Aqui esta lo que buscabamos Y ahora a trabajar!

1. 32
Inductance
CHAPTER OUTLINE
32.1 Self-Induction and Inductance
32.2 RL Circuits
32.3 Energy in a Magnetic Field
32.4 Mutual Inductance
32.5 Oscillations in an LC Circuit
32.6 The RLC Circuit
ANSWERS TO QUESTIONS
Q32.1 The coil has an inductance regardless of the nature of the
current in the circuit. Inductance depends only on the coil
geometry and its construction. Since the current is con-
stant, the self-induced emf in the coil is zero, and the coil
does not affect the steady-state current. (We assume the
resistance of the coil is negligible.)
Q32.2 The inductance of a coil is determined by (a) the geometry
of the coil and (b) the “contents” of the coil. This is similar
to the parameters that determine the capacitance of a
capacitor and the resistance of a resistor. With an inductor,
the most important factor in the geometry is the number
of turns of wire, or turns per unit length. By the “contents”
we refer to the material in which the inductor establishes
a magnetic field, notably the magnetic properties of the
core around which the wire is wrapped.
*Q32.3 The emf across an inductor is zero whenever the current is constant, large or small. Answer (d).
*Q32.4 The fine wire has considerable resistance, so a few seconds is many time constants. The final
current is not affected by the inductance of the coil. Answer (c).
*Q32.5 The inductance of a solenoid is proportional to the number of turns squared, to the cross-sectional
area, and to the reciprocal of the length. Coil A has twice as many turns with the same length of
wire, so its circumference must be half as large as that of coil B. Its radius is half as large and its
area one quarter as large. For coil A the inductance will be different by the factor 22
(1/4)(1/2) =
1/2. Answer (e).
Q32.6 When it is being opened. When the switch is initially standing open, there is no current in the
circuit. Just after the switch is then closed, the inductor tends to maintain the zero-current
condition, and there is very little chance of sparking. When the switch is standing closed, there
is current in the circuit. When the switch is then opened, the current rapidly decreases. The
induced emf is created in the inductor, and this emf tends to maintain the original current.
Sparking occurs as the current bridges the air gap between the contacts of the switch.
*Q32.7 Just before the switch is thrown, the voltage across the twelve-ohm resistor is very nearly 12 V.
Just after the switch is thrown, the current is nearly the same, maintained by the inductor. The
voltage across the 1 200-Ω resistor is then much more than 12 V. By Kirchhoff's loop rule,
the voltage across the coil is larger still: ∆VL
> ∆V1 200 Ω
> 12.0 V > ∆V12 Ω
.
225

2. 226 Chapter 32
*Q32.8 (i) (a) The bulb glows brightly right away, and then more and more faintly as the capacitor
charges up. (b) The bulb gradually gets brighter and brighter, changing rapidly at first and then
more and more slowly. (c) The bulb gradually gets brighter and brighter. (d) The bulb
glows brightly right away, and then more and more faintly as the inductor starts carrying more
and more current.
(ii) (a) The bulb goes out immediately. (b) The bulb glows for a moment as a spark jumps
across the switch. (c) The bulb stays lit for a while, gradually getting fainter and fainter.
(d) The bulb suddenly glows brightly. Then its brightness decreases to zero, changing rapidly at
first and then more and more slowly.
*Q32.9 The wire’s magnetic field goes in circles around it. We want this field to “shine” perpendicularly
through the area of the coil. Answer (c).
Q32.10 A physicist’s list of constituents of the universe in 1829 might include matter, light, heat, the stuff
of stars, charge, momentum, and several other entries. Our list today might include the quarks,
electrons, muons, tauons, and neutrinos of matter; gravitons of gravitational fields; photons of
electric and magnetic fields; W and Z particles; gluons; energy; momentum; angular momentum;
charge; baryon number; three different lepton numbers; upness; downness; strangeness; charm;
topness; and bottomness. Alternatively, the relativistic interconvertibility of mass and energy,
and of electric and magnetic fields, can be used to make the list look shorter. Some might think
of the conserved quantities energy, momentum, … bottomness as properties of matter, rather than
as things with their own existence. The idea of a field is not due to Henry, but rather to Faraday,
to whom Henry personally demonstrated self-induction. Still the thesis stated in the question has
an important germ of truth. Henry precipitated a basic change if he did not cause it. The biggest
difference between the two lists is that the 1829 list does not include fields and today’s list does.
*Q32.11 The energy stored in the magnetic field of an inductor is proportional to the square of the current.
Doubling I makes U LI=
1
2
2
get four times larger. Answer (a).
*Q32.12 Cutting the number of turns in half makes the inductance four times smaller. Doubling the
current would by itself make the stored energy four times larger, to just compensate. Answer (b).
Q32.13 The energy stored in a capacitor is proportional to the square of the electric field, and the energy stored
in an induction coil is proportional to the square of the magnetic field. The capacitor’s energy is
proportional to its capacitance, which depends on its geometry and the dielectric material inside. The
coil’s energy is proportional to its inductance, which depends on its geometry and the core mate-
rial. On the other hand, we can think of Henry’s discovery of self-inductance as fundamentally new.
Before a certain school vacation at the Albany Academy about 1830, one could visualize the universe
as consisting of only one thing, matter. All the forms of energy then known (kinetic, gravitational,
elastic, internal, electrical) belonged to chunks of matter. But the energy that temporarily maintains a
current in a coil after the battery is removed is not energy that belongs to any bit of matter. This energy
is vastly larger than the kinetic energy of the drifting electrons in the wires. This energy belongs to
the magnetic field around the coil. Beginning in 1830, Nature has forced us to admit that the universe
consists of matter and also of fields, massless and invisible, known only by their effects.

