Faraday’s Law and Inductance

635

Faraday’s Law andInductance

CHAPTER OUTLINE 23.1 Faraday’s Law of Induction 23.2 Motional emf 23.3 Lenz’s Law 23.4 Induced emfs and Electric

Fields 23.5 Self-Inductance 23.6 RL Circuits 23.7 Energy Stored in a

Magnetic Field 23.8 Context ConnectionThe

Repulsive Model for Magnetic Levitation

ANSWERS TO QUESTIONS Q23.1 The magnetic flux is ΦB BA= cosθ . Therefore the flux is maximum

when rB is perpendicular to the loop of wire and zero when there is

no component of magnetic field perpendicular to the loop. The flux is zero when the loop is turned so that the field lies in the plane of its area.

Q23.2 (a) The south pole of the magnet produces an upward magnetic field that increases as the

magnet approaches. The loop opposes change by making its own downward magnetic field; it carries current clockwise, which goes to the left through the resistor.

(b) The north pole of the magnet produces an upward magnetic field. The loop sees decreasing

upward flux as the magnet falls away, and tries to make an upward magnetic field of its own by carrying current counterclockwise, to the right in the resistor.

Q23.3 The force on positive charges in the bar is r r rF v B= ×qe j . If the bar is moving to the left, positive charge

will move downward and accumulate at the bottom end of the bar, so that an electric field will be established upward.

Q23.4 No. The magnetic force acts within the bar, but has no influence on the forward motion of the bar.

Q23.5 By the magnetic force law r r rF v B= ×qe j : the positive charges in the moving bar will flow downward

and therefore clockwise in the circuit. If the bar is moving to the left, the positive charge in the bar will flow upward and therefore counterclockwise in the circuit.

Q23.6 We ignore mechanical friction between the bar and the rails. Moving the conducting bar through the

magnetic field will force charges to move around the circuit to constitute clockwise current. The downward current in the bar feels a magnetic force to the left. Then a counterbalancing applied force to the right is required to maintain the motion.

Q23.7 As water falls, it gains speed and kinetic energy. It then pushes against turbine blades, transferring

its energy to the rotor coils of a large AC generator. The rotor of the generator turns within a strong magnetic field. Because the rotor is spinning, the magnetic flux through its turns changes in time as

ΦB BA t= cosω . Generated in the rotor is an induced emf of ε =−Nd

dtBΦ

. This induced emf is the

voltage driving the current in our electric power lines.

636 Faraday’s Law and Inductance

Q23.8 Yes. The induced eddy currents on the surface of the aluminum will slow the descent of the aluminum. It may fall very slowly.

Q23.9 The increasing counterclockwise

current in the solenoid coil produces an upward magnetic field that increases rapidly. The increasing upward flux of this field through the ring induces an emf to produce clockwise current in the ring. The magnetic field of the solenoid has a radially outward component at each point on the ring. This field component exerts upward force on the current in the ring there. The whole ring feels a total upward force larger than its weight.

rB r

Bsolenoid

rBsolenoid

rBring

rF

rF

FIG. Q23.9 Q23.10 Oscillating current in the solenoid produces an always-changing magnetic field. Vertical flux

through the ring, alternately increasing and decreasing, produces current in it with a direction that is alternately clockwise and counterclockwise. The current through the ring’s resistance produces internal energy at the rate I R2 .

Q23.11 (a) The battery makes counterclockwise current I1

in the primary coil, so its magnetic field rB1 is

to the right and increasing just after the switch is closed. The secondary coil will oppose the change with a leftward field

rB2 , which comes

from an induced clockwise current I2 that goes to the right in the resistor.

(b) At steady state the primary magnetic field is

unchanging, so no emf is induced in the secondary.

(c) The primary’s field is to the right and decreasing

as the switch is opened. The secondary coil opposes this decrease by making its own field to the right, carrying counterclockwise current to the left in the resistor.

rB1

rB1

rB2

rB2

FIG. Q23.11

Q23.12 The motional emf between the wingtips cannot be used to run a light bulb. To connect the light, an

extra insulated wire would have to be run out along each wing, making contact with the wing tip. The wings with the extra wires and the bulb constitute a single-loop circuit. As the plane flies through a uniform magnetic field, the magnetic flux through this loop is constant and zero emf is generated. On the other hand, if the magnetic field is not uniform, a large loop towed through it will generate pulses of positive and negative emf. This phenomenon has been demonstrated with a cable unreeled from the Space Shuttle.

Chapter 23 637

Q23.13 As Section 23.4 describes, the electric field lines form closed loops for an electric field created by a changing magnetic field. If a positively charged object is carried around one of these loops in the direction of

rE , the electric field does positive work on it every step of the way. The work adds up to

a nonzero positive total. This fact proves that the force is nonconservative. The work in fact is

W d q d qddt

d= ⋅ = ⋅ = − ⋅z z zr r r r r rF s E s B A . If going around once extracts energy from the changing

magnetic field, going around several lines is better. This is why generators and transformers have coils with many turns.

