt
From (1) and (2),
+10.0 − 4.00 I1 − 4.00 I1 + 4.00 I 3 = 0
Then (3) becomes
10.0 V − 4.00(0.500 I 3 + 1.25 A ) − 8.00 I 3 − (1.00)
(1.00 H)(dI 3
I1 = 0.500 I 3 + 1.25 A
and
dI 3
=0
dt
dt ) + (10.0 Ω) I 3 = 5.00 V
We solve the differential equation using Equations 32.6 and 32.7:
I 3 (t ) =
[
5.00 V
− 10.0 Ω ) t
1− e (
10.0 Ω
1.00 H
] = (0.500 A)[1 − e ]
−10t/s
I1 = 1.25 + 0.500 I 3 = 1.50 A − (0.250 A )e −10t/s
32.26
L
, we get R =
R
L
=
C
3.00 H
= 1.00 × 10 3 Ω = 1.00 kΩ
3.00 × 10 − 6 F
(a)
Using τ = RC =
(b)
τ = RC = 1.00 × 10 3 Ω 3.00 × 10 − 6 F = 3.00 × 10 −3 s = 3.00 ms
(
)(
)
© 2000 by Harcourt, Inc. All rights reserved.
250 Chapter 32 Solutions
For t ≤ 0, the current in the inductor is zero. At t = 0, it starts to
grow from zero toward 10.0 A with time constant
τ = L R = (10.0 mH) (100 Ω) = 1.00 × 10 − 4 s .
32.27
−t/τ
(
For 0 ≤ t ≤ 200 µ s,
I = I max  1 − e

 = 10.00 A 1 − e −10 000t/s
(
)

At t = 200 µ s,
I = (10.00 A ) 1 − e −2.00 = 8.65 A
(
)
Thereafter, it decays exponentially as I = I 0 e − t ′ τ ,
I = (8.65 A )e
32.28
ε
−10 000( t−200 µ s ) s
I=
(b)
Initial current is 1.00 A, :
R
so for
t ≥ 200 µ s,
(
)
= (8.65 A )e −10 000t s + 2.00 = 8.65 e 2.00 A e −10 000t s =
(63.9 A)e −10 000t s
12.0 V
= 1.00 A
12.0 Ω
(a)
=
)
∆V12 = (1.00 A)(12.00 Ω) = 12.0 V
∆V 1200 = (1.00 A)(1200 Ω) = 1.20 kV
∆V L = 1.21 kV
(c)
I = Imax e–Rt/L:
dI
R –Rt/L
dt = – Imax L e
and
dI
–L dt = ∆VL = Imax Re–Rt/L
Solving
12.0 V = (1212 V)e–1212t/2.00
so
9.90 × 10– 3 = e– 606t
Thus,
τ=
32.29
(a)
t = 7.62 ms
L 0.140
=
= 28.6 ms;
R 4.90
(
I = I max 1 − e −t/τ
e −t/τ = 0.820
)
so
I max =
ε
R
=
6.00 V
= 1.22 A
4.90 Ω
(
0.220 = 1.22 1 − e −t/τ
)
t = − τ ln(0.820) = 5.66 ms
(b)
10.0 

−
I = I max  1 − e 0.0286  = (1.22 A) 1 − e −350 = 1.22 A


(c)
I = I max e −t/τ
(
and
)
0.160 = 1.22e −t/τ
so
t = –τ ln(0.131) = 58.1 ms
Chapter 32 Solutions 251
32.30
(a)
(b)
For a series connection, both inductors carry equal currents at every instant, so dI/dt is the
same for both. The voltage across the pair is
L eq
dI
dI
dI
= L1 + L 2
dt
dt
dt
so
L eq
dI
dI
dI
= L 1 1 = L 2 2 = ∆VL
dt
dt
dt
where I = I1 + I 2
Thus,
(c)
L eq
∆VL ∆VL ∆VL
=
+
L eq
L1
L2
L eq = L 1 + L 2
and
dI dI1 dI 2
=
+
dt
dt
dt
1
1
1
=
+
L eq L 1 L 2
and
dI
dI
dI
+ R eq I = L 1 + IR 1 + L 2
+ IR 2
dt
dt
dt
Now I and dI/dt are separate quantities under our control, so functional equality requires both
L eq = L 1 + L 2
(d)
∆V = L eq
and
R eq = R 1 + R 2
dI
dI
dI
+ R eq I = L 1 1 + R 1I1 = L 2 2 + R 2 I 2
dt
dt
dt
where
I = I1 + I 2 and
dI dI1 dI 2
=
+
dt
dt
dt
We may choose to keep the currents constant in time. Then,
1
1
1
=
+
R eq R 1 R 2
We may choose to make the current swing through 0.
1
1
1
=
+
L eq L 1 L 2
Then,
This equivalent coil with resistance will be equivalent
to the pair of real inductors for all other currents as well.
L=
32.31
32.32
(a)
= 42.3 mH
1
1
U = 2 LI 2 = 2 (0.423 H)(1.75 A) 2 = 0.0648 J
so
The magnetic energy density is given by
u=
(b)
200(3.70 × 10– 4)
N ΦB
=
I
1.75
B2
(4.50 T)2
=
= 8.06 × 106 J/m3
2µ 0
2(1.26 × 10– 6 T · m/A)
The magnetic energy stored in the field equals u times the volume of the solenoid (the
volume in which B is non-zero).
[
]
U = uV = (8.06 × 106 J/m3) (0.260 m)π (0.0310 m)2 = 6.32 kJ
© 2000 by Harcourt, Inc. All rights reserved.
252 Chapter 32 Solutions
L = µ0
32.33
N 2A
(68.0)2 π (0.600 × 10 −2 )2
= µ0
= 8.21 µ H
0.0800
l
U = 21 LI 2 = 21 (8.21 × 10 −6 H)(0.770 A)2 = 2.44 µJ
32.34
(a)
2
2
(0.800)(500)2
1
1
Lε
 