3. Inductance 227
*Q32.14 (a) The instant after the switch is closed, the situation is
as shown in the circuit diagram of Figure (a).
The requested quantities are:
I I
R
I
R
V V V
L C R
L C R
= = =
= = =
0
0
0 0
0 0
, ,
, ,
ε ε
ε∆ ∆ ∆ε
(b) 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 I I
V V V
L C R
L C R
= = =
= = =
0 0 0
0 00
, ,
, ,∆ ∆ ∆ε
*Q32.15 (i) Answer (a). The mutual inductance of two loops in free space—that is, ignoring the use of
cores—is a maximum if the loops are coaxial. In this way, the maximum flux of the primary loop
will pass through the secondary loop, generating the largest possible emf given the changing
magnetic field due to the first.
(ii) Answer (c). The mutual inductance is a minimum if the magnetic field of the first coil lies in
the plane of the second coil, producing no flux through the area the second coil encloses.
Q32.16 When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance of
the coil is making the current continue to flow. At this time the magnetic field of the coil contains
all the energy that was originally stored in the charged capacitor. The current has just finished
discharging the capacitor and is proceeding to charge it up again with the opposite polarity.
Q32.17 If R
L
C
>
4
, then the oscillator is overdamped—it will not oscillate. If R
L
C
<
4
, then the
oscillator is underdamped and can go through several cycles of oscillation before the radiated
signal falls below background noise.
Q32.18 An object cannot exert a net force on itself. An object cannot create momentum out of nothing.
A coil can induce an emf in itself. When it does so, the actual forces acting on charges in
different parts of the loop add as vectors to zero. The term electromotive force does not refer to
a force, but to a voltage.
SOLUTIONS TO PROBLEMS
Section 32.1 Self-Induction and Inductance
P32.1 ε = − = −( )
−⎛
⎝⎜
⎞
⎠⎟L
I
t
∆
∆
2 00
0 0 500 1
.
.
H
A
0.010 0 s
VV s
1 H A
V
⋅
⋅
⎛
⎝
⎞
⎠
= 100
P32.2 Treating the telephone cord as a solenoid, we have:
L
N A
= =
× ⋅( )( ) ×( )− −
µ π π0
2 7 2 3
4 10 70 0 6 50 10T m A m. .
22
0 600
1 36
.
.
m
H= µ
FIG. Q32.14
+ –
+ –
IL = 0 ∆VL = e0
Q = 0
∆VC = 0
e0
∆VR = e0
IR = e0/R
IC = e0/R
Figure (a)
+ –
+ –IL = 0 ∆VL = 0
IR = 0
∆VR = 0
Q = Ce0
∆VC = e0
e0
Figure (b)

4. 228 Chapter 32
*P32.3 ε ω ω ω= − = − ( )= − = −L
dI
dt
L
d
dt
I t L I tmax maxsin cos .10 00 10 120 5 003
×( )( )( )−
π ω. cos t
ε π π= −( ) ( )= −( ) ( )6 00 120 18 8 377. cos . cost tV
P32.4 From ε = ⎛
⎝
⎞
⎠
L
I
t
∆
∆
, we have L
I t
=
( )
=
×
= ×
−
−ε
∆ ∆
24 0 10
2 40 10
3
3.
.
V
10.0 A s
H
From L
N
I
B
=
Φ
, we have ΦB
LI
N
= =
×( )( )
= ⋅
−
2 40 10 4 00
500
19 2
3
. .
.
H A
T m2
µ
P32.5 L
N A
= =
( ) ×( ) = ×
−
−µ µ0
2
0
2 4
4
420 3 00 10
0 160
4 16 10
.
.
. H
ε ε= − → =
−
=
− ×
×
= −
−
−
L
dI
dt
dI
dt L
175 10
0
6
V
4.16 10 H4
..421 A s
P32.6 ε = = ×( ) −( )−
L
dI
dt
d
dt
t t90 0 10 63 2
. V
(a) At t = 1 00. s, ε = 360 mV
(b) At t = 4 00. s, ε = 180 mV
(c) ε = ×( ) −( ) =−
90 0 10 2 6 03
. t
when t = 3 00. s
P32.7 (a) B nI= = ⎛
⎝
⎞
⎠
( )=µ µ µ0 0
450
0 120
0 040 0 188
.
. A T
(b) ΦB BA= = × ⋅−
3 33 10 8
. T m2
(c) L
N
I
B
= =
Φ
0 375. mH
(d) B LBand are proportional to current; isΦ iindependent of current.
P32.8 L
N
I
NBA
I
NA
I
NI
R
N A
R
B
= = ≈ ⋅ =
Φ µ
π
µ
π
0 0
2
2 2
P32.9 ε ε= = −−
0e L
dI
dt
kt
dI
L
e dtkt
= − −ε0
If we require I → 0 as t →∞, the solution is I
kL
e
dq
dt
kt
= =−ε0
Q Idt
kL
e dt
k L
kt
= = = −∫ ∫ −
∞
ε ε0
0
0
2
Q
k L
=
ε0
2
FIG. P32.8

5. Inductance 229
Section 32.2 RL Circuits
P32.10 Taking τ =
L
R
, I I ei
t
= − τ
:
dI
dt
I ei
t
= −
⎛
⎝
⎜
⎞
⎠
⎟
− τ
τ
1
IR L
dI
dt
+ = 0 will be true if I R e L I ei
t
i
t− −
+ ( ) −⎛
⎝
⎞
⎠
=τ τ
τ
1
0
Because τ =
L
R
, we have agreement with 0 0= .
P32.11 (a) At time t, I t
e
R
t
( )=
−( )−
ε τ
1
where τ = =
L
R
0 200. s
After a long time, I
e
R R
max =
−( ) =
−∞
ε ε1
At I t I( ) = 0 500. max 0 500
1 0 200
.
.
( ) =
−( )−
ε ε
R
e
R
t s
so 0 500 1 0 200
. .
= − −
e t s
Isolating the constants
on the right, ln ln ..
e t−
( )= ( )0 200
0 500s
and solving for t, − = −
t
0 200
0 693
.
.
s
or t = 0 139. s
(b) Similarly, to reach 90% of Imax , 0 900 1. = − −
e t τ
and t = − −( )τ ln .1 0 900
Thus, t = −( ) ( ) =0 200 0 100 0 461. ln . .s s
*P32.12 The current increases from 0 to asymptotically
approach 500 mA. In case (a) the current jumps up
essentially instantaneously. In case (b) it increases
with a longer time constant, and in case (c) the increase
is still slower.
1
0.5
0
0 0.2 0.4 0.6
t (s)
I (A)
Imax
FIG. P32.11
0
500 mA
0
t
I
(a)
(b)
(c)
FIG. P32.12