Q23.14 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.

Q23.15 The current decreases not instantaneously but over some span of time. The faster the decrease in the

current, the larger will be the emf generated in the inductor. A spark can appear at the switch as it is opened because the self-induced voltage is a maximum at this instant. The voltage can therefore briefly cause dielectric breakdown of the air between the contacts.

Q23.16 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 interconvertability 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.

Q23.17 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.

Q23.18 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 material. 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.


Q23.19 The inductance of the series combination of inductor L1 and inductor L2 is L L M1 2 12+ + , where M12 is the mutual inductance of the two coils. It can be defined as the emf induced in coil two when the current in coil one changes at one ampere per second, due to the magnetic field of coil one producing flux through coil two. The coils can be arranged to have large mutual inductance, as by winding them onto the same core. The coils can be arranged to have negligible mutual inductance, as separate toroids do.

Q23.20 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 23.1 Faraday’s Law of Induction Section 23.3 Lenz’s Law

P23.1 ε = =⋅

=− × ⋅

⋅ ⋅FHG

IKJ

⋅⋅

FHG

IKJ

−∆Φ

∆

∆

∆B

t t

r rB Ae j a fe j2 50 8 00 10

1 001 1

4. .

.

T 0.500 T m

s N s

1 T C m V C

1 N m

2

ε = 1 60. mV and IRloop

mV2.00

mA= = =ε 1 600 800

..

Ω

P23.2 ε θ πθ θ

π= − = −−F

HGIKJ = − × °− °F

HGIKJ

−NBA

tNB r

tf i∆

∆ ∆cos cos cos

. . .cos cos

.2 6 225 0 50 0 10 0 500

180 00 200

T m se j a f

ε = +9 82. mV

P23.3 Noting unit conversions from

r r rF v B= ×q and U qV= , the induced voltage is

ε

ε

θ= −⋅

= − −FHG

IKJ =

+ °

×⋅

⋅ ⋅FHG

IKJ

⋅⋅

FHG

IKJ =

= = =

−Nd

dtN

B At

IR

i

r rB Ae j a fe j0 200 1 60 0 200 0

20 0 101 1

3 200

3 200160

3cos . . cos

.∆

Ω

T m

s N s

1 T C m V CN m

V

V20.0

A

2

P23.4 ε µ µ π= = =d BA

dtnA

dIdt

n rIt

a f0 500 0 5000 0 2

2. .∆∆

(a) IR

n rR

Itring = =ε µ π0 2

2

2∆∆

(b) BI

rn rr R

It

= =µ µ π0

1

02

22

12 4∆∆

(c) The coil’s field points downward, and is increasing, so

Bring points upward .

I

FIG. P23.4

Chapter 23 639

P23.5 (a) d dIx

LdxBΦ = ⋅ =r rB A

µπ

0

2: ΦB

h

h w IL dxx

IL h wh

= = +FHGIKJ

+z µπ

µπ

0 0

2 2ln

(b) ε µπ

µπ

= − = − +FHGIKJ

LNM

OQP = − +F

HGIKJ

LNM

OQP

ddt

ddt

IL h wh

L h wh

dIdt

BΦ 0 0

2 2ln ln

επ

πµ= −

× ⋅ +FHG

IKJ = −

−4 10 1 00

21 00 10 0

1 0010 0 4 80

7 T m A m A s V

e ja f b g.ln

. ..

. .

The long wire produces magnetic flux into the page through the rectangle, shown by the first hand in the figure to the right.

As the magnetic flux increases, the rectangle produces its own magnetic field out of the page, which it does by carrying

counterclockwise current (second hand in the figure).

FIG. P23.5

P23.6 ΦB nI A= µ0b g solenoid

ε

εε

µ π

π π

= − = −

= − × ⋅ ×

= −

− −

Nd

dtN n r

dIdt

t

t

BΦ0

7 3 1 215 0 4 10 1 00 10 0 020 0 600 120

14 2 120

solenoid2

T m A m m A s

mV

e je je j b g b g a fa f

. . . cos

. cos

P23.7 ε = = FHGIKJ = +

∆Φ∆

B

tN

dBdt

A N t A0 010 0 0 080 0. .b g

At t = 5 00. s , ε π= =30 0 0 410 0 040 0 61 82

. . . . T s m mVb g b g

*P23.8 (a) Each coil has a pulse of voltage tending to produce

counterclockwise current as the projectile approaches, and then a pulse of clockwise voltage as the projectile recedes.

(b) vdt

= =×

=−1 50

106253

. m2.40 s

m s

V1 V2

t0

∆V

FIG. P23.8


P23.9 (a) Suppose, first, that the central wire is long and straight. The enclosed current of unknown amplitude creates a circular magnetic field around it, with the magnitude of the field given by Ampère’s Law.

r rB s⋅ =z d Iµ0 : B

I tR

=µ ω

π0

2max sin

at the location of the Rogowski coil, which we assume is centered on the wire. This field passes perpendicularly through each turn of the toroid, producing flux

r rB A⋅ =

µ ωπ

0

2I A t

Rmax sin

.

