234
Chapter 20
7.
The energy stored in an inductor of inductance L and carrying current I is PEL = 12 LI 2 . Thus, if
the current is doubled while the inductance is constant, the stored energy increases by a factor of
4 and the correct choice is (d).
8.
An emf is induced in the coil by any action which causes a change in the flux through the coil. The
actions described in choices (a), (b), (d), and (e) all change the flux through the coil and induce
an emf. However, moving the coil through the constant field without changing its orientation with
respect to the field will not cause a change of flux. Thus, choice (c) is the correct answer.
9.
With the current in long wire flowing in the direction shown in Figure MCQ20.9, the magnetic
flux through the rectangular loop is directed into the page. If this current is decreasing in time, the
change in the flux is directed opposite to the flux itself (or out of the page). The induced current
will then flow clockwise around the loop, producing a flux directed into the page through the loop
and opposing the change in flux due to the decreasing current in the long wire. The correct choice
is for this question is (b).
10.
A current flowing counterclockwise in the outer loop of Figure MCQ20.10 produces a magnetic
flux through the inner loop that is directed out of the page. If this current is increasing in time, the
change in the flux is in the same direction as the flux itself (or out of the page). The induced current in the inner loop will then flow clockwise around the loop, producing a flux through the loop
directed into the page, opposing the change in flux due to the increasing current in the outer loop.
The correct answer is choice (b).
11.
As the bar magnet approaches the loop from above, with its south end downward as shown
in Figure MCQ20.11, magnetic flux through the area enclosed by the loop is directed upward
and increasing in magnitude. To oppose this increasing upward flux, the induced current in the
loop will flow clockwise, as seen from above, producing a flux directed downward through the
area enclosed by the loop. After the bar magnet has passed through the plane of the loop, and is
departing with its north end upward, a decreasing flux is directed upward through the loop. To
oppose this decreasing upward flux, the induced current in the loop flows counterclockwise as
seen from above, producing flux directed upward through the area enclosed by the loop. From
this analysis, we see that (a) is the only true statement among the listed choices.
12.
With the magnetic field perpendicular to the plane of the page in Figure MCQ20.12, the flux
through the closed loop to the left of the bar is given by Φ B = BA, where B is the magnitude of the
field and A is the area enclosed by the loop. Any action which produces a change in this product,
BA, will induce a current in the loop and cause the bulb to light. Such actions include increasing or
decreasing the magnitude of the field (B), and moving the bar to the right or left and changing the
enclosed area A. Thus, the bulb will light during all of the actions in choices (a), (b), (c), and (d).
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 cos θ = 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.
56157_20_ch20_p219-251.indd 234
3/19/08 1:50:02 AM
Induced Voltages and Inductance
235
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.
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 freely-falling body. Try it for
yourself to show that an upward force also acts on the falling magnet if the south end faces the ring.
12.
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.
14.
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.
PROBLEM SOLUTIONS
20.1
The angle between the direction of the constant field and the normal to the plane of the loop is
θ = 0°, so
)
)
Φ B = BA cos θ = ( 0.50 T) ( 8.0 × 10 −2 m (12 × 10 −2 m  cos 0° = 4.8 × 10 −3 T ⋅ m 2 = 4.8 mWb
20.2
The magnetic flux through the loop is given by Φ B = BA cos θ , 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,
(a)
2

 10 −2 m  
Φ B = BA cos θ = 5.00 × 10 −5 T  20.0 cm 2 
 cos θ = 1.00 × 10 −7 T ⋅ m 2 cos θ
 1 cm  

ur
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
(c)
If θ = 90.0°, then Φ B = (1.00 × 10 −7 T ⋅ m 2
(
20.3
)
(
)
) cos 90.0° =
)
Φ B = (1.00 × 10 −7 T ⋅ m 2 cos 30.0° = 8.66 × 10 −8 T ⋅ m 2
0
Φ B = BA cos θ = B(π r 2 ) cos θ where θ is the angle between the direction of the field and the
normal to the plane of the loop.
(a)
If the field is perpendicular to the plane of the loop, θ = 0°, and
B=
(b)
ΦB
8.00 × 10 −3 T ⋅ m 2
=
= 0.177 T
(π r 2 cos θ π ( 0.12 m )2 cos 0°
)
If the field is directed parallel to the plane of the loop, θ = 90°, and
Φ B = BA cos θ = BA cos 90° = 0
56157_20_ch20_p219-251.indd 235
3/19/08 1:50:03 AM
236
Chapter 20
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 cos θ = BA cos 90.0° = 0
(a)
The magnitude of the field inside the solenoid is
20.6
 400 
 N
B = µ0 nI = µ0   I = ( 4π × 10 −7 T ⋅ m A 
( 5.00 A ) = 6.98 × 10 −3 T = 6.98 mT
 0.360 m 
 l
)
The field inside a solenoid is directed perpendicular to the cross-sectional area, so θ = 0°
and the flux through a loop of the solenoid is
(b)
Φ B = BA cos θ = B (π r 2 ) cos θ
= ( 6.98 × 10 −3 T ) π ( 3.00 × 10 −2 m ) cos 0° = 1.97 × 10 −5 T ⋅ m 2
2
20.7
(a)
The magnetic flux through an area A may be written as
Φ B = ( B cos θ ) A
= ( component of B perpendicular to A ) ⋅ A
Thus, the flux through the shaded side of the cube is
Φ B = Bx ⋅ A = ( 5.0 T) ⋅ ( 2 .5 × 10 −2 m
(b)
)
2
= 3.1 × 10 −3 T ⋅ m 2
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 .
(
)
2
−3