U = 2 LI 2 = 2 L ε
=
=
= 27.8 J
 2R 
8(30.0)2
8R 2
(b)
I=
ε
1 − e −(R/L)t
 R
[
]
R
t = ln 2
L
ε
so
2R
u = ε0
*32.36 (a)
U = 2 LI 2 = 2 (4.00 H)(0.500 A) 2 = 0.500 J
dU
= LI = (4.00 H)(1.00 A) = 4.00 J/s = 4.00 W
dt
(c)
P = (∆V)I = (22.0 V)(0.500 A) = 11.0 W
(a)
]
L
0.800
ln 2 =
ln 2 = 18.5 ms
R
30.0
1
(b)
32.37
[
B2
u = 2µ = 995 µ J/m3
0
32.35
1
1
ε
1 − e −(R/L)t → e −(R/L)t =
 R
2
t=
so
E2
3
2 = 44.2 nJ/m
=
From Equation 32.7,
I=
ε (1 − e − R t
The maximum current, after a long time t , is
I=
ε
At that time, the inductor is fully energized and
P = I(∆V) = (2.00 A)(10.0 V) = 20.0 W
(b)
P lost = I 2 R = (2.00 A)2 (5.00 Ω) = 20.0 W
(c)
P inductor = I(∆Vdrop ) = 0
(d)
U=
LI 2 (10.0 H)(2.00 A)2
=
= 20.0 J
2
2
R
R
L
)
= 2.00 A.
Chapter 32 Solutions 253
32.38
We have
u = e0
Therefore
e0
B = E e0µ 0 =
32.39
E2
2
B2
E2
=
2
2µ 0
B2
2µ 0
and
u=
so
B2 = e0µ 0E 2
6.80 × 10 5 V / m
= 2.27 × 10– 3 T
3.00 × 108 m / s
The total magnetic energy is the volume integral of the energy density, u =
Because B changes with position, u is not constant. For B = B0 ( R / r ) ,
2
B2
2µ 0
 B 2  R 4
u= 0 
 2µ 0   r 
Next, we set up an expression for the magnetic energy in a spherical shell of radius r and
thickness dr. Such a shell has a volume 4π r 2 dr, so the energy stored in it is
 2 πB0 2 R 4  dr
dU = u 4 π r 2 dr = 
µ 0  r 2

(
)
We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the
sphere. This gives
2
U=
32.40
2π B 0 R 3
µ0
=
2π (5.00 × 10–5 T)2(6.00 × 106 m)3
= 2.70 × 1018 J
(1.26 × 10– 6 T · m/A)
I1(t) = I max e − α t sin ω t with I max = 5.00 A, α = 0.0250 s −1 , and ω = 377 rad s .
dI1
= I max e − α t ( −α sin ω t + ω cos ω t )
dt
[
dI1
= 1.85 × 10 3 A s
dt
Thus,
]
dI1
= ( 5.00 A s)e − 0.0200 − (0.0250) sin(0.800( 377 )) + 377 cos(0.800( 377 ))
dt
At t = 0.800 s ,
ε 2 = −M dI1 :
dt
M=
− ε2
dI1 dt
=
+ 3.20 V
= 1.73 mH
1.85 × 10 3 A s
© 2000 by Harcourt, Inc. All rights reserved.
254 Chapter 32 Solutions
ε 2 = −M dI1 = −(1.00 × 10− 4 H)(1.00 × 10 4 A / s) cos(1000t)
32.41
dt
(ε 2 )max =
M=
32.42
32.43
1.00 V
ε2
dI1 dt
=
96.0 mV
= 80.0 mH
1.20 A / s
(a)
M=
N BΦ BA 700(90.0 × 10 − 6 )
=
= 18.0 mH
3.50
IA
(b)
LA =
Φ A 400(300 × 10 − 6 )
=
= 34.3 mH
3.50
IA
(c)
ε B = −M dI A = −(18.0 mH)(0.500 A / s) =
dt
M=
32.44
[
(
70.0  
π ( 5.00 × 10
) 0.0500
m  
B at center of (larger) loop: B1 =
(a)
M=
(b)
M=
]
N 2 Φ12 N 2 (B1A1 ) N 2 (µ 0n1I1 )A 1
=
=
= N 2 µ 0n1A 1
I1
I1
I1
M = (1.00) 4 π × 10 −7 T ⋅ m A
32.45
– 9.00 mV
µ 0 I1
2R
µ0 π r 2
Φ 2 B1A 2 (µ 0 I1 / 2R)( π r 2 )
=
=
=
I1
I1
2R
I1
µ 0 π (0.0200)2
2(0.200)
= 3.95 nH
−3
)
2
m  = 138 nH