The toroid has 2π Rn turns. As the magnetic field varies, the emf induced in it is

ε π µπ

ω µ ω ω= − ⋅ = − = −Nddt

RnI A

Rddt

t I nA tr rB A 2

20

0max

maxsin cos .

This is an alternating voltage with amplitude ε µ ωmax max= 0nA I . Measuring the amplitude determines the size Imax of the central current. Our assumptions that the central wire is long and straight and passes perpendicularly through the center of the Rogowski coil are all unnecessary.

(b) If the wire is not centered, the coil will respond to stronger magnetic fields on one side, but

to correspondingly weaker fields on the opposite side. The emf induced in the coil is proportional to the line integral of the magnetic field around the circular axis of the toroid. Ampère’s Law says that this line integral depends only on the amount of current the coil encloses. It does not depend on the shape or location of the current within the coil, or on any currents outside the coil.

*P23.10 The upper loop has area π 0 05 7 85 102 3 2. . m ma f = × − . The induced emf in it is

ε θ= − = − ° = − × = − ×− −Nddt

BA AdBdt

cos cos . .1 0 7 85 10 2 1 57 103 2 m T s V2 b g .

The minus sign indicates that it tends to produce counterclockwise current, to make its own magnetic field out of the page. Similarly, the induced emf in the lower loop is

ε θ π= − = − = − × = + ×− −NAdBdt

cos . . .0 09 2 5 09 10 5 09 102 2 2 m T s V Va f to produce

counterclockwise current in the lower loop, which becomes clockwise current in the upper loop .

The net emf for current in this sense around the figure 8 is

5 09 10 1 57 10 3 52 102 2 2. . .× − × = ×− − − V V V.

It pushes current in this sense through series resistance 2 0 05 2 0 09 3 2 64π π. . . m m m a f a f+ =Ω Ω.

The current is IR

= = × =−ε 3 52 10

13 32.

. V

2.64 mA

Ω.

Chapter 23 641

Section 23.2 Motional emf Section 23.3 Lenz’s Law P23.11 (a) For maximum induced emf, with positive charge at the top of the antenna,

r r rF v B+ += ×q e j, so the auto must move east , since east × north = up.

(b) ε = = × ×FHG

IKJ ° = ×− −B vl 5 00 10 1 20

65 0 1065 0 4 58 105

34. .

.cos . . T m

m3 600 s

Ve ja f

P23.12 IR

B vR

= =ε l

v = 1 00. m s

rFapp

FIG. P23.12 P23.13 (a)

r rl

rlF BB I I B= × =

When IR

= ε

and ε = B vl

we get FB vR

BB v

RB = = = =ll

la f a f a f a f2 2 2 22 50 1 20 2 006 00

3 00. . .

.. N .

The applied force is 3 00. N to the right .

(b) P = = =I RB v

R2

2 2 2

6 00l

. W or P = =Fv 6 00. W

rFapp

FIG. P23.13

P23.14 F I BB = l and ε = B vl

IR

B vR

= =ε l so B

IRv

=l

(a) FI R

vB =2l

l and I

F vRB= = 0 500. A

(b) I R2 2 00= . W

(c) For constant force, P = ⋅ = =

r rF v 1 00 2 00 2 00. . . N m s Wa fb g .


*P23.15 With v representing the initial speed of the bar, let u represent its speed at any later time. The

motional emf induced in the bar is ε = B ul . The induced current is IR

B uR

= =ε l. The magnetic force

on the bar is backward F I BB u

Rmdudt

= − = − =ll2 2

.

Method one: To find u as a function of time, we separate variables thus:

− =

− =

− − = − =

= = =

z z

− −

BRm

dtduu

BRm

dtduu

BRm

t u vuv

euv

u vedxdt

t

v

u

B t Rm B t Rm

2 2

2 2

02 2

0

2 2 2 2

l

l

l

l l

a f ln ln ln

.

The distance traveled is given by

dx ve dt v

RmB

eB dtRm

xRmvB

e eRmvB

xB t Rm B t Rm

0 02 2

2 2

0

2 20

2 2

2 2 2 2

0

max

max .

z z z= = −FHGIKJ − −F

HGIKJ

− = − − =

−∞

−∞

−∞ −

l l

l

l

l l

Method two: Newton’s second law is

− = − =

= −

B uR

BR

dxdt

mdudt

mduB

Rdx

2 2 2 2

2 2

l l

l.

Direct integration from the initial to the stopping point gives

mduB

Rdx

m vB

Rx

xmvRB

v

x0 2 2

02 2

2 2

0 0

z z= −

− = − −

=

l

l

l

max

max

max .

a f b g

*P23.16 Model the magnetic flux inside the metallic tube as constant as it shrinks from radius R to radius r:

2 50

2 50 2 50 12 360

2 2

22

.