∆Φ B ( ∆B ) A cos θ (1.5 T − 0 ) π 1.6 × 10 m  cos 0°
=
=
=
= 1.0 × 10 −4 V = 0.10 mV
∆t
∆t
120 × 10 −3 s
20.8
ε
20.9
With the constant field directed perpendicular to the plane of the coil, the flux through the coil is
Φ B = BA cos 0° = BA . As the enclosed area increases, the magnitude of the induced emf in the
coil is
ε
20.10
ε
=
=
56157_20_ch20_p219-251.indd 236
=
∆Φ B
 ∆A 
= B
= ( 0.30 T) ( 5.0 × 10 −3 m 2 s = 1.5 × 10 −3 V = 1.5 mV
 ∆t 
∆t
)
∆Φ B B ( ∆A ) cos θ
=
∆t
∆t
( 0.15 T) π ( 0.12 m )2 − 0  cos 0°
0.20 s
= 3.4 × 10 −2 V = 34 mV
3/19/08 4:06:35 AM
Induced Voltages and Inductance
20.11
ε
The magnitude of the induced emf is
237
∆ ( B cos θ ) A
∆Φ B
=
∆t
∆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
( 0.20 T) cos 180° − ( 0.30 T) cos 0° π ( 0.30 m )2
1.5 s
= 9.4 × 10 −2 V = 94 mV
With the field directed perpendicular to the plane of the coil, the flux through the coil is
Φ B = BA cos 0° = BA. As the magnitude of the field increases, the magnitude of the induced emf
in the coil is
ε
20.13
=
=
∆Φ B  ∆B 
2
=
A = ( 0.050 0 T s ) π ( 0.120 m )  = 2.26 × 10 −3 V = 2.26 mV
 ∆t 
∆t
The required induced emf is
From
ε
=
ε
= IR = ( 0.10 A ) ( 8.0 Ω ) = 0.80 V.
∆Φ B  ∆B 
=
NA cos θ
 ∆t 
∆t
ε
∆B
0.80 V
=
=
= 2 .7 T s
∆t NA cos θ ( 75 ) ( 0.050 m )( 0.080 m )  cos 0°
20.14
The initial magnetic field inside the solenoid is
 100 
B = µ0 nI = ( 4π × 10 −7 T ⋅ m A 
( 3.00 A ) = 1.88 × 10 −3 T
 0.200 m 
)
(a)
)
)
Φ B = BA cos θ = (1.88 × 10 −3 T (1.00 × 10 −2 m cos 0°
2
= 1.88 × 10 −7 T ⋅ m 2
(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
(a)
=
0 − 1.88 × 10 −7 T ⋅ m 2
∆Φ B
=
= 6.28 × 10 −8 V
∆t
3.00 s
The initial field inside the solenoid is
 300 
Bi = µ0 nI i = ( 4π × 10 −7 T ⋅ m A 
( 2.00 A ) = 3.77 × 10 −3 T
 0.200 m 
)
(b)
The final field inside the solenoid is
 300 
B f = µ0 nI f = ( 4π × 10 −7 T ⋅ m A 
( 5.00 A ) = 9.42 × 10 −3 T
 0.200 m 
)
)
(c)
The 4-turn coil encloses an area A = π r 2 = π (1.50 × 10 −2 m
(d)
The change in flux through each turn of the 4-turn coil during the 0.900-s period is
(
)(
2
= 7.07 × 10 −4 m 2
)
∆Φ B = ( ∆B ) A = 9.42 × 10 −3 T − 3.77 × 10 −3 T 7.07 × 10 −4 m 2 = 3.99 × 10 −6 Wb
continued on next page
56157_20_ch20_p219-251.indd 237
3/19/08 4:06:15 AM
238
Chapter 20
(e)
The average induced emf in the 4-turn coil is


