32

CHAPTER OUTLINE

32.1 Self-Inductance32.2 RL32.3 Energy in a Magnetic Field32.4 Mutual Inductance32.5 Oscillations in an LC Circuit32.6 The

Circuits

CircuitRLC

Inductance

ANSWERS TO QUESTIONS

Q32.1 The emf induced in an inductor is opposite to the direction of thechanging current. For example, in a simple RL circuit with currentflowing clockwise, if the current in the circuit increases, theinductor will generate an emf to oppose the increasing current.

Q32.2 The coil has an inductance regardless of the nature of the currentin the circuit. Inductance depends only on the coil geometry andits construction. Since the current is constant, the self-inducedemf in the coil is zero, and the coil does not affect the steady-statecurrent. (We assume the resistance of the coil is negligible.)

Q32.3 The inductance of a coil is determined by (a) the geometry of thecoil and (b) the “contents” of the coil. This is similar to theparameters that determine the capacitance of a capacitor and theresistance of a resistor. With an inductor, the most importantfactor in the geometry is the number of turns of wire, or turns perunit length. By the “contents” we refer to the material in which theinductor establishes a magnetic field, notably the magneticproperties of the core around which the wire is wrapped.

Q32.4 If the first set of turns is wrapped clockwise around a spool, wrap the second set counter-clockwise,so that the coil produces negligible magnetic field. Then the inductance of each set of turnseffectively negates the inductive effects of the other set.

Q32.5 After the switch is closed, the back emf will not exceed that of the battery. If this were the case, thenthe current in the circuit would change direction to counterclockwise. Just after the switch is opened,the back emf can be much larger than the battery emf, to temporarily maintain the clockwise currentin a spark.

Q32.6 The current decreases not instantaneously but over some span of time. The faster the decrease in thecurrent, the larger will be the emf generated in the inductor. A spark can appear at the switch as it isopened because the self-induced voltage is a maximum at this instant. The voltage can thereforebriefly cause dielectric breakdown of the air between the contacts.

Q32.7 When it is being opened. When the switch is initially standing open, there is no current in thecircuit. 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 thecircuit. When the switch is then opened, the current rapidly decreases. The induced emf is created inthe inductor, and this emf tends to maintain the original current. Sparking occurs as the currentbridges the air gap between the contacts of the switch.

239

240 Inductance

Q32.8 A physicist’s list of constituents of the universe in 1829 might include matter, light, heat, the stuff ofstars, 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 ofelectric 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 interconvertability of mass and energy, andof electric and magnetic fields, can be used to make the list look shorter. Some might think of theconserved quantities energy, momentum, … bottomness as properties of matter, rather than asthings with their own existence. The idea of a field is not due to Henry, but rather to Faraday, towhom Henry personally demonstrated self-induction. Still the thesis stated in the question has animportant germ of truth. Henry precipitated a basic change if he did not cause it. The biggestdifference between the two lists is that the 1829 list does not include fields and today’s list does.

Q32.9 The energy stored in the magnetic field of an inductor is proportional to the square of the current.

Doubling I makes U LI=12

2 get four times larger.

Q32.10 The energy stored in a capacitor is proportional to the square of the electric field, and the energystored in an induction coil is proportional to the square of the magnetic field. The capacitor’s energyis 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 corematerial. On the other hand, we can think of Henry’s discovery of self-inductance as fundamentallynew. Before a certain school vacation at the Albany Academy about 1830, one could visualize theuniverse 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 thattemporarily maintains a current in a coil after the battery is removed is not energy that belongs toany bit of matter. This energy is vastly larger than the kinetic energy of the drifting electrons in thewires. This energy belongs to the magnetic field around the coil. Beginning in 1830, Nature hasforced us to admit that the universe consists of matter and also of fields, massless and invisible,known only by their effects.

Q32.11 The inductance of the series combination of inductor L1 and inductor L2 is L L M1 2 12+ + , whereM12 is the mutual inductance of the two coils. It can be defined as the emf induced in coil two whenthe current in coil one changes at one ampere per second, due to the magnetic field of coil oneproducing flux through coil two. The coils can be arranged to have large mutual inductance, as bywinding them onto the same core. The coils can be arranged to have negligible mutual inductance,as separate toroids do.

Q32.12 The mutual inductance of two loops in free space—that is, ignoring the use of cores—is a maximumif the loops are coaxial. In this way, the maximum flux of the primary loop will pass through thesecondary loop, generating the largest possible emf given the changing magnetic field due to thefirst. The mutual inductance is a minimum if the magnetic field of the first coil lies in the plane of thesecond coil, producing no flux through the area the second coil encloses.

Q32.13 The answer depends on the orientation of the solenoids. If they are coaxial, such as two solenoidsend-to-end, then there certainty will be mutual induction. If, however, they are oriented in such away that the magnetic field of one coil does not go through turns of the second coil, then there willbe no mutual induction. Consider the case of two solenoids physically arranged in a “T” formation,but still connected electrically in series. The magnetic field lines of the first coil will not produce anynet flux in the second coil, and thus no mutual induction will be present.

Chapter 32 241

Q32.14 When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance ofthe coil is making the current continue to flow. At this time the magnetic field of the coil contains allthe energy that was originally stored in the charged capacitor. The current has just finisheddischarging the capacitor and is proceeding to charge it up again with the opposite polarity.

Q32.15 The oscillations would eventually decrease, but perhaps with very small damping. The originalpotential energy would be converted to internal energy within the wires. Such a situationconstitutes an RLC circuit. Remember that a real battery generally contains an internal resistance.

Q32.16 If RL

C>

4, then the oscillator is overdamped—it will not oscillate. If R

LC

<4

, then the oscillator is

underdamped and can go through several cycles of oscillation before the radiated signal falls belowbackground noise.

Q32.17 The condition for critical damping must be investigated to design a circuit for a particular purpose.For example, in building a radio receiver, one would want to construct the receiving circuit so that itis underdamped. Then it can oscillate in resonance and detect the desired signal. Conversely, whendesigning a probe to measure a changing signal, such free oscillations are undesirable. An electricalvibration in the probe would constitute “ringing” of the system, where the probe would measure anadditional signal—that of the probe itself! In this case, one would want to design a probe that iscritically damped or overdamped, so that the only signal measured is the one under study. Criticaldamping represents the threshold between underdamping and overdamping. One must know thecondition for it to meet the design criteria for a project.

