ID: B
2
UCSD Physics 2B
Unit Exam 4B
Magnetism & Induction
Answer Section
MULTIPLE CHOICE
1. ANS: B
Two ways to view this problem:
1. By Faraday's Law, the induced
emf
= −
d
Φ
dt
Ê
Ë
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
so curl the finger of the right hand in the direction
of the current (
emf
) and the thumb points to the right. Hence the flux is increasing to the
left
(decreasing to the right) and so the magnet (South pole to the left) must be moving to the right.
2. Lenz's Law says the current tries to oppose the change in flux. Since the current loop makes
magnetic North pole pointing to the right, the loop is trying to attract the bar magnet so it must be
moving to the right.
TOP: LENZ
2. ANS: C
Field at the center of a full circle is
B
=
μ
0
I
2
R
so the field for a semicircle must be half that. The two
semicircular currents flow in
opposite
directions, so the fields
subtract
:
B
=
μ
0
I
4
1
R
1
−
1
R
2
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
1.26
×
10
−
6
N
/
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
595
A
(
)
4
1
0.023
m
−
1
0.032
m
Ê
Ë
Á
Á
Á
Á
Á
Á
ˆ
¯
˜
˜
˜
˜
˜
˜
=
2.29
×
10
−
3
T
=
2.29m
T
TOP: FIELD SUPERPOSITION
3. ANS: B
The magnetic flux is the product of the coil area
A
, the number of turns
N
, the magnetic field
strength
B
and the
cosine
of the angle at which the field lines penetrate the coil:
Φ =
N
B
•
A
=
NBA
cos
θ
⇒
N
=
Φ
BA
cos
θ
=
0.75 Wb
0.5
T
(
)
10
×
10
−
2
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
2
cos60
(
)
=
300
TOP: FLUX
4. ANS: B
B
=
μ
0
In
=
μ
0
I
N
L
⇒
L
=
μ
0
IN
B
=
1.26
×
10
−
6
H
/
m
Ê
Ë
Á
Á
Á
ˆ
¯
˜
˜
˜
3.42
A
(
)
169
(
)
375
×
10
−
6
T
=
1.94
m
TOP: SOLENOID
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ID: B
2
5. ANS: A
First we use Faraday’s Law to get the EMF. The
flux
through the loop is the field
B
times the area
A
=
h
×
x
. The change in the flux is only due to the penetration depth
x
since the field
B
and height
h
are fixed. The power dissipated by the resistor is
P
=
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 Summer '09
 Lever
 Current, Magnetism, Magnetic Field, Lenz

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