A long horizontal wire `P` carries of `50 A`. It is rigidly fixed. Another fine wire `Q` is placed directly above and parallel to `P`. The wieght 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.

Explanation

The upper wire is in equilibrium due to weight of wire and magnetic repulsive force. Since magnetic force should be in upward direction to balance the weight of wire, for this current in upper wire should be opposite to current in lower wire.
Consider unit length of wire.
`F_(m) = W`
`(mu_(0)i_(1)i_(2))/(2 pid) = W`
`(2 xx 10^(-7) xx 50 xx 25)/(d) = 0.075`
`d = 3.3 xx 10^(-3) m`