6. 230 Chapter 32
P32.13 (a) τ = = × =−L
R
2 00 10 2 003
. .s ms
(b) I I e et
= −( )= ⎛
⎝
⎞
⎠
−− −
max
. ..
1
6 00
1 0 250 2 0τ V
4.00 Ω
00
0 176( )= . A
(c) I
R
max
.
.= = =
ε 6 00
1 50
V
4.00
A
Ω
(d) 0 800 1 2 00 0 200 32 00
. . ln . ..
= − → = −( ) ( ) =−
e tt ms
ms 222 ms
P32.14 I
R
e et
= −( )= −( )=− −ε τ
1
120
9 00
1 3 021 80 7 00
.
.. .
A
∆
∆ ∆
V IR
V V
R
L R
= = ( )( ) =
= − = −
3 02 9 00 27 2
120 27
. . .
.
V
ε 22 92 8= . V
P32.15 (a) ∆ ΩV IRR = = ( )( ) =8 00 2 00 16 0. . .A V
and ∆ ∆V VL R= − = − =ε 36 0 16 0 20 0. . .V V V
Therefore,
∆
∆
V
V
R
L
= =
16 0
0 800
.
.
V
20.0 V
(b) ∆ ΩV IRR = = ( )( ) =4 50 8 00 36 0. . .A V
∆ ∆V VL R= − =ε 0
P32.16 After a long time, 12 0 0 200. . .V A= ( )R Thus, R = 60 0. Ω. Now, τ =
L
R
gives
L R= = ×( )( )=−
τ 5 00 10 60 0 30 04
. . .s V A mH
P32.17 I I e t
= −( )−
max :1 τ dI
dt
I e t
= − ( ) −⎛
⎝
⎞
⎠
−
max
τ
τ
1
τ = = =
L
R
15 0
0 500
.
.
H
30.0
s:
Ω
dI
dt
R
L
I e t
= −
max
τ
and I
R
max =
ε
(a) t = 0:
dI
dt
R
L
I e
L
= = = =max .0 100
6 67
ε V
15.0 H
A s
(b) t = 1 50. s:
dI
dt L
e et
= = ( ) = (− − ( )ε τ
6 67 6 67
1 50 0 500
. .
. .
A s A s)) =−
e 3 00
0 332.
. A s
P32.18 Name the currents as shown. By Kirchhoff’s laws:
I I I1 2 3= + (1)
+ − − =10 0 4 00 4 00 01 2. . .V I I (2)
+ − − − ( ) =10 0 4 00 8 00 1 00 01 3
3
. . . .V I I
dI
dt
(3)
FIG. P32.18
FIG. P32.15
FIG. P32.13
continued on next page

7. Inductance 231
From (1) and (2), + − − + =10 0 4 00 4 00 4 00 01 1 3. . . .I I I
and I I1 30 500 1 25= +. . A
Then (3) becomes 10 0 4 00 0 500 1 25 8 00 1 003 3. . . . . .V A− +( )− − ( )I I
dI33
0
dt
=
1 00 10 0 5 003
3. . .H V( )⎛
⎝
⎞
⎠
+ ( ) =
dI
dt
IΩ
We solve the differential equation using equations from the chapter text:
I t e
t
3
10 0 1 005 00
1( ) =
⎛
⎝
⎜
⎞
⎠
⎟ −
−( ). . .V
10.0
H
Ω
Ω⎡⎡
⎣
⎤
⎦ = ( ) −⎡⎣ ⎤⎦
= + =
−
0 500 1
1 25 0 500
10
1 3
.
. .
A s
e
I I
t
11 50 0 250 10
. .A A s
− ( ) −
e t
P32.19 (a) Using τ = =RC
L
R
, we get R
L
C
= =
×
= × =−
3 00
1 00 10 1 006
3.
. . .
H
3.00 10 F
kΩ Ω
(b) τ = = ×( ) ×( )= × =− −
RC 1 00 10 3 00 10 3 00 103 6 3
. . .F sΩ 33 00. ms
P32.20 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
100
1 00 10 4.
.
mH
s
Ω
For 0 200≤ ≤t µs, I I e et t
= −( )= ( ) −( )− −
max . .1 10 0 1 10 000τ
A s
At t = 200 s,µ I e= ( ) −( )=−
10 00 1 8 652 00
. ..
A A.
Thereafter, it decays exponentially as I I ei
t
= − ′ τ
, so for t ≥ 200 s,µ
I e e
t
= ( ) = ( )− −( ) −
8 65 8 65
10 000 200 10 00
. .A A
s sµ 00 2 00 2 00 10 000
8 65 63 9t t
e e es s
A A+ − −
= ( ) = ( ). .
. . 110 000t s
P32.21 τ = = =
L
R
0 140
4 90
28 6
.
.
. ms
I
R
max
.
.= = =
ε 6 00
1 22
V
4.90
A
Ω
(a) I I e t
= −( )−
max 1 τ
so 0 220 1 22 1. .= −( )−
e t τ
e t−
=τ
0 820. : t = − ( ) =τ ln . .0 820 5 66 ms
(b) I I e e= −( )= ( ) −( )=− −
max
. .
. .1 1 22 1 110 0 0 028 6 350
A 222 A
(c) I I e t
= −
max
τ
and 0 160 1 22. .= −
e t τ
so t = − ( ) =τ ln . .0 131 58 1 ms
FIG. P32.20
FIG. P32.21

8. 232 Chapter 32
P32.22 (a) 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
dI
dt
L
dI
dt
L
dI
dt
eq = +1 2 so L L Leq = +1 2
(b) L
dI
dt
L
dI
dt
L
dI
dt
VLeq = = =1
1
2
2
∆ where I I I= +1 2 and
dI
dt
dI
dt
dI
dt
= +1 2
Thus,
∆ ∆ ∆V
L
V
L
V
L
L L L
eq
= +
1 2
and
1 1 1
1 2L L Leq
= +
(c) L
dI
dt
R I L
dI
dt
IR L
dI
dt
IReq eq+ = + + +1 1 2 2
Now I and
dI
dt
are separate quantities under our control, so functional equality requires
both L L L R R Req eqand= + = +1 2 1 2 .
(d) ∆V L
dI
dt
R I L
dI
dt
R I L
dI
dt
R I= + = + = +eq eq 1
1
1 1 2
2
2 2 where I I I= +1 2 and
dI
dt
dI
dt
dI
dt
= +1 2
.
We may choose to keep the currents constant in time. Then,
1 1 1
1 2R R Req
= +
We may choose to make the current swing through 0. Then,
1 1 1
1 2L L Leq
= +
This equivalent coil with resistance will bbe equivalent to the pair of real inductorss for
all other currents as well.
Section 32.3 Energy in a Magnetic Field
P32.23 L
N A
= =
( ) ×( )⎡
⎣
⎤
⎦ =
−
µ µ
π
0
2
0
2 2 2
68 0 0 600 10
0 080 0
. .
.
88 21. Hµ
U LI= = ×( )( ) =−1
2
1
2
8 21 10 0 770 2 442 6 2
. . .H A Jµ
P32.24 (a) The magnetic energy density is given by
u
B
= =
( )
× ⋅( )
= ×−
2
0
2
6
6
2
4 50
2 1 26 10
8 06 10
µ
.
.
.
T
T m A
J mm3
(b) 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 10 0 260 0 03106 2
. . .J m m m3
π == 6 32. kJ
P32.25 u
E
= ∈ =0
2
2
44 2. nJ m3
u
B
= =
2
02
995
µ
µJ m3

