Physics 112
Exam 2
Summer 2014


21. Positive charge q is moving with velocity v in the magnetic field B as shown below (with


vectors v and B on the surface of the page). What is the direction of the magnetic force?

v

q
A)
B)
C)
D)
E)

Parallel to v

Parallel to B


Exactly between v and B
Into the page
Out of the page

B
Solution: Use right hand rule: rotation form vector v to vector B gives direction of magnetic
force – into the page.
22. A positively charged particle is moving in a magnetic field. What is the angle between the
velocity of the charge and the direction of the magnetic field if the magnetic force on the particle
is zero?
A)
B)
C)
D)
E)
90°
45°
30°
0
Magnetic force cannot be zero if the speed of the particle and the magnetic field are not zero.
Solution: From equation FB  q vB sin   0 follows: sin   0  ,   0 or   180  .
23. Alpha particles of charge q  2e and mass m  6.6  10 27 kg are emitted from a radioactive
source at a speed of 1.6  10 7 m s . What magnetic field strength would be required to bend them
into a circular path of radius r  0.25 m?
A)
B)
C)
D)
E)
5.3 T
4.3 T
3.3T
2.3 T
1.3 T
v2
mv
Solution: In this problem, centripetal force is magnetic force: q vB  m  B 
.
R
qR
B
6.6  10 kg 1.6  10 m s .  1.3T
21.6  10 C.0.25m 
27
7
19
Page 1 of 8
Physics 112
Exam 2
Summer 2014
24. If a bar magnet is divided into two equal pieces,
A) the north and south poles are separated
B) two magnets result
C) the magnetic properties are destroyed
D) an electric field is created
E) none of the above
Solution:
Magnetic monopole does not exist. If a bar magnet is divided into two pieces, two magnets
result.
25. An airplane with a wing span of 60 m flies horizontally at a location where the downward
component of the Earth's magnetic field is 6.0 × 10-5 T. Find the magnitude of the induced emf
between the tips of the wings when the speed of the plane is 225 m/s.
A)
B)
C)
D)
E)
0.41 V
0.61 V
0.81 V
1.2 V
1.4 V
Solution:
From equation   Blv follows   Blv   6.0  10 5 T 60m225m / s   0.81V


26. A wire carries a current of 10 A in a direction of 30° with respect to the direction of a 0.30-T
magnetic field. Find the magnitude of the magnetic force on a 0.50-m length of the wire.
A) 0.75 N
B) 1.5 N
C) 3.0 N
D) 6.0 N
E) zero
Solution:
F  IlB sin  ; F  10 A0.5m0.3T sin 30   0.75N
Page 2 of 8
Physics 112
Exam 2
Summer 2014
27. A circular loop of wire of radius 0.50 m is in a uniform magnetic field of 0.30 T. The current
in the loop is 2.0 A. What is the magnetic torque when the plane of the loop is perpendicular to
the magnetic field?
A)
B)
C)
D)
E)
0.73 m∙N
0.52 m∙N
0.47 m∙N
0.41 m∙N
zero
Solution:
The loop is perpendicular to the magnetic field means that the magnetic dipole moment of the
loop is parallel (or antiparallel) to the magnetic field. From equation   MB sin  follows   0 .
28. What is the strength of a magnetic field 5.0 cm from a long straight wire carrying 4.0 A of
current?
A)
B)
C)
D)
E)
3.8∙10-6 T
4.9∙10-6 T
1.6 ∙10-5 T
4.7∙10-5 T
5.1∙10-5 T
Solution:
 I
4  10 7 Tm / A 4.0 A
B 0 
 1.6  10 5 T
2r
2 5.0  10 2 m




29. Two long parallel wires carry currents of 5.0 A and 8.0 A in the opposite direction. The
wires are separated by 0.30 m. Find the magnetic force per unit length between the two wires.
A)
B)
C)
D)
E)
2.7∙10-5 N repulsive
7.2∙10-5 N repulsive
2.7∙10-5 N attractive
7.2∙10-5 N attractive
zero
Solution:
 0 I 1 I 2 l 4  10 7 Tm / A 5.0 A8.0 A1m
F

