PHYSICS 212
CHAPTER 20
INDUCED VOLTAGES AND INDUCTANCE
WORKBOOK
ANSWERS
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4. Complete the sections in sequence.
5. Study and learn definitions of terms, physical quantities, units, principles, and
basic equations before attempting problems.
6. You may work on this with other students but do not copy another student’s
workbook or let a student copy this workbook. Do not copy from other sources
either.
7. Wherever possible, include diagrams in your solutions. Diagrams are required.
8. Keep this workbook after it is graded and returned to you.
9. Using the answer key, redo all questions and problems until you can answer
them all correctly by yourself without help.
10. Use the workbook to learn the general problem-solving strategy rather than
how individual problems are solved.
11. Questions marked (Basic) should be answered by students without the need
for additional assistance.
-1-
PHYSICS 212 CHAPTER 20 MULTIPLE CHOICE QUESTIONS
INDUCED VOLTAGES AND INDUCTANCE
Multiple Choice
Identify the letter of the choice that best completes the statement or answers the question.
____1.
B
A uniform 4.5-T magnetic field passes perpendicularly through the plane of a wire loop 0.10
m2 in area. What flux passes through the loop?
A. 5.0 T⋅m2
D. 0.135 T⋅m2
B. 0.45 T⋅m2
E. 0.15 T⋅m2
C. 0.25 T⋅m2
(Basic)
B____2. The units TAm2/s are equivalent to:
A. W.
B. V.
C. N/m.
D. webers.
E. F
(Basic)
____3.
B
A sensitive ammeter is connected to a wire loop and placed within the magnetic field of a
strong horseshoe magnet. The ammeter shows a deflection when:
A. the wire is moved parallel to the field.
B. the wire is moved perpendicularly to the field.
C. neither wire nor magnet is moving.
D. the wire's axis is parallel to the field.
E. the wire's axis is perpendicular to the field.
____4.
D
According to Lenz's law the direction of an induced current in a conductor will be that which
tends to produce which of the following effects?
A. enhance the effect which produces it D. oppose the effect which produces it
B. produce a greater heating effect
E. produce the greatest magnetic field
C. produce the greatest voltage
C____5. "GFI" stands for:
A. grand flux indicator.
B. ground forcing indicator.
C. ground fault interrupter.
____6.
B
D. gauss-free invention.
E. guided fault isolation.
(Basic)
The principle or law that says "an induced emf in a circuit loop produces a current whose
magnetic field opposes further change of magnetic flux" is credited to:
A. Faraday.
D. Volta.
B. Lenz.
E. Maxwell.
C. Ampere.
(Basic)
-2-
____7.
C
A coil is placed in a changing magnetic field and an emf is induced. What happens to the
induced emf if the rate of change of magnetic field quadruples?
A. There is no change.
D. The emf increases by a factor of 16.
B. The emf doubles.
E. The emf halves.
C. The emf quadruples.
B____8. The magnet moving past an object will produce eddy currents in the object if the object:
A. is magnetic material only.
D. is a liquid.
B. is a conductor.
E. is a paramagnetic material only..
C. is an insulator.
E____9. The operation of a tape player to play music depends on which of the following?
A. the Doppler effect
B. the Meissner effect
C. the photoelectric effect
D. the force acting on a current-carrying wire in a magnetic field
E. induced current from the motion of a magnet past a wire
____10. If the induced current in a wire loop were such that the flux it produces were in the same
C
direction as the change in external flux causing the current, which of the following
conservation laws would end up being violated?
A. momentum
D. angular momentum
B. charge
E. mass
C. energy
(Basic)
D___11. The operation of an electric motor depends on which of the following effects?
A. the Doppler effect
B. the Meissner effect
C. the photoelectric effect
D. the force acting on a current-carrying wire in a magnetic field
E. current from the motion of a wire in a magnetic field
A___12. The basic function of the electric generator is which of the following conversion processes?
A. mechanical energy to electrical
D. alternating current to direct
B. electrical energy to mechanical
E. direct current to alternating
C. low voltage to high or vice versa
(Basic)
B___13. The function of the electric motor is which one of the following conversion processes?
A. mechanical energy to electrical
D. alternating current to direct
B. electrical energy to mechanical
E. direct current to alternating
C. low voltage to high or vice versa
D___14. The back emf in an electric motor is its maximum value under which condition?
A. motor speed is zero
D. motor speed is a maximum
B. current is a maximum
E. minimal heating effect
C. voltage is a maximum
-3-
____15. Electricity may be generated by rotating a loop of wire between the poles of a magnet. The
A
induced current is greatest when:
A. the plane of the loop is parallel to the magnetic field.
B. the plane of the loop is perpendicular to the magnetic field.
C. the magnetic flux through the loop is a maximum.
D. the plane of the loop makes an angle of 45° with the magnetic field.
E. the plane of the loop makes an angle of 60° with the magnetic field.
D___16. The "back emf" of a motor refers to a source of voltage that:
A. occurs when the motor runs backwards.
B. occurs when the motor is used as a generator.
C. is biggest when the current through the motor is biggest.
D. is biggest when the motor turns fastest.
E. is biggest when the motor is accelerating.
____17. When a voltage is generated by rotating a coil in a magnetic field at a constant rate, the period
D
of the voltage equals the time that it takes for the coil to rotate through ____ radians.
A. 1
D. 2π
B. π/2
E. π/4
C. π
D___18. The self-inductance of a solenoid increases under which of the following conditions?
A. only the solenoid length is increased
B. only the cross sectional area is decreased
C. only the number of coils per unit length is decreased
D. only the number of coils is increased
E. only the solenoid length is decreased
____19. By what factor is the self-inductance of an air solenoid changed if only its cross-sectional area,
B
A, is tripled?
A. 1/3
D. 9
B. 3
E. 1/9
C. 6
____20. An inductor, battery, resistance, and ammeter and switch are connected in series. If the switch,
D
initially open, is now closed, what is the current's final value?
A. zero
D. battery voltage divided by resistance
B. battery voltage divided by
E. resistance times inductance
inductance
C. battery voltage times inductance
___21.
A
In a circuit made up of inductor, resistance, ammeter, battery and switch in series, at which of
the following times after the switch is closed is the rate of current increase greatest?
A. zero
D. ten time constants
B. one time constant
E. infinity
C. reciprocal of one time constant
-4-
C___22. How is the energy stored in a current-carrying inductor related to its self-inductance, L?
A. directly proportional to L2
D. inversely proportional to L
1/2
B. directly proportional to L
E. inversely proportional to L2
C. directly proportional to L
A___23. How is the energy stored in a current-carrying inductor related to the current value, I?
A. directly proportional to I2
D. inversely proportional to I
1/2
B. directly proportional to I
E. inversely proportional to I2
C. directly proportional to I
-5-
PHYSICS 212 CHAPTER 20 ADDITIONAL QUESTIONS
INDUCED VOLTAGES AND INDUCTANCE
1.
A uniform magnetic field of 15 G is oriented so that it makes a 20o angle with the plane of a square
loop of wire of area 5.00 cm2. What is the magnetic flux through this loop?
(Basic)
φ B = BA cos θ = (15×10
φ B = 2.57×10
2.
−7
−4
−4
2
0
T )(5×10 m ) cos(70 )
wb
A rectangular loop is initially held with one edge parallel to a long conducting wire carrying a 12 A
current. The loop has an area of 4.00 cm2 with one edge 3 cm from the wire. What is the induced
emf in the loop when it is pulled at .05 m/s parallel to the wire as shown?
(Basic)
As the loop pulled, neither B, A nor θ change. Thus φB does not change and the emf is 0.
3.
The bar magnet in the diagram is suddenly pulled upward away from the loop. Draw an arrow
along the front edge of the loop to indicate the direction of the induced current in the loop. (Basic)
-6-
4.
The solenoid has a length of 15 cm and is comprised of 7500 turns of wire. The current to the
solenoid is changing steadily from 0 A to 12 A in 15 seconds. The coil inside the solenoid has a
radius of 4 cm and is comprised of 250 turns of wire with a net resistance of 3 Ω. The axis of the
coil is offset 25o to the axis of the solenoid.
A. What is the induced emf in the coil?
B. What is the induced current in the coil?
C. What is the torque on the coil when the solenoid current is 6A?
E = NC
φ
φ
φ
Bf
Bf
Bi
∆φB
∆t
∆φB = φBf − φBi
µ N I 
= BS AC cos θ =  o s S  π rC2 cosθ
l