9. Inductance 233
P32.26 e dt
L
R
e
Rdt
L
L
R
eRt L Rt L−
∞
−
∞
∫ ∫= −
−⎛
⎝
⎜
⎞
⎠
⎟ = −2
0
2
02
2
2
−− ∞ −∞
= − −( )= −( )=2
0
0
2 2
0 1
2
Rt L L
R
e e
L
R
L
R
*P32.27 (a) P = I∆V = 3A 22 V = 66.0 W
(b) P = I∆VR
= I2
R = (3A)2
5 Ω = 45.0 W
(c) When the current is 3.00 A, Kirchhoff’s loop rule reads
+ −( )( )− =22 0 3 00 5 00 0. . . .V A Ω ∆VL
Then ∆VL = 7 00. V
The power being stored in the inductor is
I VL∆ = ( )( ) =3 00 7 00 21 0. . .A V W
(d) At all instants after the connection is made, the battery power is equal to the sum of the power
delivered to the resistor and the power delivered to the magnetic field. Just after t = 0 the
resistor power is nearly zero, and the battery power is nearly all going into the magnetic field.
Long after the connection is made, the magnetic field is absorbing no more power and the
battery power is going into the resistor.
P32.28 From the equation derived in the text, I
R
e Rt L
= −( )−ε 1
(a) The maximum current, after a long time t, is I
R
= =
ε 2 00. A
At that time, the inductor is fully energized and P = ( ) = ( )( ) =I V∆ 2 00 10 0 20 0. . . .A V W
(b) Plost A W= = ( ) ( ) =I R2 2
2 00 5 00 20 0. . .Ω
(c) Pinductor drop= ( )=I V∆ 0
(d) U
LI
= =
( )( ) =
2 2
2
10 0 2 00
2
20 0
. .
.
H A
J
P32.29 The total magnetic energy is the volume integral of the energy density, u
B
=
2
02µ
.
Because B changes with position, u is not constant. For B B
R
r
u
B R
r
=
⎛
⎝
⎜
⎞
⎠
⎟ =
⎛
⎝
⎜
⎞
⎠
⎟
⎛
⎝
⎜
⎞
⎠
⎟0
2
0
2
0
4
2
, .
µ
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 4 2
πr dr, so the energy stored in it is
dU u r dr
B R dr
r
= ( )=
⎛
⎝⎜
⎞
⎠⎟4
22 0
2 4
0
2
π
π
µ
We integrate this expression for r R= to r = ∞ to obtain the total magnetic energy outside the
sphere. This gives
U
B R
= =
×( ) ×( )−
2 2 5 00 10 6 00 10
1
0
2 3
0
5 2 6 3
π
µ
π . .
.
T m
226 10
2 70 106
18
× ⋅( )
= ×−
T m A
J.
FIG. P32.27

10. 234 Chapter 32
Section 32.4 Mutual Inductance
P32.30 I t I e tt
1 ( ) = −
max
α
ωsin with Imax .= 5 00 A, α = 0 025 0 1
. ,sϪ
and ω = 377 rad s
dI
dt
I e t tt1
= − +( )−
max sin cosα
α ω ω ω
At t = 0 800. s,
dI
dt
e1 0 020 0
5 00 0 025 0 0 800 377= ( ) −( ) (−
. . sin ..
A s ))( )⎡⎣
+ ( )( )⎤⎦377 0 800 377cos .
dI
dt
1 3
1 85 10= ×. A s
Thus, ε ε
2
1 2
1
3 20
1= − =
−
=
+
×
=M
dI
dt
M
dI dt
:
.
.
V
1.85 10 A s3
773 mH
P32.31 ε2
1 4 4
1 00 10 1 00 10 100= − = − ×( ) ×( )−
M
dI
dt
. . cosH A s 00t( )
ε2 1 00( ) =max
. V
P32.32 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at
distance x from the wire is B
I
x
=
µ
π
0
2
, and this field passes perpendicularly through the plane
of the loop. The flux through the loop is
ΦB d BdA B dx
I dx
x
= ⋅ = = ( ) =∫ ∫ ∫B A
µ
π
0
0 400
2 . mm
1.770 mm
∫ = ⎛
⎝
⎞
⎠
µ
π
0
2
1 70
0 400
I
ln
.
.
The mutual inductance between the wire and the loop is then
M
N
I
N I
I
N
= =
⎛
⎝
⎜
⎞
⎠
⎟ =2 12
1
2 0 2 0
2
1 70
0 400 2
Φ µ
π
µ
ln
.
. ππ
π
π
1 45
1 4 10 2 70 10
2
1 45
7 3
.
.
.( )=
× ⋅( ) ×( )( )
− −
T m A m
MM = × =−
7 81 10 78110
. H pH
P32.33 (a) M
N
I
B BA
A
= =
×( ) =
−
Φ 700 90 0 10
3 50
18 0
6
.
.
. mH
(b) L
I
A
A
A
= =
×( ) =
−
Φ 400 300 10
3 50
34 3
6
.
. mH
(c) εB
A
M
dI
dt
= − = −( )( )= −18 0 0 500 9 00. . .mH A/s mV
P32.34 (a) Solenoid 1 creates nearly uniform field everywhere inside it, given by m0
N1
I/
The flux though one turn of solenoid 2 is m0
p R2
2
N1
I/
The emf induced in solenoid 2 is –(m0
p R2
2
N1
N2
/ )(dI/dt)
The mutual inductance is m0
p R2
2
N1
N2
/
(b) Solenoid 2 creates nearly uniform field everywhere inside it, given by m0
N2
I/
and nearly zero field outside.
The flux though one turn of solenoid 1 is m0
p R2
2
N2
I/
The emf induced in solenoid 1 is –(m0
p R2
2
N1
N2
/ )(dI/dt)
The mutual inductance is m0
p R2
2
N1
N2
/
(c) The mutual inductances are the same. This is one example of von Neumann’s rule,
mentioned in the next problem.