. .

T

T T T

π πR B r

BRr

f

f

e ja f

=

= FHGIKJ = =

Chapter 23 643

P23.17 Observe that the homopolar generator has no commutator and produces a voltage constant in time: DC with no ripple. In time dt, the disk turns by angle d dtθ ω= . The outer brush slides over distance rdθ .

The radial line to the outer brush sweeps over area

dA rrd r dt= =12

12

2θ ω .

The emf generated is ε = − ⋅Nddt

r rB A

ε ω= − ° = − FHGIKJ1 0

12

2a fB dAdt

B rcos .

(We could think of this as following from the result of Example 23.3.)

The magnitude of the emf is

rB

FIG. P23.17

ε

ε

ωπ

= FHGIKJ = ⋅ ⋅ LNM

OQPFHG

IKJ

=

B r12

0 912

0 4 3 200260

24 1

2 2. .

. .

N s C m m rev min rad rev s min

V

b g a f b g

A free positive charge q shown, turning with the disk, feels a magnetic force qr rv B× radially

outward. Thus the outer contact is positive .

*P23.18 The speed of waves on the wire is

vT= = ⋅

×=−µ

2672983

N m3 10 kg

m s .

In the simplest standing-wave vibration state,

dNN = =0 642

. mλ

λ = 1 28. m

and fv= = =λ

2981 28

233m sm

Hz.

.

(a) The changing flux of magnetic field through the circuit containing the wire will drive

current to the left in the wire as it moves up and to the right as it moves down. The emf will have this same frequency of 233 Hz .

(b) The vertical coordinate of the center of the wire is described by

x A t t= =cos . cosω π1 5 2 233 cm sa f b g .

Its velocity is vdxdt

t= = − 1 5 2 2 233. sin cm 233 s sa fb g b gπ π .

Its maximum speed is 1 5 2 233 22 0. . cm s m sπa f = .

The induced emf is ε = −B vl , with amplitude

ε max max . . .= = × = ×− −B vl 4 50 10 0 02 22 1 98 103 3 T m m s Va f .


P23.19 ω π π= =2 00 2 4 00. . rev s rad rev rad sb gb g

ε ω= =12

2 832B l . mV

P23.20 (a)

rB iext extB= $ and Bext decreases; therefore, the

induced field is rB i0 0= B $ (to the right) and the

current in the resistor is directed to the right .

(b) rB iext extB= −$e j increases; therefore, the induced

field rB i0 0= +B $e j is to the right, and the current in

the resistor is directed to the right .

(c) rB kext extB= − $e j into the paper and Bext decreases;

therefore, the induced field is rB k0 0= −B $e j into the

paper, and the current in the resistor is directed to the right .

r

r

FIG. P23.20

(d) By the magnetic force law,

r r rF v BB q= ×e j . Therefore, a positive charge will move to the top of

the bar if rB is into the paper .

P23.21 (a) At terminal speed,

Mg F IwB

RwB

BwvR

wBB w v

RBT T= = = FHG

IKJ = FHG

IKJ =ε 2 2

or vMgRBT = 2 2ω

.

(b) The emf is directly proportional to vT , but the

current is inversely proportional to R. A large R means a small current at a given speed, so the loop must travel faster to get F mgB = .

rFB

rBout

rBInduced

rv

FIG. P23.21

(c) At a given speed, the current is directly proportional to the magnetic field. But the force is

proportional to the product of the current and the field. For a small B, the speed must

increase to compensate for both the small B and also the current, so vBT ∝ 1

2 .

Chapter 23 645

P23.22 (a) The force on the side of the coil entering the field (consisting of N wires) is

F N ILB N IwB= =a f a f .

The induced emf in the coil is

ε = = =Nd

dtN

d Bwxdt

NBwvBΦ a f.

so the current is IR

NBwvR

= =ε

counterclockwise.

The force on the leading side of the coil is then:

F NNBwv

RwB

N B w vR

= FHGIKJ =

2 2 2

to the left .

(b) Once the coil is entirely inside the field,

ΦB NBA= = constant ,

so ε = 0 , I = 0 , and F = 0 .

rv

rBin

rB

rF

rF

rB

FIG. P23.22

(c) As the coil starts to leave the field, the flux decreases at the rate Bwv, so the magnitude of the

current is the same as in part (a), but now the current is clockwise. Thus, the force exerted on the trailing side of the coil is:

FN B w v

R=

2 2 2

to the left again .

P23.23 (a) ε ω πmax . . .= = =NAB 1 000 0 100 0 200 120 7 54b ga fa fa f kV

(b) ε ω ω ω θt NBA t NBAa f = =sin sin

ε is maximal when sinθ = 1

or θ π= ±2

so the plane of coil is parallel to rB .

rB

FIG. P23.23

*P23.24 B nI= = × ⋅ = ×− − −µ π07 1 34 10 200 15 0 3 77 10 T m A m A Te je ja f. .

