2000 by Harcourt, Inc. All rights reserved.Chapter 32
Solutions*32.1 = L It = (3.00 10 3 H) 1. 50 A 0. 200 A0. 200 s|.`,
= 1.95 102 V = 19.5 mV 32.2 Treating the telephone cord as a
solenoid, we have:L = 0N2Al = (4 107 T m / A)(70.0)2()(6. 50 103
m)20.600 m = 1.36 H 32.3 = + L It|.`,= (2.00 H)0. 500 A0.0100
s|.
`,
= 100 V 32.4 L = 0n2A l so n =L0A l= 7.80 103 turns/ m 32.5 L =N
BI B=LIN= 240 nT m2 (through each turn) 32.6 LdIdt where L 0N2AlThu
s, 0N2Al|.
`,
dIdt 4 107 T m A( )300 ( )2 104 m2( )0.150 m10.0 A s ( ) = 2.37
mV 32.7 back = = L dIdt = L ddt (Imax sin t) = L Imax cos t = (10.0
10-3)(120 )(5.00) cos t back = (6.00 ) cos (120 t) = (18.8 V) cos
(377t) Chapter 32 Solutions 243 2000 by Harcourt, Inc. All rights
reserved.*32.8 From = L It|.`,, we have L = t|.`, = 24.0 10 3 V10.0
A/ s = 2.40 10 3 HFrom L = NBI , we have B = LIN = (2.40 10 3
H)(4.00 A)500 = 19.2 T m2 32.9 L = 0N 2A l = 0(420)2(3.00 10
4)0.160 = 4.16 10 4 H = L dIdt dIdt = L = 175 10 6 V4.16 10 4 H =
0.421 A/ s 32.10 The induced emf is LdIdt, where the
self-inductance of a solenoid is given by L 0N2Al. Thus, dIdt L
l0N2A 32.11 = L dIdt = (90.0 10-3) ddt (t 2 6t) V(a) At t = 1.00 s,
= 360 mV (b) At t = 4.00 s, = 180 mV (c) = (90.0 10-3)(2t 6) = 0
when t = 3.00 s 32.12 (a) B = 0nI = 0 .|,`4500.120 (0.0400 mA) =
188 T (b) B = BA = 3.33 10-8 T m2 (c) L = NBI = 0.375 mH 244
Chapter 32 Solutions(d) B and B are proportional to current; L is
independent of current Chapter 32 Solutions 245 2000 by Harcourt,
Inc. All rights reserved.32.13 (a) L = 0N 2A l = 0(120)2 ( 5.00
103)20.0900 = 15.8 H (b) B = m0 B L = mN 2A l = 800(1.58 10 5 H) =
12.6 mH 32.14 L = NBI = NBAI N AI 0NI2 R = 0N2A2 R 32.15 = 0ekt=
LdIdt dI = 0LektdtIf we require I 0 as t , the solution is I =
0kLekt=dqdt Q = I dt= 0kLekt0dt = 0k2L Q = 0k2L 32.16 I R(1 eRt/ L)
0. 900 R R1 eR(3.00 s)/ 2.50 H[ ] exp R(3.00 s)2. 50 H|.`, 0.100 R
2. 50 H3.00 sln 10.0 1.92 246 Chapter 32 Solutions32.17 = LR =
0.200 s: IImax = 1 et/(a) 0.500 = 1 et/ 0.200 t = ln 2.00 = 0.139 s
(b) 0.900 = 1 et/ 0.200 t = ln 10.0 = 0.461 s Figure for
GoalSolutionGoal Solution A 12.0-V battery is about to be connected
to a series circuit containing a 10.0- resistor and a
2.00-Hinductor. 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 L R 0. 2 s, which means that in 0.2 s, the current
will reach1/ e = 63% of its final value, as shown in the graph to
the right. We can see from this graph that t hetime to reach 50% of
Imax should be slightly less than the time constant, perhaps about
0.15 s, and t hetime to reach 0. 9Imax should be about 2.5 = 0.5 s.
O: The exact times can be found from the equation that describes
the rising current in the above graphand gives the current as a
function of time for a known emf, resistance, and time constant. We
settime t 0 to be the moment the circuit is first connected.A : At
time t , I t ( ) (1 et/ )Rwhere, after a long time, Imax (1 e)R RAt
I t ( ) 0. 500Imax, 0. 500 ( )R (1 et/ 0.200 s)Rso 0. 500 1 et/
0.200 sIsolating the constants on the right, ln et/ 2.00 s( ) ln 0.
