Chapter 20
Induced Voltages
and Inductance
Quick Quizzes
1.
b, c, a. At each instant, the magnitude of the induced emf is proportional to the rate of
change of the magnetic field (hence, proportional to the slope of the curve shown on the
graph).
2.
(a). All charged particles within the metal bar move straight downward with the bar.
According to right-hand rule #1, positive changes moving downward through a magnetic
field that is directed northward will experience magnetic forces toward the east. This
means that the free electrons (negative charges) within the metal will experience
westward forces and will drift toward the west end of the bar, leaving the east end with a
net positive charge.
3.
(b). According to Equation 20.3, because B and v are constant, the emf depends only on the
length of the wire moving in the magnetic field. Thus, you want the long dimension
moving through the magnetic field lines so that it is perpendicular to the velocity vector.
In this case, the short dimension is parallel to the velocity vector. From a more conceptual
point of view, you want the rate of change of area in the magnetic field to be the largest,
which you do by thrusting the long dimension into the field.
4.
(c). In order to oppose the approach of the north pole, the magnetic field generated by the
induced current must be directed upward. An induced current directed counterclockwise
around the loop will produce a field with this orientation along the axis of the loop.
5.
(b). When the iron rod is inserted into the solenoid, the inductance of the coil increases. As
a result, more potential difference appears across the coil than before. Consequently, less
potential difference appears across the bulb, and its brightness decreases.
175
176
CHAPTER 20
Answers to Conceptual Questions
2.
Consider the copper tube to be a large set of rings stacked one on top of the other. As the
magnet falls toward or falls away from each ring, a current is induced in the ring. Thus,
there is a current in the copper tube around its circumference.
4.
The flux is calculated as Φ B = BA cosθ = B⊥ A . The flux is therefore maximum when the
magnetic field vector is perpendicular to the plane of the loop. We may also deduce that
the flux is zero when there is no component of the magnetic field that is perpendicular to
the loop.
6.
No. Once the bar is in motion and the charges are separated, no external force is necessary
to maintain the motion. An applied force in the x direction will cause the bar to accelerate
in that direction.
8.
As water falls, it gains velocity and kinetic energy. It then pushes against the blades of a
turbine transferring this energy to the rotor or coil of a large alternating current generator.
The rotor moves in a strong external magnetic field and a voltage is induced in the coil.
This induced emf is the voltage source for the current in our electric power lines.
10.
The magnetic field lines around the transmission cable will be circular. If you place your
loop around the cable, there will be no field lines passing through the loop, so no emf will
be induced. The loop needs to be placed next to the cable, with the plane of the loop
containing the cable, to maximize the flux through its area.
12.
Let us assume the north pole of the magnet faces the ring. As the bar magnet falls toward
the conducting ring, a magnetic field is induced in the ring pointing upward. This upward
directed field will oppose the motion of the magnet preventing it from moving as a freelyfalling body. Try it for yourself to show that an upward force also acts on the falling
magnet if the south end faces the ring.
14.
A constant induced emf requires a magnetic field that is changing at a constant rate in one
direction — for example, always increasing or always decreasing. It is impossible for a
magnetic field to increase forever, both in terms of energy considerations and
technological concerns. In the case of a decreasing field, once it reaches zero and then
reverses direction, we again face the problem with the field increasing without bounds in
the opposite direction.
16.
As the magnet moves at high speed past the fixed coil, the magnetic flux through the coil
changes very rapidly, increasing as the magnet approaches the coil and decreasing as the
magnet moves away. The rapid change in flux through the coil induces a large emf, large
enough to cause a spark across the gap in the spark plug.
18.
(a) Clockwise. As the south end of the magnet approaches the loop from above, the flux
through the loop is directed upward and is increasing in magnitude. The induced current
will flow clockwise to produce a downward flux through the loop and oppose the
increasing flux due to the magnet. (b) Counterclockwise. When the north end of the
magnet is below the loop and is moving away, the flux through the loop is directed
upward and is decreasing in magnitude. The induced current will flow counterclockwise
to produce upward flux through the loop and oppose the decreasing flux due to the
magnet.
Induced Voltages and Inductance
177
Answers to Even Numbered Problems
1.00 × 10 −7 T ⋅ m 2
2.
(a)
4.
zero
6.
2.96 × 10 −5 T ⋅ m 2
8.
0.10 mV
(b)
8.66 × 10 −8 T ⋅ m 2
(b)
6.28 × 10 −8 V
(c)
0
10.
34 mV
12.
52.0 µ T
14.
(a)
16.
8.8 A
18.
1.00 m s
20.
2.87 mV
22.
2.8 mV
24.
Larger R makes current smaller, so the loop must travel faster to maintain equality of
magnetic force and weight.
(c) The magnetic force is proportional to the product of field and current, while the
current is itself proportional to the field. If B becomes two times smaller, the speed
must become four times larger to compensate.
26.
left to right
28.
(a)
30.
13 mV
32.
(a)
34.
(a)
36.
1.36 µ H
40.
1.92 × 10 −5 T ⋅ m 2
42.
(a)
1.00 kΩ
44.
(a)
0
1.88 × 10 −7 T ⋅ m 2
(b)
left to right
(b)
no induced current
(c)
right to left
8.0 A
(b)
3.2 A
(c)
60 V
7.5 kV
(b)
when the plane of the coil is parallel to the field
(b)
3.00 ms
(b)
3.8 V
(c)
6.0 V
(d)
2.2 V
178
CHAPTER 20
2.00 ms
(a)
50.
(a)
52.
3100 µ A, the galvanometer will definitely show the induced current
54.
(a) Amplitude doubles, period unchanged.
(b) Amplitude doubles, period cut in half.
(c) Amplitude unchanged, period cut in half.
56.
(a)
58.
0.121 A, clockwise
60.
6.8 V
62.
1.60 A
64.
(b)
66.
(a) between t = 0 and t = 2.0 s
(b) between t = 2.00 and t = 4.0 s
(c) No. When the field is increasing, the direction of the induced current is different than
when the field is decreasing.
(d) 0.24 A clockwise, current is zero, 0.12 A counterclockwise
τ = L 2R
6.25 × 1010 J
(b)
(b)
(d)
3.22 ms
48.
(b)
(c)
1.50 A
(a)
4.44 × 10 −3 H
(b)
0.176 A
46.
5.55 × 10 −4 J
τ = 2L R
2.00 kN m
0.750 mA
Induced Voltages and Inductance
179
Problem Solutions
20.1
The magnetic flux through the area enclosed by the loop is
2
Φ B = BA cosθ = B (π r 2 ) cos 0° = ( 0.30 T ) π ( 0.25 m )  = 5.9 × 10 −2 T ⋅ m 2
20.2
The magnetic flux through the loop is given by Φ B = BA cosθ where B is the magnitude
of the magnetic field, A is the area enclosed by the loop, and θ is the angle the magnetic
field makes with the normal to the plane of the loop. Thus,
Φ B = BA cosθ = ( 5.00 × 10
−5
2
-2

