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Question

A long straight wire is fixed horizontally and carries a current of 50A. A second wire having linear mass density 10×104 kg m1 is placed parallel to and directly above this wire at a separation of 50 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight ?

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Solution

μ0i1i22πd=mg (For a portion of wire of length 1m)
μ0×50×i22π×5×103=1×104×9.8
4π×107×5×i22π×5×103=9.8×104
2×i2×103=9.3×103×101
i2=9.82×101=0.49A

1554029_1093703_ans_64e29c56069b4bbeb97ba37c6460e365.JPG

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