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Physics 112
Exam 2
Summer 2010
21. A proton has a speed of 3.0 × 106 m/s in a direction perpendicular to a uniform
magnetic field, and the proton moves in a circle of radius 0.20 m. What is the magnitude
of the magnetic field? (Proton mass 1.67× 10-27kg)
A) 0.08 T
B) 0.16 T
C) 0.24 T
D) 0.32 T
E) 0.64 T
Solution:
q  e  1.6  10 19 C
m  1.67  10  27 kg
v  3.0  10 6 m/s
qvB sin  
  90 
r  0.20m
B
B=?
mv 2

r
B
1.67 10 kg3.0 10 m / s  
1.6 10 C 0.2m
27
6
19
mv
er
B  0.16T
22. 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


B
A) out of the page
B) into the page

C) parallel to the vector B
D) force is zero
E) it is not enough information to answer the question
Solution:
From the right-hand rule follows that magnetic force is directed into the page. (Rotate
from v to B.)
Page 1 of 9
Physics 112
Exam 2
Summer 2010
23. 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) Zero
B) 0.25 N
C) 0.75 N
D) 1.5 N
E) 3.0 N
Solution:
I  10 A
F  IlB sin 
  30 
F  10 A0.5m 0.3T sin 30   0.75 N
B  0.3T
l  0 .5 m
F ?
24. A doubly charged helium atom whose mass is 6.6 1027 kg is accelerated by a voltage
of 2100 V. What will be its radius of curvature if it moves in a plane perpendicular to a
uniform 0.340-T field?
A) 1.1  10 2 m
B) 1.8  10 2 m
C) 2.2  10 2 m
D) 2.7  10 2 m
E) 3.3  10 2 m
Solution
The velocity of the ion can be found using energy conservation. The electrical potential
energy of the ion becomes kinetic energy as it is accelerated. Then, since the ion is
moving perpendicular to the magnetic field, the magnetic force will be a maximum. That
force will cause the ion to move in a circular path.
Einitial  Efinal  qV  12 mv 2  v 
Fmax  qvB  m
r
mv
qB
m

v
2qV
m
2
r

2qV
1
m

qB
B
2mV
q

1
0.340 T


2 6.6  1027 kg  2100 V 

19
2 1.60  10 C

 2.7  102 m
Page 2 of 9
Physics 112
Exam 2
Summer 2010
25. 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 parallel to the magnetic field?
A) zero
B) 0.21 N∙m
C) 0.41 N∙m
D) 0.47 N∙m
E) 0.52 N∙m
Solution:
r  0.50m
B  0.30T
I  2 .0 A
  90
 ?

M  IA  Ir 2
  MB sin   Ir 2 B sin 
  2.0 A 0.50m 2 0.30T sin 90   0.47 N·m
26. A current-caring circular coil is in uniform magnetic field. The maximum magnitude
of the torque on the coil occurs when the angle between the field and magnetic moment is
A) 0
B) 45°
C) 90°
D) 135°
E) 180°
Solution:
  MB sin 
   max 
  90 
27. Two long parallel wires carry equal currents. The magnitude of the force between
the wires is F. The current in each wire is now doubled. What is the magnitude of the
new force between the two wires?
A) F/4
B) F/2
C) F
D) 2F
E) 4F
Solution:
IIl
F 0 1 2 .
The force is proportional to the current in each wire, because of that
2r
 2I 2I 2 l
Fnew  0 1
 4F
2r
Page 3 of 9
Physics 112
Exam 2
Summer 2010
28. Two long parallel wires are 4.0 cm apart. Each wire carries the current 10 A in the
opposite direction. Find the magnetic field halfway between the wires.
A) 10-4 T
B) 2*10-4 T
C) 3*10-4 T
D) 2*10-2 T
E) zero
Solution:


B1  B2
 


B  B1  B2  2 B1
B  2 B1 
0 I
r
B
4 10

T  m / A  10 A
 2  10 4 T
  0.04m / 2
7
29. 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) The induced emf is 4 times as much.
B) The induced emf is twice times as much.
C) The induced emf is half as much.
D) The induced emf is one qorter as much.
E) There is no change in the induced emf.
Solution:
Faraday’s law:  N  N 1   N
 B
t
30. The magnetic field inside an air filled solenoid is B. The area of the solenoid is
doubled, keeping the current flowing through the solenoid and the number of turns per
unit length unchanged. Find the magnetic field inside the new solenoid.
A) B
B) 2B
C) 4B
D) B/2
E) B/4
Solution:
Magnetic field inside solenoid is independent from the area of the solenoid.
Page 4 of 9
Physics 112
Exam 2
Summer 2010
31. A circular coil lies flat on a horizontal table. A bar magnet is held above its center
with its north pole pointing down. The stationary magnet induces (when viewed from
above)
A) a clockwise current in the coil.
B) a counterclockwise current in the coil.
C) no current in the coil.
D) a current whose direction cannot be determined from the information given.
E) alternating current
Solution:
Faraday’s law:   
 B
. Magnet is stationary – flax is not changing – emf is zero –
t
no current in the coil.
32. A uniform magnetic field B is perpendicular to the area bounded by the U shaped
conductor and a movable metal rod of length l. The rod is moving along the conductor at
a speed v. The total resistance of the loop is R. What is power needed to move the rod?
B 2l 2 v 2
R
2 2
Bl v
P
R
2 2
B lv
P
R
2 2
B l v
P
R
Blv
P
R
A) P 
B)
C)
D)
E)
Solution:
Given : B, l , v, R
Find : P
  Blv
 Blv
I
R

