Nguyen, Thanh – Homework 14 – Due: Mar 14 2008, 7:00 pm – Inst: Weathers
1
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printout
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have
20
questions.
Multiplechoice questions may continue on
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
A solenoid 6
.
1 cm in radius and 31 m in length
has 12900 uniformly spaced turns and carries
a current of 6
.
7 A. Consider a plane circular
surface of radius 2
.
9 cm located at the center
of the solenoid with its axis coincident with
the axis of the solenoid.
What is the magnetic flux Φ through this
circular surface?(1 Wb = 1 T m
2
)
Correct answer: 9
.
25675
×
10

6
Wb.
Explanation:
Let :
‘
= 31 m
,
N
= 12900
,
I
= 6
.
7 A
,
and
r
= 2
.
9 cm = 0
.
029 m
.
Basic Concepts:
B
=
μ
0
N I
l
I
=
μ
0
n I
Φ
B
=
Z
~
B
·
d
~
A
The magnetic field in the solenoid is given by
B
=
μ
0
I N
‘
Call
A
the area of the plane circular surface.
The the magnetic flux through this surface is
Φ =
B A
=
μ
0
I N
‘
π r
2
=
μ
0
(6
.
7 A)(12900)
31 m
π
(0
.
029 m)
2
=
9
.
25675
×
10

6
Wb
.
The radius of the solenoid is not required since
it is smaller that the plane circular surface.
keywords:
002
(part 1 of 1) 10 points
A rectangular loop located a distance from a
long wire carrying a current is shown in the
figure. The wire is parallel to the longest side
of the loop.
4
.
8 cm
3
.
71 cm
20
.
3 cm
0
.
0289 A
Find the total magnetic flux through the
loop.
Correct answer: 6
.
71885
×
10

10
Wb.
Explanation:
Let :
c
= 4
.
8 cm
,
a
= 3
.
71 cm
,
b
= 20
.
3 cm
,
and
I
= 0
.
0289 A
.
c
a
b
r
dr
I
From Amp`
ere’s law, the strength of the
magnetic field created by the currentcarrying
wire at a distance
r
from the wire is (see
figure.)
B
=
μ
0
I
2
π r
,
so the field varies over the loop and is directed
perpendicular to the page. Since
~
B
is parallel
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Nguyen, Thanh – Homework 14 – Due: Mar 14 2008, 7:00 pm – Inst: Weathers
2
to
d
~
A
, the magnetic flux through an area
element
dA
is
Φ
≡
Z
B dA
=
Z
μ
0
I
2
π r
dA .
Note:
~
B
is not uniform but rather depends on
r
, so it cannot be removed from the integral.
In order to integrate, we express the area
element shaded in the figure as
dA
=
b dr
.
Since
r
is the only variable that now appears
in the integral, we obtain for the magnetic
flux
Φ
B
=
μ
0
I
2
π
b
Z
a
+
c
c
d r
r
=
μ
0
I b
2
π
ln
r
fl
fl
fl
a
+
c
c
=
μ
0
I b
2
π
ln
a
+
c
c
¶
=
μ
0
(0
.
0289 A)(0
.
203 m)
2
π
ln
a
+
c
c
¶
=
μ
0
(0
.
0289 A)(0
.
203 m)
2
π
(0
.
572626)
=
6
.
71885
×
10

10
Wb
.
keywords:
003
(part 1 of 2) 10 points
A 0
.
1304 A current is charging a capacitor
that has circular plates, 8
.
84 cm in radius.
The plate separation is 2
.
62 mm.
The permitivity or free space is 8
.
85
×
10

12
and the permeability of free space is 1
.
25664
×
10

6
T
·
m
/
A.
What is the time rate of increase of electric
field between the plates?
Correct answer: 6
.
00178
×
10
11
V
/
m s.
Explanation:
Let :
I
= 0
.
1304 A
,
r
= 8
.
84 cm = 0
.
0884 m
,
A
=
π r
2
= 0
.
0245502 m
2
,
and
²
0
= 8
.
85
×
10

12
.
Let Φ
E
be the flux of the electric field, defined
as Φ
E
=
Z
~
E
·
d
~
A
. Thus
d
Φ
E
dt
=
d
dt
(
E A
) =
I
²
0
Time rate of increase of electric field between
the plates is
P
=
d E
dt
=
I
²
0
A
=
0
.
1304 A
(8
.
85
×
10

12
)(0
.
0245502 m
2
)
=
6
.
00178
×
10
11
V
/
m s
.
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 Work, Magnetic Field, Thanh

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