Chapter 32 Solutions 255
*32.46
Assume the long wire carries current I.
Then the magnitude of the magnetic field it
generates at distance x from the wire is B = µ 0 I 2 π x, and this field passes perpendicularly
through the plane of the loop. The flux through the loop is
Φ B = ∫ B ⋅ dA = ∫ B dA = ∫ B (ldx ) =
µ 0 I l 1.70 mm dx
µ Il  1.70 
= 0 ln
 0.400 
2 π ∫0.400 mm x
2π
The mutual inductance between the wire and the loop is then
M=
1(4 π × 10 −7 T ⋅ m A)(2.70 × 10 −3 m)
N 2 Φ12 N 2 µ 0 I l  1.70  N 2 µ 0 l
=
=
ln
1.45) =
(
(1.45)
 0.400 
I1
2π I
2π
2π
M = 7.81 × 10 −10 H = 781 pH
32.47
With I = I1 + I 2 , the voltage across the pair is:
∆V = − L 1
dI
dI
dI1
dI
dI
− M 2 = − L 2 2 − M 1 = − L eq
dt
dt
dt
dt
dt
dI1 ∆V M dI 2
=
+
L 1 L 1 dt
dt
So,
−
and
− L2
(−L1L2 + M 2 )
32.48
[1]
dI 2 ∆V M dI1
=
+
L2
L 2 dt
dt
−
leads to
(− L1L 2 + M 2 )
dI1
= ∆V (L 2 − M)
dt
Adding [1] to [2],
(− L 1L 2 + M 2 )
dI
= ∆V (L 1 + L 2 − 2M)
dt
So,
L eq = −
I max =
(b)
dI 2
= ∆V(L1 − M)
dt
By substitution,
At different times,
(a)
dI 2 M( ∆V ) M 2 dI 2
+
+
= ∆V
dt
L1
L 1 dt
[2]
L 1L 2 − M 2
∆V
=
dI / dt
L 1 + L 2 − 2M
(UC )max = (U L )max
so
1.00 × 10 −6 F
C
( ∆V )max =
( 40.0 V) = 0.400 A
L
10.0 × 10 −3 H
© 2000 by Harcourt, Inc. All rights reserved.
[
]
1 C ∆V )2
2 (
max
(
= 21 LI 2
)
max
256 Chapter 32 Solutions
32.49
32.50
[
]
1 C ∆V )2
2 (
max
(
= 21 LI 2
)
( ∆VC )max =
so
max
L
I max =
C
20.0 × 10 −3 H
(0.100 A) = 20.0 V
0.500 × 10 − 6 F
When the switch has been closed for a long time, battery, resistor,
and coil carry constant current I max = ε / R . When the switch is
opened, current in battery and resistor drops to zero, but the coil
carries this same current for a moment as oscillations begin in the
LC loop.
We interpret the problem to mean that the voltage amplitude of
2
2
.
these oscillations is ∆V, in 21 C ( ∆V ) = 21 LI max
Then, L =
32.51
C =
(
)
2
2
0.500 × 10 − 6 F (150 V ) ( 250 Ω)
C ( ∆V )
C ( ∆V ) R 2
=
=
= 0.281 H
2
I max
ε2
( 50.0 V )2
2
2
1
1
=
= 608 pF
2
(2 π f ) L (2 π ⋅ 6.30 × 106 )2 (1.05 × 106 )
Goal Solution
A fixed inductance L = 1.05 µ H is used in series with a variable capacitor in the tuning section of a radio.
What capacitance tunes the circuit to the signal from a station broadcasting at 6.30 MHz?
G: It is difficult to predict a value for the capacitance without doing the calculations, but we might expect
a typical value in the µF or pF range.
O: We want the resonance frequency of the circuit to match the broadcasting frequency, and for a simple
RLC circuit, the resonance frequency only depends on the magnitudes of the inductance and
capacitance.
A : The resonance frequency is f 0 =
Thus,
C=
1
2 π LC
1
1
=
= 608 pF
2
(2 π f 0 ) L (2 π )(6.30 × 106 Hz) 2 (1.05 × 10 −6 H)
[
]
L : This is indeed a typical capacitance, so our calculation appears reasonable. However, you probably
would not hear any familiar music on this broadcast frequency. The frequency range for FM radio
broadcasting is 88.0 – 108.0 MHz, and AM radio is 535 – 1605 kHz. The 6.30 MHz frequency falls in the
Maritime Mobile SSB Radiotelephone range, so you might hear a ship captain instead of Top 40
tunes! This and other information about the radio frequency spectrum can be found on the National
Telecommunications and Information Administration (NTIA) website, which at the time of this
printing was at
http://www.ntia.doc.gov/osmhome/allochrt.html
Chapter 32 Solutions 257
f=
1
:
2 π LC
(a)
f=
1
1
=
= 135 Hz
2 π LC 2 π (0.0820 H)(17.0 × 10 − 6 F)
(b)
Q = Qmax cos ω t = (180 µ C) cos(847 × 0.00100) = 119 µC
(c)
I=
(a)
f=
(b)
Q = C ε = (1.00 × 10 − 6 F)(12.0 V) = 12.0 µC
(c)
1C
2
32.52
32.53
32.54
L=
1
1
=
= 0.220 H
(2 π f )2 C (2 π ⋅ 120)2 (8.00 × 10 − 6 )
dQ
= −ω Qmax sin ω t = −(847)(180) sin(0.847) = – 114 mA
dt
1
2 π LC
ε2 =
=
1
2 π (0.100 H)(1.00 × 10 − 6 F)
= 503 Hz
1 LI 2
2 max
I max = ε
C
= 12 V
L
1.00 × 10 − 6 F
= 37.9 mA
0.100 H
(d) At all times U = 21 C ε 2 = 21 (1.00 × 10 − 6 F)(12.0 V)2 = 72.0 µ J
ω=
32.55
1
=
LC
1
(3.30 H)(840 × 10
Q = Qmax cos ω t, I =
−12
F
dQ
= −ω Qmax sin ω t
dt
( [105 × 10 ]cos[(1.899 × 10
−6
(a)
Q2
UC =
=
2C
(b)
2
U L = 21 LI 2 = 21 Lω 2Qmax
sin 2 (ω t ) =
UL
(c)
)
= 1.899 × 10 4 rad s
(105 × 10
=
4
(
2 840 × 10 −12
−6
)
2
[(
)(
rad s 2.00 × 10 −3 s
)
2
= 6.03 J
2
Qmax
sin 2 (ω t )
2C
)(
C sin 2 1.899 × 10 4 rad s 2.00 × 10 −3 s
(
)] )
2 840 × 10 −12 F
)
)] =
0.529 J
U total = UC + U L = 6.56 J
© 2000 by Harcourt, Inc. All rights reserved.
258 Chapter 32 Solutions
32.56
(a)
ωd =
2
1  R
−
=
LC  2L 
fd =
Therefore,
32.57
(2.20 × 10 )(1.80 × 10 )
−3
−6