−7 TM
(4π ×10 A )(7500)(12 A)π (.04m) 2 cos(25)
=
= 3.434 ×10−3 wb
.15m
= 0 because I s is zero
(3.435 ×10−3 wb)
E = (250)
= 5.725 × 10−2 V
15s
E 5.725 × 10−2
IC = =
= 1.908 ×10−2 A
R
15s
µ o N S I S (4π × 10−7 TmA )(7500)(6 A)
BS =
=
= 3.77 ×10−1T
l
.15m
mC = N C I C AC = (250)(1.908 ×10−2 A)π (.04m) 2 = 2.398 ×10−2 Am 2
τ = mC BS sin θ = (2.398 ×10−2 Am 2 )(3.77 ×10−1T ) sin(250 ) = 3.820 ×10−3 Nm
-7-
5.
An inductor is 2 cm long, has a radius of .5 cm and is comprised of 75 turns of wire. What is its
inductance?
(Basic)
L=
µo N 2 A
l
=
(4π × 10 −7
)(75) 2 π (.005m) 2
.02m
Tm
A
L = 2.776 × 10−5 H
6.
The inductor in problem 5 is connected in series to a 0.5 Ω and a battery whose emf is 24 V. What
current flows in this circuit 0.1 seconds after the switch is closed?
I=
E
 24V
1 − e − Rt / L  = 
R
 .5Ω

