P30_030.PDF

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Chapter 30 - 30.30
30. A close look at the path reveals that only currents 1, 3, 6 and 7 are enclosed. Thus, noting the different
current directions described in the problem, we obtain
B
ds = µ 0 (7 i
−
6 i +3 i + i )=5 µ 0 i.
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