 2.7  10 7 N
2r
2 0.3m


Page 3 of 8
Physics 112
Exam 2
Summer 2014
30. A long solenoid has 100 turns/cm and carries a current of 1.00A. A sphere of radius 1.00 cm
is placed inside the solenoid. Find the magnetic flux through the sphere.
A)
B)
C)
D)
E)
1.6∙10-6 T∙m2
1.6∙10-7 T∙m2
4.0∙10-8 T∙m2
8.0∙10-8 T∙m2
zero
Solution:
Magnetic flux through any closed surface is zero.
31. What is the direction of the induced current in the circular loop due to the increasing current
shown below?
I increasing
A)
B)
C)
D)
E)
Counterclockwise
Clockwise
The direction of the current is periodically changing
Induced current is equal to zero
There is not enough information to answer the quastion
Solution:
The increasing current in the wire creates increasing magnetic field and flux trough the loop.
According to the right hand rule, this field is directed out of page. Because this field is
increasing, the induced field should have opposite direction (into the page). According to right
hand rule, this induced field is created by the clockwise current through the loop.
32. An AC generator consists of 100 turns (loops) of wire with total resistance 12 Ω. The area of
each loop is 0.090 m2. The loops rotate in a magnetic field of 0.50 T at a constant angular speed
of 60 revolutions per second. Find the maximum induced emf.
A)
B)
C)
D)
E)
0.27 kV
0.54 kV
1.7 kV
3.4 kV
5.1 kV
Solution:
 N max  N 1 max  NBA  1000.50T 0.090m 2 2  60 Hz   1.7kV
Page 4 of 8
Physics 112
Exam 2
Summer 2014
33. Doubling the number of loops of wire in a coil produces what kind of change on the induced
emf, assuming all other factors remain constant?
A)
B)
C)
D)
E)
The induced emf increases 4 times
The induced emf increases 2 times
The induced emf decreases 2 times
The induced emf decreases 4 times
The induced emf remains the same
Solution:
 N  N 1   2  2 1
34. A simple RL circuit contains a 6.0-Ω resistor and an 18-H inductor. What is this circuit's
time constant?
A)
B)
C)
D)
E)
108 s
3.0 s
0.33 s
0.20 s
0.11 s
Solution:
  L / R  18H  6.0  3.0s
35. A 6 V battery is connected to coil 1 of the transformer shown in the figure below. Coil 2 has
twice as many turns as coil 1. What is the voltage drop across coil 2?
A)
B)
C)
D)
E)
12 V
6V
3V
2V
0
1
2
Solution:
Batteries provide DC current. Only a changing magnetic flux induces an EMF. Therefore, the
voltage across coil 2 is zero.
Page 5 of 8
Physics 112
Exam 2
Summer 2014
36. The magnetic field inside an air-filled solenoid 36 cm long and 2.0 cm in diameter is 0.80 T.
Approximately how much energy is stored in this field?
A)
B)
C)
D)
E)
5J
11 J
18 J
21 J
29 J
Solution:
UB
B2
B2
B2

UB 
V
lr 2 ;
V
2 0
2 0
2 0
UB 

0.80T 2
2 4  10 Tm / A
7

0.36m  1.0  10  2 m2  28.8 J
37. What is the current through a 2.50-mH coil due to a 110-V, 60.0 Hz source?
A)
B)
C)
D)
E)
0.94 A
2.5 A
104 A
117 A
154 A
Solution:
V
V
110V
I

 117 A
; I 
X L 2fL
2 60.0 Hz  2.50  10 3 Hz


38. A circular parallel-plate capacitor with plates 4.0 cm in diameter is accumulating charge at
the rate of 3.0 mC/s at some instant in time. What is the magnitude of the induced magnetic field
at the distance 1.0 cm measured radially outward from the center of the plates at this instant?
A)
B)
C)
D)
E)
2.0 T
0.5 T
2.0  10 8 T
1.5  10 8 T
0.5  10 8 T
Solution:
 I
B  0 2 r;
2R
B
4  10
7

 1.0  10
Tm / A 3.0  10 3 A

2 2.0  10 m
2

2
2

m  1.5  10 8 T
Page 6 of 8
Physics 112
Exam 2
Summer 2014
39. An electromagnetic wave in vacuum is moving in +y direction. At time t=0 and at position
(x,y,z)=(0,0,0), the electric field is pointing in the +z direction. In what direction is the magnetic
field pointing at that time and position?
 y 
A) +x
v
B) –x
B
C) +y
x
D) –y

E) +z
E
z
Solution:
Use right hand rule: rotation from vector E to vector B to gives direction of vector v.
Also rotation from x direction to y direction leads to z direction.
40. If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT, what is the
peak magnitude of the electric field?
A)
B)
C)
D)
E)
zero
1.75 V/m
3.50 V/m
5.25 V/m
7.00 V/m
Solution:
B  E / c  E  Bc ; E  17.5  10 9 T 3  10 8 m / s



Page 7 of 8
Physics 112
Exam 2
Summer 2014
Record Sheet
You may fill in this sheet with your choices, detach it and take it with you after the exam for
comparison with the posted answers
21
31
22
32
23
33
24
34
25
35
26
36
27
37
28
38
29
39
30
40
Page 8 of 8