For the small coil, ΦB N NBA t NB r t= ⋅ = =r rB A cos cosω π ω2e j .

Thus, ε π ω ω= − =d

dtNB r tBΦ 2 sin

ε π π π π= × =− −30 0 3 77 10 0 080 0 4 00 4 00 28 6 4 003 2 1. . . . sin . . sin .a fe j b g e j b g a f b g T m s mVt t .


Section 23.4 Induced emfs and Electric Fields

P23.25 dBdt

t= 0 060 0. ε π π= = =d

dtr

dBdt

r EBΦ12

12

At t = 3 00. s ,

Err

dBdt

=FHGIKJ = ⋅

⋅ ⋅FHG

IKJ

ππ

12

120 020 0

0 060 0 3 001.

. . m

2 T s s

N s1 T C m

2e jb g

rE = × −1 80 10 3

1. N C perpendicular to and counterclockwiser

rE

rBin

FIG. P23.25

P23.26 (a) dBdt

t t= −6 00 8 002. . ε =d

dtBΦ

At t = 2 00. s , ER dB dt

r= =

ππ

ππ

2

2

2

2

8 00 0 025 0

2 0 050 0b g b g

b g. .

.

F qE= = × −8 00 10 21. N

(b) When 6 00 8 00 02. .t t− = , at t = 1 33. s and at t = 0 .

rE

rBin

FIG. P23.26

Section 23.5 Self-Inductance

P23.27 ε = = × −FHG

IKJ = × =− −L

It

∆∆

3 00 101 50 0 200

1 95 10 19 53 2.. .

. . H A A0.200 s

V mVe j

P23.28 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

l

T m A m

m H

e ja f e j. .

.. .

P23.29 ε ω ω ω π ω= − = − = − = − × −LdIdt

Lddt

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

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

P23.30 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 .

Chapter 23 647

P23.31 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 .

P23.32 LN

INBA

INA

INIR

N AR

B= = ≈ ⋅ =Φ µ

πµ

π0 0

2

2 2

FIG. P23.32

Section 23.6 RL Circuits

P23.33 (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 εR

e

R

t

=− − s

1

0.5

00 0.2 0.4 0.6

t (s)

I (A)Imax

FIG. P23.33

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 f

and 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 .


P23.34 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= .

P23.35 (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 00

1 50 V

4.00 A

Ω

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

FIG. P23.35

P23.36 (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 0

0 800.

. V

20.0 V.

(b) ∆ ΩV IRR = = =4 50 8 00 36 0. . . A Va fa f

∆ ∆V VL R= − =ε 0

FIG. P23.36

*P23.37 For the increasing current IR

e Lt R= − −ε1e j . The final value is

εR

, so the condition on ∆t is

0 8 1

0 2

5

5

5

.

.

ln

ln

ε εR R

e

e

eL tR

tR

L

L t R

L t R

L t R

= −

=

=

=

=

−

−

+

∆

∆

∆

∆

∆

e j

At the moment when the battery is removed, the current in the coil is quite precisely εR

. During the

decrease, I I eR

eiLt R Lt R= =− −ε

.

(a) at t tR

L= =∆ ln5

, II

ei

L t R= = =− ∆ 0 200 20 0%. .

(b) at t t= 2∆ , II

e ei

L t R L t R= = = = =− −2 2 20 2 0 04 4 00%∆ ∆e j a f. . .

Chapter 23 649

P23.38 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 mH

FIG. P23.38

P23.39 Name the currents as shown. By Kirchhoff’s laws:

+ − − =I I I1 2 3 0 (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. P23.39

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 23.13 and 23.14:

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

P23.40 (a) IR

= = =ε 12 01 00

..

V12.0

AΩ

(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. P23.40

(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 .


P23.41 τ = = =LR

0 1404 90

28 6..

. ms

IRmax

..= = =ε 6 00

1 22 V

4.90 A

Ω

(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. P23.41

(c) I I e t= −

maxτ and 0 160 1 22. .= −e t τ

so t = − =τ ln . .0 131 58 1a f ms .

Section 23.7 Energy Stored in a Magnetic Field P23.42 (a) The magnetic energy density is given by

uB= =

× ⋅= ×

−

2

0

2

66

24 50

2 1 26 108 06 10

µ.

..

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 (the

volume 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

P23.43 LN A= =

×LNM

OQP =

−

µ µπ

µ0

2

0

2 2 268 0 0 600 10

0 080 08 21

l

. .

..

a f e j H

U LI= = × =−12

12

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

P23.44 (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. P23.44

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

Chapter 23 651

P23.45 uE=∈ =0

2

244 2. nJ m3 u

B= =2

02995

µµ J m3

Section 23.8 Context ConnectionThe Repulsive Model for Magnetic Levitation P23.46 The induced emf in the leading edge of the loop is

ε = = FHG

IKJFHG

IKJ =B v B Bl 0 2 400

1 00022 2. . m km h

mkm

h3 600 s

m s2a fb g e j

The induced current is IR

B= =

⋅ε 22 2

25

. m

s

2e jΩ

The magnetic force on the lower section of one loop is r r

lr

F BB I= × :

rFB B B=

⋅FHG

IKJ °22 2

0 1 90.

