Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih
1
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The due time is Central
time.
001
(part 1 of 1) 10 points
The figure below shows a coaxial cable of
radii
a
,
b
, and
c
in which equal, uniformly
distributed, but antiparallel currents
i
exist
in the two conductors.
O
i
out
fl
i
in
⊗
F
E
D
C
r
1
r
2
r
3
r
4
c
b
a
Which
expression
gives
the
magnitude
B
(
r
3
) at
D
of the magnetic field in the re
gion
b < r
3
< a
?
1.
B
(
r
3
) =
μ
0
i
(
a
2

r
2
3
)
2
π r
3
(
a
2

b
2
)
correct
2.
B
(
r
3
) = 0
3.
B
(
r
3
) =
μ
0
i r
3
2
π b
2
4.
B
(
r
3
) =
μ
0
i
(
a
2
+
r
2
3

2
b
2
)
2
π r
3
(
a
2

b
2
)
5.
B
(
r
3
) =
μ
0
i r
3
2
π c
2
6.
B
(
r
3
) =
μ
0
i
(
r
2
3

b
2
)
2
π r
3
(
a
2

b
2
)
7.
B
(
r
3
) =
μ
0
i r
3
2
π a
2
8.
B
(
r
3
) =
μ
0
i
π r
3
9.
B
(
r
3
) =
μ
0
i
(
a
2

b
2
)
2
π r
3
(
r
2
3

b
2
)
10.
B
(
r
3
) =
μ
0
i
2
π r
3
Explanation:
Ampere’s Law
states that the line inte
gral
I
~
B
·
d
~
‘
around any closed path equals
μ
0
I
, where
I
is the total steady current pass
ing through any surface bounded by the closed
path.
Considering the symmetry of this problem,
we choose a circular path, so
Ampere’s Law
is simplified to
B
(2
π r
3
) =
μ
0
i
in
,
where
r
3
is the radius of the circle and
i
in
is
the current enclosed.
Since, when
b < r
3
< a
, for the cylinder,
A
in
A
cylinder
=
π
(
r
2
3

b
2
)
π
(
a
2

b
2
)
,
we have
B
=
μ
0
I
in
2
π r
3
=
μ
0
•
i

i
π
(
r
2
3

b
2
)
π
(
a
2

b
2
)
‚
2
π r
3
=
μ
0
i
a
2

r
2
3
a
2

b
2
¶
2
π r
3
=
μ
0
i
(
a
2

r
2
3
)
2
π r
3
(
a
2

b
2
)
.
002
(part 1 of 1) 10 points
Calculate the resonance frequency of a se
ries
RLC
circuit for which the capacitance is
99
μ
F
,
the resistance is 33 kΩ
,
and the induc
tance is 56 mH
.
Correct answer: 67
.
5941 Hz.
Explanation:
Let :
R
= 33 kΩ = 33000 Ω
,
L
= 56 mH = 0
.
056 H
,
and
C
= 99
μ
F = 9
.
9
×
10

5
F
.
The resonance frequency is the frequency at
which the current becomes maximum, or the
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Platt, David – Quiz 3 – Due: Nov 15 2005, 10:00 pm – Inst: Ken Shih
2
impedance becomes minimum.
This occurs
when
X
L
=
X
C
ω L
=
1
ω C
.
From this condition, the resonance frequency
is given by
f
=
1
2
π
√
L C
=
1
2
π
p
(0
.
056 H) (9
.
9
×
10

5
F)
=
67
.
5941 Hz
.
003
(part 1 of 1) 10 points
Consider an electromagnetic wave pattern as
shown in the figure below.
E
B
The wave is
1.
traveling left to right.
correct
2.
traveling right to left.
3.
a standing wave and is stationary.
Explanation:
The
~
E
vector and
~
B
vector are not at the
same point on the velocity axis.
Pick an instant in time, where the
E
and
B
fields are at the same point on the velocity
axis.
z
v
x
y
E
B
For instance, let us choose the point where
the
~
E
vector is along the
x
axis, as shown in
the above figures. At this same instant, the
~
B
vector is along the negative
y
axis (at a point
with a phase difference of 360
◦
from the place
on the velocity (
z
) axis where the
~
E
vector is
drawn).
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 Fall '08
 Turner
 Physics, Magnetic Field, Platt, Ken Shih, Bnet Bnet Bnet, David Quiz

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