Q32.18 An object cannot exert a net force on itself. An object cannot create momentum out of nothing. A coilcan induce an emf in itself. When it does so, the actual forces acting on charges in different parts ofthe loop add as vectors to zero. The term electromotive force does not refer to a force, but to avoltage.

SOLUTIONS TO PROBLEMS

Section 32.1 Self-Inductance

P32.1 ε = = ×−F

HGIKJ = × =− −L

It

∆∆

3 00 101 50 0 200

1 95 10 19 53 2.. .

. . H A A0.200 s

V mVe j

P32.2 Treating the telephone cord as a solenoid, we have:

LN A

= =× ⋅ ×

=− −

µ π πµ0

2 7 2 3 24 10 70 0 6 50 10

0 6001 36

T m A m

m H

e ja f e j. .

.. .

P32.3 ε = − = −−FHG

IKJ

⋅⋅

FHG

IKJ =L

It

∆∆

2 000 0 500 1

100..

H A

0.010 0 s V s

1 H A Va f

P32.4 LN

ILIN

BB= → = = ⋅

ΦΦ 240 nT m2 through each turn

P32.5 ε ε ω ω ω π ωback = − = = = = × −LdIdt

Lddt

I t L I t tmax maxsin cos . . cosb g e ja fa f10 0 10 120 5 003

ε π πback V= =6 00 120 18 8 377. cos . cosa f b g a f a ft t

242 Inductance

P32.6 From ε = FHGIKJL

It

∆∆

, we have LI t

= =×

= ×−

−ε∆ ∆b g

24 0 102 40 10

33.

. V

10.0 A s H .

From LN

IB=

Φ, we have ΦB

LIN

= =×

= ⋅−2 40 10 4 00

50019 2

3. ..

H A T m2e ja fµ .

P32.7 LN A

= =×

= ×−

−µ µ0

2 02 4

4420 3 00 10

0 1604 16 10

a f e j.

.. H

εε

= − → =−

=− ×

×= −

−

−LdIdt

dIdt L

175 100 421

6 V4.16 10 H

A s4 .

P32.8 ε = = × −−LdIdt

ddt

t t90 0 10 63 2.e j e j 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.e ja ft

when t = 3 00. s .

P32.9 (a) B nI= = FHGIKJ =µ µ µ0 0

4500 120

0 040 0 188.

. A Tb g

(b) ΦB BA= = × ⋅−3 33 10 8. T m2

(c) LN

IB= =

Φ0 375. mH

(d) B LB and are proportional to current; is independent of currentΦ

P32.10 (a) LN A

= =×

=−

µ µ πµ0

2 02 3 2

120 5 00 10

0 090 015 8

a f e j.

.. H

(b) ′ = → = = × =−Φ ΦBm

BmL

N Aµµ

µ

0

25800 1 58 10 12 6. . H mHe j

*P32.11 We can directly find the self inductance of the solenoid:

ε = −LdIdt

+ = −−

0 080 1 8

..

V A

0.12 sL L

N A= × =−5 33 10 3 0

2

. Vs Aµ

.

Here A r=π 2 , 200 2 m= N rπ , and = −N 10 3 me j . Eliminating extra unknowns step by step, we have

5 33 10200 40 000

4

10 40 000

4 105 33 10

0 750

3 02 2

02 2

07

3

3

.

..

× = =FHG

IKJ = =

=×

×=

−−

−

−

Vs A m

2 m m Tm

A

WbmA AVs

m

2 2µ π µ π

πµ

πN r N

N

e j

Chapter 32 243

P32.12 LN

INBA

INA

INIR

N AR

B= = ≈ ⋅ =Φ µ

πµ

π0 0

2

2 2

FIG. P32.12

P32.13 ε ε= = −−0e L

dIdt

kt

dIL

e dtkt= − −ε 0

If we require I → 0 as t → ∞ , the solution is IkL

edqdt

kt= =−ε 0

Q IdtkL

e dtk L

kt= = = −z z −∞ ε ε0

0

02 Q

k L=

ε 02 .

Section 32.2 RL Circuits

P32.14 IR

e Rt L= − −ε1e j : 0 900 1 3 00 2 50. . .ε ε

R Re R= − − s Ha f

exp.

..

.ln . .

−FHG

IKJ =

= =

R

R

3 002 50

0 100

2 5010 0 1 92

s H

H3.00 s

a f

Ω

P32.15 (a) At time t, I te

R

t

a f e j=

− −ε τ1

where τ = =LR

0 200. s .

After a long time, Ie

R Rmax =−

=−∞ε ε1e j

.

At I t Ia f = 0 500. max 0 5001 0 200

..

a f e jε ε

R

e

R

t

=− − s

1

0.5

00 0.2 0.4 0.6

t (s)

I (A)Imax

FIG. P32.15

so 0 500 1 0 200. .= − −e t s .

Isolating the constants on the right, ln ln ..e t− =0 200 0 500 se j a fand solving for t, − = −

t0 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 900a f .Thus, t = − =0 200 0 100 0 461. ln . . s sa f a f .

244 Inductance

P32.16 Taking τ =LR

, I I e t= −0

τ :dIdt

I e t= −FHGIKJ

−0

1τ

τ

IR LdIdt

+ = 0 will be true if I L I et t0 0

10Re− −+ −FHGIKJ =

τ τ

τe j .

Because τ =LR

, we have agreement with 0 0= .

P32.17 (a) τ = = × =−LR

2 00 10 2 003. . s ms

(b) I I e et= − = FHGIKJ − =− −

max. ..

.16 00

1 0 1760 250 2 00τe j e j V4.00

AΩ

(c) IRmax

..= = =

ε 6 001 50

V4.00

AΩ

(d) 0 800 1 2 00 0 200 3 222 00. . ln . ..= − → = − =−e tt ms ms msa f a fFIG. P32.17

P32.18 IR

e et= − = − =− −ε τ11209 00

1 3 021.80 7 00e j e j... A

∆

∆ ∆

V IR

V VR

L R

= = =

= − = − =

3 02 9 00 27 2

120 27 2 92 8

. . .

. .

a fa f V

Vε

P32.19 Note: It may not be correct to call the voltage or emf across a coil a “potentialdifference.” Electric potential can only be defined for a conservative electricfield, and not for the electric field around an inductor.(a) ∆ ΩV IRR = = =8 00 2 00 16 0. . . A Va fa f

and ∆ ∆V VL R= − = − =ε 36 0 16 0 20 0. . . V V V .