. sin m A

25 V s m

2

a f = ⋅ ⋅0 0889 2. m C J s 3 2 2e jB up

We require Σ =Fy 0: 100 0F mgB − =

B2 48 89 5 10 9 80. . m C N s N2 2 2⋅ ⋅ = ×e j a f

B = ⋅ ⋅ = ×235 2 102 N s C m T

P23.47 (a) BNI

= =× ⋅

= ×−

−µ π0

73

4 10 1 400 2 00

1 202 93 10

l

T m A A

m T upward

e jb ga f b g.

..

(b) uB= =

×

× ⋅= ⋅F

HGIKJ = =

−

−

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

rB

(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

superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges and has a radially outward horizontal component. This component exerts upward force on the clockwise superconductor current. The total force on the core is upward . You

can think of it as a force of repulsion between the solenoid with its north end pointing up, and the core, with its north end pointing down.

(e) 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 16.13 shows that the pressure is two-thirds of the translational energy density in an ideal gas.


Additional Problems P23.48 (a) Doubling the number of turns.

Amplitude doubles: period unchanged

(b) Doubling the angular velocity.

doubles the amplitude: cuts the period in half

(c) Doubling the angular velocity while reducing the

number of turns to one half the original value.

Amplitude unchanged: cuts the period in half

FIG. P23.48

P23.49 ε θ π= − = − °FHGIKJN

ddt

BA N rdBdt

cos cosa f e j2 0

ε

ε

ε

π π

π π π

π

= − ×LNM

OQP +

= − ×LNM

OQP ×

= − ×

− −

− − − −

− −

30 0 2 70 10 1 50 0 3 20 2 523

30 0 2 70 10 3 20 10 2 523 2 523

7 22 10 2 523

3 2 1

3 2 3 1 1

3 1

. . . . sin

. . . cos

. cos

a f e j a f a f e ja f e j e j e j e j

e j e j

m mT mT s

m T s s

V s

ddt

t

t

t

*P23.50 ε θ π= − = − ° = − −×

FHG

IKJ =−N

tBA N r

Bt

∆∆

∆∆

cos cos .. .

..a f e j e ja f2

30 1 0 005 00 11 50 5 0020 0 10

0 875 mT T

s V2

(a) IR

= = =ε 0 87543 8

..

V0.020 0

AΩ

(b) P = = =ε I 0 875 43 8 38 3. . . V A Wa fa f

*P23.51 Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we

use a bar magnet to produce field 10 3− T through the coil in one direction along its axis. Suppose we then flip the magnet to reverse the flux in 10 1− s . The average induced emf is then

ε

ε

θπ

π

= − = − = − °− °FHG

IKJ

= − −FHG

IKJ

−−

−

Nt

NBA

tNB r

tB∆Φ

∆∆

∆ ∆cos cos cos

. ~

2

3 21

4

180 0

20 10 0 015 02

1010

e j

a fe j b g T ms

V

Chapter 23 653

P23.52 IR

=+ε ε induced and ε induced = − d

dtBAa f

F mdvdt

IBd

dvdt

IBdm

BdmR

dvdt

BdmR

Bvd

= =

= = +

= −

ε ε

εinducedb g

a f

To solve the differential equation, let u Bvd= −ε

dudt

Bddvdt

Bddudt

BdmR

u

= −

− =1

so duu

BdmR

dtu

u t

0

2

0z z= −

a f.

rB

FIG. P23.52

Integrating from t = 0 to t t= , lnu

uBdmR

t0

2

= −a f

or u

ue B d t mR

0

2 2= − .

Since v = 0 when t = 0 , u0 = ε

and u Bvd= −ε

ε ε− = −Bvd e B d t mR2 2.

Therefore, vBd

e B d t mR= − −ε1

2 2e j .

P23.53 The enclosed flux is ΦB BA B r= = π 2 .

The particle moves according to r rF a∑ = m : qvB

mvr

sin902

° =

rmvqB

= .

Then ΦBB m v

q B=

π 2 2

2 2 .

(a) vq Bm

B= =× ⋅ ×

×= ×

− −

−

Φ 2

2

6 9 2

16 25

15 10 30 10 0 6

2 102 54 10

π π

T m C T

kg m s

2e je j a fe j

..

continued on next page


(b) Energy for the particle-electric field system is conserved in the firing process:

U Ki f= : q V mv∆ = 12

2

∆Vmv

q= =

× ×

×=

−

−

2 16 5 2

92

2 10 2 54 10

2 30 10215

kg m s

C V

e je je j

..

P23.54 (a) Consider an annulus of radius r, width dr, height b, and resistivity ρ. Around its

circumference, a voltage is induced according to

ε ω π π ω ω= − ⋅ = − = +Nddt

ddt

B t r B r tr rB A 1 2 2

max maxcos sinb g .