7.60
−
 2 2.20 × 10 −3

(
2
)

 = 1.58 × 10 4 rad / s


ωd
= 2.51 kHz
2π
(b)
Rc =
4L
= 69.9 Ω
C
(a)
ω0 =
1
1
=
= 4.47 krad/s
LC
(0.500)(0.100 × 10 −6 )
(b)
ωd =
1  R
−
= 4.36 krad/s
LC  2L 
(c)
∆ω
= 2.53% lower
ω0
2
Choose to call positive current clockwise in Figure 32.19. It drains charge from the capacitor
according to I = – dQ/dt. A clockwise trip around the circuit then gives
32.58
32.59
1
(a)
+
Q
dI
− IR − L = 0
C
dt
+
Q dQ
d dQ
+
R+L
= 0, identical with Equation 32.29.
C
dt
dt dt
Q = Qmax e
0.500 = e
t=−
(b)
−
−
cos ω dt
Rt
2L
−
Rt
2L
so
I max ∝ e
and
Rt
= − ln(0.500)
2L
so
Q = 0.500 Qmax = 0.707Qmax
2L
 2L 
ln(0.500) = 0.693
 R
R
2
U 0 ∝ Qmax
t=−
Rt
2L
and
U = 0.500U 0
2L
 2L 
ln(0.707) = 0.347
 R
R
(half as long)
Chapter 32 Solutions 259
With Q = Qmax at t = 0, the charge on the capacitor at any time is Q = Qmax cos ω t where
ω = 1 LC . The energy stored in the capacitor at time t is then
32.60
U=
2
Q 2 Qmax
=
cos 2 ω t = U 0 cos 2 ω t.
2C
2C
1
4
cos ω t =
When U = U 0 ,
1
2
and
t
π
=
LC 3
Therefore,
π2
t2
=
LC
9
or
L=
The inductance is then:
32.61
(a)
(b)
d 20.0t )
ε L = − L dI = −(1.00 mH) (
=
dt
dt
9t 2
π 2C
– 20.0 mV
Q = ∫ I dt = ∫ ( 20.0t )dt = 10.0t 2
t
t
0
0
∆VC =
(c)
1
3
ω t = π rad
(
)
−Q
−10.0t 2
=
= − 10.0 MV s 2 t 2
C
1.00 × 10 −6 F
(−10.0t ) ≥ 1 (1.00 × 10 )(20.0t) ,
2
2(1.00 × 10 )
2 2
Q2
1
When
≥ LI 2 ,
2
2C
or
(
−3
−6
)
then 100t 4 ≥ 400 × 10 − 9 t 2 . The earliest time this is true is at
32.62
(a)
ε L = − L dI = − L d
(b)
I=
dt
dQ
,
dt
dt
When
t
t
Q = ∫ I dt = ∫ Kt dt = 21 Kt 2
so
0
∆VC =
Thus
1C
2
( ∆VC )2 =
t = 4.00 × 10 − 9 s = 63.2 µs
(Kt) = –LK
and
(c)
2
1 LI 2 ,
2
1C
2
0
−Q
Kt2
= −
2C
C
 K2 t 4  1
2 2
 4C 2  = 2 L K t


t = 2 LC
© 2000 by Harcourt, Inc. All rights reserved.
(
)
260 Chapter 32 Solutions
2
1  Q
1
1 Q2
=
+ LI 2
2C  2 
2
2 C
32.63
so
The flux through each turn of the coil is
I=
3Q 2
4CL
ΦB =
LI
=
N
Q
2N
3L
C
where N is the number of turns.
Equation 30.16: B =
32.64
µ 0 NI
2π r
b
(a)
b
µ 0 NI
µ NIh dr µ 0 NIh  b 
h dr = 0
=
ln
 a
2π r
2 π ∫a r
2π
a
Φ B = ∫ B dA = ∫
L=
µ 0N 2h  b 
NΦ B
=
ln
 a
2π
I
(b)
L=
µ 0 (500)2 (0.0100)  12.0 
= 91.2 µH
ln
 10.0 
2π
(c)
Lappx =
*32.65 (a)
µ 0 N 2  A  µ 0 (500)2  2.00 × 10 − 4 m 2 
=

 = 90.9 µH
2π  R 
2π
0.110


Nµ 0 IR 2
Nµ 0 I
=
2
2 3/2
2R
2(R + 0 )
At the center,
B=
So the coil creates flux through itself
Φ B ≈ BA cos θ =
When the current it carries changes,
ε L = −N d ΦB ≈ − N π Nµ 0 R dI =
dt
L≈
so
Nµ 0 I
π
π R 2 cos 0° = Nµ 0 IR
2R
2
2
dt
−L
dI
dt
π 2
N µ 0R
2
π
T·m
L ≈ 2 12  4π × 10–7 A  0.14 m = 2.8 × 10–7 H ~ 100 nH


(b)
2π r ≈ 3(0.3 m), so r ≈ 0.14 m;
(c)
2.8 × 10–7 V · s/A
L
≈
= 1.0 × 10– 9 s ~ 1 ns
270 V/A
R
Chapter 32 Solutions 261
32.66
(a)
If unrolled, the wire forms the diagonal of a 0.100 m
(10.0 cm) rectangle as shown. The length of this rectangle
is
L′ =
9.80 m
L′
(9.80 m)2 − (0.100 m)2
The mean circumference of each turn is C = 2π r ′ , where r ′ =
radius of each turn.
The number of turns is then:
(9.80 m)2 − (0.100 m)2
N=
L′
=
C
(b)
R=
−8
ρ l 1.70 × 10 Ω ⋅ m (10.0 m )
=
= 0.522 Ω
2
A
π 0.322 × 10 −3 m
(c)
L=
2
µ N 2 A 800µ 0  L′ 
=
π ( r ′ )2
l′
l′  C 
2π
(
(
 24.0 + 0.644 
× 10 − 3 m


