P30_024.PDF
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60 KB
)
Pobierz
Chapter 30 - 30.24
24. We label these wires 1 through 5, left to right, and use Eq. 30-15 (divided by length). Then,
F
1
=
µ
0
i
2
2
π
1
d
+
1
2
d
+
1
3
d
+
1
4
d
j=
25
µ
0
i
2
24
πd
j
=
(13)(4
π
×
10
−
7
T
m
/
A)(3
.
00 A)
2
(1
.
00 m)j
24
π
(8
.
00
×
10
−
2
m)
·
=4
.
69
×
10
−
5
N
/
m
ˆ
j;
F
2
=
µ
0
i
2
2
π
1
2
d
+
1
3
d
j=
5
µ
0
i
2
12
πd
j=1
.
88
×
10
−
5
N
/
m
ˆ
j;
F
3
= 0 (because of symmetry);
F
4
=
−
F
2
;and
F
5
=
−
F
1
.
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