Therefore,∆∆

VV

R

L= =

16 00 800

..

V20.0 V

.

(b) ∆ ΩV IRR = = =4 50 8 00 36 0. . . A Va fa f∆ ∆V VL R= − =ε 0

FIG. P32.19

P32.20 After a long time, 12 0 0 200. . V A= a fR . Thus, R = 60 0. Ω . Now, τ =LR

gives

L R= = × =−τ 5 00 10 60 0 30 04. . . s V A mHe jb g .

P32.21 I I e t= − −max 1 τe j : dI

dtI e t= − −FHG

IKJ

−max

τ

τe j 1

τ = = =LR

15 00 500

..

H30.0

sΩ

:dIdt

RL

I e t= −max

τ and IRmax =ε

(a) t = 0 : dIdt

RL

I eL

= = = =max .0 1006 67

ε V15.0 H

A s

(b) t = 1 50. s : dIdt L

e e et= = = =− − −ε τ 6 67 6 67 0 3321.50 0 500 3 00. . .. . A s A s A sb g b ga f

Chapter 32 245

P32.22 I I e t= − −max 1 τe j : 0 980 1 3 00 10 3

. .= − − × −e τ

0 020 0

3 00 100 020 0

7 67 10

3 00 10

34

3.

.ln .

.

.=

= −×

= ×

− ×

−−

−e τ

τ b g s

τ =LR

, so L R= = × =−τ 7 67 10 10 0 7 674. . .e ja f mHFIG. P32.22

P32.23 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 33. . . . V I I

dIdt

a f (3)

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 .

FIG. P32.23

Then (3) becomes 10 0 4 00 0 500 1 25 8 00 1 00 03 33. . . . . . V A− + − − =I I

dIdt

b g a f

1 00 10 0 5 0033. . . H Va f a fdI

dtIF

HGIKJ + =Ω .

We solve the differential equation using Equations 32.6 and 32.7:

I t e e

I I e

t t

t

310 0 1.00 10

1 310

5 001 0 500 1

1 25 0 500 1 50 0 250

a f a fa f

a f= FHGIKJ − = −

= + = −

− −

−

..

. . . .

. V10.0

A

A A

H s

s

ΩΩ

P32.24 (a) Using τ = =RCLR

, we get RLC

= =×

= × =−3 00

1 00 10 1 0063.

. . H

3.00 10 F kΩ Ω .

(b) τ = = × × = × =− −RC 1 00 10 3 00 10 3 00 10 3 003 6 3. . . . F s msΩe je jP32.25 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 = = = × −LR

10 0100

1 00 10 4..

mH

sa fa fΩ

.

For 0 200≤ ≤t sµ , I I e et t= − = −− −max .1 10 0 1 10 000τe j a fe j A s .

At t = 200 sµ , I e= − =−10 00 1 8 652 00. .. A Aa fe j .

Thereafter, it decays exponentially as I I e t= − ′0

τ , so for t ≥ 200 sµ ,

FIG. P32.25

I e e e e et t t t= = = =− − − + − −8 65 8 65 8 65 63 910 000 200 10 000 2 00 2 00 10 000 10 000. . . .. . A A A A s s s s sa f a f e j a fb gµ .

246 Inductance

P32.26 (a) IR

= = =ε 12 0

1 00.

. V

12.0 A

Ω

(b) Initial current is 1.00 A: ∆ ΩV12 1 00 12 00 12 0= =. . . A Va fa f∆ ΩV1 200 1 00 1 200 1 20= =. . A kVa fb g∆VL = 1 21. kV . FIG. P32.26

(c) I I e Rt L= −max :

dIdt

IRL

e Rt L= − −max

and − = = −LdIdt

V I R eLRt L∆ max .

Solving 12 0 1 212 1 212 2 00. . V V= −b ge t

so 9 90 10 3 606. × =− −e t .

Thus, t = 7 62. ms .

P32.27 τ = = =LR

0 1404 90

28 6..

. ms

IRmax

..= = =

ε 6 001 22

V4.90

AΩ

(a) I I e t= − −max 1 τe j so 0 220 1 22 1. .= − −e t τe j

e t− =τ 0 820. : t = − =τ ln . .0 820 5 66a f ms

(b) I I e e= − = − =− −max

. . . .1 1 22 1 1 2210 0 0 028 6 350e j a fe j A A

FIG. P32.27

(c) I I e t= −max

τ and 0 160 1 22. .= −e t τ

so t = − =τ ln . .0 131 58 1a f ms .

P32.28 (a) For a series connection, both inductors carry equal currents at every instant, so dIdt

is the

same for both. The voltage across the pair is

LdIdt

LdIdt

LdIdteq = +1 2 so L L Leq = +1 2 .

(b) LdIdt

LdIdt

LdIdt

VLeq = = =11

22 ∆ where I I I= +1 2 and

dIdt

dIdt

dIdt

= +1 2 .

Thus, ∆ ∆ ∆VL

VL

VL

L L L

eq= +

1 2and

1 1 1

1 2L L Leq= + .

(c) LdIdt

R I LdIdt

IR LdIdt

IReq eq+ = + + +1 1 2 2

Now I and dIdt

are separate quantities under our control, so functional equality requires

both L L L R R Req eqand= + = +1 2 1 2 .

continued on next page

Chapter 32 247

(d) ∆V LdIdt

R I LdIdt

R I LdIdt

R I= + = + = +eq eq 11

1 1 22

2 2 where I I I= +1 2 and dIdt

dIdt

dIdt

= +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 be equivalent to the pair of real inductors forall other currents as well.

Section 32.3 Energy in a Magnetic Field

P32.29 LN

IB= =

×=

−Φ 200 3 70 10

1 7542 3

4.

..

e j mH so U LI= = =

12

12

0 423 1 75 0 064 82 2. . . H A Ja fa f .

P32.30 (a) The magnetic energy density is given by

µµ

= =× ⋅

= ×−

B2

0

2

66

24 50

2 1 26 108 06 10

.

..

T

T m A J m3a f

e j.

(b) The magnetic energy stored in the field equals u times the volume of the solenoid (thevolume in which B is non-zero).