2
)
(
24.0 + 0.644
mm is the mean
2
= 127
)
)
2
2
800 4 π × 10 −7  (9.80 m )2 − (0.100 m )2   24.0 + 0.644 

−3
×
10
L=
π
m




 π ( 24.0 + 0.644) × 10 −3 m  
2
0.100 m



L = 7.68 × 10 −2 H = 76.8 mH
32.67
From Ampere’s law, the magnetic field at distance r ≤ R is found as:
 I 
B( 2 π r ) = µ 0 J π r 2 = µ 0 
π r 2 , or
2
 πR 
( )
( )
B=
µ 0 Ir
2π R2
The magnetic energy per unit length within the wire is then
µ0 I 2
R B2
U
=∫
2 π r dr ) =
(
0 2µ
l
4π R4
0
0.100 m
R
∫0
r 3 dr =
µ0 I 2  R4 
  =
4π R4  4 
µ0 I 2
16 π
This is independent of the radius of the wire.
© 2000 by Harcourt, Inc. All rights reserved.
262 Chapter 32 Solutions
The primary circuit (containing the battery and solenoid) is an
RL circuit with R = 14.0 Ω , and
32.68
(
)
(
)
4 π × 10 −7 (12 500) 1.00 × 10 − 4
µ N 2A
L= 0
=
= 0.280 H
l
0.0700
(a)
The time for the current to reach 63.2% of the maximum
value is the time constant of the circuit:
τ=
(b)
(c)
2
L 0.280 H
=
= 0.0200 s = 20.0 ms
R 14.0 Ω
 I f − 0
 ∆I 
= L

