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Question

A long,straight wire is fixed horizontally and carries a current of 50.0 A.A second wire having linear mass density 1.0×104 kg m1 is placed parallel to and directly above this wire at separation of 5.0 mm.What current should this second wire carry such that the magnetic repulsion can balance its weight ?

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Solution

According to question μ0i1i22πd=dFdl

2×107×50×i25×103=1×104×9.8

i2=9.8×10720×107A

0.49 A in opposite direction.


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