500 ( )and solving for t , t0. 200 s 0.693 or t 0.139 s(b)
Similarly, to reach 90% of Imax, 0. 900 1 et/ and t ln 1 0. 900 (
)Thus, t 0. 200 s ( )ln 0.100 ( ) 0. 461 sL: The calculated times
agree reasonably well with our predictions. We must be careful to
avoidconfusing 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 2000 by Harcourt, Inc. All rights reserved.32.18
Taking L R, I I0et/ : dIdt I0et/ 1|.`, IR + LdIdt 0 will be true if
I0Ret/ + L I0et/ ( ) 1|.`, 0Because L R, we have agreement with 0 =
0*32.19 (a) L R = 2.00 10 3 s = 2.00 ms (b) I Imax1 et/ ( ) 6.00
V4.00 |.`,1 e0.250/ 2.00( ) 0.176 A (c) Imax = R = 6.00 V4.00 =
1.50 A (d) 0.800 = 1 et/ 2.00 ms t = (2.00 ms) ln(0.200) = 3.22 ms
*32.20 I = R (1 et/ ) = 1209.00 (1 e1.80/ 7.00) = 3.02 AVR = IR =
(3.02)(9.00) = 27.2 VVL = VR = 120 27.2 = 92.8 V 32.21 (a) VR IR
(8.00 )(2.00 A) 16.0 V and VL VR 36.0 V 16.0 V 20.0 VTherefore,
VRVL16.0 V20.0 V 0.800 (b) VR IR (4. 50 A)(8.00 ) 36.0 V VL VR 0
Figure for GoalSolution248 Chapter 32 SolutionsGoal Solution For
the RL circuit shown in Figure P32.19, let L = 3.00 H, R = 8.00 ,
and = 36.0 V. (a) Calculate the ratioof the potential difference
across the resistor to that across the inductor when I = 2.00 A.
(b) Calculate t hevoltage across the inductor when I = 4.50 A.G:
The voltage across the resistor is proportional to the current, IR
VR , while the voltage across t heinductor is proportional to the
rate of change in the current, L LdI dt . When the switch is
firstclosed, the voltage across the inductor will be large as it
opposes the sudden change in current. As t hecurrent approaches its
steady state value, the voltage across the resistor increases and
the inductorsemf decreases. The maximum current will be / R = 4.50
A, so when I = 2.00 A, the resistor andinductor will share similar
voltages at this mid-range current, but when I = 4.50 A, the entire
circuitvoltage 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 : (a) When I 2.00 A, the voltage across the
resistor is VR IR 2.00 A ( ) 8.00 ( ) 16.0 VKirchhoff's loop rule
tells us that the sum of the changes in potential around the loop
must be zero: VR L 36.0 V 16.0 V L 0 so L 20.0 V and VRL16.0 V20.0
V 0.800(b) Similarly, for I 4. 50 A, VR IR 4. 50 A ( ) 8.00 ( )
36.0 V VR L 36.0 V 36.0 V L 0 so L 0L: We see that when I 2.00 A,
VR < L, but they are similar in magnitude as expected. Also
aspredicted, 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
afterthe switch is reopened.*32.22 After a long time, 12.0 V =
(0.200 A)R Thus, R = 60.0 . Now, = LR givesL = R = (5.00 10 4
s)(60.0 V/ A) = 30.0 mH 32.23 I Imax1 et/ ( ): dIdt Imaxet/ ( )
1|.`, = LR = 15.0 H30.0 = 0.500 s :dIdt = RL Imax et/ and Imax =
R(a) t = 0: dIdt = RL Imax e0 = L = 100 V15.0 H = 6.67 A/ s (b) t =
1.50 s: dIdt = L et/ = (6.67 A/ s)e 1.50/ (0.500) = (6.67 A/
s)e3.00 = 0.332 A/ s Chapter 32 Solutions 249 2000 by Harcourt,
Inc. All rights reserved.32.24 I Imax1 et/ ( ) 0. 980 1 e3.0010 3/
0.0200 e3.0010 3/ 3.00 103ln(0.0200) 7.67 104s L R, so L R (7.67
104)(10.0) 7.67 mH 32.25 Name the currents as shown. By Kirchhoffs
laws: I1 I2 + I3(1) +10.0 V 4.00I1 4.00I2 0 (2) +10.0 V 4.00I1
8.00I3 1.00 ( )dI3dt 0 (3)From (1) and (2), +10.0 4.00I1 4.00I1 +
4.00I3 0 and I1 0. 500I3 + 1. 25 AThen (3) becomes 10.0 V 4.00 0.
