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 $\frac{\mathrm{dF}}{\mathrm{dl}}=\frac{{ }_{{\mu }_0}}{4\mathrm{\pi }}\frac{2{\mathrm{i}}_1{\mathrm{i}}_2}{\mathrm{d}}$

It 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.

$\begin{array}{l}\mathrm{So},{\mathrm{F}}_{\mathrm{m}}=\mathrm{Mg}\\ \frac{\mathrm{F}}{\mathrm{L}}=\frac{\mathrm{Mg}}{\mathrm{L}}\Rightarrow \frac{{\mathrm{\mu }}_0{\mathrm{i}}_1{\mathrm{i}}_2}{2\mathrm{\pi d}}=\frac{\mathrm{Mg}}{\mathrm{L}}\\ \mathrm{d}=\frac{{\mathrm{\mu }}_0{\mathrm{i}}_1{\mathrm{i}}_2}{2\mathrm{\pi }\frac{\mathrm{Mg}}{\mathrm{L}}}=\frac{4\mathrm{\pi }\times {10}^{-7}\times 50\times 25}{2\mathrm{\pi }\times 0.075}\text{m}\\ \mathrm{d}=\frac{{10}^{-2}}{3}m\end{array}$