U uV= = × =8 06 10 0 260 0 031 0 6 326 2. . . . J m m m kJ3e j a f b gπ

P32.31 LN A

= =×L

NMOQP =

−

µ µπ

µ0

2

0

2 2 268 0 0 600 10

0 080 08 21

. .

..

a f e j H

U LI= = × =−12

12

8 21 10 0 770 2 442 6 2. . . H A Je ja f µ

P32.32 (a) U LI LR

LR

= = FHGIKJ = = =

12

12 2 8

0 800 500

8 30 027 82

2 2

2

2

2ε ε .

..

a fa fa f J

(b) IR

e R L t= FHGIKJ − −ε

1 b g soε ε

21

12R R

e eR L t R L t= FHGIKJ − → =− −b g b g

RL

t = ln 2 so tLR

= = =ln.

.ln .2

0 80030 0

2 18 5 ms

P32.33 uE

=∈ =0

2

244 2. nJ m3 u

B= =

2

02995

µµ J m3

*P32.34 e dtLR

eRdtL

LR

eLR

e eLR

LR

Rt L Rt L Rt L−∞

−∞

− ∞ −∞z z= −−FHG

IKJ = − = − − = − =2

0

2

0

20

0

22

2 2 20 1

2e j a f

248 Inductance

P32.35 (a) U LI= =12

12

4 00 0 5002 2. . H Aa fa f U = 0 500. J

(b) When the current is 1.00 A,Kirchhoff’s loop rule reads + − − =22 0 1 00 5 00 0. . . V A a fa fΩ ∆VL .Then ∆VL = 17 0. V .The power being stored in the inductor is

I VL∆ = =1 00 17 0 17 0. . . A V Wa fa f .FIG. P32.35

(c) P = =I V∆ 0 500 22 0. . A Va fa f P = 11 0. W

P32.36 From Equation 32.7, IR

e Rt L= − −ε1e j .

(a) The maximum current, after a long time t , is IR

= =ε

2 00. A .

At that time, the inductor is fully energized and P = = =I V∆a f a fa f2 00 10 0 20 0. . . A V W .

(b) Plost A W= = =I R2 22 00 5 00 20 0. . .a f a fΩ

(c) Pinductor drop= =I V∆e j 0

(d) ULI

= = =2 2

210 0 2 00

220 0

. ..

H A J

a fa f

P32.37 We have uE

=∈0

2

2and u

B=

2

02µ.

Therefore ∈ =0

2 2

02 2E B

µso B E2

0 0=∈ µ 2

B E= ∈ =×

×= × −

0 0

5

836 80 10

3 00 102 27 10µ

..

. V m m s

T .

P32.38 The total magnetic energy is the volume integral of the energy density, uB

=2

02µ.

Because B changes with position, u is not constant. For B BRr

= FHGIKJ0

2

, uB R

r=FHGIKJFHGIKJ

02

0

4

2µ.

Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thicknessdr. Such a shell has a volume 4 2π r dr , so the energy stored in it is

dU u r drB R dr

r= =

FHG

IKJ4

22 02 4

02π

πµe j .

We integrate this expression for r R= to r = ∞ to obtain the total magnetic energy outside thesphere. This gives

UB R

= =× ×

× ⋅= ×

−

−

2 2 5 00 10 6 00 10

1 26 102 70 100

2 3

0

5 2 6 3

618π

µ

π . .

..

T m

T m A J

e j e je j

.

Chapter 32 249

Section 32.4 Mutual Inductance

P32.39 I t I e tt1 a f = −

maxα ωsin with Imax .= 5 00 A , α = −0 025 0 1. s , and ω = 377 rad s

dIdt

I e t tt1 = − +−max sin cosα α ω ω ωb g .

At t = 0 800. s ,dIdt

e1 0 020 05 00 0 025 0 0 800 377 377 0 800 377= − +−. . sin . cos .. A sb g b g a fc h a fc hdIdt

1 31 85 10= ×. A s .

Thus, ε 21= −M

dIdt

: MdI dt

=−

=+×

=ε 2

1

3 201 73

..

V1.85 10 A s

mH3 .

P32.40 ε 21 4 41 00 10 1 00 10 1 000= − = − × ×−M

dIdt

t. . cos H A se je j b g

ε 2 1 00b gmax.= V

P32.41 MdI dt

= = =ε 2

1

96 080 0

..

mV1.20 A s

mH

P32.42 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at

distance x from the wire is BIx

=µπ

0

2, and this field passes perpendicularly through the plane of the

loop. The flux through the loop is

ΦB d BdA B dxI dx

xI

= ⋅ = = = = FHGIKJz z z zB A a f µ

πµ

π0

0 400

0

2 21 70

0 400.

ln.. mm

1.70 mm

.

The mutual inductance between the wire and the loop is then

MN

IN I

IN

M

= = FHGIKJ = =

× ⋅ ×

= × =

− −

−

2 12

1

2 0 2 07 3

10

21 70

0 400 21 45

1 4 10 2 70 10

21 45

7 81 10 781

Φ µπ

µπ

π

πln

..

..

.

.

a f e je j a f T m A m

H pH

P32.43 (a) MN

IB BA

A= =

×=

−Φ 700 90 0 10

3 5018 0

6.

..

e j mH

(b) LIA

A

A= =

×=

−Φ 400 300 10

3 5034 3

6e j.

. mH

(c) ε BAM

dIdt

= − = − = −18 0 0 500 9 00. . . mH A s mVa fb g

250 Inductance

*P32.44 The large coil produces this field at the center of the small coil: N I R

x R

1 0 1 12

212 3 2

2

µ

+e j. The field is normal to

the area of the small coil and nearly uniform over this area, so it produces flux

Φ121 0 1 1

2

212 3 2 2

2

2=

+

N I R

x RR

µπ

e j through the face area of the small coil. When current I1 varies, this is the

emf induced in the small coil:

εµ π πµ

2 21 0 1

222

212 3 2 1

1 2 0 12

22

212 3 2

1 1

2 2= −

+= −

+= −N

ddt

N R R

x RI

N N R R

x R

dIdt

MdIdte j e j

so MN N R R

x R=

+

1 2 0 12

22

212 3 2

2

πµ

e j.

P32.45 With I I I= +1 2 , the voltage across the pair is:

∆V LdIdt

MdIdt

LdIdt

MdIdt

LdIdt

= − − = − − = −11 2

22 1

eq .