R
Blv
B 2l 2 v
F  IlB 
lB 
R
R
2 2 2
B l v
Pext  Fv 
R
2
B 2l 2 v 2
 Blv 
Pdis  I R  
 R
R
 R 
2
Page 5 of 9
Physics 112
Exam 2
Summer 2010
33. The coil of a generator has 50 loops and a cross-sectional area of 0.25 m2. What is
the maximum emf generated by this generator if it is spinning with an angular velocity of
4.0 rad/s in a 2.0 T magnetic field?
A) 20 V
C) 40 V
D) 60 V
E) 80 V
E) 100 V
Solution:
 N max  N 1 max  NBA  502.0T 0.25m 2 4.0rad / s   100V
34. A wire loop is in a uniform magnetic field. Current flows in the wire loop, as shown.
What does the loop do?
A)
B)
C)
D)
E)
moves to the right
moves up
remains motionless
rotates
moves out of the page
Solution:
There is no magnetic force on the top and bottom legs, since they are parallel to the B
field. However, the magnetic force on the right side is into the page, and the magnetic
force on the left side is out of the page. Therefore, the entire loop will tend to rotate.
This is how a motor works!
35. The inductive reactance in an ac circuit changes by what factor when the frequency is
tripled?
A) 1/3
B) 1/9
C) 3
D) 9
E) remains unchanged
Solution:
The inductive reactance in an ac circuit is proportional to the frequency: X L  L  2fL .
Because of that the inductive reactance is tripled when the frequency is tripled.
Page 6 of 9
Physics 112
Exam 2
Summer 2010
36. A 4.0-mH coil carries a current of 5.0 A. How much energy is stored in the coil's
magnetic field?
A) 2.0 mJ
B) 10 mJ
C) 20 mJ
D) 50 mJ
E) 80 mJ
Solution:
U  12 LI 2
U
1
2
4.0 10
3

H 5.0 A  50  10 3 J  50mJ
2
37. An electric field can be produced by a
A) changing magnetic field
B) constant magnetic field
C) either a constant or a changing magnetic field
D) changing gravitational field
E) none of the given answers
Solution:
 E cos   
 BA cos  
t
An electric field can be produced by a changing magnetic field.
38. An inductance coil operates at 240 V and 60.0 Hz. It draws 12.8 A. What is the
coil’s inductance?
A) 1.0  10 2 H
B) 2.0  10 2 H
C) 3.0  10 2 H
D) 4.0  10 2 H
E) 5.0  10 2 H
Solution:
We find the reactance from Ohm’s law, and the inductance by Eq. 21-11b.
V  IX L  X L 
V
I
X L  2 fL  L 
XL
2 f

V
2 fI

240 V
2  60.0 Hz 12.8 A
 4.97  102 H
Page 7 of 9
Physics 112
Exam 2
Summer 2010
39. Unpolarized light passes through five successive ideal polarizers, each of whose axis
makes a 45° angle with the previous one. What fraction of the light intensity is
transmitted?
A) 1/2
B) 1/4
C) 1/8
D) 1/16
E) 1/32
Solution:
I 1  12 I 0
I 2  I 1 cos 2 45   12 I 1  14 I 0
I 3  I 2 cos 2 45   12 I 2  18 I 0
I 4  I 3 cos 2 45   12 I 3  161 I 0
I 5  I 4 cos 2 45   12 I 4 
1
32
I0
40. An electromagnetic wave in vacuum has a frequency of 1.00 MHz. What is the
wavelength of the wave?
A) 1.00 cm
B) 3.00 cm
C) 1.00 m
D) 3.00 m
E) 300 m
Solution:

c 3.00  108 m / s

 300m
f
1.00  10 6 s 1
Page 8 of 9
Physics 112
Exam 2
Summer 2010
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
B) 0.16 T
31
C) no current in the coil.
22
B) into the page
32
23
C) 0.75 N
33
E) 100 V
24
D) 2.7  10 2 m
34
D) rotates
25
D) 0.47 N∙m
35
C) 3
26
C) 90°
36
D) 50 mJ
27
E) 4F
37
A) changing magnetic
field
28
B) 2*10-4 T
38
E) 5.0  10 2 H
29
B) The induced emf is
twice times as much.
39
E) 1/32
30
A) B
40
E) 300 m
B 2l 2 v 2
A) P 
R
Page 9 of 9