635 Faraday’s Law and Inductance 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 Φ B BA = cos θ . Therefore the flux is maximum when r B 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 r F v B = × q e 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 r F v B = × q e 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 Φ B BA t = cos ω . Generated in the rotor is an induced emf of ε = − Nd dt B Φ . This induced emf is the voltage driving the current in our electric power lines.
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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.
638 Faraday’s Law and Inductance
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
640 Faraday’s Law and Inductance
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 .
642 Faraday’s Law and Inductance
*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 .
644 Faraday’s Law and Inductance
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 .
646 Faraday’s Law and Inductance
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 .
648 Faraday’s Law and Inductance
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 .
650 Faraday’s Law and Inductance
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.
652 Faraday’s Law and Inductance
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
654 Faraday’s Law and Inductance
(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
656 Faraday’s Law and Inductance
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
continued on next page
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
658 Faraday’s Law and Inductance
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
.
.. . .
continued on next page
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
660 Faraday’s Law and Inductance
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