m 
2  10
2
−7
T )  20.0 cm 
  cosθ = ( 1.00 × 10 T ⋅ m ) cosθ
1
cm

 

JG
(a) When B is perpendicular to the plane of the loop, θ = 0° and
Φ B = 1.00 × 10 −7 T ⋅ m 2
(b) If θ = 30.0°, then Φ B = ( 1.00 × 10 −7 T ⋅ m 2 ) cos 30.0° = 8.66 × 10 −8 T ⋅ m 2
(c) If θ = 90.0°, then Φ B = ( 1.00 × 10 −7 T ⋅ m 2 ) cos90.0° = 0
20.3
The magnetic flux through the loop is given by Φ B = BA cosθ where B is the magnitude
of the magnetic field, A is the area enclosed by the loop, and θ is the angle the magnetic
field makes with the normal to the plane of the loop. Thus,
Φ B = BA cosθ = ( 0.300 T )( 2.00 m ) cos 50.0° = 7.71 × 10 −1 T ⋅ m 2
2
20.4
The magnetic field lines are tangent to the surface of the cylinder, so that no magnetic
field lines penetrate the cylindrical surface. The total flux through the cylinder is zero
20.5
(a) Every field line that comes up through the area A on one side of the wire goes back
down through area A on the other side of the wire. Thus, the net flux through the
coil is zero
(b) The magnetic field is parallel to the plane of the coil , so θ = 90.0° . Therefore,
Φ B = BA cosθ = BA cos 90.0° = 0
180
CHAPTER 20
20.6
The magnetic field generated by the current in the solenoid is
 250 
−2
B = µ0 nI = ( 4π × 10 −7 T ⋅ m A ) 
 ( 15.0 A ) = 2.36 × 10 T
 0.200 m 
and the flux through each turn on the solenoid is
Φ B = BA cosθ
= ( 2.36 × 10
20.7
−2
 π ( 4.00 × 10 −2 m )2 
 cos0° = 2.96 × 10 −5 T ⋅ m 2
T) 


4


(a) The magnetic flux through an area A may be written as
y
Φ B = ( B cosθ ) A
ur
B
= ( component of B perpendicular to A ) ⋅ A
x
Thus, the flux through the shaded side of the cube is
Φ B = Bx ⋅ A = ( 5.0 T ) ⋅ ( 2.5 × 10 −2 m ) = 3.1 × 10 −3 T ⋅ m 2
2
z
(b) Unlike electric field lines, magnetic field lines always form closed loops, without
beginning or end. Therefore, no magnetic field lines originate or terminate within
the cube and any line entering the cube at one point must emerge from the cube at
some other point. The net flux through the cube, and indeed through any closed
surface, is zero .
20.8
ε
20.9
From
=
∆Φ B
∆t
ε
=
B=
=
( ∆B ) A cosθ
∆t
∆Φ B
∆t
( ∆A
=
( 1.5 T − 0 ) π ( 1.6 × 10 −3 m )  cos 0°
2
=
120 × 10 −3 s
B ∆A cosθ
, we find that
∆t
ε
∆t ) cosθ
=
18 × 10 −3 V
= 0.18 T
( 0.10 m 2 s ) cos 0°
= 1.0 × 10 −4 V = 0.10 mV
Induced Voltages and Inductance
20.10
ε
=
181
∆Φ B B ( ∆A ) cosθ
=
∆t
∆t
2
( 0.15 T ) π ( 0.12 m ) − 0  cos 0°
=
= 3.4 × 10 −2 V = 34 mV
0.20 s
20.11
The magnitude of the induced emf is
ε
=
∆Φ B
∆t
=
∆ ( B cosθ ) A
∆t
If the normal to the plane of the loop is considered to point in the original direction of
the magnetic field, then θ i = 0° and θ f = 180° . Thus, we find
ε
20.12
ε
=
1.5 s
2
= 9.4 × 10 −2 V = 94 mV
ε ⋅ ∆t
∆Φ B NBA  ∆ ( cosθ ) 
=
, so B =
∆t
∆t
NA  ∆ ( cosθ ) 
( 0.166 V ) ( 2.77 × 10 −3 s )
B=
= 5.20 × 10 −5 T = 52.0 µ T
2
500 π ( 0.150 m ) 4  [ cos0° − cos 90°]
or
20.13
=
( 0.20 T ) cos 180° − ( 0.30 T ) cos 0° π ( 0.30 m )
The required induced emf is
From
ε
=
ε
= IR = ( 0.10 A )( 8.0 Ω ) = 0.80 V .
∆Φ B  ∆B 
=
 NA cosθ
∆t
 ∆t 
ε =
∆B
0.80 V
=
= 2.7 T s
∆t NA cosθ ( 75 ) ( 0.050 m )( 0.080 m )  cos 0°
20.14
The initial magnetic field inside the solenoid is
 100 
−3
B = µ0 nI = ( 4π × 10 −7 T ⋅ m A ) 
 ( 3.00 A ) = 1.88 × 10 T
0.200
m


