Review Questions
Multiple Choice
Identify the choice that best completes the statement or answers the question.
____
1. How much current is required to produce a force of 0.96 N on a 75 cm long wire oriented at right angles to a
0.24 T magnetic field?
a. 0.050 A
c. 5.3 A
b. 0.17 A
d. 0.19 A
____
2. A 0.050-m-long wire is carrying a current of 2.0 A at right angles to a 0.15-T magnetic field. What is the
magnitude of the force exerted by the magnetic field on the wire?
a. 0.015 N
c. 0.67 N
b. 0.0037 N
d. 6.0 N
____
3. Which device is used to measure very small currents?
a. armature
c. voltmeter
b. galvanometer
d. potentiometer
____
4. Which rule is used to find the direction of force acting on a current-carrying wire in a magnetic field?
a. the first right-hand rule
b. the third right-hand rule
c. the second right-hand rule
d. the fourth right-hand rule
____
5. The _____ is a method used to determine the direction of the force on a current carrying wire in a magnetic
field.
a. first right hand rule
c. third right hand rule
b. second right hand rule
d. none of the above
____
6. A/An _____ is a magnet created when current flows though a wire coil.
a. solenoid
c. electric motor
b. electromagnet
d. armature
____
7. A/An _____ is a device that is used to measure very small currents.
a. solenoid
c. galvanometer
b. electromagnet
d. armature
____
8. The _____ is a method used to determine the direction of a magnetic field relative to the direction of
conventional current.
a. first right hand rule
c. third right hand rule
b. second right hand rule
d. none of the above
____
9. A/An _____ is formed when the magnetic fields of electrons are aligned in the same direction.
a. magnetic field
c. domain
b. magnetic flux
d. polarization
____ 10. A/An _____ is an apparatus that converts electric energy to rotational kinetic energy.
a. solenoid
c. electric motor
b. electromagnet
d. armature
____ 11. A/An _____ is the number of magnetic field lines that pass through a surface.
a. magnetic field
c. solenoid
b. magnetic flux
d. armature
____ 12. A/An _____ is the wire coil of an electric motor.
a. solenoid
c. domain
b. electromagnet
d. armature
____ 13. A/An _____ is described as the area around a magnet where other magnets would be accelerated.
a. magnetic field
c. domain
b. magnetic flux
d. solenoid
____ 14. A straight wire carrying a current of 7.2 A has a field of 8.9 10-3 T at right angles to it. What length of
wire in the field will experience a force of 2.1 N?
a. 2.6 10-3 m
c. 1.3 10-1 m
-1
b. 3.1 10 m
d. 3.3 101 m
____ 15. Assume that a 19 cm length of wire carries a current at right angles to a 4.1 T magnetic field. The force on the
wire is 3.10 10-2 N. What is the current in the wire?
a. 3.4 10-7 A
c. 1.9 10-2 A
b. 39.8 10-3 A
d. 9.8 A
____ 16. A 7.12 µC charge is moving at one half the speed of light perpendicular to a magnetic field of 4.02 mT.
What is the force on the charge?
a. 4.30 N
c. 4.30 1012 N
1
b. 32.90 10 N
d. 1.00 1016 N
____ 17. A uniform magnetic field of 0.25 T points vertically downward. A proton enters the field with a horizontal
velocity of 4.0 10 m/s north. What are the magnitude and direction of the instantaneous force exerted on
the proton as it enters the magnetic field?
a. 1.6 10-13 N west
c. 1.6 10-13 N east
-13
b. 1.6 10 N downward
d. 1.6 10-13 N to the right
____ 18. A wire that is 0.50 m long and carrying a current of 8.0 A is at right angles to a uniform magnetic field. The
force on the wire is 0.40 N. What is the strength of the magnetic field?
a. 0.010 T
c. 1.0 T
b. 0.10 T
d. 10 T
____ 19. The current through a wire that is 0.80 m long is 5.0 A. The wire is at right angles to a 0.60 T magnetic field.
What is the magnitude of the force on the wire?
a. 0.24 N
c. 24 N
b. 2.4 N
d. 240 N
____ 20. A wire that is 25 cm long is at right angles to a 0.30 T uniform magnetic field. The current through the wire
is 6.0 A. What is the magnitude of the force on the wire?
a. 0.45 N
c. 45 N
b. 4.5 N
d. 450 N
Completion
Complete each statement.