ε = N 2  ∆Φ B  = 4  3.99 × 10
∆t
−6
0.900 s
Wb 
−5
 = 1.77 × 10 V
Since the current increases at a constant rate during this time interval, the induced emf at
any instant during the interval is the same as the average value given above.
(f)
20.16
The induced emf is small, so the current in the 4-turn coil will also be very small.
This means that the magnetic field generatedd by this current will be negligibly small
in comparison to the field generated by thee solenoid.
The magnitude of the average emf is
ε
=
=
N ( ∆Φ B ) NBA  ∆ ( cos θ ) 
=
∆t
∆t
)
200 (1.1 T) (100 × 10 −4 m 2 ( cos 0° − cos 180°)
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 ) NB ( ∆A ) cos θ
=
∆t
∆t
(
)(
)(
)
200 50.0 × 10 −6 T  39.0 cm 2 1 m 2 10 4 cm 2  cos 62.0°
=
= 1.002 × 10 −5 V = 10.2 µV
1.80 s
20.18
With the magnetic field perpendicular to the plane of the coil, the flux through each turn of the
coil is Φ B = BA = B (π r 2 . Since the area remains constant, the change in flux due to the changing magnitude of the magnetic field is ∆Φ B = ( ∆B ) π r 2 .
)
The induced emf is:
(b)
When looking down on the coil from a location on the positive z-axis, the magnetic field (in
the positive z-direction) is directed up toward you and increasing in magnitude. This means
the change in the flux through the coil is directed upward. In order to oppose this change in
flux, the current must produce a flux directed downward through the area enclosed by the
coil. Thus, the current must flow clockwise as seen from your viewing location.
(c)
Since the turns of the coil are connected in series, the total resistance of the coil is
Req = NR . Thus, the magnitude of the induced current is
I=
56157_20_ch20_p219-251.indd 238
 B0 − 0 ) π r 2 
NB0π r 2
= −
t
t−0


ε = − N  ∆Φ  = − N  (
(a)
ε
Req
=
∆t
NB0π r 2 t
B π r2
= 0
NR
tR
3/19/08 1:50:05 AM
Induced Voltages and Inductance
20.19
The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of
the metallic truck body. Thus, the motional emf induced across the width of the truck is
ε = B⊥ lv = ( 35 × 10 −6
20.20
239

 1m 
T ( 79.8 in ) 
( 37 m s ) = 2.6 × 10 −3 V = 2.6 mV
 39.37 in  

)
The vertical component of the Earth’s magnetic field is perpendicular to the horizontal velocity of
the wire. Thus, the magnitude of the motional emf induced in the wire is
ε = B⊥ lv = ( 40.0 × 10 −6
)
T ( 2.00 m ) (15.0 m s ) = 1.20 × 10 −3 V = 1.20 mV
Imagine holding your right hand horizontal with the fingers pointing north (the direction of
the wire’s velocity), such that when you close your hand the fingers curl downward (in the
direction of B⊥). Your thumb will then be pointing westward. By right-hand rule 1, the magnetic force on charges in the wire would tend to move positive charges westward. Thus,
the west end of the wire will be positive reelative to the east end .
20.21
(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 q = 90°), the car should move toward the east through the northward horizontal component of the magnetic field.
(b)
ε = Bh lv, where
Bh is the horizontal component of the magnetic field.
ε = ( 50.0 × 10 −6