The resistance around the loop is ρ ρ πl

A

r

bdrx=

2b g.

The eddy current in the ring is dIB r t bdr

rB rb dr t

= = =ε π ω ωρ π

ω ωρresistance

max maxsin sin2

2 2b gb g .

The instantaneous power is d dIB r b dr t

iP = =ε π ω ωρ

max sin2 3 2 2

2.

The time average of the function sin cos2 12

12

2ω ωt t= − is 12

012

− =

so the time-averaged power delivered to the annulus is

dB r b dr

P = max2 3 2

4π ω

ρ.

The power delivered to the disk is P P= =z zdB b

r drR

max2 2

3

0 4π ωρ

P = −FHG

IKJ =

B b R B R bmax max2 2 4 2 4 2

4 40

16π ωρ

π ωρ

.

(b) When Bmax gets two times larger, Bmax

2 and P get 4 times larger.

(c) When f and ω π= 2 f double, ω 2 and P get 4 times larger.

(d) When R doubles, R4 and P become 2 164 = times larger.

P23.55 We are given ΦB t t= − ⋅6 00 18 03 2. .e j T m2

and ε = − = − +d

dtt tBΦ

18 0 36 02. . .

Maximum ε occurs when ddt

tε = − + =36 0 36 0 0. .

which gives t = 1 00. s .

Therefore, the maximum current (at t = 1 00. s ) is IR

= =− +

=ε 18 0 36 06 00

. ..

a f V3.00

AΩ

.

Chapter 23 655

*P23.56 (a)

ε θθ θ ω= − = − ° = − = − = −

= − =

Nddt

BAddt

Ba Ba d

dtBacos cos . .

. .

12

02

12

12

0 5 0 5 2

0 125 0 125

2 22 2 T m rad s

V V clockwise

a fa f

The – sign indicates that the induced emf produces clockwise current, to make its own magnetic field into the page.

(b) At this instant θ ω= = =t 2 0 25 0 5 rad s s rad. .a f . The arc PQ has length

rθ = =0 5 0 5 0 25. . . rad m ma fa f . The length of the circuit is 0 5 0 5 0 25 1 25. . . .m m m m+ + = its

resistance is 1 25 6 25. . m 5 m Ω Ωb g = . The current is 0 125

0 020 0.

.V

6.25 A clockwise

Ω= .

P23.57 Suppose the field is vertically down. When an electron is moving away

from you, the force on it is in the direction given by

q cr rv B× as − × = −away downb g = − =left right .

Therefore, the electrons circulate clockwise.

rB

rv

rBc

rF

FIG. P23.57 (a) As the downward field increases, an emf is induced to produce some current that in turn

produces an upward field. This current is directed counterclockwise, carried by

negative electrons moving clockwise. Therefore the original electron motion speeds up.

(b) At the circumference, we have F mac c∑ = : q vBmv

rc sin902

° =

mv q rBc= .

The increasing magnetic field rBav in the area enclosed by the orbit produces a tangential

electric field according to

r r r rE s B A⋅ = − ⋅z d

ddt av E r r

dBdt

av2 2π πb g = Er dB

dtav=

2.

An electron feels a tangential force according to F mat t∑ = : q E mdvdt

= .

Then qr dB

dtm

dvdt

av

2= q

rB mv q rBav c2

= =

and B Bav c= 2 .

P23.58 ε θ= − = −Ndd t

Ndd t

BABΦcosa f

ε θ µ= −FHGIKJ = − × ° ×F

HGIKJ = −−

−NB

At

cos . cos ..

.∆∆

200 50 0 10 62 039 0 10

10 264

T m

1.80 s V

2

e ja f


P23.59 The magnetic field produced by the current in the straight wire is perpendicular to the plane of the coil at all points within the coil. The

magnitude of the field is BIr

=µπ0

2. Thus, the flux linkage is

NNIL dr

rNI L h w

htB

h

h w

Φ = = +FHGIKJ +

+zµπ

µπ

ω φ0 0

2 2max ln sinb g .

Finally, the induced emf is

FIG. P23.59

ε

ε

ε

µ ωπ

ω φ

π ππ

ω φ

π φ

= − +FHGIKJ +

= −×

+FHGIKJ +

= − +

− −

0

7 1

21

4 10 100 50 0 0 200 200

21

5 00

87 1 200

NI L wh

t

t

t

max ln cos

. .ln

.cos

. cos .

b g

e ja fa fa fe j b ga f b g

m s cm5.00 cm

mV

The term sin ω φt +b g in the expression for the current in the straight wire does not change

appreciably when ω t changes by 0.10 rad or less. Thus, the current does not change appreciably during a time interval

∆t < = ×−

−0 10

2001 6 10

14.

.π s

se j

.

