30.2
If we have a current
I
2
going through the inner solenoid, it produces a field:
B
=
μ
0
n
2
I
2
The flux through the outer solenoid is
φ
12
=
BA
= (
μ
0
n
2
I
2
)(
πr
2
2
)
Therefore the mutual inductance is
M
=
N
1
φ
12
I
2
=
N
1
μ
0
n
2
I
2
πr
2
2
I
2
=
N
1
μ
0
n
2
πr
2
2
Finally, we want the mutual inductance
per unit length
so we divide by the length.
We notice that
N
1
/l
=
n
1
(total number of turns divided by the length is number of turns
per unit length). So our final answer is:
m
=
N
1
μ
0
n
2
πr
2
2
/l
=
μ
0
n
1
n
2
πr
2
2
30.4
The magnetic field due to a long wire is
B
=
μ
0
I
2
πr
So the flux through the rectangular loop is:
φ
=
Z
BdA
=
w
Z
l
2
l
1
μ
0
I
2
πr
dr
=
w
μ
0
I
2
π
ln
l
2
l
1
The mutual inductance is
M
=
φ
I
=
μ
0
w
2
π
ln
l
2
l
1
30.9
E
=

L
dI
dt
=

(440
mH
)(3
.
60
A/s
) =

1
.
584
V
1
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30.10
The inductance per unit length for a coaxial cable is (see example 305 for the derivation
of the formula):
L
l
=
μ
0
2
π
ln
r
2
r
1
Plug in
r
2
= 3
.
0mm and
L/l
= 55nH, we will be able to find
r
1
.
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 Spring '12
 welsch
 Inductance, Inductor, RL circuit, LC circuit

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