So, − = +dIdt

VL

ML

dIdt

1

1 1

2∆

and − + + =LdIdt

M VL

ML

dIdt

V22

1

2

1

2∆∆

a f(a) (b)

FIG. P32.45

− + = −L L MdIdt

V L M1 22 2

1e j b g∆ . [1]

By substitution, − = +dIdt

VL

ML

dIdt

2

2 2

1∆

leads to − + = −L L MdIdt

V L M1 22 1

2e j b g∆ . [2]

Adding [1] to [2], − + = + −L L MdIdt

V L L M1 22

1 2 2e j b g∆ .

So, LV

dI dtL L M

L L Meq = − =−

+ −∆ 1 2

2

1 2 2.

Section 32.5 Oscillations in an LC Circuit

P32.46 At different times, U UC Lb g b gmax max= so

12

12

2 2C V LI∆a fLNM

OQP = FHG

IKJmax max

ICL

Vmax max.

. .= =××

=−

−∆a f a f1 00 1040 0 0 400

6 F10.0 10 H

V A3 .

Chapter 32 251

P32.4712

12

2 2C V LI∆a fLNM

OQP = FHG

IKJmax max

so ∆VLC

ICb g a fmax max.

.. .= =

××

=−

−20 0 100 500 10

0 100 20 03

6 H F

A V

P32.48 When the switch has been closed for a long time, battery, resistor, and

coil carry constant current IRmax =ε

. When the switch is opened,

current in battery and resistor drops to zero, but the coil carries thissame 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 12

12

2 2C V LI∆a f = max . FIG. P32.50

Then, LC V

I

C V R= = =

×=

−∆ ∆ Ωa f a f e ja f a f

a f2

2

2 2

2

6 2 2

2

0 500 10 150 250

50 00 281

max

.

..

ε

F V

V H .

P32.49 This radio is a radiotelephone on a ship, according to frequency assignments made by internationaltreaties, laws, and decisions of the National Telecommunications and Information Administration.

The resonance frequency is fLC0

12

=π

.

Thus, Cf L

= =× ×

=−

1

2

1

2 6 30 10 1 05 10608

02 6 2 6π πb g e j e j. . Hz H

pF .

P32.50 fLC

=1

2π: L

f C= =

×=

−

1

2

1

2 120 8 00 100 2202 2 6π πb g a f e j.. H

P32.51 (a) fLC

= =×

=−

12

1

2 0 082 0 17 0 10135

6π π . . H F Hz

b ge j

(b) Q Q t= = × =max cos cos .ω µ µ180 847 0 001 00 119 C Cb g b g

(c) IdQdt

Q t= = − = − = −ω ωmax sin sin .847 180 0 847 114a fa f a f mA

P32.52 (a) fLC

= =×

=−

12

1

2 0 100 1 00 10503

6π π . . H F Hz

a fe j

(b) Q C= = × =−ε µ1 00 10 12 0 12 06. . . F V Ce ja f

(c)12

12

2 2C LIε = max

ICLmax .= =

×=

−

ε 12 37 96

V1.00 10 F

0.100 H mA

FIG. P32.52

(d) At all times U C= = × =−12

12

1 00 10 12 0 72 02 6 2ε µ. . . F V Je ja f .

252 Inductance

P32.53 ω = =×

= ×−

1 1

3 30 840 101 899 10

12

4

LC ..

H F rad s

a fe jQ Q t= max cosω , I

dQdt

Q t= = −ω ωmax sin

(a) UQ

CC = =× × ×

×=

− −

−

26 4 3

2

122

105 10 1 899 10 2 00 10

2 840 106 03

cos . ..

rad s s J

e je je je j

(b) U LI L Q tQ t

CL = = =12

12 2

2 2 2 22 2

ω ωω

maxmaxsin

sinb g b g

UL =× × ×

×=

− −

−

105 10 1 899 10 2 00 10

2 840 100 529

6 2 2 4 3

12

C rad s s

F J

e j e je je j

sin . ..

(c) U U UC Ltotal J= + = 6 56.

Section 32.6 The RLC Circuit

P32.54 (a) ω d LCRL

= − FHGIKJ =

× ×−

×

FHGG

IKJJ = ×

− − −

12

1

2 20 10 1 80 10

7 60

2 2 20 101 58 10

2

3 6 3

2

4

. .

.

..

e je j e j rad s

Therefore, fdd= =

ωπ2

2 51. kHz .

(b) RL

Cc = =4

69 9. Ω

P32.55 (a) ω 06

1 1

0 500 0 100 104 47= =

×=

−LC . ..

a fe j krad s

(b) ω d LCRL

= − FHGIKJ =

12

4 362

. krad s

(c)∆ωω 0

2 53%= . lower

P32.56 Choose to call positive current clockwise in Figure 32.21. It drains charge from the capacitor

according to IdQdt

= − . A clockwise trip around the circuit then gives

+ − − =QC

IR LdIdt

0

+ + + =QC

dQdt

R Lddt

dQdt

0 , identical with Equation 32.28.

Chapter 32 253

*P32.57 The period of damped oscillation is Td

=2πω

. After one oscillation the charge returning to the

capacitor is Q Q e Q eRT L R L d= =− −max max

2 2 2π ω . The energy is proportional to the charge squared, so

after one oscillation it is U U e UR L d= =−0

200 99π ω . . Then

e

L

L LLC

RL

LC

LC

R L

d

d

d2

2

2

1 2

62

6

10 99

2 21 010 1 0 001 005

2 21 250

14

1 563 102

4

1 563 10

π ω

πω

ωπ

=

= =

= = = −FHG

IKJ

× = −

= ×

.

ln . .

.

.

0.001 005

2

2

Ω

ΩΩ

ΩΩ

Ω

b g

a f

We are also given

ω π

π

= × =

=×

= × −

2 101

1

2 102 533 10

3

3 28

s

s s2

LC

LCe j

.

Solving simultaneously,

C L

LL

C C

= ×

×= × =

=×

= =

−

−

−

2 533 10

2 533 101 563 10 0 199

2 533 100 199

127

8

2

86

8

.

.. .

