Version 104 – Exam 2 – chiu – (57460)
1
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printout
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have
16
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before answering.
001
10.0points
Consider the setup in Figure above. What
is the change in potential energy Δ
U
=
U
C

U
D
, in moving an electron from D to C?
1.
2
k e q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
2.
k q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
3.
k q s
parenleftbigg
1
a

1
b
parenrightbigg
4.

k e q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
5.

k e q s
parenleftbigg
1
a

1
b
parenrightbigg
6.
2
k e q s
parenleftbigg
1
a

1
b
parenrightbigg
7.
2
k e q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
8.
k e q s
parenleftbigg
1
a

1
b
parenrightbigg
9.
k e q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
correct
10.
2
k e q s
parenleftbigg
1
a

1
b
parenrightbigg
Explanation:
V
C

V
D
=

integraldisplay parenleftbigg

2
k q s
x
3
parenrightbigg
dx
= 2
k q s
integraldisplay
d
parenleftbigg

1
2
x
2
parenrightbigg
= 2
k q s
parenleftbigg

1
2
a
2
+
1
2
b
2
parenrightbigg
.
(1)
Multiplying eq(1) by the electronic charge
e, we arrive at the potential energy difference
from D to C is given by
U
C

U
D
=

e
(
V
(
a
)

V
(
b
)) =
k e q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
.
Intuitive reasoning on the sign of Δ
U
: Nat
ural tendency of the motion is from high po
tential energy to lower potential energy. Since
when the electron is released it should move
from C to D, so
U
C
> U
D
.
Alternative explanation:
Let the center of the dipole be at the origin.
At a distance
x
along the +ˆ
x
direction
V
dipole
(
x
) =
V
q
parenleftBig
x
+
s
2
parenrightBig
+
V
−
q
parenleftBig
x

s
2
parenrightBig
V
dipole
(
x
) =
kq
x
+
s/
2
+
k
(

q
)
x

s/
2
V
dipole
(
x
) =
(
x

s/
2)

(
x
+
s/
2)
x
2

(
s/
2)
2
≈ 
kqs
x
2
So, we have
V
C
(
x
)

V
D
(
x
) =

k q s
parenleftbigg
1
a
2

1
b
2
parenrightbigg
Check:
E is along the ˆ
x
direction,
V
C
is
expected to be lower than
V
D
.
002
10.0points
A long currentcarrying wire, oriented North
South, lies on a table (it is connected to bat
teries which are not shown). A compass lies
on top of the wire, with the compass needle
about 3 mm above the wire.
With the cur
rent running, the compass deflects 16
◦
to the
West.
At this location, the horizontal com
ponent of the Earth’s magnetic field is about
2
×
10
−
5
T.
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Version 104 – Exam 2 – chiu – (57460)
2
What is the magnitude of the magnetic field
at location A, on the table top, a distance
2
.
5 cm to the East of the wire, due only to the
current in the wire?
1. 6.88e07
2. 5.54e07
3. 4.02e07
4. 3.66e07
5. 6.43e07
6. 3.8e07
7. 5.33e07
8. 3.39e07
9. 3.92e07
10. 5.13e07
Correct answer: 6
.
88
×
10
−
7
T.
Explanation:
let :
B
earth
= 2
×
10
−
5
T
,
r
compass
= 3 mm = 0
.
003 m
,
r
A
= 2
.
5 cm = 0
.
025 m
,
and
θ
= 16
◦
.
The magnetic field due to a wire is
B
=
μ
0
4
π
2
I
r
Since
tan
(
θ
)
=
B
compass
B
earth
,
B
compass
=
B
earth
tan
(
θ
).
By writing the equation for
the magnetic field of a wire for the compass
and for point A and dividing the two equa
tions, we obtain
B
A
=
B
compass
r
compass
r
A
=
B
earth
tan
(
θ
)
r
compass
r
A
= (2
×
10
−
5
T)
tan
(16
◦
)
0
.
003 m
0
.
025 m
= 6
.
88
×
10
−
7
T
.
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 Spring '11
 Chiu
 Physics, Magnetic Field, Electric charge

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