11. Inductance 235
P32.35 The large coil produces this field at the center of the small coil:
N I R
x R
1 0 1 1
2
2
1
2 3 2
2
µ
+( )
. The field is
normal to the area of the small coil and nearly uniform over this area, so it produces flux
Φ12
1 0 1 1
2
2
1
2 3 2 2
2
2
=
+( )
N I R
x R
R
µ
π through the face area of the small coil. When current I1 varies,
this is the emf induced in the small coil:
ε µ π πµ
2 2
1 0 1
2
2
2
2
1
2 3 2 1
1 2 0
2
= −
+( )
= −N
d
dt
N R R
x R
I
N N R11
2
2
2
2
1
2 3 2
1 1
2
R
x R
dI
dt
M
dI
dt+( )
= − so M
N N R R
x R
=
+( )
1 2 0 1
2
2
2
2
1
2 3 2
2
πµ
P32.36 With I I I= +1 2 , the voltage across the pair is:
∆V L
dI
dt
M
dI
dt
L
dI
dt
M
dI
dt
L
dI
dt
= − − = − − = −1
1 2
2
2 1
eq
So, − = +
dI
dt
V
L
M
L
dI
dt
1
1 1
2∆
and − +
( ) + =L
dI
dt
M V
L
M
L
dI
dt
V2
2
1
2
1
2
∆
∆
− +( ) = −( )L L M
dI
dt
V L M1 2
2 2
1∆
By substitution, − = +
dI
dt
V
L
M
L
dI
dt
2
2 2
1∆
leads to − +( ) = −( )L L M
dI
dt
V L M1 2
2 1
2∆
Adding [1] to [2], − +( ) = + −( )L L M
dI
dt
V L L M1 2
2
1 2 2∆
So, L
V
dI dt
L L M
L L M
eq = − =
−
+ −
∆ 1 2
2
1 2 2
Section 32.5 Oscillations in an LC Circuit
P32.37 At different times, U UC L( ) = ( )max max
so
1
2
1
2
2 2
C V LI∆( )⎡
⎣⎢
⎤
⎦⎥ = ⎛
⎝
⎞
⎠max max
I
C
L
Vmax max
.
.= ( ) =
×
×
−
−
∆
1 00 10
40 0
6
F
10.0 10 H
V3
(( ) = 0 400. A
P32.38
1
2
1
2
2 2
C V LI∆( )⎡
⎣⎢
⎤
⎦⎥ = ⎛
⎝
⎞
⎠max max
so ∆V
L
C
IC( ) = =
×
×
−
−max max
.
.
.
20 0 10
0 500 10
0 10
3
6
H
F
00 20 0A V( )= .
FIG. P32.36

12. 236 Chapter 32
P32.39 When the switch has been closed for a long time, battery, resistor,
and coil carry constant current I
R
max .=
ε 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 these
oscillations is ∆V, in
1
2
1
2
2 2
C V LI∆( ) = max .
Then, L
C V
I
C V R
=
( ) =
( ) =
×( )−
∆ ∆
2
2
2 2
2
6
0 500 10 150
max
.
ε
F V(( ) ( )
( )
=
2 2
2
250
50 0
0 281
V
H
Ω
.
. .
P32.40 (a) f
LC
= =
( ) ×( )
=
−
1
2
1
2 0 082 0 17 0 10
135
6π π . .H F
Hz
(b) Q Q t= = ( ) ×( )=max cos cos .ω µ µ180 847 0 00100 119C C
(c) I
dQ
dt
Q t= = − = −( )( ) ( )= −ω ωmax sin sin .847 180 0 847 1114 mA
P32.41 This radio is a radiotelephone on a ship, according to frequency assignments made by
international treaties, laws, and decisions of the National Telecommunications and Information
Administration.
The resonance frequency is f
LC
0
1
2
=
π
Thus, C
f L
=
( )
=
×( )⎡⎣ ⎤⎦ × −
1
2
1
2 6 30 10 1 05 100
2
6
2
6
π π . .Hz HH
pF
( )
= 608
P32.42 (a) f
LC
= =
( ) ×( )
=
−
1
2
1
2 0 100 1 00 10
503
6π π . .H F
Hz
(b) Q C= = ×( )( )=−
ε µ1 00 10 12 0 12 06
. . .F V C
(c)
1
2
1
2
2 2
C LIε = max
I
C
L
max .= =
×
=
−
ε 12 37 9
6
V
1.00 10 F
0.100 H
mA
(d) At all times U C= = ×( )( ) =−1
2
1
2
1 00 10 12 0 72 02 6 2
ε µ. . .F V J
P32.43 ω = =
( ) ×( )
= ×
−
1 1
3 30 840 10
1 899 10
12
4
LC .
.
H F
rad s
Q Q t= max cos ,ω I
dQ
dt
Q t= = −ω ωmax sin
(a) U
Q
C
C = =
×⎡⎣ ⎤⎦ ×( ) ×−2 6 4
2
105 10 1 899 10 2 00cos . .rad s 110
2 840 10
6 03
3
2
12
−
−
( )⎡⎣ ⎤⎦( )
×( )
=
s
J.
FIG. P32.42
FIG. P32.39
continued on next page

13. Inductance 237
(b) U LI L Q t
Q t
C
L = = ( )=
( )1
2
1
2 2
2 2 2 2
2 2
ω ω
ω
max
max
sin
sin
UL =
×( ) ×( ) ×− −
105 10 1 899 10 2 00 106 2 2 4 3
C rad ssin . . ss
F
J
( )⎡⎣ ⎤⎦
×( )
=−
2 840 10
0 52912
.
(c) U U UC Ltotal J= + = 6 56.
Section 32.6 The RLC Circuit
P32.44 (a) ωd
LC
R
L
= − ⎛
⎝
⎞
⎠
=
×( ) ×( )
−− −
1
2
1
2 20 10 1 80 10
7 6
2
3 6
. .
. 00
2 2 20 10
1 58 103
2
4
.
.
×( )
⎛
⎝
⎜
⎞
⎠
⎟ = ×−
rad s
Therefore, fd
d
= =
ω
π2
2 51. kHz
(b) R
L
C
c = =
4
69 9. Ω
P32.45 (a) ω0
6
1 1
0 500 0 100 10
4 47= =
( ) ×( )
=
−LC . .
. krad s
(b) ωd
LC
R
L
= −
⎛
⎝
⎞
⎠
=
1
2
4 36
2
. krad s
(c)
∆ω
ω0
2 53= . % lower
P32.46 Choose to call positive current clockwise in Figure 32.15. It drains charge from the capacitor
according to I
dQ
dt
= − . A clockwise trip around the circuit then gives
+ − − =
Q
C
IR L
dI
dt
0
+ + + =
Q
C
dQ
dt
R L
d
dt
dQ
dt
0, identical with Equation 32.28.
P32.47 (a) Q Q e tRt L
d= −
max cos2
ω so I e Rt L
max ∝ − 2
0 500 2
. = −
e Rt L
and
Rt
L2
0 500= − ( )ln .
t
L
R
L
R
= − ( )=
⎛
⎝
⎜
⎞
⎠
⎟
2
0 500 0 693
2
ln . .
(b) U Q0
2
∝ max and U U= 0 500 0. so Q Q Q= =0 500 0 707. .max max
t
L
R
L
R
= − ( )=
⎛
⎝
⎜
⎞
⎠
⎟
2
0 707 0 347
2
ln . . (half as long)