..

s

s H

s H

nF

2

22

2

Ω

P32.58 (a) Q Q e tRt Ld= −

max cos2 ω so I e Rt Lmax ∝ − 2

0 500 2. = −e Rt L andRtL2

0 500= − ln .a f

tL

RL

R= − = F

HGIKJ

20 500 0 693

2ln . .a f

(b) U Q02∝ max and U U= 0 500 0. so Q Q Q= =0 500 0 707. .max max

tL

RL

R= − = F

HGIKJ

20 707 0 347

2ln . .a f (half as long)

254 Inductance

Additional Problems

*P32.59 (a) Let Q represent the magnitude of the opposite charges on the plates of a parallel platecapacitor, 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 EQ

AQ

AE = ∈ = ∈∈

=∈

12

12 20

20

2

02 2

2

02 . Modeling this as a negative or inward pressure, we

have for the force on one plate F PAQ

A= =

∈

2

022

, in agreement with our first analysis.

(b) The lower of the two current sheets shown creates

above it magnetic field B k= −µ 0

2Js e j . Let and w

represent the length and width of each sheet. Theupper sheet carries current J ws and feels force

F B i k j= × = × − =I J wJ w J

ss sµ µ0 0

2

2 2e j .

The force per area is PFw

Js= =µ 0

2

2.

y

x

z

Js

Js

FIG. P32.59(b)

(c) Between the two sheets the total magnetic field is µ µ

µ0 002 2

J JJs s

s− + − =k k ke j e j , with

magnitude B Js= µ 0 . Outside the space they enclose, the fields of the separate sheets are

in opposite directions and add to zero .

(d) u BJ J

Bs s= = =

12 2 20

2 02 2

0

02

µµ

µµ

(e) This energy density agrees with the magnetic pressure found in part (b).

P32.60 With Q Q= max at t = 0 , the charge on the capacitor at any time is Q Q t= max cosω where ω =1LC

.

The energy stored in the capacitor at time t is then

UQ

CQ

Ct U t= = =

2 22

02

2 2max cos cosω ω .

When U U=14 0 , cosω t =

12

and ω πt =13

rad .

Therefore,tLC

=π3

ortLC

2 2

9=

π.

The inductance is then: LtC

=9 2

2π.

Chapter 32 255

P32.61 (a) ε L LdIdt

d tdt

= − = − = −1 0020 0

20 0..

. mH mVa f a f

(b) Q Idt t dt tt t

= = =z z0 0

220 0 10 0. .a f

∆VQ

Ct

tC =−

=−

×= −−

10 01 00 10

10 02

62.

..

F MV s2e j

(c) When Q

CLI

22

212

≥ , or −

×≥ ×

−−

10 0

2 1 00 10

12

1 00 10 20 02 2

63 2.

.. .

tt

e je j e ja f ,

then 100 400 104 9 2t t≥ × −e j . The earliest time this is true is at t = × =−4 00 10 63 29. . s sµ .

P32.62 (a) ε L LdIdt

Lddt

Kt LK= − = − = −a f

(b) IdQdt

= , so Q Idt Ktdt Ktt t

= = =z z0 0

212

and ∆VQ

CKt

CC =−

= −2

2

(c) When 12

12

2 2C V LIC∆b g = ,12 4

12

2 4

22 2C

K tC

L K tFHGIKJ = e j

Thus t LC= 2

P32.6312

12 2

12

2 22Q

C CQ

LI= FHGIKJ + so I

QCL

=34

2

.

The flux through each turn of the coil is ΦBLIN

QN

LC

= =2

3

where N is the number of turns.

P32.64 BNIr

=µ

π0

2

(a) ΦBa

b

a

b

BdANIr

hdrNIh dr

rNIh b

a= = = = F

HGIKJz z zµ

πµ

πµ

π0 0 0

2 2 2ln

LN

IN h b

aB= = F

HGIKJ

Φ µπ

02

2ln

(b) L = FHGIKJ =

µπ

µ02500 0 010 0

212 010 0

91 2a f b g.

ln..

. H

FIG. P32.64

(c) LN A

Rappx

2 m H= F

HGIKJ =

×FHG

IKJ =

−µπ

µπ

µ02

02 4

25002

2 00 100 110

90 9a f .

.. , only 0.3% different.

256 Inductance

P32.65 (a) At the center, BN IR

R

N IR

=+

=µ µ0

2

2 2 3 20

2 0 2e j.

So the coil creates flux through itself ΦB BAN I

RR N IR= = ° =cos cosθ

µπ

πµ0 2

020

2.

When the current it carries changes, επ

µLBN

ddt

N N RdIdt

LdIdt

= − ≈ − FHGIKJ = −

Φ2 0

so L N R≈π

µ2

20 .

(b) 2 3 0 3π r = . ma f so r ≈ 0 14. m

L

L

≈ × ⋅ = ×− −ππ

21 4 10 0 14 2 8 10

100

2 7 7e je ja f T m A m H

nH

. .

~

(c)LR

=× ⋅

= ×−

−2 8 10270

1 0 107

9..

V s A V A

sLR

~1 ns

P32.66 (a) If unrolled, the wire forms the diagonal of a0.100 m (10.0 cm) rectangle as shown. The length ofthis rectangle is

′ = −L 9 80 0 1002 2. . m ma f a f . ′ L

0.100 m9.80 m

FIG. P32.66(a)

The mean circumference of each turn is C r= ′2π , where ′ =+

r24 0 0 644

2. .

mm is the mean

radius of each turn. The number of turns is then:

NLC

=′

=−

+ ×=−

9 80 0 100

2 24 0 0 644 2 10127

2 2

3

. .

. .

m m

m

a f a fa fπ

.

(b) RA

= =× ⋅

×=

−

−

ρ

π

1 70 10 10 0

0 322 100 522

8

3 2

. .

..

m m

m

ΩΩ

e ja fe j

(c) LN A L

Cr=

′=

′′FHGIKJ ′

µ µπ

20

22800 a f

L

L

=× −

+ ×

F

HGG

I

KJJ

+FHG

IKJ ×

LNM

OQP

= × =

−

−−

−

800 4 10

0 100

9 80 0 100

24 0 0 644 1024 0 0 644

210

7 68 10 76 8

7 2 2

3

2

32

2

π

ππ

e j a f a fa f.

. .

. .. .

. .

m

m m

m m

H mH

Chapter 32 257

P32.67 From Ampere’s law, the magnetic field at distance r R≤ is found as:

B r J rIR

r2 02

0 22π µ π µ

ππb g e j e j= =

FHGIKJ , or B

IrR

=µπ

022

.