km   0.278 m s  
T cos 65.0°  (1.20 m )  65.0


h   1 km h  

)
= 4.58 × 10 − 4 V
20.22
(a)
Since ε = B⊥ lv, the magnitude of the vertical component of the Earth’s magnetic field at
this location is
ε
Bvertical = B⊥ =
(b)
20.23
lv
=
0.45 V
= 6.0 × 10 −6 T = 6.0 µ T
3
×
25
m
3
.
0
10
m
s
(
)(
)
Yes. The magnitude and direction of the Earth’s field varies from one location to the other,
so the induced voltage in the wire changes. Further, the voltage will change if the tether
cord changes its orientation relative to the Earth’s field.
r
ε = B⊥ lv, where B⊥ is the component of the magnetic field perpendicular to the velocity v.
Thus,
ε = ( 50.0 × 10 −6
20.24
56157_20_ch20_p219-251.indd 239
From
)
T sin 58.0°  ( 60.0 m ) ( 300 m s ) = 0.763 V
ε = B lv, the required speed is
ε = IR = ( 0.500 A ) ( 6.00 Ω ) =
v=
Bl Bl
( 2.50 T) (1.20 m )
1.00 m s
3/19/08 4:44:37 AM
240
20.25
Chapter 20
(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.
20.26
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.
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.
20.29
56157_20_ch20_p219-251.indd 240
(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.
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.
3/19/08 1:50:06 AM
Induced Voltages and Inductance
20.30
(a)
241
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 ε = Bwv in this conductor,
with the left end at the higher potential. As a result, an
induced current of magnitude
I=
ε
R
=
Bwv
R
flows clockwise around the loop. The field then exerts
an upward force of magnitude
B2 w2 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
B 2 w 2 vt
= Mg
R
20.31
or
vT =
MgR
B2 w2
(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.
(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
ε
=
∆Φ B NB ( ∆A )
=
= NBwv
∆t
∆t
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 = ε 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 w2 v
 NBwv 
F = BI ( Nw ) = B 
Nw
=
directedd toward the left
(
)
 R 
R
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.
continued on next page
56157_20_ch20_p219-251.indd 241
3/19/08 4:05:53 AM
242
Chapter 20
(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 ε = 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 = ε 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 w2 v
 NBwv 
F = BI ( Nw ) = B 
Nw ) =
(

 R 
R
20.32
(a)
directedd toward the left
The motional emf induced in the bar must be ε = IR , where I is the current in this series
circuit. Since ε = B⊥ lv , the speed of the moving bar must be
v=
ε
B⊥ l
=
)
−3
IR ( 8.5 × 10 A ( 9.0 Ω )
=
= 0.73 m s
B⊥ l
( 0.30 T) ( 0.35 m )
The flux through the closed loop formed by the rails, the bar, and the resistor is directed
into the page and is increasing in magnitude. To oppose this change in flux, the current
must flow in a manner so as to produce flux out of the page through the area enclosed by
the loop. This means the current will flow counterclockwise .
(b)
The rate at which energy is delivered to the resistor is
P = I 2 R = ( 8.5 × 10 −3 A ) ( 9.0 Ω ) = 6.5 × 10 −4 W = 0.65 mW = 0.65 mJ s
2
(c)
20.33
An external force directed to the right acts on the bar to balance the magnetic force to the
left. Hence, work is being done by the external force , which is transformed into the resistor’s thermal energy.
The emf induced in a rotating coil is directly proportional to the angular frequency of the coil.
Thus,
ε2
ε1
20.34
=
ω2
ω1
or
 
 ω1 

rev min 
( 24.0 V) = 13.3 V
 900 rev min 
ε 2 =  ω 2  ε1 =  500
ε max = NBhorizontal Aω = 100 ( 2 .0 × 10 −5
rev   2π rad   1 min  
2 
T ( 0.20 m )  1500


 

min
1 rev   60 s  

)
= 1.3 × 10 −2 V = 13 mV
20.35
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 ε = ε max sin ω t
where ε max = 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
ε max
56157_20_ch20_p219-251.indd 242
)
(1.0 × 10−3 T π ( 8.0 × 10−6 m
 πd2 
= NBAω = B 
ω
=
4
 4 
) (120π s ) =
2
−1
1.9 × 10 −11 V
3/19/08 1:50:08 AM
Induced Voltages and Inductance
20.36
(a)
Using
ε max = NBAω ,
ε max = 1 000 ( 0.20 T) ( 0.10 m 2 )  60

(b)
20.37
rev   1 min   2π rad 
rad

ω =  120
 
 
 = 4π

min
60 s
1 revv
s
T=
2π
= 0.50 s
ω
rad s ) = 60 V
(a)
ε max = NBAω = 500 ( 0.60 T) ( 0.080 m ) ( 0.20 m ) ( 4π
(b)
ε = ε max sin (ω t ) = ( 60 V) sin ( 4π
(c)
The emf is first maximum at t =
(a)
Immediately after the switch is closed, the motor coils are still stationary and the back emf
ε 240 V = 8.0 A .
is zero. Thus, I = =
R 30 Ω
(b)
At maximum speed,

I=
20.39
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 .
and the period is
20.38
243
ε − εback
R
π
rad s ) 
 32

s  = 57 V

T 0.50 s
=
= 0.13 s .
4
4
ε back = 145 V and
=
240 V − 145 V
= 3.2 A
30 Ω
(c)
εback = ε − IR = 240 V − ( 6.0 A ) ( 30 Ω ) =
(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
ε = ε max sin ω t
where
60 V
ε max = 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
rev   1 min 