− (.5 Ω )(.1s ) /(2776×10−5 H ) 
 1 − e


I = 48.0 A
-8-
7.
There are two conducting, parallel rods spaced 10 cm apart. The bottom ends of the rods are
connected by a wire with a 2 Ω resistor. A uniform and horizontal magnetic field of 2.25 T is
oriented perpendicularly to the plane of the two rods. A small conducting bar of mass 15g is placed
across the two vertical rods. This bar is horizontal and slides against the vertical rods without
friction. When released the bar falls and reaches a constant speed. What is that speed?
FB = mg
IlB sin θ = mg
E
  lB = mg
 R
 vBl 

 lB = mg
 R 
vB 2l 2
= mg
R
Rmg (2Ω)(.015kg )(9.8 sm2 )
v= 2 2 =
= 5.81 ms
2
2
Bl
(2.25T ) (.1m)
-9-
8.
A conducting loop is pulled away from a current, carrying wire. Show in which direction the
induced current will flow around the loop.
The current in the long wire produces an outward magnetic field in the vicinity of the loop. As the loop
is moved to the right it moves into weaker field strength. This causes the flux to decrease. By Lenz’s
law an induced current must produce additional flux to offset this decrease. This requires the induced
current’s magnetic field to point in the same direction as the field from the long wire, outward. To do
this the current must flow around the loop in a counter-clockwise direction.
-10-
PHYSICS 212 CHAPTER 20 PROBLEMS
INDUCED VOLTAGES AND INDUCTANCE
PRACTICE PROBLEMS:
HOMEWORK PROBLEMS:
6, 8, 16, 19, 27, 29, 36, 41, 55, 59
1, 5, 9, 14, 15, 21, 23, 25, 30, 35, 39, 45, 49
[NOTE: Problem numbers in italic are included in the student study guide; Problem numbers underlined
are online at www.cp7e.com.]
1.
A magnetic field of strength 0.30 T is directed perpendicular to a plane circular loop of wire of
radius 25 cm. Find the magnetic flux through the area enclosed by this loop.
(Basic)
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


