Kapoor (mk9499) – oldhomework 23 – Turner – (60230)
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001
(part 1 oF 4) 10.0 points
A long solenoid carries a current oF 30 A
.
Another coil (oF larger diameter than the
solenoid) is coaxial with the center oF the
solenoid, as in the fgure below.
The permeability oF Free space is 4
π
×
10
7
T m
/
A.
5 m
1 m
Outside solenoid has 150 turns
Inside solenoid has 7100 turns
12 cm
2 cm
±ind the magnetic feld in the solenoid.
Correct answer: 0
.
0535327 T.
Explanation:
ℓ
2
ℓ
1
Outside solenoid has
N
1
turns
Inside solenoid has
N
2
turns
A
1
A
2
Let :
N
2
= 7100
,
ℓ
2
= 5 m
,
I
2
= 30 A
,
and
μ
0
= 4
π
×
10
7
T m
/
A
.
Basic Concepts:
Mutual Inductance:
M
21
=
N
1
Φ
21
I
2
=
N
2
Φ
12
I
1
=
M
12
.
(1)
Solution:
As in the fgure, iF the solenoid
carries a current
I
2
, then the magnetic feld is
given by
B
2
=
μ
0
N
2
I
2
ℓ
2
(2)
= (4
π
×
10
7
T m
/
A)
(7100) (30 A)
(5 m)
=
0
.
0535327 T
.
002
(part 2 oF 4) 10.0 points
±ind the magnetic ²ux Φ
12
through the cross
sectional area oF the large circular coil due to
the 30 A current in the solenoid.
Correct answer: 6
.
72712
×
10

5
T m
2
.
Explanation:
Let :
r
1
= 12 cm = 0
.
12 m
and
r
2
= 2 cm = 0
.
02 m
,
A
1
=
π r
1
2
=
π
(0
.
12 m)
2
= 0
.
0452389 m
2
and
A
2
=
π r
2
2
=
π
(0
.
02 m)
2
= 0
.
00125664 m
2
.
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 Fall '08
 Turner
 Physics, Work, Magnetic Field, Solenoid, Correct Answer, Inductor, Kapoor

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