21. The number of magnetic field lines passing through a surface is called ____________________.
22. A(n) ____________________is an apparatus that converts electric energy into rotational kinetic energy.
Matching
Define the following terms by matching them with the correct definition.
a. polarized
g. second right hand rule
b. magnetic fields
h. domain
c. magnetic flux
i. third right hand rule
d. first right hand rule
j. galvanometer
e. solenoid
k. electric motor
f. electromagnet
l. armature
____ 23. formed when the magnetic fields of electrons are aligned in the same direction
____ 24. a method used to determine the direction of the field produced by an electromagnet relative to the flow of
conventional current
____ 25. the area around a magnet where other magnets may be attracted or repelled.
____ 26. the wire coil of an electric motor
____ 27. an apparatus that converts electric energy to rotational kinetic energy
____ 28. the number of magnetic field lines that pass through a surface
____ 29. a magnet created when current flows through a wire coil
____ 30. a method used to determine the direction of the force on a current carrying wire in a magnetic field
____ 31. a method used to determine the direction of a magnetic field relative to the direction of conventional current
____ 32. a device used to measure very small currents
____ 33. a long coil of wire with many loops; combined fields create a greater total strength
____ 34. the property of having two distinct poles or opposite ends
Problem
35. What is the force on a 12 cm straight wire in a 1.9 T magnetic field when the current in the wire is 25 A?
36. A wire that is 0.50 m long and carrying a current of 8.0 A is at a right angle to a 0.40 T magnetic field. How
strong is the force that acts on the wire?
37. How much current will be required to produce a force of 0.38 N on a 10.0 cm length of wire at right angles to
a 0.49 T field?
38. A 40.0 cm long copper wire carries a current of 6.0 A and weighs 0.35 N. A certain magnetic field is strong
enough to balance the force of gravity on the wire. What is the strength of the magnetic field?
39. A wire that is 75 cm long, carrying a current of 6.0 A is at right angles to a uniform magnetic field. The
magnitude of the force acting on the wire is 0.60 N. What is the strength of the magnetic field?
Review Questions
Answer Section
MULTIPLE CHOICE
1. ANS: C
Force on the wire, F = 0.96 N
Length of the wire, L = 75 cm = 0.75 m
Strength of the magnetic field, B = 0.24 T
Amount of current required,
Feedback
A
B
C
D
Convert the length of wire to meters and then calculate the current.
Force should be divided and not multiplied by the product of length of wire and strength
of field.
Correct!
The current is equal to the force divided by the product of length of wire and strength of
magnetic field.
PTS: 1
DIF: Bloom's Level 2
REF: Page 653
OBJ: 24.2.2 Solve problems involving magnetic field strength and the forces on current carrying wires¸ and
on moving charged particles in magnetic fields.
TOP: Solve problems involving magnetic field strength and the forces on current carrying wires, and on
moving charged particles in magnetic fields.;
KEY: Third right-hand rule
MSC: 3
NOT: /a/ Convert the length of wire to meters and then calculate the current. /b/ Force should be divided and
not multiplied by the product of length of wire and strength of field. /c/ Correct! /d/ The current is equal to the
force divided by the product of length of wire and strength of magnetic field.
2. ANS: A
The amount of current flowing through the conductor, I = 2.0 A
The length of the conductor, L = 0.050 m
The strength of the magnetic field, B = 0.15 T
The expression for force is F = ILB
Therefore, the magnitude of force is =
= 0.015 N
Feedback
A
B
C
D
Correct!
The current should be multiplied with the product of strength of the magnetic field and
the length of the conductor.
The strength of the magnetic field should be multiplied with the product of length of the
conductor and the current.
The length should be multiplied with the product of strength of the magnetic field and
current.
PTS: 1
DIF: Bloom's Level 1
REF: Page 653
OBJ: 24.2.2 Solve problems involving magnetic field strength and the forces on current carrying wires¸ and
on moving charged particles in magnetic fields.
TOP: Solve problems involving magnetic field strength and the forces on current carrying wires, and on
moving charged particles in magnetic fields.;
KEY: Third right-hand rule
MSC: 3
NOT: /a/ Correct! /b/ The current should be multiplied with the product of strength of the magnetic field and
the length of the conductor. /c/ The strength of the magnetic field should be multiplied with the product of
length of the conductor and the current. /d/ The length should be multiplied with the product of strength of the
magnetic field and current.
3. ANS: B
A galvanometer is used to measure very small currents.
Feedback
A
B
C
D
The armature is the wire coil of a motor that is mounted in the magnetic field.
Correct!
A voltmeter is used to measure potential difference.
A potentiometer is used to vary the amount of current flowing through the circuit.
PTS: 1
DIF: Bloom's Level 1
REF: Page 655
OBJ: 24.2.2 Solve problems involving magnetic field strength and the forces on current carrying wires¸ and
on moving charged particles in magnetic fields.
TOP: Solve problems involving magnetic field strength and the forces on current carrying wires, and on
moving charged particles in magnetic fields.;
KEY: Galvanometer
MSC: 1
NOT: /a/ The armature is the wire coil of a motor that is mounted in the magnetic field. /b/ Correct! /c/ A
voltmeter is used to measure potential difference. /d/ A potentiometer is used to vary the amount of current
flowing through the circuit.
4. ANS: B
The third right-hand rule gives the method of finding the direction of the force acting on a current carrying
conductor in a magnetic field.
Feedback
A
B
C
D
The first right-hand rule is used to find the direction of the magnetic field relative to the
direction of the conventional current.
Correct!
The second right-hand rule is used to find the polarity of an electromagnet.
The fourth right-hand rule is to find the direction of force on charges inside a conductor
moving through a magnetic field.
PTS: 1
DIF: Bloom's Level 1
REF: Page 652
OBJ: 24.2.1 Relate magnetic induction to the direction of the force on a current carrying wire in a magnetic
field.