500I3 + 1. 25 A ( ) 8.00I3 1.00 ( )dI3dt 0 1.00 H ( ) dI3dt ( ) +
10.0 ( ) I3 5.00 VWe solve the differential equation using
Equations 32.6 and 32.7: I3t ( ) 5.00 V10.0 1 e 10.0 ( )t 1.00 H[ ]
= 0. 500 A ( ) 1 e10t/ s[ ] I1 1. 25 + 0. 500I3 1. 50 A 0. 250 A (
)e10t/ s 32.26 (a) Using RC LR, we get R LC 3.00 H3.00 106 F 1.00
103 1.00 k (b) RC 1.00 103 ( )3.00 106 F( ) 3.00 103 s 3.00 ms 250
Chapter 32 Solutions32.27 For t 0, the current in the inductor is
zero. At t 0, it starts t ogrow from zero toward 10.0 A with time
constant L R 10.0 mH ( ) 100 ( ) 1.00 104 s .For 0 t 200 s, I Imax1
et/ |.`,= 10.00 A ( ) 1 e10000t/ s( ) At t 200 s, I 10.00 A ( ) 1
e2.00( ) 8.65 AThereafter, it decays exponentially as I I0e t , so
for t 200 s , I 8.65 A ( )e10000 t 200 s ( ) s 8.65 A ( )e10000t s
+2.00 8.65e2.00 A( )e10000t s = 63. 9 A ( )e10000t s 32.28 (a) I =
R = 12.0 V12.0 = 1.00 A (b) Initial current is 1.00 A, : V12 =
(1.00 A)(12.00 ) = 12.0 V V1200 = (1.00 A)(1200 ) = 1.20 kV VL =
1.21 kV (c) I = Imax eRt/ L:dIdt = Imax RL eRt / Land L dIdt = VL =
Imax ReRt/ LSolving 12.0 V = (1212 V)e1212t/ 2.00so 9.90 10 3 = e
606tThu s, t = 7.62 ms 32.29 LR 0.1404. 90 28.6 ms; Imax R 6.00 V4.
90 1. 22 A(a) I Imax1 et/ ( )so 0. 220 1. 22 1 et/ ( ) et/ 0.820 t
ln(0.820) 5.66 ms (b) I Imax1 e10.00.0286|.
`,
(1. 22 A) 1 e350( ) 1.22 A (c) I Imaxet/ and 0.160 1. 22et/ so t
= ln(0.131) = 58.1 ms Chapter 32 Solutions 251 2000 by Harcourt,
Inc. All rights reserved.32.30 (a) For a series connection, both
inductors carry equal currents at every instant, so dI/ dt is t
hesame for both. The voltage across the pair is LeqdIdt= L1dIdt+
L2dIdtso Leq= L1+ L2 (b) LeqdIdt= L1dI1dt= L2dI2dt= VLwhere I = I1+
I2 and dIdt=dI1dt+dI2dtThus, VLLeq= VLL1+ VLL2 and 1Leq=1L1+1L2 (c)
LeqdIdt+ ReqI = L1dIdt+ IR1+ L2dIdt+ IR2Now I and dI/dt are
separate quantities under our control, so functional equality
requires both Leq= L1+ L2 and Req= R 1+ R 2 (d) V = LeqdIdt+ ReqI =
L1dI1dt+ R1I1= L2dI2dt+ R2I2 where I = I1+ I2 and
dIdt=dI1dt+dI2dtWe may choose to keep the currents constant in
time. Then, 1Req=1R 1+1R 2We may choose to make the current swing
through 0. Then, 1Leq=1L1+1L2 This equivalent coil with resistance
will be equivalentto the pair of real inductors for all other
currents as well. 32.31 L = N BI = 200(3.70 10 4)1.75 = 42.3 mH so
U = 12 LI 2 = 12 (0.423 H)(1.75 A) 2 = 0.0648 J 32.32 (a) The
magnetic energy density is given byu = B 220 = (4.50 T)22(1.26 10 6
T m/ A) = 8.06 106 J/ m3 (b) The magnetic energy stored in the
field equals u times the volume of the solenoid (t hevolume in
which B is non-zero).U = uV = (8.06 106 J/ m3) (0. 260 m)(0.0310
m)2[ ] 6.32 kJ 252 Chapter 32 Solutions32.33 L 0N2Al 0(68.0)2(0.600
102)20.0800 8. 21 H U 12LI212(8. 21 106H)(0.770 A)2 2.44 J 32.34
(a) U = 12 LI 2 = 12 L 2R|.`,2L28R2 (0.800)(500)28(30.0)2 = 27.8 J
(b) I = R|.`,1 e(R/ L)t[ ]so 2R R|.`,1 e(R/ L)t[ ]e(R/ L)t12 RLt ln
2 so t LRln 2 0.80030.0ln 2 18.5 ms 32.35 u = 0 E 22 = 44.2 nJ/ m3
u = B 220 = 995 J/ m3 *32.36 (a) U = 12 LI 2 = 12 (4.00 H)(0.500 A)
2 = 0.500 J (b)dUdt = LI = (4.00 H)(1.00 A) = 4.00 J/ s = 4.00 W
(c) P = (V)I = (22.0 V)(0.500 A) = 11.0 W 32.37 From Equation 32.7,
I R1 e Rt L( )(a) The maximum current, after a long time t , is I R
2.00 A.At that time, the inductor is fully energized and P I(V)
(2.00 A)(10.0 V) = 20.0 W (b) Plost I2R (2.00 A)2(5.00 ) = 20.0 W
(c) Pinductor I(Vdrop) 0 (d) U LI22 (10.0 H)(2.00 A)22 20.0 J
Chapter 32 Solutions 253 2000 by Harcourt, Inc. All rights
reserved.32.38 We have u e0E22and u B220Therefore e0E22 B220so B2
e00E2 B E e00 6.80 105 V/ m3.00 108 m / s 2.27 10 3 T 32.39 The
total magnetic energy is the volume integral of the energy density,
u B220Because B changes with position, u is not constant. For B B0R
/ r ( )2, u B0220|.