5.
A long, straight wire lies in the plane of a circular coil with a radius of 0.010 m. The wire carries
a current of 2.0 A and is placed along a diameter of the coil. (a) What is the net flux through the
coil? (b) If the wire passes through the center of the coil and is perpendicular to the plane of the
coil, find the net flux through the coil.
(Basic)
(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
-11-
9.
A square, single-turn coil 0.20 m on a side is placed with its plane perpendicular to a constant
magnetic field. An emf of 18 mV is induced in the coil winding when the area of the coil
decreases at the rate of 0.10 m2/s. What is the magnitude of the magnetic field?(Basic)
From
B=
14.
ε
=
( ∆A
∆Φ B
∆t
ε
=
∆t ) cosθ
B ∆A cosθ
, we find that
∆t
=
18 × 10 −3 V
( 0.10 m s ) cos 0°
2
= 0.18 T
A square, single-turn wire loop 1.00 cm on a side is placed inside a solenoid that has a circular
cross section of radius 3.00 cm, as shown in Figure P20.14. The solenoid is 20.0 cm long and
wound with 100 turns of wire. (a) If the current in the solenoid is 3.00 A, find the flux through the
loop. (b) If the current in the solenoid is reduced to zero in 3.00 s, find the magnitude of the
average induced emf in the loop.
Figure P20.14
The initial magnetic field inside the solenoid is
 100 
−3
B = µ 0 nI = 4π × 10 −7 T ⋅ m A 
 ( 3.00 A ) = 1.88 × 10 T
 0.200 m 
(
(a)
)
(
)(
)
2
Φ B = BA cosθ = 1.88 × 10−3 T 1.00 × 10 −2 m cos 0°
= 1.88 × 10−7 T ⋅ m 2
-12-
(b)
When the current is zero, the flux through the loop is Φ B = 0 and the average induced emf
has been
ε
15.
=
∆Φ B 1.88 × 10 −7 T ⋅ m 2 − 0
=
= 6.28 × 10−8 V
∆t
3.00 s
A 300-turn solenoid with a length of 20 cm and a radius of 1.5 cm carries a current of 2.0 A. A
second coil of four turns is wrapped tightly about this solenoid so that it can be considered to have
the same radius as the solenoid. Find (a) the change in the magnetic flux through the coil and (b)
the magnitude of the average induced emf in the coil when the current in the solenoid increases to
5.0 A in a period of 0.90 s.
If the solenoid has current I, the magnetic field inside it is
 300 
−4
B = µ 0 nI = 4π × 10−7 T ⋅ m A 
 I = 6.00 π × 10 T A ⋅ I
 0.200 m 
∆Φ B = ( ∆B ) A cosθ
(
)
(
(
)
)
(
= 4.0 × 10 −6 T ⋅ m 2
(b)
)
2
=  6.00 π × 10 −4 T A ( 5.0 A − 2.0 A )  π 1.5 × 10 −2 m  cos 0°

 

(a)
ε
=
N ( ∆Φ B )
∆t
=
(
4 4.0 × 10−6 T ⋅ m 2
0.90 s
-13-
) = 1.8 × 10
−5
V = 18 µ V
21.
An automobile has a vertical radio antenna 1.20 m long. The automobile travels at 65.0 km/h on a
horizontal road where the Earth’s magnetic field is 50.0 :T, directed toward the north and
downwards at an angle of 65.0/ below the horizontal. (a) Specify the direction that the automobile
should move in order to generate the maximum motional emf in the antenna, with the top of the
antenna positive relative to the bottom. (b) Calculate the magnitude of this induced emf.
(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 B is the horizontal component of the magnetic field.


