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.

As force per unit length between two parallel current carrying wires separated by a distance dis given by dFdl= μ04π 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  ⇒  μ0i1i22πd = MgLd= μ0i1i22π MgL  =   4π × 10−7 ×50 × 252π × 0.075md=10−23m