Wong, Lemuel – Homework 11 – Due: Apr 16 2007, 11:00 pm – Inst: Matt Mackie
1
This
printout
should
have
16
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
The due time is Central
time.
NOTE the unusual due TIME: 5pm central,
which is 6pm our time. this is meant to allow
more time with the solutions before the exam.
001
(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.
5
.
5 cm
6
.
88 cm
23
.
2 cm
0
.
0661 A
Find the total magnetic flux through the
loop.
Correct answer: 2
.
48839
×
10

9
Wb.
Explanation:
Let :
c
= 5
.
5 cm
,
a
= 6
.
88 cm
,
b
= 23
.
2 cm
,
and
I
= 0
.
0661 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
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
.
0661 A)(0
.
232 m)
2
π
ln
a
+
c
c
¶
=
μ
0
(0
.
0661 A)(0
.
232 m)
2
π
(0
.
811334)
=
2
.
48839
×
10

9
Wb
.
keywords:
002
(part 1 of 1) 10 points
A solenoid 8
.
3 cm in radius and 20 m in length
has 14000 uniformly spaced turns and carries
a current of 9
.
4 A. Consider a plane circular
surface of radius 2 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: 1
.
03907
×
10

5
Wb.
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Wong, Lemuel – Homework 11 – Due: Apr 16 2007, 11:00 pm – Inst: Matt Mackie
2
Explanation:
Let :
‘
= 20 m
,
N
= 14000
,
I
= 9
.
4 A
,
and
r
= 2 cm = 0
.
02 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
(9
.
4 A)(14000)
20 m
π
(0
.
02 m)
2
=
1
.
03907
×
10

5
Wb
.
The radius of the solenoid is not required since
it is smaller that the plane circular surface.
keywords:
003
(part 1 of 1) 10 points
A coil is wrapped with 119 turns of wire on
the perimeter of a circular frame (of radius
32 cm). Each turn has the same area, equal
to that of the frame.
A uniform magnetic
field is directed perpendicular to the plane of
the coil. This field changes at a constant rate
from 23 mT to 66 mT in 22 ms.
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 Spring '08
 Turner
 Physics, Inductance, Work, Electromotive Force, Magnetic Field, Correct Answer, Inductor, Faraday's law of induction

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