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Question

Three long straight conductors are kept perpendicular to the plane of paper. Currents 2A, 3A are passing through the two conductors into the plane of paper in first two conductors and 5A current passes through third conductor, directed out of the paper. A closed loop encloses the conductors, then the value of B.dl over the closed loop is (assume current into the paper as negative and out of the paper as positive)

A
2μ
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B
0
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C
10μ
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D
+μ
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Solution

The correct option is B 0
Using ampere's circuital law,
The line integral of magnetic field in a closed loop equals μ0i , where i is the current enclosed by the loop
B.dl=μ0i
=μ0(2+35)
=0

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