`,
Rr|.`,4Next, we set up an expression for the magnetic energy in
a spherical shell of radius r andthickness dr. Such a shell has a
volume 4 r 2 dr, so the energy stored in it is dU u 4 r2dr( )
2B02R40|.
`,
drr2We integrate this expression for r = R to r = to obtain the
total magnetic energy outside t hesphere. This gives U 2 B20 R 30 =
2 (5.00 105 T)2(6.00 106 m)3(1.26 10 6 T m/ A) = 2.70 1018 J 32.40
I1(t ) Imaxetsin t with Imax 5.00 A, 0.0250 s1, and 377 rad s .
dI1dt Imaxet sin t + cos t ( )At t 0.800 s , dI1dt 5.00 A s (
)e0.0200 0.0250 ( )sin 0.800 377 ( ) ( ) + 377cos 0.800 377 ( ) ( )
[ ] dI1dt 1.85 103 A sThus, 2 MdI1dt: M 2dI1dt +3. 20 V1.85 103 A s
1.73 mH 254 Chapter 32 Solutions32.41 2 MdI1dt (1.00 104 H)(1.00
104 A/ s) cos(1000t ) 2( )max 1.00 V 32.42 M = 2dI1dt=96.0 mV1. 20
A/ s= 80.0 mH 32.43 (a) M NBBAIA700(90.0 106)3. 50 18.0 mH (b) LA
AIA400(300 106)3. 50 34.3 mH (c) B MdIAdt (18.0 mH)(0. 500 A/ s)
9.00 mV 32.44 M N212I1N2B1A1( )I1N2 0n1I1( )A1 [ ]I1 N2 0n1A1 M
(1.00) 4 107 T m A( )70.00.0500 m|.`, 5.00 103 m( )2
]]] 138 nH 32.45 B at center of (larger) loop: B1 0I12R(a) M
2I1B1A2I1(0I1/ 2R)(r2)I1 022rR (b) M 0(0.0200)22(0. 200) 3.95 nH
Chapter 32 Solutions 255 2000 by Harcourt, Inc. All rights
reserved.*32.46 Assume the long wire carries current I. Then the
magnitude of the magnetic field itgenerates at distance x from the
wire is B 0I 2x, and this field passes perpendicularlythrough the
plane of the loop. The flux through the loop is B B dA BdA B ldx (
) 0I l2dxx0.400 mm1.70 mm 0Il2ln1.700. 400|.`,The mutual inductance
between the wire and the loop is then M N212I1N20I l2Iln1.700.