TOP: Relate magnetic induction to the direction of the force on a current carrying wire in a magnetic field.
KEY: Third right-hand rule
MSC: 2
NOT: /a/ The first right-hand rule is used to find the direction of the magnetic field relative to the direction of
the conventional current. /b/ Correct! /c/The second right-hand rule is used to find the polarity of an
electromagnet. /d/ The fourth right-hand rule is to find the direction of force on charges inside a conductor
moving through a magnetic field.
5. ANS: C
The third right hand rule is used to determine the direction of force on a current carrying wire in a magnetic
field.
PTS: 1
DIF: Bloom's Level 2
REF: p. 652
NAT: B.4
6. ANS: B
An electromagnet is simply a coil of wire with electric current flowing through it.
PTS: 1
DIF: Bloom's Level 2
NAT: E.2
7. ANS: C
Galvanometers can detect very small currents.
REF: p. 649
PTS: 1
DIF: Bloom's Level 2
REF: p. 655
NAT: E.2
8. ANS: A
Using the first right hand rule, one can determine the direction of the magnetic field relative to the direction of
conventional current.
PTS: 1
DIF: Bloom's Level 2
REF: pp. 648-649
NAT: B.4
9. ANS: C
When the magnetic fields of electrons are aligned in the same direction, a domain is formed.
PTS: 1
DIF: Bloom's Level 2
REF: p. 650
NAT: B.4
10. ANS: C
An electric motor converts electric energy into rotational kinetic energy.
PTS: 1
DIF: Bloom's Level 2
REF: p. 656
NAT: E.2
11. ANS: B
The number of magnetic field lines that pass through a surface is the magnetic flux.
PTS: 1
DIF: Bloom's Level 2
NAT: B.4
12. ANS: D
The wire coil of an electromagnet is called an armature.
REF: p. 646
PTS: 1
DIF: Bloom's Level 2
NAT: B.4
13. ANS: A
Magnetic fields exist around magnets and affect other magnets.
REF: p. 656
PTS: 1
NAT: B.4
14. ANS: D
DIF: Bloom's Level 2
REF: p. 646
PTS: 1
NAT: UCP.3
15. ANS: B
DIF: Bloom's Level 3
REF: p. 652
PTS: 1
NAT: UCP.3
16. ANS: A
DIF: Bloom's Level 3
REF: p. 652
PTS: 1
DIF: Bloom's Level 3
REF: p. 657
NAT: UCP.3
17. ANS: A
The direction of the force is perpendicular to both the velocity of the particle and the magnetic field.
PTS: 1
NAT: UCP.3
18. ANS: B
DIF: Bloom's Level 5
REF: p. 657
PTS: 1
NAT: UCP.3
19. ANS: B
5.0 A 0.80 m
DIF: Bloom's Level 5
REF: p. 653
PTS: 1
NAT: UCP.3
20. ANS: A
6.0 A 0.25 m
0.60 T = 2.4 N
DIF: Bloom's Level 5
REF: p. 653
0.30 T = 0.45 N
PTS: 1
NAT: UCP.3
DIF: Bloom's Level 5
REF: p. 653
COMPLETION
21. ANS: magnetic flux
PTS:
OBJ:
NAT:
KEY:
22. ANS:
PTS:
OBJ:
TOP:
MSC:
1
DIF: Bloom's Level 1
REF: Page 646
24.1.1 Describe the properties of magnets and the origin of magnetism in materials.
G.3
TOP: Describe the properties of magnets and the origin of magnetism in materials.
Magnetic Flux
MSC: 1
electric motor
1
DIF: Bloom's Level 1
REF: Page 656
24.2.3 Describe the design and operation of an electric motor.
Describe the design and operation of an electric motor.
KEY: Electric motor
1
MATCHING
23. ANS: H
PTS: 1
REF: pp. 644-646, 648-650, 652, 655-656
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
NAT:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
ANS:
NAT:
ANS:
ANS:
ANS:
ANS:
B.4
G
B
L
K
C
F
I
B.4
D
J
E
A
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
PTS:
1
1
1
1
1
1
1
NAT:
NAT:
NAT:
NAT:
NAT:
NAT:
DIF:
B.4
B.4
B.4
E.2
B.4
E.2
Bloom's Level 1
PTS:
PTS:
PTS:
PTS:
1
1
1
1
NAT:
NAT:
NAT:
NAT:
B.4
E.2
E.2
B.4
PROBLEM
35. ANS:
F = ILB = 5.7 N
PTS: 1
NAT: UCP.3
36. ANS:
1.6 N
DIF: Bloom's Level 3
REF: p. 653
PTS: 1
NAT: UCP.3
37. ANS:
7.8 A
DIF: Bloom's Level 3
REF: p. 653
PTS: 1
NAT: UCP.3
38. ANS:
0.15 T
DIF: Bloom's Level 3
REF: p. 653
PTS: 1
NAT: UCP.3
39. ANS:
0.13 T
DIF: Bloom's Level 3
REF: p. 653
PTS: 1
NAT: UCP.3
DIF: Bloom's Level 3
REF: p. 653