(a)
Φ B = BA cosθ = ( 1.88 × 10 −3 T )( 1.00 × 10 −2 m ) cos 0°
2
= 1.88 × 10 −7 T ⋅ m 2
182
CHAPTER 20
(b) When the current is zero, the flux through the loop is Φ B = 0 and the average
induced emf has been
ε
20.15
=
∆Φ B 1.88 × 10 −7 T ⋅ m 2 − 0
=
= 6.28 × 10 −8 V
3.00 s
∆t
If the solenoid has current I, the magnetic field inside it is
 300 
−4
B = µ0 nI = ( 4π × 10 −7 T ⋅ m A ) 
 I = ( 6.00 π × 10 T A ) ⋅ I
 0.200 m 
∆Φ B = ( ∆B ) A cosθ
(a)
2
= ( 6.00 π × 10 −4 T A ) ( 5.0 A − 2.0 A )  π ( 1.5 × 10 −2 m )  cos 0°


= 4.0 × 10 −6 T ⋅ m 2
ε
(b)
20.16
=
N ( ∆Φ B )
∆t
=
4 ( 4.0 × 10 −6 T ⋅ m 2 )
0.90 s
= 1.8 × 10 −5 V = 18 µ V
The magnitude of the average emf is
ε
=
=
N ( ∆Φ B )
∆t
=
NBA  ∆ ( cosθ ) 
∆t
200 ( 1.1 T ) ( 100 × 10 −4 m 2 ) ( cos 0° − cos180° )
0.10 s
Therefore, the average induced current is I =
20.17
ε
R
=
= 44 V
44 V
= 8.8 A
5.0 Ω
If the magnetic field makes an angle of 28.0° with the plane of the coil, the angle it makes
with the normal to the plane of the coil is θ = 62.0° . Thus,
ε
=
N ( ∆Φ B )
∆t
=
NB ( ∆A ) cosθ
∆t
200 ( 50.0 × 10 -6 T ) ( 39.0 cm 2 )( 1 m 2 10 4 cm 2 )  cos 62.0°
=
= 1.02 × 10 −5 V = 10.2 µ V
1.80 s
Induced Voltages and Inductance
20.18
From ε = B Av , the required speed is
v=
20.19
ε
BA
=
IR ( 0.500 A )( 6.00 Ω )
=
= 1.00 m s
B A ( 2.50 T )( 1.20 m )
ε = B⊥ Av , where B⊥
G
v . Thus,
is the component of the magnetic field perpendicular to the velocity
ε = ( 50.0 × 10−6 T ) sin 58.0° ( 60.0 m ) ( 300
20.20
m s ) = 0.763 V
(
)
The speed of the beam after falling freely for 9.00 m, starting from rest v0 y = 0 , is
vy = v02y + 2ay ( ∆y ) = 0 + 2 ( −9.80 m s 2 ) ( −9.00 m ) = 13.3 m s
Since the induced emf is ε = B⊥ Av , where B⊥ is the component of the magnetic field
G
perpendicular to the velocity v , we find
ε = (18.0 × 10−6 T ) (12.0 m ) ( 13.3 m s ) = 2.87 × 10−3 V =
20.21
183
2.87 mV
(a) Observe that only the horizontal component, Bh , of Earth’s magnetic field is
effective in exerting a vertical force on charged particles in the antenna. For the
magnetic force, Fm = qvBh sin θ , on positive charges in the antenna to be directed
upward and have maximum magnitude (when θ =90°), the car should move
toward the east through the northward horizontal component of the magnetic
field.
(b)
ε = BhAv , where Bh
is the horizontal component of the magnetic field.