400|.`, N20l21. 45 ( ) 1(4 107T m A)(2.70 103m)21. 45 ( ) M 7.81
1010 H 781 pH 32.47 With I I1 + I2, the voltage across the pair is:
V L1dI1dt MdI2dt L2dI2dt MdI1dt LeqdIdtSo, dI1dt= VL1+ML1dI2dtand
L2dI2dt+M V ( )L1+M2L1dI2dt= V(a) (b) (L1L2 + M2)dI2dt V(L1 M)
[1]By substitution, dI2dt= VL2+ML2dI1dtleads to ( L1L2+ M2)dI1dt= V
(L2 M) [2]Adding [1] to [2], ( L1L2+ M2)dIdt= V (L1+ L2 2M)So, Leq=
VdI / dt= L1L2 M2L1+ L2 2M 32.48 At different times, UC( )max UL(
)max so 12C V ( )2[ ]max12LI2( )max Imax CL V ( )max 1.00 106F10.0
103H40.0 V ( ) 0.400 A 256 Chapter 32 Solutions32.49 12C V ( )2[
]max12LI2( )maxso VC( )max LCImax 20.0 103H0. 500 106F0.100 A ( )
20.0 V 32.50 When the switch has been closed for a long time,
battery, resistor,and coil carry constant current Imax= / R. When
the switch isopened, current in battery and resistor drops to zero,
but the coilcarries this same current for a moment as oscillations
begin in t heLC loop.We interpret the problem to mean that the
voltage amplitude ofthese oscillations is V, in 12C V (
)212LImax2.Then, L C V ( )2Imax2 C V ( )2R22 0. 500 106 F( )150 V (
)2250 ( )250.0 V ( )2 0.281 H 32.51 C =1(2 f )2L=1(2 6. 30
106)2(1.05 106)= 608 pF Goal Solution A fixed inductance L = 1.05 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 expecta typical value in the F or pF range.O: We want the
resonance frequency of the circuit to match the broadcasting
frequency, and for a simpleRLC circuit, the resonance frequency
only depends on the magnitudes of the inductance andcapacitance.A :
The resonance frequency is f0 12 LCThu s, C 1(2 f0)2L 1(2)(6. 30
106 Hz)[ ]2(1.05 106 H) 608 pFL: This is indeed a typical
capacitance, so our calculation appears reasonable. However, you
probablywould not hear any familiar music on this broadcast
frequency. The frequency range for FM radiobroadcasting is 88.0
108.0 MHz, and AM radio is 535 1605 kHz. The 6.30 MHz frequency
falls in t heMaritime Mobile SSB Radiotelephone range, so you might
hear a ship captain instead of Top 40tunes! This and other
information about the radio frequency spectrum can be found on the
Nat ionalTelecommunications and Information Administration (NTIA)
website, which at the time of t hisprinting was at http:/ /
www.ntia.doc.gov/ osmhome/ allochrt.html Chapter 32 Solutions 257
2000 by Harcourt, Inc. All rights reserved.32.52 f =12 LC: L =1(2 f
)2C=1(2 120)2(8.00 106)= 0.220 H 32.53 (a) f 12 LC 12 (0.0820
H)(17.0 106F) 135 Hz (b) Q Qmaxcos t (180 C) cos(847 0.00100) 119 C
(c) I dQdt Qmaxsin t (847)(180) sin (0.847) 114 mA 32.54 (a) f =12
LC=12 (0.100 H)(1.00 106F)= 503 Hz (b) Q = C = (1.00 106F)(12.0 V)
= 12.0 C (c) 12C2=12LImax2 Imax= CL = 12 V1.00 106F0.100 H= 37.9 mA
(d) At all times U = 12C2=12(1.00 106F)(12.0 V)2= 72.0 J 32.55 1LC
13. 30 H ( ) 840 1012 F( ) 1.899 104 rad s Q Qmaxcos t , I dQdt
Qmaxsin t(a) UC Q22C 105 106[ ]cos 1.899 104 rad s( )2.00 103 s( )[
] ( )22 840 1012( ) 6.03 J (b) UL 12LI212L2Qmax2sin2t ( )
Qmax2sin2t ( )2C UL 105 106 C( )2sin21.899 104 rad s( )2.00 103 s(
)[ ]2 840 1012 F( ) 0.529 J (c) Utotal UC +UL 6.56 J 258 Chapter 32
Solutions32.56 (a) d 1LC R2L|.`,212. 20 103( )1.80 106( ) 7.602 2.
20 103( )|.