ω = 2π f = 2π  100
 = 10.5 rad s


min   60.0 s 
(b)
Thus,
ε max = (10.0 ) ( 55.0 × 10 −6
or
ε max = 1.81 × 10 −5
π

T  ( 0.100 m ) ( 0.040 0 m )  (10.5 rad s )
4

)
V = 18.1 µV
When the rotation axis is parallel to the field, then B⊥ = 0 giving
ε max =
0
It is easily understood that the induced SSM 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.
56157_20_ch20_p219-251.indd 243
3/19/08 1:50:08 AM
244
20.40
Chapter 20
(a)
In terms of its cross-sectional area (A), length ( l ), and number of turns (N), the self
inductance of a solenoid is given as L = µ0 N 2 A l. Thus, for the given solenoid,
L=
20.41
µ0 N 2 (π d 2 4 )
l
−7
)
T ⋅ m A ( 580 ) π ( 8.0 × 10 −2 m
2
4 ( 0.36 m )
(b)
ε = − L  ∆I  = − ( 5.9 × 10 −3 H ) ( +4.0
From
ε
∆t
)
2
= 5.9 × 10 −3 H = 5.9 mH
A s ) = −24 × 10 −3 V = −24 mV
= L ∆I ∆t , we have
L=
20.42
( 4π × 10
=
ε
∆I ∆t
=
ε ( ∆t ) (12 × 10 −3 V) ( 0.50 s )
∆I
=
2.0 A − 3.55 A
= 4.0 × 10 −3 H = 4.0 mH
NΦB
T ⋅ m2
are
I
A
The units of
F
From the force on a moving charged particle, F = qvB , the magnetic field is B =
and
qv
we find that
1T=1
N
N ⋅s
=1
C ⋅ ( m s)
C⋅m
 N ⋅ s  2 (N ⋅ m) ⋅ s  J 
⋅m =
Thus, T ⋅ m 2 = 
=   ⋅s = V⋅s
 C ⋅ m 
 C
C
and
20.43
ε
T ⋅ m2 V ⋅ s
which is the same as the units of
=
∆
I
∆t
A
A
(a)
2
2
−7
−2