ε = ( 50.0 × 10−6 T ) cos 65.0° (1.20 m )  65.0

= 4.58 × 10 − 4 V
km   0.278 m s  


h   1 km h  
184
CHAPTER 20
20.22
During each revolution, one of the rotor blades sweeps out a horizontal circular area of
radius A, A = π A 2 . The number of magnetic field lines cut per revolution is
∆Φ B = B⊥ A = Bvertical A . The induced emf is then
ε
20.23
∆Φ B Bvertical (π A
=
=
1 f
∆t
2
) = ( 5.0 × 10
−5
2
T ) π ( 3.0 m ) 
= 2.8 × 10 −3 V = 2.8 mV
0.50 s
(a) To oppose the motion of the magnet, the magnetic field generated by the induced
current should be directed to the right along the axis of the coil. The current must
then be left to right through the resistor.
(b) The magnetic field produced by the current should be directed to the left along the
axis of the coil, so the current must be right to left through the resistor.
20.24
(a) As the bottom conductor of the loop falls, it cuts across
the magnetic field lines coming out of the page. This
induces an emf of magnitude E = Bwv in this conductor,
with the left end at the higher potential. As a result, an
induced current of magnitude
I=
E Bwv
=
R
R
flows clockwise around the loop. The field then exerts
an upward force of magnitude
w
I
ur
Fm
l
ur
Bout
ur
v
B2 w 2 v
 Bwv 
Fm = BIw = B 
w =
R
 R 
on this current-carrying conductor forming the bottom of the loop. If the loop is
falling at terminal speed, the magnitude of this force must equal the downward
gravitational force acting on the loop. That is, when v = vt , we must have
B2 w 2 vt
= Mg
R
or
vt =
MgR
B2 w 2
(b) A larger resistance would make the current smaller, so the loop must reach higher
speed before the magnitude of the magnetic force will equal the gravitational force.
(c) The magnetic force is proportional to the product of the field and the current, while
the current itself is proportional to the field. If B is cut in half, the speed must
become four times larger to compensate and yield a magnetic force with magnitude
equal to the that of the gravitational force.
185
Induced Voltages and Inductance
20.25
(a) After the right end of the coil has
entered the field, but the left end
has not, the flux through the area
enclosed by the coil is directed into
the page and is increasing in
magnitude. This increasing flux
induces an emf of magnitude
E=
ur
v
w
∆Φ B NB ( ∆A )
=
= NBwv
∆t
∆t
l
×
×
×
×
×
×
ur
Bin
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
in the loop. Note that in the above equation, ∆A is the area enclosed by the coil that
enters the field in time ∆t . This emf produces a counterclockwise current in the
loop to oppose the increasing inward flux. The magnitude of this current is
I = E R = NBwv R . The right end of the loop is now a conductor, of length Nw,
carrying a current toward the top of the page through a field directed into the page.
The field exerts a magnetic force of magnitude
N 2 B2 w 2 v
 NBwv 
F = BI ( Nw ) = B 
Nw
=
(
)

R
 R 
directed toward the left
on this conductor, and hence, on the loop.
(b) When the loop is entirely within the magnetic field, the flux through the area
enclosed by the loop is constant. Hence, there is no induced emf or current in the
loop, and the field exerts zero force on the loop.
(c) After the right end of the loop emerges from the field, and before the left end
emerges, the flux through the loop is directed into the page and decreasing. This
decreasing flux induces an emf of magnitude E = NBwv in the loop, which
produces an induced current directed clockwise around the loop so as to oppose the
decreasing flux. The current has magnitude I = E R = NBwv R . This current flowing
upward, through conductors of total length Nw, in the left end of the loop,
experiences a magnetic force given by
N 2 B2 w 2 v
 NBwv 
F = BI ( Nw ) = B 
Nw
=
(
)

R
 R 
20.26
directed toward the left
When the switch is closed, the magnetic field due to the current from the battery will be
directed to the left along the axis of the cylinder. To oppose this increasing leftward flux,
the induced current in the other loop must produce a field directed to the right through
the area it encloses. Thus, the induced current is left to right through the resistor.
186
CHAPTER 20
20.27
Since the magnetic force, Fm = qvB sin θ , on a positive charge is directed toward the top
of the bar when the velocity is to the right, the right hand rule says that the magnetic
field is directed into the page .
20.28
When the switch is closed, the current from the battery produces a magnetic field
directed toward the right along the axis of both coils.
(a) As the battery current is growing in magnitude, the induced current in the
rightmost coil opposes the increasing rightward directed field by generating a field
toward to the left along the axis. Thus, the induced current must be left to right
through the resistor.
(b) Once the battery current, and the field it produces, have stabilized, the flux through
the rightmost coil is constant and there is no induced current .
(c) As the switch is opened, the battery current and the field it produces rapidly
decrease in magnitude. To oppose this decrease in the rightward directed field, the
induced current must produce a field toward the right along the axis, so the
induced current is right to left through the resistor.
20.29
When the switch is closed, the current from the battery produces a magnetic field
directed toward the left along the axis of both coils.
(a) As the current from the battery, and the leftward field it produces, increase in
magnitude, the induced current in the leftmost coil opposes the increased leftward
field by flowing right to left through R and producing a field directed toward the
right along the axis.
(b) As the variable resistance is decreased, the battery current and the leftward field
generated by it increase in magnitude. To oppose this, the induced current is
right to left through R, producing a field directed toward the right along the axis.
(c) Moving the circuit containing R to the left decreases the leftward field (due to the
battery current) along its axis. To oppose this decrease, the induced current is
left to right through R, producing an additional field directed toward the left
along the axis.
(d) As the switch is opened, the battery current and the leftward field it produces
decrease rapidly in magnitude. To oppose this decrease, the induced current is
left to right through R, generating additional magnetic field directed toward the
left along the axis.
Induced Voltages and Inductance
20.30
ε max = NBhorizontal Aω = 100 ( 2.0 × 10−5 T ) ( 0.20 m )2  1500

187
rev  2π rad  1 min  



min  1 rev  60 s  
= 1.3 × 10 −2 V = 13 mV
20.31
Note the similarity between the situation in this problem and a generator. In a generator,
one normally has a loop rotating in a constant magnetic field so the flux through the
loop varies sinusoidally in time. In this problem, we have a stationary loop in an
oscillating magnetic field, and the flux through the loop varies sinusoidally in time. In
both cases, a sinusoidal emf E = Emax sin ωt where Emax = NBAω is induced in the loop.
The loop in this case consists of a single band ( N = 1) around the perimeter of a red
blood cell with diameter d = 8.0 × 10 −6 m . The angular frequency of the oscillating flux
through the area of this loop is ω = 2π f = 2π ( 60 Hz ) = 120π rad s . The maximum
induced emf is then
Emax
20.32
 π d2
= NBAω = B 
 4
(1.0 × 10