`,
2 1. 58 104rad / sTherefore, fd d2 2.51 kHz (b) Rc 4LC 69.9
32.57 (a) 0 1LC 1(0. 500)(0.100 106) 4.47 krad/ s (b) d 1LC
R2L|.`,2 4.36 krad/ s (c) 0 = 2.53% lower 32.58 Choose to call
positive current clockwise in Figure 32.19. It drains charge from
the capacitoraccording to I = dQ/dt. A clockwise trip around the
circuit then gives +QC IR LdIdt= 0 +QC+dQdtR + LddtdQdt= 0,
identical with Equation 32.29.32.59 (a) Q QmaxeRt2L cos dt so Imax
eRt2L 0. 500 eRt2Land Rt2L ln(0. 500) t 2LRln 0. 500 ( )
0.6932LR|.`, (b) U0 Qmax2 and U 0. 500U0so Q = 0.500 Qmax =
0.707Qmax t 2LRln(0.707) = 0. 3472LR|.`, (half as long) Chapter 32
Solutions 259 2000 by Harcourt, Inc. All rights reserved.32.60 With
Q Qmax at t 0, the charge on the capacitor at any time is Q Qmaxcos
t where 1 LC . The energy stored in the capacitor at time t is then
U Q22C Qmax22Ccos2t U0cos2t .When U 14U0, cos t 12and t 13 rad
Therefore, tLC 3or t2LC 29The inductance is then: L 9t22C 32.61 (a)
L LdIdt 1.00 mH ( )d 20.0t ( )dt 20.0 mV (b) Q I dt0t 20.0t ( )dt0t
10.0t2 VC QC 10.0t21.00 106 F 10.0 MV s2( )t2 (c) When Q22C 12LI2,
or 10.0t2( )22 1.00 106( ) 121.00 103( )20.0t ( )2,then 100t4 400
109( )t2. The earliest time this is true is at t 4.00 109 s 63.2 s
32.62 (a) L= LdIdt= Lddt(Kt ) = LK (b) I =dQdt, so Q I dt0t Kt dt0t
12Kt2and VC QC Kt22C (c) When 12C VC( )2=12LI2, 12CK2t44C2|.
`,
=12L K2t2( )Thu s t = 2 LC 260 Chapter 32 Solutions32.63
12Q2C=12CQ2|.`,2+12LI2so I =3Q24CLThe flux through each turn of the
coil is B=LIN= Q2N3LC where N is the number of turns.32.64 Equation
30.16: B 0NI2r(a) B BdA 0NI2rhdrab 0NIh2drrab 0NIh2lnba|.`, L NBI
0N2h2lnba|.`, (b) L 0(500)2(0.0100)2ln12.010.0|.`, 91.2 H (c) Lappx
0N22AR|.`, 0(500)222.00 104 m20.110|.
`,
90.9 H *32.65 (a) At the center, B N0IR22(R2+ 02)3/ 2 N0I2RSo
the coil creates flux through itself B BA cos N0I2R R2cos 0
2N0IRWhen the current it carries changes, L NdBdt N 2N0RdIdt
LdIdtso L 2N20R (b) 2 r 3(0.3 m), so r 0.14 m; L 2 12 .|,`4 107 T
mA 0.14 m = 2.8 107 H ~ 100 nH (c)LR 2.8 107 V s/ A270 V/ A = 1.0
10 9 s ~ 1 ns Chapter 32 Solutions 261 2000 by Harcourt, Inc. All
rights reserved.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
rectangleis L 9.80 m ( )2 0.100 m ( )2 L 0.100 m9.80 mThe mean
circumference of each turn is C r 2 , where r 24.0 + 0.6442 mm is
the meanradius of each turn. The number of turns is then: N LC 9.80
m ( )2 0.100 m ( )2224.0 + 0.6442|.`, 103 m 127 (b) R lA 1.70 108
m( )10.0 m ( ) 0. 322 103 m( )2 0.522 (c) L N2A l 8000 l LC|.`,2 r
( )2 L 800 4 107( )0.100 m9.80 m ( )2 0.100 m ( )2 24.0 + 0.644 ( )
103 m|.
`,
224.0 + 0.6442|.`, 103 m
]]]2 L 7.68 102 H 76.8 mH 32.67 From Amperes law, the magnetic
field at distance r R is found as: B 2r ( ) 0J r2( ) 0IR2|.
`,
r2( ), or B 0Ir2R2The magnetic energy per unit length within the
wire is then Ul B2202r dr ( )0R 0I24R4r3dr0R 0I24R4R44|.
`,
= 0I216 This is independent of the radius of the wire.262
Chapter 32 Solutions32.68 The primary circuit (containing the
battery and solenoid) is a nRL circuit with R 14.0 , and L 0N2Al 4
107( )12 500 ( )21.00 104( )0.0700 0. 280 H(a) The time for the
current to reach 63.2% of the maxi mu mvalue is the time constant
of the circuit: LR 0. 280 H14.0 0.0200 s 20.0 ms (b) The solenoid's
average back emf is L L It|.`, LIf 0t|.