 ∆t 
 ∆t 
The solenoid's average back emf is
εL
where
I f = 0.632 I max = 0.632
Thus,
ε L = (0.280 H) 2.71 A  =
0.0200 s
=L
 ∆V 
 60.0 V 
= 0.632
= 2.71 A
 R 
 14.0 Ω 
37.9 V
The average rate of change of flux through each turn of the overwrapped concentric coil is the
same as that through a turn on the solenoid:
(
)
(
)
4 π × 10 −7 T ⋅ m A (12 500 0.0700 m )( 2.71 A ) 1.00 × 10 − 4 m 2
∆Φ B µ 0n( ∆I )A
=
=
= 3.04 mV
∆t
0.0200 s
∆t
(d) The magnitude of the average induced emf in the coil is
the average induced current is
I=
32.69
εL
R
=
(
εL
= N ( ∆Φ B ∆t ) and magnitude of
)
820
N  ∆Φ B 
=
3.04 × 10 −3 V = 0.104 A = 104 mA
R  ∆t  24.0 Ω
Left-hand loop:
E − (I + I 2 )R 1 − I 2 R2 = 0
Outside loop:
E − (I + I 2 )R 1 − L
Eliminating I 2 gives
E ′ − IR′ − L
dI
=0
dt
dI
=0
dt
E′
(1 − e − R ′t L )
R′
This is of the same form as Equation 32.6,
so its solution is of the same form as Equation 32.7:
I (t ) =
But R′ = R1R2 / ( R1 + R2 ) and E ′ = R2E / ( R1 + R2 ), so
E R 2 / (R 1 + R2 )
E
E′
=
=
R′ R 1R2 / (R 1 + R2 ) R 1
Thus
I(t) =
E
(1 − e − R ′ t L )
R1
Chapter 32 Solutions 263
When switch is closed, steady current I 0 = 1.20 A. W h e n
the switch is opened after being closed a long time, the
current in the right loop is
32.70
I = I0e
32.71
(a)
− R2 t L
=
I0
I
Rt
I 
= ln 0
 I
L
so
e Rt
L
Therefore,
L=
R2 t
(1.00 Ω)(0.150 s) = 0.0956 H = 95.6 mH
=
ln( I 0 I ) ln(1.20 A 0.250 A )
and
While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the
circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer
loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:
[
](
)
+ ε 0 − ( 2.00 + 6.00) × 10 3 Ω 9.00 × 10 −3 A = 0
+ ε 0 = 72 .0 V with end b at the higher potential
(b)
(c)
After the switch is opened, the current around the outer loop decays as
I = I max e − R t
L
with
I max = 9.00 mA ,
R = 8.00 kΩ ,
and
Thus, when the current has reached a value I = 2.00 mA , the elapsed time is:
t=
 L   I max   0.400 H   9.00 
=
ln
= 7.52 × 10 −5 s = 75.2 µs
ln
 R   I   8.00 × 10 3 Ω   2.00 
© 2000 by Harcourt, Inc. All rights reserved.
L = 0.400 H
264 Chapter 32 Solutions
32.72
(a)
The instant after the switch is closed, the situation is as shown i n
the circuit diagram of Figure (a). The requested quantities are:
∆VC = 0,
-
+
IR = ε0/R
I L = 0, IC = ε 0 R , I R = ε 0 R
∆VL = ε 0 ,
∆VL = ε0
IL = 0
Q=0
∆VC = 0
∆V R = ε 0
∆VR = ε0
+ -
IC = ε0/R
ε0
Figure (a)
(b)
IL = 0
After the switch has been closed a long time, the steady-state
conditions shown in Figure (b) will exist. The currents and
voltages are:
I L = 0,
∆VL = 0,
IC = 0,
∆VL = 0
-
IR = 0
Q = Cε0
IR = 0
∆VC = ε 0 ,
+
∆VR = 0
∆VC = ε0
+ -
∆V R = 0
ε0
Figure (b)
32.73
When the switch is closed, as
shown in Figure (a), the current
in the inductor is I :
12.0 – 7.50I – 10.0 = 0 → I = 0.267 A
When the switch is opened, the
initial current in the inductor
remains at 0.267 A.
IR = ∆V:
(0.267 A)R ≤ 80.0 V
(a)
(b)
R ≤ 300 Ω
Goal Solution
To prevent damage from arcing in an electric motor, a discharge resistor is sometimes placed in parallel