We define a critical length, c t∆ = × × = ×−3 00 10 1 6 10 4 8 108 4 4. . . m s s me je j equal to the distance to

which field changes could be propagated during an interval of 1 6 10 4. × − s . This length is so much larger than any dimension of the coil or its distance from the wire that, although we consider the straight wire to be infinitely long, we can also safely ignore the field propagation effects in the vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the vicinity of the coil.

If the angular frequency ω were much larger, say, 200 105 1π × − s , the corresponding critical length would be only 48 cm. In this situation propagation effects would be important and the above expression for ε would require modification. As a “rule of thumb” we can consider field

propagation effects for circuits of laboratory size to be negligible for frequencies, f = ωπ2

, that are

less than about 106 Hz.

P23.60 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

FIG. P23.60


Chapter 23 657

(b) L = FHGIKJ =

µπ

µ02500 0 010 0

212 010 0

91 2a f b g.

ln..

. H

(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.

P23.61 (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 µ π π µ0 2

020

2.

When the current it carries changes, ε π 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

≈ × ⋅ = ×− −π π2

1 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

s LR

~1 ns

P23.62 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 Va fR ≤

R ≤ 300 Ω

(a) (b)

FIG. P23.62


P23.63 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 resistance would be at least 10 6− Ω .)

P23.64 (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 loops

have 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= l sinθ

giving F IIr

Ir

= ° =llµ

πµ

π0 0

2

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

. .

..

P23.65 P = I V∆ IV

= = ××

= ×P∆

1 00 105 00 10

93.

. W

200 10 V A3

From Ampère’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. P23.65

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

.

.. . .


Chapter 23 659

(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

l l lln

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 cm

2.00 cm J MJ

e je j e j.ln

.. .

(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 l 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

= =×

× °−l lsin.

.. 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

= = =× ×

=−

l

5 00 10 20 0 10

2 0 050 0318

3 3. .

.

A T

m Pa

e je jb gπ

.

ANSWERS TO EVEN PROBLEMS P23.2 9.82 mV

P23.4 (a) µ π0 2

2

2n rR

It

∆∆FHGIKJ ; (b)

µ π02

22

14n rr R

It

∆∆FHGIKJ ;

(c) upward P23.6 − 14 2 120. cos mVa f a ft P23.8 (a) see the solution; (b) 625 m/s P23.10 13.3 mA counterclockwise in the lower

loop and clockwise in the upper loop P23.12 1.00 m/s P23.14 (a) 0.500 A; (b) 2.00 W; (c) 2.00 W P23.16 360 T P23.18 (a) 233 Hz; (b) 1.98 mV P23.20 (a) to the right; (b) to the right; (c) to the right; (d) into the paper

P23.22 (a) FN B w v

R=

2 2 2

to the left;

(b) 0;

(c) FN B w v

R=

2 2 2

to the left

P23.24 28 6 4. sin mVa f b gπ t P23.26 (a) 8 00 10 21. × − N, clockwise; (b) 1.33 s and 0 P23.28 1 36. Hµ P23.30 19 2. T m2µ ⋅ P23.32 see the solution P23.34 see the solution P23.36 (a) 0.800; (b) 0 P23.38 7.67 mH


P23.40 (a) 1.00 A; (b) 12.0 V, 1.20 kV, 1.21 kV; (c) 7.62 ms P23.42 (a) 8 06 106. × J m3 ; (b) 6.32 kJ P23.44 (a) 0.500 J; (b) 17.0 W; (c) 11.0 W P23.46 2 102× T P23.48 see the solution (a) doubling N doubles amplitude; (b) doubling ω doubles amplitude and

halves period; (c) doubling ω and halving N leaves the

amplitude the same and cuts the period in half

P23.50 (a) 43.8 A; (b) 38.3 W

P23.52 see the solution

P23.54 (a) π ω

ρB R bmax

2 4 2

16; (b) 4; (c) 4; (d) 16

P23.56 (a) 0.125 V to produce clockwise current; (b) 20.0 mA clockwise P23.58 −10 2. Vµ P23.60 (a) see the solution; (b) 91 2. Hµ ; (c) 90 9. Hµ P23.62 300 Ω P23.64 (a) 62.5 GJ; (b) 2 000 N

CONTEXT 7 CONCLUSION SOLUTIONS TO PROBLEMS CC7.1 The alternating current must complete a cycle of reversing and reversing again as the vehicle travels

forward by 2 10 20 m ma f = . The period of alternation must be

Tx

v= = ⋅ F

HGIKJFHG

IKJ =

∆ 201 000

3 6000 180

m h400 km

km m

sh

s. .

So the frequency is fT

= =15 56. Hz .

CC7.2 The current in the magnet creates an upward magnetic field, so the N and S poles on the

solenoid core are shown correctly. On the rail in front of the brake, the upward flux of rB increases as

the coil approaches, so a current is induced here to create a downward magnetic field. This is

clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic

field by being counterclockwise as the picture correctly shows.

ANSWERS TO EVEN CONTEXT 7 CONCLUSION PROBLEMS CC7.2 both are correct

Faraday’s Law and Inductance

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