`,
where If 0.632Imax 0.632 VR|.`, 0.63260.0 V14.0 |.`, 2.71 AThus,
L 0. 280 H ( )2.71 A0.0200 s|.`, 37.9 V (c) The average rate of
change of flux through each turn of the overwrapped concentric coil
is t hesame as that through a turn on the solenoid: Bt 0n I ( )At 4
107 T m A( )12500 0.0700 m ( ) 2.71 A ( ) 1.00 104 m2( )0.0200 s =
3.04 mV (d) The magnitude of the average induced emf in the coil is
L N B t ( ) and magnitude ofthe average induced current is I LR
NRBt|.`, 82024.0 3.04 103 V( ) 0.104 A 104 mA 32.69 Left-hand loop:
E (I + I2)R 1 I2R2 0Outside loop: E (I + I2)R 1 LdIdt 0Eliminating
I 2 gives E I R LdIdt 0This is of the same form as Equation 32.6,
so its solution is of the same form as Equation 32.7: I t ( ) E R(1
e R t L)But R R1R2/ R1 + R2( ) and E R2E/ R1 + R2( ), so E R ER 2/
(R 1 + R2)R 1R2/ (R1 + R2) ER 1Thu s I(t ) = ER 1(1 e R t L)Chapter
32 Solutions 263 2000 by Harcourt, Inc. All rights reserved.32.70
When switch is closed, steady current I0 1. 20 A. Wh e nthe switch
is opened after being closed a long time, t hecurrent in the right
loop is I I0eR2t Lso eRt LI0Iand RtL lnI0I|.`,Therefore, L R2tln
I0I ( ) 1.00 ( ) 0.150 s ( )ln 1. 20 A 0. 250 A ( ) 0.0956 H 95.6
mH 32.71 (a) While steady-state conditions exist, a 9.00 mA flows
clockwise around the right loop of t hecircuit. Immediately after
the switch is opened, a 9.00 mA current will flow around the out
erloop of the circuit. Applying Kirchhoffs loop rule to this loop
gives: +0 2.00 + 6.00 ( ) 103 [ ]9.00 103 A( ) 0 +0 72 0 . V with
end at the higher potential b (b) (c) After the switch is opened,
the current around the outer loop decays as I ImaxeRt L with Imax
9.00 mA, R 8.00 k, and L 0. 400 HThus, when the current has reached
a value I 2.00 mA, the elapsed time is: t LR|.`,lnImaxI|.`, 0. 400
H8.00 103 |.`,ln9.002.00|.
`,
7. 52 105 s 75.2 s 264 Chapter 32 Solutions32.72 (a) The instant
after the switch is closed, the situation is as shown i nthe
circuit diagram of Figure (a). The requested quantities are: IL 0,
IC 0R, IR 0R VL 0, VC 0, VR 0 (b) After the switch has been closed
a long time, the steady-stateconditions shown in Figure (b) will
exist. The currents andvoltages are: IL 0, IC 0, IR 0 VL 0, VC 0,
VR 0 IR = 0+-0Q = 0VC = 0IR = 0/ R+-IL = 0 VL = 0VR = 0Figure (a)
+-0Q = C0VC = 0+-IL = 0 VL = 0VR = 0Figure (b) IC = 0/ R32.73 When
the switch is closed, asshown in Figure (a), the currentin the
inductor is I :12.0 7.50I 10.0 = 0 I = 0.267 AWhen the switch is
opened, t heinitial current in the inductorremains at 0.267 A.IR =
V: (0.267 A)R 80.0 VR 300 (a) (b)Goal Solution To prevent damage
from arcing in an electric motor, a discharge resistor is sometimes
placed in parallelwith the armature. If the motor is suddenly
unplugged while running, this resistor limits the voltagethat
appears across the armature coils. Consider a 12.0-V dc motor with
an armature that has a resistanceof 7.50 and an inductance of 450
mH. Assume that the back emf in the armature coils is 10.0 V
whenthe motor is running at normal speed. (The equivalent circuit
for the armature is shown in FigureP32.73.) Calculate the maximum
resistance R that limits the voltage across the armature to 80.0 V
whenthe motor is unplugged.Chapter 32 Solutions 265 2000 by
Harcourt, Inc. All rights reserved.G: We should expect R to be
significantly greater than the resistance of the armature coil, for
otherwise alarge portion of the source current would be diverted
through R and much of the total power wouldbe 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
motorsinertia 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 flowthrough the discharge resistor, which will create a
voltage across R that we wish to limit to 80 V. Astime passes, the
current will be reduced by the opposing back emf, and as the motor
slows down, t heback 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 t othe outer
loop: + 12.0 V I 7. 50 ( ) 10.0 V 0so I 2.00 V7. 50 0. 267 AWe then
require that VR 80.0 V 0. 267 A ( )Rso R VRI 80.0 V0. 267 A 300 L:
As we expected, this discharge resistance is considerably greater
than the coils resistance. Note thatwhile the motor is running, the
discharge resistor turns P (12 V)2300 0. 48 W of power int oheat
(or wastes 0.48 W). The source delivers power at the rate of about
P IV 0. 267 A + 12 V/ 300 ( ) [ ]12 V ( ) 3.68 W, so the discharge
resistor wastes about 13% of t hetotal 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 0N12Al14 107 T m A( )1000 ( )21.00 104 m2(
)0. 500 m 2. 51 104 H 251 H (b) M N22I1N21I1N2BAI1N2 0N1 l1( )I1 [
]AI1 0N1N2Al1 M 4 107 T m A( )1000 ( ) 100 ( ) 1.00 104 m2( )0. 500
m 2. 51 105 H 25.1 H (c) 1 MdI2dt, or I1R1 MdI2dt and I1 dQ1dt
MR1dI2dt Q1 MR1dI20tf MR1I2f I2i ( ) MR10 I2i( ) M I2iR1 Q1 2. 51
105 H( )1.00 A ( )1000 2. 51 108 C 25.1 nC 266 Chapter 32
Solutions32.75 (a) It has a magnetic field, and it stores energy,
so L = 2UI 2 is non-zero.(b) Every field line goes through the
rectangle between the conductors.(c) = LI so L I 1IBday awa L 1Ix
dyawa 0I2 y + 0I2 w y ( )|.