with the armature. If the motor is suddenly unplugged while running, this resistor limits the voltage
that appears across the armature coils. Consider a 12.0-V dc motor with an armature that has a resistance
of 7.50 Ω and an inductance of 450 mH. Assume that the back emf in the armature coils is 10.0 V when
the motor is running at normal speed. (The equivalent circuit for the armature is shown in Figure
P32.73.) Calculate the maximum resistance R that limits the voltage across the armature to 80.0 V when
the motor is unplugged.
Chapter 32 Solutions 265
G: We should expect R to be significantly greater than the resistance of the armature coil, for otherwise a
large portion of the source current would be diverted through R and much of the total power would
be wasted on heating this discharge resistor.
O: When the motor is unplugged, the 10-V back emf will still exist for a short while because the motor’s
inertia will tend to keep it spinning. Now the circuit is reduced to a simple series loop with an emf,
inductor, and two resistors. The current that was flowing through the armature coil must now flow
through the discharge resistor, which will create a voltage across R that we wish to limit to 80 V. As
time passes, the current will be reduced by the opposing back emf, and as the motor slows down, the
back emf will be reduced to zero, and the current will stop.
A : The steady-state coil current when the switch is closed is found from applying Kirchhoff's loop rule to
the outer loop:
+ 12.0 V − I (7.50 Ω) − 10.0 V = 0
2.00 V
= 0.267 A
7.50 Ω
so
I=
We then require that
∆V R = 80.0 V = (0.267 A )R
so
R=
∆V R
80.0 V
=
= 300 Ω
0.267 A
I
L : As we expected, this discharge resistance is considerably greater than the coil’s resistance. Note that
while the motor is running, the discharge resistor turns P = (12 V)2 300 Ω = 0.48 W of power into
heat (or wastes 0.48 W).
The source delivers power at the rate of about
P = IV = [0.267 A + (12 V / 300 Ω)](12 V ) = 3.68 W, so the discharge resistor wastes about 13% of the
total power. For a sense of perspective, this 4-W motor could lift a 40-N weight at a rate of 0.1 m/s.
32.74
(
)
(
)
(a)
L1 =
2
4 π × 10 −7 T ⋅ m A (1000) 1.00 × 10 − 4 m 2
µ 0 N12 A
=
= 2.51 × 10 − 4 H = 251 µH
l1
0.500 m
(b)
M=
N 2 Φ 2 N 2 Φ1 N 2BA N 2 µ 0 ( N1 l1 ) I1 A µ 0 N1N 2 A
=
=
=
=
l1
I1
I1
I1
I1
[
(4π × 10
M=
(c)
−7
dt
Q1 =
(
T ⋅ m A (1000)(100) 1.00 × 10 − 4 m 2
0.500 m
ε 1 = − M dI 2 ,
Q1 = −
)
]
or I1R1 = − M
(
dI 2
dt
and I1 =
)
= 2.51 × 10 − 5 H = 25.1 µH
dQ1
M dI 2
=−
R1 dt
dt
)
M I 2i
M tf
M
M
dI 2 = −
I 2 f − I 2i = − (0 − I 2i ) =
R1 ∫0
R1
R1
R1
(2.51 × 10
−5
)
H (1.00 A )
1000 Ω
= 2.51 × 10 −8 C = 25.1 nC
© 2000 by Harcourt, Inc. All rights reserved.
266 Chapter 32 Solutions
32.75
2U
is non-zero.
I2
(a)
It has a magnetic field, and it stores energy, so L =
(b)
Every field line goes through the rectangle between the conductors.
(c)
Φ = LI
so
Thus
For an RL circuit,
32.76
L=
Φ 1 w−a
=
I I ∫y=a
L=
w−a
 µ0I
µ 0 I  2 µ 0 Ix
2µ 0 x
1 w−a
x
dy
+
=
dy
=
ln
y
 2 π y 2 π w −y  I ∫ 2 π y
I ∫a
2π
( )