ω =

−3
T ) π ( 8.0 × 10 −6 m ) ( 120π s -1 )
2
4
= 1.9 × 10 −11 V
(a) Immediately after the switch is closed, the motor coils are still stationary and the
ε 240 V = 8.0 A
back emf is zero. Thus, I = =
R 30 Ω
(b) At maximum speed,
I=
(c)
ε − ε back
R
=
ε back = 145 V and
240 V − 145 V
= 3.2 A
30 Ω
ε back = ε − IR = 240 V − ( 6.0 A )( 30 Ω ) =
60 V
188
CHAPTER 20
20.33
(a) When a coil having N turns and enclosing area A rotates at angular frequency ω in
a constant magnetic field, the emf induced in the coil is
E = Emax sin ωt where Emax = NB⊥ Aω
Here, B⊥ is the magnitude of the magnetic field perpendicular to the rotation axis of
the coil. In the given case, B⊥ = 55.0 µ T ; A = π ab where a = ( 10.0 cm ) 2 and
b = ( 4.00 cm ) 2 ; and
ω = 2π f = 2π  100

rev   1 min 

 = 10.5 rad s
min   60.0 s 
Thus,
π

Emax = ( 10.0 ) ( 55.0 × 10 −6 T )  ( 0.100 m ) ( 0.040 0 m )  ( 10.5 rad s )
4

or
Emax = 1.81 × 10 −5 V = 18.1 µ V
(b) When the rotation axis is parallel to the field, then B⊥ = 0 giving
Emax = 0
It is easily understood that the induced emf is always zero in this case if you
recognize that the magnetic field lines are always parallel to the plane of the coil,
and the flux through the coil has a constant value of zero.
20.34
(a) Using
ε max = NBAω ,
ε max = 1 000 ( 0.20 T ) ( 0.10 m2 )  60

(b)
rev  2π rad  
3

 = 7.5 × 10 = 7.5 kV
s  1 rev  
ε max occurs when the flux through the loop is changing the most rapidly. This is
when the plane of the loop is parallel to the magnetic field .
20.35
ω =  120

rev  1 min  2π rad 
rad


 = 4π
min  60 s  1 rev 
s
and the period is
T=
2π
ω
= 0.50 s
(a)
ε max = NBAω = 500 ( 0.60 T ) ( 0.080 m )( 0.20 m ) ( 4π
(b)
ε = ε max sin (ωt ) = ( 60 V ) sin ( 4π

π
rad s ) 
 32
rad s ) = 60 V

s   = 57 V

Induced Voltages and Inductance
(c) The emf is first maximum at t =
20.36
189
T 0.50 s
=
= 0.13 s
4
4
Treating the coiled telephone cord as a solenoid,
L=
µ0 N A
2
A
=
( 4π × 10
−7
2
2 π
T ⋅ m A ) ( 70.0 )  ( 1.30 × 10 −2 ) 
4

0.600 m
= 1.36 × 10 −6 H = 1.36 µ H
20.37
20.38
Eav = L
∆I
 1.50 A − 0.20 A 
−2
= ( 3.00 × 10 -3 H ) 
 = 2.0 × 10 V = 20 mV
∆t
0.20 s


The units of
NΦB
T ⋅ m2
are
A
I
From the force on a moving charged particle, F = qvB , the magnetic field is B =
we find that
1 T =1
N
N⋅s
=1
C ⋅ (m s)
C⋅m
 N ⋅ s  2 (N ⋅ m) ⋅ s  J 
Thus, T ⋅ m 2 = 
=  ⋅s = V⋅s
⋅m =
C
 C⋅m 
C
and
20.39
(a)
ε
T ⋅ m2 V ⋅ s
=
which is the same as the units of
A
A
∆I ∆t
L=
µ0 N 2 A
A
=
( 4π × 10
−7
2
2
T ⋅ m A ) ( 400 ) π ( 2.5 × 10 −2 m ) 