ε = ( 50.0 × 10−6 T ) cos 65.0° (1.20 m )  65.0 km   0.278 m s  
h

h   1 km h  
= 4.58 × 10− 4 V
23.
A bar magnet is positioned near a coil of wire as shown in Figure P20.23. What is the direction of
the current in the resistor when the magnet is moved (a) to the left? (b) to the right?(Basic)
Figure P20.23
(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
-14-
right to left through the resistor.
25.
A rectangular coil with resistance R has N turns, each of length R and width w as shown in Figure
P20.25. The coil moves into a uniform magnetic field B with constant velocity v. What are the
magnitude and direction of the total magnetic force on the coil (a) as it enters the magnetic field,
(b) as it moves within the field, and (c) as it leaves the field?
Figure P20.25
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=
∆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 = 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 B 2 w2 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.
-15-
(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 B 2 w2 v
 NBwv 
F = BI ( Nw ) = B 
Nw
=
(
)

R
 R 
30.
directed toward the left
A 100-turn square wire coil of area 0.040 m2 rotates about a vertical axis at 1 500 rev/min, as
indicated in Figure P20.30. The horizontal component of the Earth’s magnetic field at the location
of the loop is 2.0 × 10!5 T. Calculate the maximum emf induced in the coil by the Earth’s field.
Figure P20.30
ε max = NBhorizontal Aω = 100 ( 2.0 × 10−5 T ) ( 0.20 m )2 1500

rev  2π rad  1 min  


min 
 1 rev  60 s  
= 1.3 × 10−2 V = 13 mV
35.
In a model AC generator, a 500-turn rectangular coil 8.0 cm by 20 cm rotates at 120 rev/min in a
uniform magnetic field of 0.60 T. (a) What is the maximum emf induced in the coil? (b) What is
the instantaneous value of the emf in the coil at t = (B/32) s? Assume that the emf is zero at t = 0.
(c) What is the smallest value of t for which the emf will have its maximum value?
-16-


ω =  120
rev  1 min  2π rad 
rad


 = 4π
min  60 s  1 rev 
s
T=
and the period is
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 (ω t ) = ( 60 V ) sin ( 4π

(c)
39.
rad s ) = 60 V
 π 
rad s ) 
s   = 57 V
 32  
The emf is first maximum at t =
T 0.50 s
=
= 0.13 s
4
4
A solenoid of radius 2.5 cm has 400 turns and a length of 20 cm. Find (a) its inductance and (b) the
rate at which current must change through it to produce an emf of 75 mV. (Basic)
(a)
L=
µ0 N A
2
ℓ
=
( 4π ×10
−7
)
(
= 2.0 × 10−3 H = 2.0 mH
(b)
From
ε
)
2
2
T ⋅ m A ( 400 ) π 2.5 × 10 −2 m 


0.20 m
∆I ε
75 × 10−3 V
=
=
= 38 A s
= L ( ∆I ∆t ) ,
∆t
L 2.0 × 10−3 H
-17-
45.
Calculate the resistance in an RL circuit in which L = 2.50 H and the current increases to 90.0% of
its final value in 3.00 s.
(
)
−t τ
−t τ
=1−
From I = I max 1 − e
, e
I
I max
If at t = 3.00 s , then
−3.00 s
= 1.30 s
e −3.00 s τ = 0.100 or τ =
ln ( 0.100 )
Since the time constant of an RL circuit is τ = L R , the resistance is
R=
49.
L
τ
=
2.50 H
= 1.92 Ω
1.30 s
A 24-V battery is connected in series with a resistor and an inductor, with R = 8.0 S and L = 4.0 H,
respectively. Find the energy stored in the inductor (a) when the current reaches its maximum value
and (b) one time constant after the switch is closed.
The current in the circuit at time t is I =
ε
(1 − e ) , and the energy stored in the inductor is
R
−t τ
PEL =
1 2
LI
2
(a)
As t → ∞ , I → I max =
(b)
−1
At t = τ , I = I max 1 − e = ( 3.0 A )(1 − 0.368 ) = 1.9 A
and
(
PEL =
ε
R
=
24 V
= 3.0 A , and
8.0 Ω
)
1
2
( 4.0 H )(1.9 A ) = 7.2 J
2
-18-