µ0 N 2 A ( 4π × 10 T ⋅ m A ) ( 400 ) π ( 2 .5 × 10 m ) 
L=
=
l
0.20 m
= 2 .0 × 10 −3 H = 2 .0 mH
(b)
20.44
From
From
ε
ε
= L ( ∆I ∆t ),
∆I ε
75 × 10 −3 V
=
=
= 38 A s
∆t
L 2 .0 × 10 −3 H
= 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
ΦB =
20.45
56157_20_ch20_p219-251.indd 244
)
−3
L ⋅ I ( 2 .40 × 10 H ( 4.00 A )
=
= 1.92 × 10 −5 T ⋅ m 2
500
N
L 12 × 10 −3 H
=
= 4.0 × 10 −3 s = 4.0 ms
R
3.0 Ω
(a)
τ=
(b)
I max =
ε,
R
so
ε = I max R = (150 × 10 −3 A ) ( 3.0 Ω ) = 4.5 × 10 −1 V =
0.45 V
3/19/08 1:50:09 AM
Induced Voltages and Inductance
20.46
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
(a)
RC =
(b)
L
,
R
L
=
C
R=
or
3.00 H
= 1.00 × 10 3 Ω = 1.00 kΩ
3.00 × 10 −6 F
The common value of the two time constants is
τ=
20.47
245
ε,
L
3.00 H
=
= 3.00 × 10 −3 s = 3.00 ms
R 1.00 × 10 3 Ω
ε = I max R = ( 8.0 A ) ( 0.30 Ω ) =
(a)
I max =
(b)
The time constant is τ =
R
so
2.4 V
L
, giving
R
L = τ R = ( 0.25 s ) ( 0.30 Ω ) = 7.5 × 10 −2 H = 75 mH
)
The current as a function of time is I = I max (1 − e−t τ , so at t = τ ,
(c)
(
)
I = I max 1 − e−1 = 0.632 I max = 0.632 ( 8.0 A ) = 5.1 A
At t = τ , I = 5.1 A and the voltage drop across the resistor is
(d)
∆VR = − IR = − ( 5.1 A ) ( 0.30 Ω ) = −1.5 V
Applying Kirchhoff’s loop rule to the circuit shown in Figure 20.27 gives
ε + ∆VR + ∆VL = 0. Thus, at t = τ , we have
∆VL = − (ε + ∆VR ) = − ( 2.4 V − 1.5 V) = −0.90 V
20.48
resistor is ∆VR = RI = ε (1 − e
the inductor is
−t τ
ε
)
(1 − e−t τ . The potential difference across the
R
, and from Kirchhoff’s loop rule, the potential difference across
The current in the RL circuit at time t is I =
)
)
∆VL = ε − ∆VR = ε 1 − (1 − e−t τ  = ε e−t τ
20.49
)
) = ( 6.0 V) (1 − 0.368 ) =
(a)
At t = 0, ∆VR = ε (1 − e−0 = ε (1 − 1) = 0
(b)
At t = τ , ∆VR = ε (1 − e−1
(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
)
From I = I max (1 − e−t τ , e−t τ = 1 −
If
I
I max
3.8 V
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=
56157_20_ch20_p219-251.indd 245
L 2.50 H
=
= 1.92 Ω
τ
1.30 s
3/19/08 1:50:10 AM
246
20.50
Chapter 20
τ=
(b)
I=
(c)
I max =
(d)
I = I max (1 − e−t τ
and
20.51
L 8.00 mH
=
= 2 .00 ms
R
4.00 Ω
(a)
(a)
ε
R
.00 V 
(1 − e ) =  64.00
 (1 − e
Ω
−t τ
ε
R
=
(a)
= 1 − I I max ,
)
If I = 3.0 A, the stored energy will be
1 2 1
2
LI = ( 2.08 × 10 −4 H ( 3.0 A ) = 9.36 × 10 −4 J = 0.936 mJ
2
2
)
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,
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
When the solenoid described above carries a current of I = 0.500 A, the stored energy is
1 2 1
2
LI = ( 4.44 × 10 −3 H ( 0.500 A ) = 5.55 × 10 −4 J
2
2
)
ε
(
)
The current in the circuit at time t is I = 1 − e−t τ , and the energy stored in the inductor is
R
1
PEL = LI 2
2
(a)
As t → ∞, I → I max =
PEL →
(b)
(a)
ε
R
=
24 V
= 3.0 A , and
8.0 Ω
1 2
1
2
LI max = ( 4.0 H ) ( 3.0 A ) = 18 J
2
2
)
At t = τ , I = I max (1 − e−1 = ( 3.0 A ) (1 − 0.368 ) = 1.9 A
and
20.54
−t τ
−3
2 ( PEL ) 2 ( 0.300 × 10 J
=
= 2.08 × 10 −4 H = 0.208 mH
2
I2
(1.70 A )
PEL =
20.53
) yields e
The energy stored by an inductor is PEL = 12 LI 2 , so the self inductance is
L=
(b)
0.176 A
6.00 V
= 1.50 A
4.00 Ω
PEL =
20.52
)=
t = − τ ln (1 − I I max ) = − ( 2 .00 ms ) ln (1 − 0.800 ) = 3.22 ms
L=
(b)
− 250 ×10 −6 s 2.00 ×10 −3 s
PEL =
1
( 4.0 H ) (1.9 A )2 = 7.2 J
2
Use Table 17.1 to obtain the resistivity of the copper wire and find
Rwire =
−8
ρCu L ρCu L (1.7 × 10 Ω ⋅ m ) ( 60.0 m )
= 2 =
= 1.3 Ω
2
Awire π rwire
π ( 0.50 × 10 −3 m )
continued on next page
56157_20_ch20_p219-251.indd 246
3/19/08 4:05:27 AM
Induced Voltages and Inductance
247
L
L
60.00 m
=
=
= 4.8 × 10 2 turns
Circumference of a loop 2π rsolenoid 2π ( 2.0 × 10 −2 m
(b)
N=
(c)
The length of the solenoid is
)
)
l = N ( diameter of wire ) = N ( 2 rwire ) = ( 480 ) 2 ( 0.50 × 10 −3 m = 0.48 m
2
( 4π × 10−7 T ⋅ m A) ( 480 ) π ( 2.0 × 10−2 m )
µ N 2 Asolenoid µ0 N 2π rsolenoid
L= 0
=
=
l
l
0.48 m
2
(d)
giving
2
L = 7.6 × 10 −4 H = 0.76 mH
L
L
7.6 × 10 −4 H
=
=
= 4.6 × 10 −4 s = 0.46 ms
Rtotal Rwire + rinternal 1.3 Ω + 0.350 Ω
ε =
6.0 V
I max =
= 3.6 A
Rtotal 1.3 Ω + 0.350 Ω
τ=
(e)
(f)
)
I = I max (1 − e−t τ , so when I = 0.999 I max, we have 1 − e−t τ = 0.999 and
t
or
e−t τ = 1 − 0.999 = 0.001. Thus, − = ln ( 0.001)
t = −τ ⋅ ln ( 0.001)
τ
(g)
giving
t = − ( 0.46 ms ) ⋅ ln ( 0.001) = 3.2 ms
( PEL )max =
(h)
1 2
1
2
LI max = ( 7.6 × 10 −4 H ( 3.6 A ) = 4.9 × 10 −3 J = 4.9 mJ
2
2
)
20.55
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 .
20.56
ε
20.57
=
ε ⋅ ∆t
∆Φ B NBA  ∆ ( cos θ ) 
=
, so B =
∆t
∆t
NA  ∆ ( cos θ ) 
( 0.166 V) ( 2 .77 × 10 −3 s )
= 5.20 × 10 −5
2