0.20 m
= 2.0 × 10 −3 H = 2.0 mH
(b) From
ε
= L ( ∆I ∆ t ) ,
∆I ε
75 × 10 −3 V
=
=
= 38 A s
∆t L 2.0 × 10 −3 H
F
and
qv
190
CHAPTER 20
20.40
From
ε
= L ( ∆I ∆t ) , the self-inductance is
L=
ε
∆I ∆t
=
24.0 × 10 −3 V
= 2.40 × 10 −3 H
10.0 A s
Then, from L = N Φ B I , the magnetic flux through each turn is
−3
L ⋅ I ( 2.40 × 10 H ) ( 4.00 A )
ΦB =
=
= 1.92 × 10 −5 T ⋅ m 2
500
N
20.41
The inductive time constant is τ = L R . From
L=
ε
∆I ∆t
Ω⋅s
=s.
Ω
20.42
with units of
ε
= L ( ∆I ∆t ) , the self-inductance is
V V
=   ⋅ s = Ω ⋅ s . Thus, the units of the time constant are
A s A
(a) The time constant of the RL circuit is τ = L R , and that of the RC circuit is τ = RC . If
the two time constants have the same value, then
RC =
L
L
3.00 H
, or R =
=
= 1.00 × 10 3 Ω = 1.00 kΩ
-6
R
C
3.00 × 10 F
(b) The common value of the two time constants is
τ=
20.43
L
3.00 H
=
= 3.00 × 10 -3 s = 3.00 ms
R 1.00 × 10 3 Ω
The maximum current in a RL circuit I max = ε R , so the resistance is
R=
ε
I max
=
6.0 V
= 20 Ω
0.300 A
The inductive time constant is τ = L R , so
L = τ ⋅ R = ( 600 × 10 −6 s ) ( 20 Ω ) = 1.2 × 10 −2 H = 12 mH
Induced Voltages and Inductance
20.44
The current in the RL circuit at time t is I =
ε
(1 − e ) . The potential difference across
R
−t τ
the resistor is ∆VR = RI = ε ( 1 − e −t τ ) , and from Kirchhoff’s loop rule, the potential
difference across the inductor is
∆VL = ε − ∆VR = ε 1 − ( 1 − e −t τ )  = ε e −t τ
(a) At t = 0 , ∆VR = ε ( 1 − e −0 ) = ε ( 1 − 1) = 0
(b) At t = τ , ∆VR = ε ( 1 − e −1 ) = ( 6.0 V )( 1 − 0.368 ) = 3.8 V
(c) At t = 0 , ∆VL = ε e −0 = ε = 6.0 V
(d) At t = τ , ∆VL = ε e −1 = ( 6.0 V )( 0.368 ) = 2.2 V
20.45
From I = I max ( 1 − e −t τ ) , e −t τ = 1 −
If
I
I max
I
I max
= 0.900 at t = 3.00 s , then
e − 3.00 s τ = 0.100 or τ =
−3.00 s
= 1.30 s
ln ( 0.100 )
Since the time constant of an RL circuit is τ = L R , the resistance is
R=
20.46
(a)
τ=
(b) I =
(c)
L
τ
=
2.50 H
= 1.92 Ω
1.30 s
L 8.00 mH
=
= 2.00 ms
4.00 Ω
R
ε
 6.00 V 
1− e ) = 
(
 (1 − e
R
 4.00 Ω 
I max =
−t τ
ε
R
=
− 250×10 −6 s 2.00×10 -3 s
)=
0.176 A
6.00 V
= 1.50 A
4.00 Ω
(d) I = I max ( 1 − e −t τ ) yields e −t τ = 1 − I I max ,
and
191
t = −τ ln ( 1 − I I max ) = − ( 2.00 ms ) ln ( 1 − 0.800 ) = 3.22 ms
192
CHAPTER 20
1 2 1
2
LI = ( 70.0 × 10 −3 H ) ( 2.00 A ) = 0.140 J
2
2
20.47
PEL =
20.48
(a) The inductance of a solenoid is given by L = µ0 N 2 A l , where N is the number of
turns on the solenoid, A is its cross-sectional area, and l is its length. For the given
solenoid,
L=
µ0 N 2 (π r 2 )
l
( 4π × 10
=
−7
T ⋅ m A ) ( 300 ) π ( 5.00 × 10 −2 m )
2
0.200 m
2
= 4.44 × 10 −3 H
(b) When the solenoid described above carries a current of I = 0.500 A , the stored
energy is
PEL =
20.49
1 2 1
2
LI = ( 4.44 × 10 −3 H ) ( 0.500 A ) = 5.55 × 10 −4 J
2
2
The current in the circuit at time t is I =
inductor is PEL =
ε
R
(1 − e ) , and the energy stored in the
−t τ
1 2
LI
2
(a) As t →∞ , I → I max =
ε
R
=
24 V
= 3.0 A , and
8.0 Ω
1 2
1
2
PEL → LI max
= ( 4.0 H )( 3.0 A ) = 18 J
2
2
(b) At t = τ , I = I max ( 1 − e −1 ) = ( 3.0 A )( 1 − 0.368 ) = 1.9 A
and
20.50
PEL =
1
2
( 4.0 H )(1.9 A ) = 7.2 J
2
(a) When the two resistors are in series, the total resistance is
Req = R + R = 2R , and the time constant of the circuit is τ =
L
L
=
Req
2R
(b) With the resistors now connected in parallel, the total resistance is
Req =
( R )( R ) R
R+R
=
2
, and the time constant is τ =
2L
L
=
Req
R
Induced Voltages and Inductance
20.51
According to Lenz’s law, a current will be induced in the coil to oppose the change in
magnetic flux due to the magnet. Therefore, current must be directed from b to a
through the resistor, and Va − Vb will be negative .
20.52
While the coil is between the poles of the magnet, the component of the field
perpendicular to the plane of the coil is Bi = 0.10 T . After the coil is pulled out of the
field, B f ≈ 0 .
193
The magnitude of the average induced emf as the coil is moved is
ε
=N
∆ ( BA )
∆Φ B
( ∆B ) A
=N
=N
∆t
∆t
∆t
and the average induced current in the galvanometer is
ε
I=
2
N ( ∆B ) A 10 ( 0.10 T − 0 ) π ( 0.020 m ) 
=
=
R
R ( ∆t )
( 2.0 Ω ) ( 0.20 s )
= 3.1 × 10 −3 A = 3 100 µ A
This means the galvanometer will definitely show the induced current and even be
overloaded.
20.53
(a) The current in the solenoid reaches I = 0.632 I max in a time of t = τ = L R , where
L=
Thus,
µ0 N 2 A
A
t=
( 4π × 10
=
−7
T ⋅ m A ) ( 12 500 ) ( 1.00 × 10 −4 m 2 )
2
7.00 × 10 -2 m
0.280 H
= 2.00 × 10 −2 s = 20.0 ms
14.0 Ω
(b) The change in the solenoid current during this time is
 ∆V 
 60.0 V 
∆I = 0.632 I max − 0 = 0.632 
 = 0.632 
 = 2.71 A
 R 
 14.0 Ω 
so the average back emf is