14. 238 Chapter 32
Additional Problems
P32.48 (a) Let Q represent the magnitude of the opposite charges on the plates of a parallel plate
capacitor, the two plates having area A and separation d. The negative plate creates electric
field E =
∈
Q
A2 0
toward itself. It exerts on the positive plate force F =
∈
Q
A
2
02
toward the
negative plate. The total field between the plates is
Q
A∈0
. The energy density is
u E
Q
A
Q
A
E = ∈ = ∈
∈
=
∈
1
2
1
2 2
0
2
0
2
0
2 2
2
0
2
. Modeling this as a negative or inward pressure,
we have for the force on one plate F PA
Q
A
= =
∈
2
0
2
2
, in agreement with our first analysis.
(b) The lower of the two current sheets shown creates
above it magnetic field B k= −( )µ0
2
Js ˆ . Let and w
represent the length and width of each sheet. The
upper sheet carries current J ws and feels force
F B i k j= × = × −( )=I J w
J w J
s
s sµ µ0 0
2
2 2
ˆ ˆ ˆ .
The force per area is P
F
w
Js
= =
µ0
2
2
.
(c) Between the two sheets the total magnetic field is
µ µ
µ0 0
0
2 2
J J
Js s
s−( )+ −( )=ˆ ˆ ˆk k k , with
magnitude B Js= µ0 .Outside the space they enclose, the fields of the separate sheets are
in opposite directions and add to zero .
(d) u B
J J
B
s s
= = =
1
2 2 20
2 0
2 2
0
0
2
µ
µ
µ
µ
(e) This energy density agrees with the magnetic pressure found in part (b).
P32.49 (a) εL L
dI
dt
d t
dt
= − = −( )
( ) = −1 00
20 0
20 0.
.
.mH mV
(b) Q Idt t dt t
t t
= = ( ) =∫ ∫
0 0
2
20 0 10 0. .
∆V
Q
C
t
tC =
−
=
−
×
= −( )−
10 0
1 00 10
10 0
2
6
2.
.
.
F
MV s2
(c) When
Q
C
LI
2
2
2
1
2
≥ , or
−( )
×( )
≥ ×( )( )−
−
10 0
2 1 00 10
1
2
1 00 10 20 0
2 2
6
3
.
.
. .
t
t
22
, then
100 400 104 9 2
t t≥ ×( )−
. The earliest time this is true is at t = × =−
4 00 10 63 29
. . .s sµ
y
x
z
Js
Js
FIG. P32.48(b)

15. Inductance 239
P32.50 (a) εL L
dI
dt
L
d
dt
Kt LK= − = − ( )= −
(b) I
dQ
dt
= , so Q Idt Ktdt Kt
t t
= = =∫ ∫
0 0
21
2
and ∆V
Q
C
Kt
C
C =
−
= −
2
2
(c) When
1
2
1
2
2 2
C V LIC∆( ) = ,
1
2 4
1
2
2 4
2
2 2
C
K t
C
L K t
⎛
⎝
⎜
⎞
⎠
⎟ = ( )
Thus t LC= 2
P32.51
1
2
1
2 2
1
2
2 2
2Q
C C
Q
LI=
⎛
⎝
⎜
⎞
⎠
⎟ + so I
Q
CL
=
3
4
2
The flux through each turn of the coil is ΦB
LI
N
Q
N
L
C
= =
2
3
where N is the number of turns.
*P32.52 (a) The inductor has no voltage across it. It behaves as a short circuit . The battery sees
equivalent resistance 4 Ω + (1/4 Ω + 1/8 Ω)–1
= 6.67 Ω. The battery current is
10 V/6.67 Ω = 1.50 A. The voltage across the parallel combination of resistors is
10 V – 1.50 A 4 Ω = 4 V. The current in the 8-Ω resistor and the inductor is
4V/8Ω = 500 mA .
(b) U = (1/2) LI 2
= (1/2) 1 H(0.5 A)2
= 125 mJ
(c) The energy becomes 125 mJ of additional internal energy in the 8-Ω resistor and the 4-Ω
resistor in the middle branch.
(d) The current decreases from 500 mA toward zero,
showing exponential decay with a time constant of
L/R = 1 H/12 Ω = 83.3 ms.
*P32.53 (a) Just after the circuit is connected, the potential difference across the resistor is 0 and the
emf across the coil is 24.0 V.
(b) After several seconds, the potential difference across the resistor is 24.0 V and that across
the coil is 0.
(c) The resistor voltage and inductor voltage always add to 24 V. The resistor voltage increases
monotonically, so the two voltages are equal to each other, both being 12.0 V, just once.
The time is given by 12 V = IR = Rε/R(1 – e–Rt/L
) = 24 V(1 – e–6Ωt/0.005 H
).
This is 0.5 = e–1 200 t
or 1 200 t = ln 2 giving t = 0.578 ms after the circuit is connected .
(d) As the current decays the potential difference across the resistor is always equal to the emf
across the coil. It decreases from 24.0 V to zero.
0
t
83.3 ms
0
500 mA
I
FIG. P32.52(d)