500 π ( 0.150 m ) 4  [ cos 0° − cos 90°]
T = 52 .0 µ T
or
B=
(a)
The current in the solenoid reaches I = 0.632 I max in a time of t = τ = L R, where
−7
−4
2
µ N 2 A ( 4π × 10 T ⋅ m A ) (12 500 ) (1.00 × 10 m )
L= 0
=
= 0.280 H
l
7.00 × 10 −2 m
2
Thus,
(b)
t=
0.280 H
= 2 .00 × 10 −2 s = 20.0 ms
14.0 Ω
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 .00 × 10 −2 s 
ε back = L  ∆I  = ( 0.280 H ) 
∆t
continued on next page
56157_20_ch20_p219-251.indd 247
3/19/08 4:05:33 AM
248
Chapter 20
(c)
∆Φ B ( ∆B ) A 12  µ0 n ( ∆I )  A µ0 N ( ∆I ) A
=
=
=
∆t
∆t
∆t
2 l ⋅ ( ∆t )
=
20.58
( 4π × 10
ε coil
)
T ⋅ m A (12 500 ) ( 2.71 A ) (1.00 × 10 −4 m 2
−7
N coil ( ∆Φ B
Rcoil
−3
V
s
)
)=
1.52 × 10 −3 V
) = 0.051 9 A =
I=
(a)
The gravitational force exerted on the ship by the pulsar supplies the centripetal acceleraGM pulsar mship mship v 2
tion needed to hold the ship in orbit. Thus, Fg =
, giving
=
2
rorbit
rorbit
Rcoil
=
m ( 2 .00 × 10
−2
(d)
GM pulsar
v=
24.0 Ω
( 6.67 × 10
=
rorbit
−11
)
N ⋅ m kg 2 ( 2.0 × 10 30 kg
3.0 × 10 m
7
)=
51.9 mA
2.1 × 10 6 m s
(b)
The magnetic force acting on charged particles moving through a magnetic field is perpendicular to both the magnetic field and the velocity of the particles (and therefore perpendicular to the ship’s length). Thus, the charged particles in the materials making up the
spacecraft experience magnetic forces directed from one side of the ship to the other, meaning that the induced emf is directed from side to side within the ship.
(c)
ε = B⊥ lv , where ! = 2rship = 0.080 km = 80 m is the side to side dimension of the ship.
This yields
)
ε = (1.0 × 10 2
20.59
)
∆t ) ( 820 ) (1.52 × 10
=
2 ( 7.00 × 10
−2
)
T ( 80 m ) ( 2.1 × 10 6 m s = 1.7 × 1010 V
(d)
The very large induced emf would lead to powerful spontaneous electric discharges. The
strong electric and magnetic fields would disrupt the flow of ions in their bodies.
(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 l. The magnitude of
the induced current is
I=
ε
R
=
( ∆Φ B
R
so the field strength is
∆t )
=
B ( ∆A ∆t ) B lv
=
R
R
B=
IR
,
lv
giving
Fapp = I 2 R v
Thus, the current is
I=
56157_20_ch20_p219-251.indd 248
Fapp ⋅ v
R
=
(1.00 N ) ( 2 .00
8.00 Ω
m s)
= 0.500 A
(b)
P = I 2 R = ( 0.500 A )2 ( 8.00 Ω ) = 2 .00 W
(c)
Pinput = Fapp ⋅ v = (1.00 N ) ( 2 .00 m s ) = 2 .00 W
3/19/08 1:50:13 AM
Induced Voltages and Inductance
20.60
(a)
When the motor is first turned on, the coil is not rotating so the back emf is zero. Thus, the
current is a maximum with only the resistance of the windings limiting its value. This gives
I max = ε R, or
R=
(b)
ε
I max
=
120 V
= 11 Ω
11 A
When the motor has reached maximum speed, the steady state value of the current is
I = (ε − ε back ) R, giving the back emf as
ε back = ε − IR = 120 V − ( 4.0 A ) (11 Ω ) =
20.61
249
76 V
If d is the distance from the lightning bolt to the center of the coil, then
ε av
=
N ( ∆Φ B ) N ( ∆B ) A N  µ0 ( ∆I ) 2π d  A N µ0 ( ∆I ) A
=
=
=
2π d ( ∆t )
∆t
∆t
∆t
)
)
)
2
100 ( 4π × 10 −7 T ⋅ m A ( 6.02 × 10 6 A − 0 π ( 0.800 m ) 
=
2π ( 200 m ) (10.5 × 10 −6 s
= 1.15 × 10 5 V = 115 kV
20.62
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
( 3.00 m )2 − (1.50 m )2
= 2.60 m
The decrease in the enclosed area has been
2