2.71 A 
 = 37.9 V
-2
 2.00 × 10 s 
ε back = L  ∆I  = ( 0.280 H ) 
 ∆t 
= 0.280 H
194
CHAPTER 20
(c)
∆Φ B ( ∆B ) A 12  µ0 n ( ∆I )  A µ0 N ( ∆I ) A
=
=
=
2 A ⋅ ( ∆t )
∆t
∆t
∆t
( 4π × 10
=
(d) I =
ε coil
Rcoil
=
−7
T ⋅ m A ) ( 12 500 ) ( 2.71 A ) ( 1.00 × 10 −4 m 2 )
2 ( 7.00 × 10 -2 m )( 2.00 × 10 -2 s )
N coil ( ∆Φ B ∆t )
Rcoil
=
( 820 ) (1.52 × 10 −3 V )
24.0 Ω
= 1.52 × 10 −3 V
= 0.051 9 A = 51.9 mA
20.54
e(t)
(a)
(b)
10 mV
Original Curve
(c)
5 mV
t (ms)
0
0.5
0.25
0.75
1.0
1.25
1.5
1.75
2.0
–5 mV
–10 mV
(a) Doubling the number of turns doubles the amplitude but does not alter the period.
(b) Doubling the angular velocity doubles the amplitude and also cuts the period in
half.
(c) Doubling the angular velocity while reducing the number of turns to one half the
original value leaves the amplitude unchanged but does cut the period in half.
20.55
Q = I av ( ∆t ) =
or
Q=
ε av
R
B ( ∆A )
1 ∆Φ
( ∆t ) =  B  ( ∆t ) =
(15.0 × 10
R  ∆t 
−6
R
2
T ) ( 0.200 m ) − 0 
= 1.20 × 10 −6 C = 1.20 µ C
0.500 Ω
Induced Voltages and Inductance
20.56
(a)
(b)
PEL =
2
1 2 1
LI = ( 50.0 H ) ( 50.0 × 10 3 A ) = 6.25 × 1010 J
2
2
( 4π × 10
µII
= 0 1 2 =
A 2π d
F
= 2.00 × 10 3
20.57
195
−7
T ⋅ m A )( 50.0 × 10 3 A )
2
2π ( 0.250 m )
N
kN
= 2.00
m
m
(a) To move the bar at uniform speed, the magnitude of the applied force must equal
that of the magnetic force retarding the motion of the bar. Therefore, Fapp = B I A . The
magnitude of the induced current is
I=
ε
R
=
( ∆Φ B
∆t )
R
=
so the field strength is
B ( ∆A ∆t )
R
B=
=
B Av
R
IR
, giving Fapp = I 2 R v
Av
Thus, the current is
I=
Fapp ⋅ v
R
=
( 1.00 N ) ( 2.00 m s )
8.00 Ω
= 0.500 A
(b) √ = I 2 R = ( 0.500 A ) ( 8.00 Ω ) = 2.00 W
2
(c)
√input = Fapp ⋅ v = ( 1.00 N ) ( 2.00 m s ) = 2.00 W
196
CHAPTER 20
20.58
When A and B are 3.00 m apart, the area enclosed
by the loop consists of four triangular sections,
each having hypotenuse of 3.00 m, altitude of
1.50 m, and base of
2
2
( 3.00 m ) − (1.50 m ) = 2.60 m
The decrease in the enclosed area has been
A
3.00 m
3.00 m
3.00 m
3.00 m
B
2
1

∆A = Ai − A f = ( 3.00 m ) − 4  ( 1.50 m )( 2.60 m )  = 1.21 m 2
2


The average induced current has been
I av =
ε av ( ∆Φ B
=
R
R
∆t )
=
B ( ∆A ∆t )
R
=
( 0.100 T ) ( 1.21 m 2 0.100 s )
10.0 Ω
= 0.121 A
As the enclosed area decreases, the flux (directed into the page) through this area also
decreases. Thus, the induced current will be directed clockwise around the loop to
create additional flux directed into the page through the enclosed area.
20.59
If d is the distance from the lightning bolt to the center of the coil, then
ε av
=
N ( ∆Φ B )
∆t
=
N ( ∆B ) A N  µ0 ( ∆I ) 2π d  A N µ0 ( ∆I ) A
=
=
2π d ( ∆t )
∆t
∆t
2
100 ( 4π × 10 −7 T ⋅ m A )( 6.02 × 106 A − 0 ) π ( 0.800 m ) 
=
2π ( 200 m ) ( 10.5 × 10 −6 s )
= 1.15 × 10 5 V= 115 kV
20.60
The flux through the surface area of the tent is the same as that through the tent base.
Thus, as the tent is flattened, the change is flux is
∆Φ B = B ( ∆Abase ) = B  L ( 2L ) − L ( 2 ⋅ L cosθ )  = 2 L2 B ( 1 − cosθ )
The magnitude of the average induced emf is then
ε av
2
∆Φ B 2 L B ( 1 − cosθ ) 2 ( 1.5 m ) ( 0.30 T )( 1 − cos 60° )
=
=
=
= 6.8 V
0.10 s
∆t
∆t
2
Induced Voltages and Inductance
20.61
(a)
ε av
=
=
197
2
∆Φ B B ( ∆A ) B (π d 4 ) − 0 
=
=
∆t
∆t
∆t
( 25.0 mT ) π ( 2.00 × 10 -2 m )
4 ( 50.0 × 10 −3 s )
2
= 0.157 mV
As the inward directed flux through the loop decreases, the induced current goes
clockwise around the loop in an attempt to create additional inward flux through
the enclosed area. With positive charges accumulating at B,
point B is at a higher potential than A
(b)
ε av
-2
∆Φ B ( ∆B ) A ( 100 − 25.0 ) mT  π ( 2.00 × 10 m )
=
=
=
= 5.89 mV
∆t
∆t
4 ( 4.00 × 10 −3 s )
2
As the inward directed flux through the enclosed area increases, the induced
current goes counterclockwise around the loop in an attempt to create flux directed
outward through the enclosed area.
With positive charges now accumulating at A,
point A is at a higher potential than B
20.62
The induced emf in the ring is
ε av
=
∆Φ B ( ∆B ) Asolenoid ( ∆Bsolenoid 2 ) Asolenoid 1 
 ∆I

=
=
=  µ0 n  solenoid   Asolenoid
2
∆t
∆t
∆t
 ∆t  
=
2
1
4π × 10 −7 T ⋅ m A ) ( 1 000 )( 270 A s ) π  3.00 × 10 −2 m   = 4.80 × 10 −4 V
(