The magnetic energy per unit length within the wire is then

U Brdr

IR

r drIR

R IR R

= = =FHGIKJ =z z2

00

02

43

0

02

4

40

2

22

4 4 4 16µπ

µπ

µπ

µπ

b g .

This is independent of the radius of the wire.

P32.68 The primary circuit (containing the battery and solenoid) is an RLcircuit with R = 14 0. Ω , and

LN A

= =× ×

=− −

µ π0

2 7 2 44 10 12 500 1 00 10

0 070 00 280

e jb g e j.

.. H .

(a) The time for the current to reach 63.2% of the maximum valueis the time constant of the circuit:

τ = = = =LR

0 2800 020 0 20 0

.. .

H14.0

s msΩ

.

(b) The solenoid’s average back emf is ε LfL

It

LI

t= FHGIKJ =

−FHGIKJ

∆∆ ∆

0 FIG. P32.68

where I IV

Rf = = FHGIKJ =

FHG

IKJ =0 632 0 632 0 632

60 02 71. . .

..max

∆Ω V

14.0 A .

Thus, ε L =FHG

IKJ =0 280

2 7137 9.

.. H

A0.020 0 s

Va f .

(c) The average rate of change of flux through each turn of the overwrapped concentric coil isthe same as that through a turn on the solenoid:

∆Φ∆

∆∆

B

tn I A

t= =

× ⋅ ×

=

− −µ π

07 44 10 12 500 0 070 0 2 71 1 00 10

0 020

3 04

a f e jb ga fe j T m A m A m

0 s

mV

2. . .

.

.

(d) The magnitude of the average induced emf in the coil is ε LBN

t= FHG

IKJ

∆Φ∆

and magnitude of

the average induced current is

IR

NR t

L B= = FHGIKJ = × = =−ε ∆Φ

∆ Ω820

24 03 04 10 0 104 1043

.. .

V A mAe j .

258 Inductance

P32.69 Left-hand loop: ε − + − =I I R I R2 1 2 2 0b g .

Outside loop: ε − + − =I I R LdIdt2 1 0b g .

Eliminating I2 gives ′ − ′ − =ε IR LdIdt

0 .

This is of the same form as Equation 32.6, so its solution is of the sameform as Equation 32.7:

FIG. P32.69

I tR

e R t La f e j=′′

− − ′ε1 .

But ′ =+

RR R

R R1 2

1 2 and ′ =

+ε

εRR R

2

1 2, so

′′

=+

+=

ε ε εR

R R RR R R R R

2 1 2

1 2 1 2 1

b gb g .

Thus I tR

e R t La f e j= − − ′ε

11 .

P32.70 When switch is closed, steady current I0 = 1.20 A . When the switch isopened after being closed a long time, the current in the right loop is

I I e R t L= −0

2

so eII

Rt L = 0 andRtL

II

= FHGIKJln 0 .

Therefore, LR tI I

= = = =2

0

1 00 0 1501 20

0 095 6 95 6ln

. .ln .

. .b ga fa fb g

s A 0.250 A

H mHΩ

.FIG. P32.70

P32.71 (a) While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of thecircuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outerloop of the circuit. Applying Kirchhoff’s loop rule to this loop gives:

+ − + × × =

+ =

−ε

ε

03 3

0

2 00 6 00 10 9 00 10 0

72 0

. . .

.

a f e j A

V with end at the higher potential

Ω

b

(b)

FIG. P32.71(b)

(c) After the switch is opened, the current around the outer loop decays as

I I e Rt L= −max 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:

tLR

II

= FHGIKJFHGIKJ = ×FHG

IKJFHGIKJ = × =−ln

.ln

.

.. .max 0 400 9 00

2 007 52 10 75 25 H

8.00 10 s s3 Ω

µ .

Chapter 32 259

P32.72 (a) The instant after the switch is closed, the situationis as shown in the circuit diagram of Figure (a). Therequested quantities are:

I IR

IR

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, thesteady-state conditions shown in Figure (b) willexist. The currents and voltages are:

I I I

V V V

L C R

L C R

= = =

= = =

0 0 0

0 00

, ,

, ,∆ ∆ ∆ε+ –

+ –IL = 0 VL=∆ 0

IR = 0

VR =∆ 0Q = ε 0C

VC=∆ ε 0

ε 0

Figure (b)

+ –

+ –IL = 0 VL=∆ ε 0

VR =∆Q = 0

VC=∆ 0

ε 0

Figure (a)

ε 0

IR =ε 0 /R

IC =ε 0 /R

FIG. P32.72

P32.73 When the switch is closed, as shown infigure (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 initialcurrent in the inductor remains at 0.267 A.

IR V= ∆ : 0 267 80 0. . A Va fR ≤

R ≤ 300 Ω(a) (b)

FIG. P32.73

P32.74 (a) LN A

10 1

2

1

7 2 44

4 10 1 000 1 00 10

0 5002 51 10 251= =

× ⋅ ×= × =

− −−µ π

µ T m A m

m H H

2e jb g e j.

..

(b) MN

IN

IN BA

I

N N I A

IN N A

= = = = =2 2

1

2 1

1

2

1

2 0 1 1 1

1

0 1 2

1

Φ Φ µ µb g

M =× ⋅ ×

= × =− −

−4 10 1 000 100 1 00 10

0 5002 51 10 25 1

7 45

πµ

T m A m

m H H

2e jb ga fe j.

.. .

(c) ε 12= −M

dIdt

, or I R MdIdt1 1

2= − and IdQdt

MR

dIdt1

1

1

2= = −

QMR

dIMR

I IMR

IMIR

t

f i ii

f

11

20 1

2 21

22

10= − = − − = − − =z d i b g

Q1

58

2 51 10 1 002 51 10 25 1=

×= × =

−−

. .. .

H A

1 000 C nC

e ja fΩ

260 Inductance

P32.75 (a) It has a magnetic field, and it stores energy, so LU

I=

22 is non-zero.

(b) Every field line goes through the rectangle between the conductors.

(c) Φ = LI so LI I

BdAy a

w a

= ==

−

zΦ 1

LI

xdyIy

Iw y I

Ixy

dyx

ya

w a

a

w a

= +−

FHG

IKJ = =

− −

z z12 2

22

22

0 0 0 0µπ

µπ

µπ

µπb g ln .

Thus Lx w a

a=

−FHGIKJ

µπ0 ln .