`,
2I0Ix2 ydy 20x2ln yawaThu s L 0xlnw aa|.`,32.76 For an RL
circuit, I(t ) ImaxeRLt: I(t )Imax 1 109 eRLt 1RLt RLt 109so Rmax
(3.14 108)(109)(2. 50 yr)(3.16 107 s / yr) 3. 97 1025 (If the ring
were of purest copper, of diameter 1 cm, and cross-sectional area 1
mm2, itsresistance would be at least 10 6 ).32.77 (a) UB = 12 LI 2
= 12 (50.0 H)(50.0 10 3 A) 2 = 6.25 1010 J (b) Two adjacent turns
are parallel wires carrying current in the same direction. Since
the loopshave such large radius, a one-meter section can be
regarded as straight. Then one wire creates a field of B = 0I2 r
This causes a force on the next wire of F = IlB sin giving F =
Il0I2 rsin 90 0lI22 rSolving for the force, F = (4 107 N/ A2) (1.00
m)(50.0 10 3 A) 2(2)(0.250 m) = 2000 N Chapter 32 Solutions 267
2000 by Harcourt, Inc. All rights reserved.32.78 P I V ( ) I PV
1.00 109 W200 103 V 5.00 103 AFrom Amperes law, B 2r ( ) 0Ienclosed
or B 0Ienclosed2r(a) At r a 0.0200 m, Ienclosed 5.00 103 A and B 4
107 T m A( )5.00 103 A( )2 0.0200 m ( ) 0.0500 T 50.0 mT (b) At r b
0.0500 m, Ienclosed I 5.00 103 A and B 4 107 T m A( )5.00 103 A( )2
0.0500 m ( ) 0.0200 T 20.0 mT (c) U u dV B r ( ) [ ]22 rldr (
)20rarb 0I2l4drrab 0I2l4lnba|.`, U 4 107 T m A( )5.00 103 A( )21000
103 m( )4ln5.00 cm2.00 cm|.
`,
2. 29 106 J 2.29 MJ (d) The magnetic field created by the inner
conductor exerts a force of repulsion on the current i nthe outer
sheath. The strength of this field, from part (b), is 20.0 mT.
Consider a smallrectangular section of the outer cylinder of length
l and width w. It carries a current of 5.00 103 A( )w2 0.0500 m (
)|.
`,
and experiences an outward force F IlBsin 5.00 103 A( )w2 0.0500
m ( ) l 20.0 103 T( )sin 90.0The pressure on it is P FA Fwl5.00 103
A( )20.0 103 T( )2 0.0500 m ( ) 318 Pa 268 Chapter 32
Solutions*32.79 (a) B 0NIl 4 107 T m A( )1400 ( ) 2.00 A ( )1. 20 m
2. 93 103 T (upward) (b) u B2202. 93 103 T( )22 4 107 T m A( ) 3.
42 Jm31 N m1 J|.
`,
3. 42 Nm2 3.42 Pa (c) To produce a downward magnetic field, the
surface of the super conductormust carry a clockwise current.(d)
The vertical component of the field of the solenoid exerts an
inward force on t hesuperconductor. The total horizontal force is
zero. Over the top end of the solenoid, its fielddiverges and has a
radially outward horizontal component. This component exerts
upwardforce on the clockwise superconductor current. The total
force on the core is upward . Youcan 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 Pa ( )
1.10 102 m( )2
]]] 1. 30 103 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.