2 
(
)
Thus, the induced current in the ring is
I ring =
20.63
ε av
R
=
4.80 × 10 −4 V
= 1.60 A
3.00 × 10 -4 Ω
(a) As the rolling axle (of length A = 1.50 m ) moves perpendicularly to the uniform
= BAv will exist between its ends.
magnetic field, an induced emf of magnitude
The current produced in the closed-loop circuit by this induced emf has magnitude
ε
I=
ε av ( ∆Φ B
=
R
R
∆t )
=
B ( ∆A ∆t )
R
=
B Av ( 0.800 T )( 1.50 m ) ( 3.00 m s )
=
= 9.00 A
0.400 Ω
R
198
CHAPTER 20
(b) The induced current through the axle will cause the magnetic field to exert a
retarding force of magnitude Fr = BI A on the axle. The direction of this force will be
G
opposite to that of the velocity v so as to oppose the motion of the axle. If the axle
G
is to continue moving at constant speed, an applied force in the direction of v and
having magnitude Fapp = Fr must be exerted on the axle.
Fapp = BI A = ( 0.800 T ) ( 9.00 A )( 1.50 m ) = 10.8 N
(c) Using the right-hand rule, observe that positive charges within the moving axle
experience a magnetic force toward the rail containing point b, and negative charges
experience a force directed toward the rail containing point a. Thus, the rail
containing b will be positive relative to the other rail. Point
b is then at a higher potential than a , and the current goes from b to a through the
resistor R.
(d)
G
G
No . Both the velocity v of the rolling axle and the magnetic field B are
unchanged. Thus, the polarity of the induced emf in the moving axle is unchanged,
and the current continues to be directed from b to a through the resistor R.
20.64
(a) The flowing water is a conductor moving through Earth’s magnetic field. A
motional emf given by ε = B ( w ) v will exist in the water between the plates and
the induced current in the load resistor is
I=
ε
Rtotal = B ( w ) v Rtotal
where
Rtotal = Rwater + R = ρ
Thus,
I=
A
A
+R=ρ
w
+R
ab
B(w)v
abvB
=
ρ ( w ab ) + R
ρ + abR w
(b) If R = 0 ,
I=
( 100 m )( 5.00 m ) ( 3.00 m s ) ( 50.0 × 10−6 T )
100 Ω ⋅ m + 0
= 7.50 × 10 −4 A = 0.750 mA
199
Induced Voltages and Inductance
20.65
Consider the closed conducting path made up by
the horizontal wire, the vertical rails and the path
containing the resistance R at the bottom of the
figure. As the wire slides down the rails, the
outward directed flux through the area enclosed
by this path is decreasing as the area decreases.
This decreasing flux produces an induced current
which flows counterclockwise around the
conducting path (and hence, right to left through
the horizontal wire) to oppose the decrease in flux.
ur
Bout
l
m
The wire is now carrying a current toward the left
through a magnetic field directed out of the page.
The field then exerts an upward magnetic force on
the wire of magnitude
R
B ( ∆Φ B ∆t ) l B ( B ∆A ∆t ) l B2 ( lv ) l B2 l2 v
 E
=
=
=
F = BI lsin 90° = B   l =
R
R
R
R
R
Observe that this upward directed magnetic force opposes the weight of the wire, and
its magnitude is proportional to the speed of the falling wire. The wire will be at its
terminal speed ( v = vt ) when the magnitude of the magnetic force equals the weight of
the wire. That is, when
B2 l2 vt
= mg
R
20.66
which gives
vt =
The flux through the area enclosed by the coil is
given by Φ B = BA , where B is the magnetic field
perpendicular to the plane of the coil, and A is the
enclosed area. Compare Figures (a) and (b) at the
right and observe that when B > 0 , Φ B is directed
into the page through the interior of the coil. The
induced current in the coil will have a magnitude
of
I=
mgR
B2 l2
B(T)
0.6
0.3
0
2
4
(a)
E ∆Φ B ∆t  ∆B  A
=
=

R
R
 ∆t  R
where R is the resistance of the coil.
6
Coil
ur
Bin
(b)
t(s)
200
CHAPTER 20
(a) From the above result, note that the induced current has greatest magnitude when
∆B ∆t is the greatest (that is, when the graph of B vs. t has the steepest slope).
From Figure (a) above, this is seen to be in the interval between t = 0 and t = 2.0 s .
(b) I = 0 when ∆B ∆t = 0 (that is, when B is constant). This is true in the interval
between t = 2.0 s and t = 4.0 s .
(c) No. The direction of the induced current will always be such as to oppose the
change that is occurring in the flux through the coil. When B is decreasing the
change in the flux (and hence the induced current) is directed opposite to what it is
when B is increasing.
(d) Between t = 0 and t = 2.0 s , the flux through the coil is directed into the page and
decreasing in magnitude. The induced current must flow clockwise around the coil
so that the flux it generates through the interior of the coil is into the page opposing
the change that is occurring in the primary flux. The magnitude is
 ∆B  A  0.60 T  0.20 m 2
I =
= 0.24 A
 =

 ∆t  R  2.0 s  0.25 Ω
Between t = 2.0 s and t = 4.0 s , the induced current is zero because the flux is
constant.
Between t = 4.0 and t = 6.0 s , the flux through the coil is directed into the page and
increasing in magnitude. The induced current must flow counterclockwise around
the coil so that the flux it generates through the interior of the coil is out of the page
opposing the change that is occurring in the primary flux. The current magnitude is
 ∆B  A  0.30 T  0.20 m 2
I =
= 0.12 A
 =

 ∆t  R  2.0 s  0.25 Ω