PHY2049 Fall 2009
Profs. A. Petkova, A. Rinzler, S. Hershfield
Exam 3 Solution
1. In the figure
R
represents a variable resistor and the battery supplies an emf that
drives current clockwise around the loop.
The following is done.
The resistance
is steadily increased and then steadily decreased. For each stage of this sequence
(in order), what is the direction of the induced emf by the inductor? CW denotes
clockwise, CCW denotes counterclockwise, and NC means no change.
Answer:
CW, CCW
Solution:
When the resistance is increased, the current in the clockwise direction is decreased.
To oppose that the
induced emf will be in the clockwise direction. When the resistance is decreased, the current in the clockwise direction
will increase. To oppose that the induced emf will be in the counterclockwise direction.
2. A solenoid that is 85.0 cm long has a crosssectional area of 17.0 cm
2
. There are 950 turns of wire carrying current of
10.0 A.
What is the total energy stored in the magnetic field inside the solenoid (neglect end effects)?
Answer:
0.113 J
Solution:
The magnetic field in a solenoid is
B
=
μ
o
nI
, where
n
is the number of terms per unit length. The energy
density is
u
=
B
2
/
(2
μ
o
), and the net energy is
U
=
u
Vol., where the volume inside the solenoid is its length times its
crosssectional area.
3. A square wire loop with 2.00 m sides is perpendicular to a uniform magnetic field,
with half the area of the loop in the field as shown.
The loop contains an ideal
battery with emf
E
= 15
.
0V. If the magnitude of the field varies with time according
to
B
= 0
.
0620

0
.
870
t
, with
B
in Tesla and
t
in seconds, what is the net emf in
the circuit, and its direction?
Answer:
16.7 V, CCW
Solution:
The area in the magnetic field is 0
.
5(2
m
)
2
= 2
m
2
. The rate of change of the magnetic flux is (2
m
2
)
dB/dt
=
2
×
0
.
87 = 1
.
74
V
, which is also the induced emf. Because the flux is out of the page and decreasing in magnitude, via
Lenz’s law the induced current is in the CCW direction. Thus, the induced emf and the emf of the battery are in the
same direction. The net emf is 15
.
0
V
+ 1
.
7
V
.
4. The magnetic field of a cylindrical magnet that has a poleface diameter of 3.3 cm can be varied sinusoidally between 28.0
and 30.0 T at a frequency of 15 Hz. At a radial distance of 1.4 cm, what is the amplitude of the electric field induced by
the variation? (Note that the magnetic field is directed along the axis of the cylinder.)
Answer:
0.66 V/m
Solution:
The amplitude of the magnetic field has the form
B
(
t
) = 29
T
+ 1
T
sin(2
πft
), so the rate of change of flux is
d
Φ
B
/dt
= 2
πf
(1
T
)
A
cos(2
πt
). For a path of integration of radius
r
= 1
.
4 cm, Faraday’s law implies that
contintegraldisplay
vector
E
·
dvectors
= 2
πrE
=

d
Φ
B
dt
==

2
πf
(1
T
)
πr
2
cos(2
πt
)
.
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 Spring '08
 Any
 Physics, Current, Resistance, Snell's Law, Magnetic Field, Inductor

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