gilvin (jg47854) – 19. Ampere’s law – meyers – (21235)
1
This
printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – find all choices
before answering.
This homework is due Tuesday, april19, at
midnight Tucson time.
001
10.0points
A lightening bolt may carry a current of
4200 A for a short period of time.
The permeability of free space is 1
.
25664
×
10
−
6
T
·
m
/
A.
What is the resulting magnetic field 68 m
from the bolt?
Suppose that the bolt extends far above
and below the point of observation.
Correct answer: 12
.
3529
μ
T.
Explanation:
Given :
L
= 68 m
and
I
= 4200 A
.
Treat the lighting bolt as a long, straight
conductor. Then the magnetic field is
B
=
μ
0
I
2
π L
=
(
1
.
25664
×
10
−
6
T
·
m
/
A
)
(4200 A)
2
π
(68 m)
×
parenleftbigg
1
.
0
×
10
6
μ
T
1 T
parenrightbigg
=
12
.
3529
μ
T
.
002
10.0points
A long, straight, thinwalled cylindrical shell
of radius
R
carries a current
I
.
Find the magnetic field inside the cylinder
and outside the cylinder.
1.
B
=
5
μ
0
I
2
π r
, r > R
;
B
= 0
, r < R
2.
B
=
μ
0
I
2
π r
, r > R
;
B
= 0
, r < R
correct
3.
B
=
3
μ
0
I
2
π r
, r > R
;
B
= 0
, r < R
4.
B
= 0
, r > R
;
B
=
3
μ
0
I
2
π r
, r < R
5.
B
= 0
, r > R
;
B
=
μ
0
I
2
π r
, r < R
6.
B
= 0
, r > R
;
B
=
5
μ
0
I
2
π r
, r < R
Explanation:
Applying Amp`
ere’s law for
r > R
,
contintegraldisplay
C
vector
B
·
dl
=
B
(2
π r
) =
μ
0
I
B
r>R
=
μ
0
I
2
π r
.
applying Amp`
ere’s law for
r < R
,
contintegraldisplay
C
vector
B
·
dl
=
B
(2
π r
) = 0
B
r<R
=
0
.
003(part1of3)10.0points
A wire of radius 1 cm carries a current of
90 A that is uniformly distributed over its
crosssectional area.
The permeability of free space is 4
π
×
10
−
7
T
·
m
/
A.
Find the magnetic field at a distance of
0
.
1 cm from the center of the wire.
Correct answer: 0
.
00018 T.
Explanation:
Let :
I
= 90 A
,
R
= 1 cm = 0
.
01 m
,
r
= 0
.
1 cm = 0
.
001 m
,
and
μ
0
= 4
π
×
10
−
7
T
·
m
/
A
.
Using Amp`
ere’s law, we have
contintegraldisplay
C
vector
B
·
d
vector
l
=
μ
0
I
C
.
Because
r < R
,
I
C
=
r
2
R
2
I
, so
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gilvin (jg47854) – 19. Ampere’s law – meyers – (21235)
2
B
(2
π r
) =
μ
0
r
2
R
2
I
B
=
μ
0
r
2
π R
2
I
=
(4
π
×
10
−
7
T
·
m
/
A) (0
.
001 m)
2
π
(0
.
01 m)
2
×
(90 A)
=
0
.
00018 T
.
004(part2of3)10.0points
Find the magnetic field at the surface of the
wire.
Correct answer: 0
.
0018 T.
Explanation:
Using Amp`
ere’s law, we have
contintegraldisplay
C
vector
B
·
d
vector
l
=
μ
0
I
C
.
Because
r
=
R
,
I
C
=
I
, so
B
(2
π R
) =
μ
0
I
B
=
μ
0
I
2
π R
= 2
μ
0
I
4
π R
=
(4
π
×
10
−
7
T
·
m
/
A) (90 A)
2
π
(0
.
01 m)
=
0
.
0018 T
.
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