P32.76 For an RL circuit,

I t I e R L ta f b g= −max :

I tI

eRL

tR L ta f b gmax

= − = ≅ −− −1 10 19

RL

t = −10 9 so Rmax

.

. ..=

×

×= ×

− −−

3 14 10 10

2 50 3 16 103 97 10

8 9

725e je j

b ge j yr s yr Ω .

(If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm2 , its resistancewould be at least 10 6− Ω ).

P32.77 (a) U LIB = = × = ×12

12

50 0 50 0 10 6 25 102 3 2 10. . . H A Ja fe 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 BIr

=µπ0

2.

This causes a force on the next wire of F I B= sinθ

giving F IIr

Ir

= ° =µπ

µπ

0 02

290

2sin .

Evaluating the force, F = ××

=−4 101 00 50 0 10

2 0 2502 0007

3 2

ππ

N A m A

m N2e j

a fe ja f

. .

..

Chapter 32 261

P32.78 P = I V∆ IV

= =××

= ×P

∆1 00 10

5 00 109

3..

W200 10 V

A3

From Ampere’s law, B r I2 0π µb g = enclosed or BI

r=

µπ

0

2enclosed .

(a) At r a= = 0 020 0. m , Ienclosed A= ×5 00 103.FIG. P32.73

and B =× ⋅ ×

= =−4 10 5 00 10

2 0 020 00 050 0 50 0

7 3π

π

T m A A

m T mT

e je jb g

.

.. . .

(b) At r b= = 0 050 0. m , I Ienclosed A= = ×5 00 103.

and B =× ⋅ ×

= =−4 10 5 00 10

2 0 050 00 020 0 20 0

7 3π

π

T m A A

m T mT

e je jb g

.

.. . .

(c) U udVB r r dr I dr

rI b

aa

b

a

b

= = = = FHGIKJz z za f b g2

0

02

022

2 4 4

πµ

µπ

µπ

ln

U =× ⋅ × × F

HGIKJ = × =

−4 10 5 00 10 1 000 10

45 00

2 29 10 2 297 3 2 3

6π

π

T m A A m cm2.00 cm

J MJe je j e j.

ln.

. .

(d) The magnetic field created by the inner conductor exerts a force of repulsion on the currentin the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a smallrectangular section of the outer cylinder of length and width w.

It carries a current of 5 00 102 0 050 0

3..

×FHG

IKJ A

me j b gw

π

and experiences an outward force

F I Bw

= =×

× °−sin.

.. sin .θ

π

5 00 10

2 0 050 020 0 10 90 0

33

A

m T

e jb g e j .

The pressure on it is PFA

Fw

= = =× ×

=−5 00 10 20 0 10

2 0 050 0318

3 3. .

.

A T

m Pa

e je jb gπ

.

262 Inductance

P32.79 (a) BNI

= =× ⋅

= ×−

−µ π0

73

4 10 1 400 2 00

1 202 93 10

T m A A

m T upward

e jb ga f b g.

..

(b) uB

= =×

× ⋅=

⋅FHG

IKJ = =

−

−

2

0

3 2

72

2 93 10

2 4 103 42

13 42 3 42

µ π

.. . .

T

T m A J m

N m1 J

N m Pa3 2e je j e j

(c) To produce a downward magnetic field, the surface of the superconductor must carry aclockwise current.

(d) The vertical component of the field of the solenoid exerts an inward force on thesuperconductor. 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 . 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= = ×LNM

OQP = ×− −3 42 1 10 10 1 30 102 2 3. . . Pa m Na f e jπ

Note that we have not proven that energy density is pressure. In fact, it is not in some cases;Equation 21.2 shows that the pressure is two-thirds of the translational energy density in anideal gas.

ANSWERS TO EVEN PROBLEMS

P32.2 1 36. Hµ P32.26 (a) 1 00. A; (b) ∆V12 12 0= . V ,∆V1 200 1 20= . kV, ∆VL = 1 21. kV ; (c)7 62. msP32.4 240 nWb

P32.28 (a), (b), and (c) see the solution;P32.6 19 2. Wbµ(d) yes; see the solution

P32.8 (a) 360 mV ; (b) 180 mV; (c) t = 3 00. sP32.30 (a) 8 06. MJ m3 ; (b) 6 32. kJ

P32.10 (a) 15 8. Hµ ; (b) 12 6. mHP32.32 (a) 27 8. J; (b) 18 5. ms

P32.12 see the solution P32.34 see the solution

P32.14 1 92. Ω P32.36 (a) 20 0. W; (b) 20 0. W; (c) 0;(d) 20 0. J

P32.382

2 70 1002 3

0

18πµB R

= ×. JP32.16 see the solution

P32.18 92 8. VP32.40 1 00. V

P32.20 30 0. mHP32.42 781 pH

P32.22 7 67. mH

P32.44 MN N R R

x R=

+

1 2 0 12

22

212 3 2

2

πµ

e jP32.24 (a) 1 00. kΩ ; (b) 3 00. ms

Chapter 32 263

P32.46 400 mA P32.64 (a) see the solution; (b) 91 2. Hµ ;(c) 90.9 µH , 0.3% smaller

P32.48 281 mH

P32.66 (a) 127; (b) 0 522. Ω; (c) 76 8. mHP32.50 220 mH

P32.68 (a) 20.0 ms; (b) 37.9 V; (c) 3.04 mV;P32.52 (a) 503 Hz ; (b) 12 0. Cµ ; (c) 37 9. mA ; (d) 104 mA

(d) 72 0. JµP32.70 95.6 mH

P32.54 (a) 2 51. kHz; (b) 69 9. Ω

P32.72 (a) I IR

IRL C R= = =0 0 0, ,

ε ε,

∆ ∆ ∆V V VL C R= = =ε ε0 00, , ;P32.56 see the solution

P32.58 (a) 0 6932

.L

RFHGIKJ ; (b) 0 347

2.

LRFHGIKJ

(b) I I IL C R= = =0 0 0, , ,∆ ∆ ∆V V VL C R= = =0 00, ,ε

P32.609 2

2tCπ

P32.74 (a) 251 Hµ ; (b) 25 1. Hµ ; (c) 25 1. nC

P32.76 3 97 10 25. × − Ω

P32.62 (a) ε L LK= − ; (b) ∆VKtCc =

− 2

2; P32.78 (a) 50.0 mT; (b) 20.0 mT; (c) 2.29 MJ;

(d) 318 Pa(c) t LC= 2