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Physics 112
Exam 2
Summer 2015


21. Negative 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?


B
A) Parallel to v

B) Parallel to B


C) Exactly between v and B


q
D) Into the page
v
E) Out of the page
Solution:
If the charge would be positive then from the right-hand rule follows that magnetic force is
directed out of the page (rotate from v to B). On the negative charge the force is into the page.
22. A positively charged particle (q = + 60 μC) moves at 2.0  10 5 m/s in a direction that is 30o to
a uniform magnetic field of 0.50 T. What is the magnetic force on the charge?
A)
B)
C)
D)
E)
1.0 N
1.5 N
2.0 N
2.5 N
3.0 N
Solution:
F  qvB sin θ  60  10 -6 C 2.0  105 m / s 0.50T sin 30  3.0 N



23. An electron is moving at speed v = 800 m/s. It enters a region with uniform electric field
 

E = 400 V/m and uniform magnetic B field. Vectors E , B and v are perpendicular to each other.
The electron continues on the straight path. What is the magnitude of the magnetic field B?
A)
B)
C)
D)
E)
0.5 T
1.0 T
1.5 T
2.5 T
3.0 T
Solution:
The electron moves in a straight line so the electric and magnetic forces on the electron must
cancel each other.
FE  FB  qE  qvB  B  E / v
B
400 V m
 0.5T
800 m s
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Physics 112
Exam 2
Summer 2015
24. The force on a current-carrying wire in a magnetic field is the strongest when
A)
B)
C)
D)
E)
The current is parallel to the field lines
The current is at a 30° angle with respect to the field lines
The current is at a 45° angle with respect to the field lines
The current is at a 60° angle with respect to the field lines
The current is perpendicular to the field lines
Solution:
F  IlB sin  
For   90  :
F  Fmax  IlB .
25. A coil of wire has 100 turns and a cross sectional area of 0.10 m2. The coil is free to rotate
and sits in a uniform 0.50 T magnetic field that points at 30° to the normal of the coil. If a 4.0A
current passes through the coil, determine the torque on the coil?
A)
B)
C)
D)
E)
2.0N  m
5.0N m
10N m
20N  m
There is no torque because the current in the coil is steady
Solution:
The torque is given by equation   NIAB sin  .
  100  4.0 A   0.10m2   0.50T  sin 30o  10N  m
26. 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!
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Physics 112
Exam 2
Summer 2015
27. Two long parallel wires are 4.0 cm apart. Each wire carries the current 3.0A in the same
direction. Find the magnetic field at a point that is 3.0 cm from one wire and 1.0 cm from the
other wire.
A)
B)
C)
D)
E)
4 10 5 T
6  10 5 T
4 10 5 T
6  10 7 T
zero
out of the page
out of the page
into the page
into the page
I
I
3.0 cm
1.0 cm
Solution:

According to the right hand rule, at the point of interest, the magnetic field BL due to the left wire

is directed into the page, and the field B R due to the right wire is directed out of the page.
 I
 I
BL  0 , and BR  0 .
2rL
2rR
 

The total field is B  B L  B R .

Because rL  3.0cm  rR  1.0cm , we have BL  BR , and vector B R is directed out of the page.
B  BR  BL 

4 10
7
0 I 0 I 0 I  1 1 
  


2rR 2rL
2  rR rL 

T  m / A  3.0 A 
1
1
1 

 1
5

 2  10 5  3.0


T  4  10 T
2
2 
2
1
.
0
3
.
0
1
.
0

10
3
.
0

10




28. If current in each of two long parallel wires is doubled, what happened with force of
interaction between these wires?
A)
B)
C)
D)
E)
Double
Quadruple
Decrease by one-half
Decrease by one-fourth
Remain the same
Solution:
 IIl
F 0 1 2
2r
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Physics 112
Exam 2
Summer 2015
29. If current in a solenoid is doubled, what happened with the field inside the solenoid?
A)
B)
C)
D)
E)
Double
Quadruple
Decrease by one-half
Decrease by one-fourth
Remain the same
Solution: B  0 NI / l  0 nI
30. A bar magnet falls through a loop of wire with the north pole entering first. As the north pole
enters the wire, the induced current will be (as viewed from above):
A)
B)
C)
D)
E)
Zero
Clockwise
Counterclockwise
To top of loop
A current whose direction cannot be determined from the information given

v
Solution: According to the Lenz’s law the induced current creates magnetic
field that opposes the original change in the flax. As the north pole of the
falling magnet enters the wire the down directed flux is increasing. The
induced counterclockwise current creates magnetic field directed up.
S