1
∆A = Ai − A f = ( 3.00 m ) − 4  (1.50 m ) ( 2.60 m )  = 1.20 m 2

2
The average induced current has been
I av =
ε av
R
=
( ∆Φ B
R
∆t )
=
)
2
B ( ∆A ∆t ) ( 0.100 T) (1.20 m 0.100 s
=
= 0.120 A
R
10.0 Ω
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.
20.63
(a)
ε av
=
=
)
2
∆Φ B B ( ∆A ) B (π d 4 − 0 
=
=
∆t
∆t
∆t
( 25.0 mT) π ( 2 .00 × 10 −2
4 ( 50.0 × 10 −3 s
)
m
)
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 .
continued on next page
56157_20_ch20_p219-251.indd 249
3/19/08 4:45:01 AM
250
Chapter 20
(b)
ε av
(
−2
∆Φ B ( ∆B ) A (100 − 25.0 ) mT π 2 .00 × 10 m
=
=
=
∆t
∆t
4 4.00 × 10 −3 s
(
)
)
2
= 5.89 mV
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 .
20.64
The induced emf in the ring is
ε av
=
∆Φ B ( ∆B ) Asolenoid ( ∆Bsolenoid 2 ) Asolenoid 1 
 ∆I

=
=
=  µ0 n  solenoid   Asolenoid
 ∆t  
∆t
∆t
∆t
2
=
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 =
20.65
(a)
ε av
R
4.80 × 10 −4 V
= 1.60 A
3.00 × 10 −4 Ω
As the rolling axle (of length l = 1.50 m) moves perpendicularly to the uniform magnetic
field, an induced emf of magnitude ε = Blv will exist between its ends. The current produced in the closed-loop circuit by this induced emf has magnitude
I=
(b)
=
ε av
R
=
( ∆Φ B
R
∆t )
=
B ( ∆A ∆t ) B lv ( 0.800 T) (1.50 m ) ( 3.00 m s )
=
=
= 9.00 A
R
R
0.400 Ω
The induced current through the axle will cause the magnetic field to exert a retarding force
of magnitude Fr = BI l on the axle. The direction of this force will be opposite to that of
r
the velocity v so as to oppose the motion of the axle. If the axle is to continue moving at
r
constant speed, an applied force in the direction of v and having magnitude Fapp = Fr must
be exerted on the axle.
Fapp = BI l = ( 0.800 T) ( 9.00 A ) (1.50 m ) = 10.8 N
(c)
(d)
56157_20_ch20_p219-251.indd 250
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.
r
r
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.
3/19/08 4:04:31 AM
Induced Voltages and Inductance
20.66
(a)
251
The time required for the coil to move distance l and exit the field is t = l v, where v is
the constant speed of the coil. Since the speed of the coil is constant, the flux through the
area enclosed by the coil decreases at a constant rate. Thus, the instantaneous induced emf
is the same as the average emf over the interval t seconds in duration, or
ε = − N ∆Φ = − N (
∆t
0 − BA )
Bl 2 NBl 2
=N
=
= NBlv
t−0
t
lv
I=
ε
=
NBlv
R
(b)
The current induced in the coil is
(c)
The power delivered to the coil is given by P = I 2 R , or
R
 N 2 B 2 l2 v 2 
N 2 B 2 l2 v 2
=
R

R2
R
P =

(d)
The rate that the applied force does work must equal the power delivered to the coil, so
Fapp ⋅ v = P or
Fapp =
(e)
P
v
=
N 2 B 2 l2 v 2 R N 2 B 2 l2 v
=
v
R
As the coil is emerging from the field, the flux through the area it encloses is directed into
the page and decreasing in magnitude. Thus, the change in the flux through the coil is
directed out of the page. The induced current must then flow around the coil in such a direction as to produce flux into the page through the enclosed area, opposing the change that is
occurring. This means that the current must flow clockwise around the coil.
As the coil is emerging from the field, the left side of the coil is carrying an induced
current directed toward the top of the page through a magnetic field that is directed
into the page. Right-hand rule 1, then shows that this side of the coil will experience
a magnetic force directed to the left , opposing the motion of the coil.
56157_20_ch20_p219-251.indd 251
3/19/08 1:50:14 AM