λ
22
6.089
−
10
5
−
×
V sec
⋅
m
=
λ
22
µ
0
2
π
⋅
I
−
ln
1
21
⋅
⋅
−
:=
λ
21
5.991
10
5
−
×
V sec
⋅
m
=
λ
12
6.516
10
5
−
×
V sec
⋅
m
=
λ
11
6.438
−
10
5
−
×
V sec
⋅
m
=
λ
21
µ
0
2
π
⋅
I
−
ln
1
20
⋅
⋅
:=
λ
12
µ
0
2
π
⋅
I ln
1
26
⋅
⋅
−
:=
λ
11
µ
0
2
π
⋅
I ln
1
25
⋅
⋅
:=
That's how the author's solution manual solves it.
We really could use some more details.
We seek the flux that links the telephone wires and electrical wires.
This relates to a mutual
inducance behavior, not a self inductance behavior.
Therefore, we should identify the currents that
cause the flux and calculate the amount of that flux that links the other lines.
There will be 4 terms
here, a mutual flux generated by each of two currents (going and returning) linked to each of two
telephone wires.
Make every effort to get the algebraic signs consistent.
V
t
0.116
V
mi
=
V
t
7.215
10
5
−
×
V
m
=
V
t
j
ω
⋅
λ
⋅
:=
Find the voltage per unit length
λ
1.914
−
10
7
−
×
Wb
m
=
λ
µ
0
2
π
⋅
I ln
d
p
d
s
+
d
t
+
d
p
d
s
+
⋅
I
−
(
) ln
d
s
d
t
+
d
s
⋅
+
⋅
:=
Find the flux that links the telephone line.
j
1
−
:=
d
t
1 ft
⋅
:=
d
s
20 ft
⋅
:=
d
p
5 ft
⋅
:=
ω
2
π
⋅
60
⋅
rad
sec
⋅
:=
I
100 A
⋅
:=
µ
0
4
π
⋅
10
7
−
⋅
henry
m
⋅
:=
State the given
5 ft
1 ft
20 ft
+I
 I
Suppose that a 60 Hz single phase power line and an open wire telephone line are parallel to
each other and in the same horizontal plane.
The power line spacing is 5 feet and the
telephone wire spacing is 12 inches.
The nearest conductors of the two lines are 20 feet apart.
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 Spring '11
 gully
 Inch, Electrical resistance, Threephase electric power, Telephone line, GMR

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