16. 240 Chapter 32
*P32.54 We have 9 V = 2 A R + L (0.5 A/s) and 5 V = 2 A R + L (–0.5 A/s)
Solving simultaneously, 9 V – 5 V = L(1 A/s) so L = 4.00 H and 7 V = 2 A R so R = 3.50 Ω
*P32.55 Between t = 0 and t = 1 ms, the rate of change of current
is 2 A/s, so the induced voltage
∆Vab
= –L dI/dt is –100 mV. Between t = 1 ms and
t = 2 ms, the induced voltage is zero. Between t = 2 ms
and t = 3 ms the induced voltage is –50 mV. Between
t = 3 ms and t = 5 ms, the rate of change of current is
(–3/2) A/s, and the induced voltage is +75 mV.
*P32.56 (a) ω = (LC)–1/2
= (0.032 V ⋅ s/A 0.000 5 C/V)–1/2
= 250 rad/s
(b) ω = −
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
×
−−
1
2
1
1 6 10
4
2 1 2
5
LC
R
L
/
. s2
ΩΩ
2 0.032 V s/A⋅ ⋅
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
2 1 2/
242 rad/s
(c) ω = −
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
×
−−
1
2
1
1 6 10
15
2 1 2
5
LC
R
L
/
. s2
2 0.032 V s/A
Ω
⋅ ⋅
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
2 1 2/
87.0 rad/s
(d) ω = −
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟ =
×
−−
1
2
1
1 6 10
17
2 1 2
5
LC
R
L
/
. s2
2 0.032 V s/A
Ω
⋅ ⋅
⎡
⎣⎢
⎤
⎦⎥
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟
2 1 2/
gives an imaginary answer. In
parts (a), (b), and (c) the calculated angular frequency is experimentally verifiable.
Experimentally, in part (d) no oscillations occur . The circuit is overdamped.
*P32.57 B
NI
r
=
µ
π
0
2
(a) ΦB
a
b
a
b
BdA
NI
r
hdr
NIh dr
r
NIh
= = =
=
∫ ∫ ∫
µ
π
µ
π
µ
π
0 0
0
2 2
2
lln
b
a
⎛
⎝
⎜
⎞
⎠
⎟
L
N
I
N h b
a
B
= =
⎛
⎝
⎜
⎞
⎠
⎟
Φ µ
π
0
2
2
ln
(b) L =
( ) ( ) ⎛
⎝
⎜
⎞
⎠
⎟ =
µ
π
µ0
2
500 0 010 0
2
12 0
10 0
91 2
.
ln
.
.
. HH
(c) L
N A
R
appx
2
m
=
⎛
⎝
⎜
⎞
⎠
⎟ =
( ) × −
µ
π
µ
π
0
2
0
2 4
2
500
2
2 00 10.
00 110
90 9
.
. ,
⎛
⎝
⎜
⎞
⎠
⎟ = Hµ This approximate result is
only 0.3% different from the precise result.
FIG. P32.57
t (ms)
∆Vab (mV)
−100
0 0 2 4 6
100
FIG. P32.55

17. Inductance 241
P32.58 (a) At the center, B
N IR
R
N I
R
=
+( )
=
µ µ0
2
2 2 3 2
0
2 0 2
So the coil creates flux through itself ΦB BA
N I
R
R N IR= = ° =cos cosθ
µ
π
π
µ0 2
0
2
0
2
When the current it carries changes, ε π
µL
B
N
d
dt
N N R
dI
dt
L
dI
dt
= − ≈ −
⎛
⎝
⎜
⎞
⎠
⎟ = −
Φ
2
0
so L N R≈
π
µ
2
2
0
(b) 2 3 0 3πr = ( ). m so r ≈ 0 14. m
L
L
≈ ( ) × ⋅( )( )= ×− −π
π
2
1 4 10 0 14 2 8 10
10
2 7 7
T m A m H. .
~ 00 nH
(c)
L
R
=
× ⋅
= ×
−
−2 8 10
270
1 0 10
7
9.
.
V s A
V A
s
L
R
~1 ns
P32.59 Left-hand loop: ε − +( ) − =I I R I R2 1 2 2 0
Outside loop: ε − +( ) − =I I R L
dI
dt
2 1 0
Eliminating I2 gives ′− ′ − =ε IR L
dI
dt
0
This is of the same form as the differential equation 32.6 in
the chapter text for a simple RL circuit, so its solution is of the
same form as the equation 32.7 for the current in the circuit:
I t
R
e R t L
( ) =
′
−( )′ − ′ε 1
But ′ =
+
R
R R
R R
1 2
1 2
and ε′ =
+
R
R R
2
1 2
ε , so ε ε ε′
′
=
+( )
+( )
=
R
R R R
R R R R R
2 1 2
1 2 1 2 1
Thus I t
R
e R t L
( ) = −( )− ′ε
1
1
P32.60 From Ampère’s law, the magnetic field at distance r R≤ is found as:
B r J r
I
R
r2 0
2
0 2
2
π µ π µ
π
π( )= ( )=
⎛
⎝
⎜
⎞
⎠
⎟( ), or B
Ir
R
=
µ
π
0
2
2
The magnetic energy per unit length within the wire is then
U B
rdr
I
R
r dr
I
R
R R
= ( ) = =∫ ∫
2
00
0
2
4
3
0
0
2
4
2
2
4 4µ
π
µ
π
µ
π
RR I4
0
2
4 16
⎛
⎝⎜
⎞
⎠⎟ =
µ
π
This is independent of the radius of the wire.
FIG. P32.59

18. 242 Chapter 32
P32.61 (a) 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
ε
0
3 3
0
2 00 6 00 10 9 00 10 0. . . AΩ
== 72 0. V with end at the higher potentialb
(b)
FIG. P32.61(b)
(c) After the switch is opened, the current around the outer loop decays as
I I ei
Rt L
= −
with Imax .= 9 00 mA, R = 8 00. ,kΩ and L = 0 400. H.
Thus, when the current has reached a value I = 2 00. mA, the elapsed time is:
t
L
R
I
I
i
= ⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
=
×
⎛
⎝
⎞
⎠
ln
.
ln
0 400 H
8.00 103
Ω
99 00
2 00
7 52 10 75 25.
.
. .⎛
⎝
⎞
⎠
= × =−
s sµ
P32.62 (a) It has a magnetic field, and it stores energy, so L
U
I
=
2
2
is non-zero.
(b) Every field line goes through the rectangle between the conductors.
(c) Φ = LI so L
I I
BdA
y a
w a
= =
=
−
∫
Φ 1
L
I
xdy
I
y
I
w y I
Ix
a
w a
= +
−( )
⎛
⎝⎜
⎞
⎠⎟ =
−
∫
1
2 2
2
2
0 0 0µ
π
µ
π
µ
ππ
µ
πy
dy
x
y
a
w a
∫ =
−
2
2
0
ln
Thus L
x w a
a
=
−⎛
⎝
⎞
⎠
µ
π
0
ln
P32.63 When the switch is closed, as shown
in Figure (a), the current in the
inductor is I:
12 0 7 50 10 0 0 0 267. . . .− − = → =I I A
When the switch is opened, the initial
current in the inductor remains at
0.267 A.
IR V= ∆ :
0 267 80 0. .A V( ) ≤R
R ≤ 300 Ω
P32.64 For an RL circuit,
I t I ei
R L t
( ) =
−( )
:
I t
I
e
R
L
t
i
R L t( ) = − = ≅ −− −( )
1 10 19
R
L
t = −
10 9
so Rmax
.
. .
=
×( )( )
( ) ×(
− −
3 14 10 10
2 50 3 16 10
8 9
7
yr s yr))
= × −
3 97 10 25
. Ω
(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm2
, its
resistance would be at least 10 6−
Ω.)
FIG. P32.63