B
N
I
31. A coil of 600 turns with area 100 cm2 is placed in a uniform magnetic field. The angle
between the direction of the field and the perpendicular to the loop is 60°. The field changes
at the rate of 0.010 T/s. What is the magnitude of induced emf in the coil?
A)
B)
C)
D)
E)
0.01 V
0.02 V
0.03 V
0.10 V
0.20 V
Solution:
 B   BA cos   NBA cos 
N  600
  60

A  100cm
2
B
 0.010T / s
t
 ?
 B
B
N
A cos 
t
t
B
B
  
N
A cos 
t
t



 600  0.010T / s   100  10  4 m 2 cos 60   0.03V
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Physics 112
Exam 2
Summer 2015
32. An airplane with a wing span of 50 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 200 m/s.
A)
B)
C)
D)
E)
0.4 V
0.5 V
0.6 V
1.2 V
1.8 V
Solution:
  Blv 
  6.0  10 5 T 50m200m / s   0.60V
33. A 6 V battery is connected to the primary coil of a transformer that is designed to convert
120 V to 240 V. What is voltage drop across the secondary coil?
A)
B)
C)
D)
E)
6V
12 V
120 V
240 V
Zero
Solution:
Batteries provide DC current. Only a changing magnetic flux induces an EMF. Therefore, the
voltage across the secondary coil is zero.
34. What is the direction of the induced current in the circular loop due to an increasing current
that flows to the left in the straite wire as shown below?
A)
B)
C)
D)
E)
Clockwise
Counterclockwise
Alternating current
No current induced
Not enough information to answer
Solution:
The increasing current in the wire will cause an increasing field out of the page through the loop.
To oppose this increase, the induced current in the loop will produce a flux into the page, so the
direction of the induced current will be clockwise.
5
Physics 112
Exam 2
Summer 2015
35. A 4.0-mH coil carries a current of 5.0 A. How much energy is stored in the coil's magnetic
field?
A)
B)
C)
D)
E)
2.0 mJ
10 mJ
20 mJ
50 mJ
80 mJ
Solution:
U  12 LI 2
U
1
2
4.0 10
3

H 5.0 A  50  10 3 J  50mJ
2
36. After how many time constants does the current in a LR circuit reach 90% of its maximum
value?
A)
B)
C)
D)
E)
1.2
2.3
3.4
4.5
never
Solution:

I  1  e t /    I max 1  e t /   
R
t /    ln 1  I / I max   t /    ln 1  0.9  2.3
37. An inductance coil operates at 240 V and 60.0 Hz. It draws 12.8 A. What is the coil’s
inductance?
A)
B)
C)
D)
E)
1.0  10 2 H
2.0  10 2 H
3.0  10 2 H
4.0  10 2 H
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
6
Physics 112
Exam 2
Summer 2015
38. The amplitude of magnetic field in a sinusoidal electromagnetic wave is 4.0 µT. What is the
amplitude of the electric field in this wave?
A)
B)
C)
D)
E)
2400 V/m
1200 V/m
600 V/m
300 V/m
150 V/m
Solution:
Speed of an electromagnetic wave in vacuum is given by, c = E/B.
So, E  cB  3  108 m / s 4.0  10 6 T  1200V / m



39. Unpolarized light passes through three successive ideal polarizers, each of whose axis makes
a 30° angle with the previous one. What fraction of the light intensity is transmitted?
A)
B)
C)
D)
E)
1/4
1/8
3/8
9/32
27/64
Solution:
I 1  12 I 0
I 2  I 1 cos 2 30   34 I 1  83 I 0
I 3  I 2 cos 2 30   34 I 2 
9
32
I0
40. An electromagnetic wave in vacuum has a frequency of 1.00 MHz. What is the wavelength
of the wave?
A)
B)
C)
D)
E)
1.00 cm
3.00 cm
1.00 m
3.00 m
300 m
c 3.00  108 m / s
Solution:   
 300m
f
1.00  10 6 s 1
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Physics 112
Exam 2
21
31
D) Into the page
C) 0.03 V
22
32
E) 3.0 N
C) 0.6 V
23
33
A) 0.5 T
E) Zero
24
34
E) The current is
perpendicular to the field
lines
A) Clockwise
25
35
C) 10N m
D) 50 mJ
26
36
D) Rotates
B) 2.3
27
37
A) 4 10 T out of the
page
E) 5.0  10 2 H
28
38
B) Quadruple
B) 1200 V/m
29
39
A) Double
D) 9/32
30
40
C) Counterclockwise
E) 300 m
5
Summer 2015
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