A long horizontal wire P carries a current of $$50$$ A. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P. The weight of wire Q is $$0.075$$ $$N/m$$ and carries a current of $$25$$ A. Find the position of wire Q from P so that the wire Q remains suspended due to magnetic repulsion.

Question

Infinity Learn · Accepted Answer

As force per unit length between two parallel current carrying wires separated by a distance dis given by $$dFdl=$$ μ$$04$$π $$2i1i2dIt$$ is repulsive if the current in the wires are in opposite direction.In order that wire Q remains suspended, the magnetic force must be equal to its weight.So,  Fm $$=MgFL$$ $$=$$ MgL  ⇒  μ$$0i1i22$$$$π$$d $$=$$ $$MgLd=$$ μ$$0i1i22$$π MgL  $$=$$   $$4$$π × $$10$$−$$7$$ ×$$50$$ × $$252$$π × $$0.075md=10$$−$$23m$$

A long horizontal wire P carries a current of 50 A. It is rigidly fixed. Another fine wire Q is placed directly above and parallel to P. The weight of wire Q is 0.075 N/m and carries a current of 25 A. Find the position of wire Q from P so that the wire Q remains suspended due to magnetic repulsion.

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