a
L=
µ 0x  w − a 
ln
 a 
π
I(t) = I max e
−
R
t = 10 − 9
L
B da
R
t
L :
R
− t
R
I(t)
= 1 − 10 − 9 = e L ≅ 1 − t
L
I max
Rmax =
so
(3.14 × 10 − 8 )(10 − 9 )
= 3.97 × 10 − 25 Ω
(2.50 yr)(3.16 × 107 s / yr)
(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its
resistance would be at least 10– 6 Ω).
32.77
1
1
(a)
UB = 2 LI 2 = 2 (50.0 H)(50.0 × 10 3 A)
(b)
Two adjacent turns are parallel wires carrying current in the same direction. Since the loops
have such large radius, a one-meter section can be regarded as straight.
2
= 6.25 × 1010 J
µ 0I
2π r
Then one wire creates a field of
B=
This causes a force on the next wire of
F = IlB sin θ
giving
F = Il
Solving for the force,
µ lI 2
µ0I
sin 90° = 0
2π r
2π r
F = (4π × 10–7 N/A2)
(1.00 m)(50.0 × 10 3 A) 2
= 2000 N
(2π)(0.250 m)
Chapter 32 Solutions 267
P = I ( ∆V )
32.78
I=
P
1.00 × 10 9 W
=
= 5.00 × 10 3 A
∆V
200 × 10 3 V
B( 2 π r ) = µ 0 I enclosed
From Ampere’s law,
(a)
(b)
At r = a = 0.0200 m,
I enclosed = 5.00 × 10 3 A
(4π × 10
B=
)(
T ⋅ m A 5.00 × 10 3 A
2 π (0.0200 m )
At r = b = 0.0500 m,
B=
(c)
−7
(4π × 10
−7
)(
2 π (0.0500 m )
U = ∫ u dV = ∫
r=a
(4π × 10
U=
−7
) = 0.0500 T =
b dr
∫a
µ 0 I enclosed
2π r
50.0 mT
and
) = 0.0200 T =
[B(r )]2 (2π rldr ) = µ 0 I 2l
B=
and
I enclosed = I = 5.00 × 10 3 A
T ⋅ m A 5.00 × 10 3 A
r=b
or
=
20.0 mT
µ 0 I 2l  b 
ln
 a
4π
2µ 0
4π
)(
) (1000 × 10 m) ln  5.00 cm  = 2.29 × 10
T ⋅ m A 5.00 × 10 3 A
4π
2
r
3
 2.00 cm 


6
J = 2.29 MJ
(d) The magnetic field created by the inner conductor exerts a force of repulsion on the current i n
the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small
rectangular section of the outer cylinder of length l and width w . It carries a current of

w
(5.00 × 10 A) 2π (0.0500
m ) 
3
(5.00 × 10 A)w l(20.0 × 10 T) sin 90.0˚
2 π (0.0500 m )
3
and experiences an outward force
F = IlBsin θ =
The pressure on it is
P=
(
−3
)(
)
5.00 × 10 3 A 20.0 × 10 −3 T
F
F
=
=
= 318 Pa
A wl
2 π (0.0500 m )
© 2000 by Harcourt, Inc. All rights reserved.
268 Chapter 32 Solutions
(
)
*32.79 (a)
4 π × 10 − 7 T ⋅ m A (1400)( 2.00 A )
µ 0 NI
B=
=
= 2.93 × 10 − 3 T (upward)
l
1.20 m
(b)
2.93 × 10 − 3 T
N
B2
J  1 N ⋅ m
u=
=
= 3.42 3 
 = 3.42 2 = 3.42 Pa


1
J
2µ 0 2 4 π × 10 −7 T ⋅ m A
m
m
(c)
To produce a downward magnetic field, the surface of the super conductor
(
(
)
2
)
must carry a clockwise current.
(d) The vertical component of the field of the solenoid exerts an inward force on the
superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field
diverges and has a radially outward horizontal component. This component exerts upward
force on the clockwise superconductor current. The total force on the core is upward . You
can think of it as a force of repulsion between the solenoid with its north end pointing up,
and the core, with its north end pointing down.
(e)
(
)
2
F = PA = ( 3.42 Pa) π 1.10 × 10 −2 m  = 1.30 × 10 − 3 N


Note that we have not proven that energy density is pressure. In fact, it is not in some cases;
see problem 12 in Chapter 21.