19. Inductance 243
P32.65 (a) U LIB = = ( ) ×( ) = ×
1
2
1
2
50 0 50 0 10 6 25 102 3 2 10
. . .H A JJ
(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.
Then one wire creates a field of B
I
r
=
µ
π
0
2
This causes a force on the next wire of F I B= sinθ
giving F I
I
r
I
r
= =
µ
π
µ
π
0 0
2
2
90
2
sin º
Evaluating the force, F = ×( )
( ) ×( )−
4 10
1 00 50 0 10
2 0 250
7
3 2
π
π
N A
m A
m
2
. .
.(( )
= 2 000 N
P32.66 P = I V∆ I
V
= =
×
×
= ×
P
∆
1 00 10
5 00 10
9
3.
.
W
200 10 V
A3
From Ampère’s law, B r I2 0π µ( )= enclosed or B
I
r
=
µ
π
0
2
enclosed
(a) At r a= = 0 020 0. m, Ienclosed A= ×5 00 103
.
and
T m A A
m
B =
× ⋅( ) ×( )
( )
−
4 10 5 00 10
2 0 020 0
7 3
π
π
.
.
== =0 050 0 50 0. .T mT
(b) At r b= = 0 050 0. m , I Ienclosed A= = ×5 00 103
.
and B =
× ⋅( ) ×( )
( )
=
−
4 10 5 00 10
2 0 050 0
0
7 3
π
π
T m A A
m
.
.
.0020 0 20 0T mT= .
(c) U udV
B r r dr I dr
ra
b
a
b
= =
( )[ ] ( )
= =∫ ∫ ∫
2
0
0
2
2
2 4
π
µ
µ
π
µµ
π
0
2
4
I b
a
ln⎛
⎝
⎞
⎠
U =
× ⋅( ) ×( ) ×( )−
4 10 5 00 10 1000 10
4
7 3 2 3
π
π
T m A A m.
lln
.
. .
5 00
2 29 10 2 296
cm
2.00 cm
J MJ
⎛
⎝
⎞
⎠
= × =
(d) The magnetic field created by the inner conductor exerts a force of repulsion on the current
in 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 and width w.
It carries a current of 5 00 10
2 0 050 0
3
.
.
×( ) ( )
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟A
m
w
π
and experiences an outward force
F I B
w
= =
×( )
( )
×sin
.
.
.θ
π
5 00 10
2 0 050 0
20 0 10
3
A
m
−−
( )3
90 0T °sin .
The pressure on it is P
F
A
F
w
= = =
×( ) ×( )−
5 00 10 20 0 10
2 0 050 0
3 3
. .
.
A T
π mm
Pa
( )
= 318
FIG. P32.66

20. 244 Chapter 32
P32.67 (a) B
NI
= =
× ⋅( )( )( )
=
−
µ π0
7
4 10 1 400 2 00
1 20
2
T m A A
m
.
.
..93 10 3
× ( )−
T upward
(b) u
B
= =
×( )
× ⋅( )
=
−
−
2
0
3 2
7
2
2 93 10
2 4 10
3 42
µ π
.
.
T
T m A
J m33
2
N m
1 J
N m Pa
( ) ⋅⎛
⎝
⎜
⎞
⎠
⎟
= =
1
3 42 3 42. .
(c) To produce a downward magnetic field, the surface of the superconductor must carry a
clockwise current.
(d) The vertical component of the field of the solenoid exerts an inward force on the supercon-
ductor. 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) F PA= = ( ) ×( )⎡
⎣⎢
⎤
⎦⎥ = ×−
3 42 1 10 10 1 30 102 2
. . .Pa mπ −−3
N
Note that we have not proved that energy density is pressure. In fact, it is not in some cases.
Chapter 21 proved that the pressure is two-thirds of the translational energy density in an
ideal gas.
ANSWERS TO EVEN PROBLEMS
P32.2 1 36. Hµ
P32.4 19 2. Wbµ
P32.6 (a) 360 mV (b) 180 mV (c) t = 3 00. s
P32.8 See the solution.
P32.10 See the solution.
P32.12 See the solution.
P32.14 92 8. V
P32.16 30 0. mH
P32.18 (500 mA)(1 – e−10t/s
), 1.50 A – (0.250 A) e−10t/s
P32.20 0 for t < 0; (10 A)(1 – e−10 000t
) for 0 < t < 200 µs; (63.9 A) e−10 000t
for t > 200 µs
P32.22 (a), (b), and (c) See the solution. (d) Yes; see the solution.
P32.24 (a) 8 06. MJ m3
(b) 6 32. kJ
P32.26 See the solution.

21. Inductance 245
P32.28 (a) 20 0. W (b) 20 0. W (c) 0 (d) 20 0. J
P32.30 1.73 mH
P32.32 781 pH
P32.34 (a) and (b) m0
p R2
2
N1
N2
/ (c) They are the same.
P32.36 (L1
L2
− M2
)/(L1
+ L2
− 2M)
P32.38 20.0 V
P32.40 (a) 135 Hz (b) 119 mC (c) –114 mA
P32.42 (a) 503 Hz (b) 12 0. Cµ (c) 37 9. mA (d) 72 0. Jµ
P32.44 (a) 2 51. kHz (b) 69 9. Ω
P32.46 See the solution.
P32.48 (b) m0
Js
2
/2 away from the other sheet (c) m0
Js
and zero (d) m0
Js
2
/2
P32.50 (a) εL LK= − (b) ∆V
Kt
C
c =
− 2
2
(c) t LC= 2
P32.52 (a) a short circuit; 500 mA (b) 125 mJ (c) The energy becomes 125 mJ of additional internal
energy in the 8-Ω resistor and the 4-Ω resistor in the middle branch. (d) See the solution. The
current decreases from 500 mA toward zero, showing exponential decay with a time constant of
83.3 ms.
P32.54 L = 4.00 H and R = 3.50 Ω
P32.56 (a) 250 rad/s (b) 242 rad/s (c) 87.0 rad/s (d) In parts (a), (b), and (c) the calculated angular
frequency is experimentally verifiable. In part (d) the equation for w gives an imaginary answer.
Experimentally, no oscillations occur when the circuit is overdamped.
P32.58 (a) L ≈ (p /2)N 2
m0
R (b) ~100 nH (c) ~1 ns
P32.60 See the solution.
P32.62 (a) It creates a magnetic field. (b) The long narrow rectangular area between the conductors
encloses all of the magnetic flux.
P32.64 3 97 10 25
. × −
Ω
P32.66 (a) 50.0 mT (b) 20.0 mT (c) 2.29 MJ (d) 318 Pa