A long horizontally fixed wire carries a current of 100 A directly above and parallel to it is a fine wire that carries a current of 20 A and weight 0.04 N/m The distance between the two wires for which the upper wire is just supported by magnetic repulsion is

Explanation

The upper wire is acted on by
(i) Its weight per unit length directed vertically downward =`0.073 N//m`
(ii) The magnetic force on it per unit length by the current in the lower wire = `vec F_(mag)`
Then for equilibrium`vec F_(mag)` per metre = `0.073 N//m`
`mu_(0)/4pi 2i_(1)i_(2)/r=mu_(0)/4pi (2)(100)(20)/r=0.073`
`rArr r=(10^(-7))(4000)/0.073=5.48xx10^(-3)m`
Sınce the weight of the upper wire is directed towards the lower wire, the `F_(m)` on the upper wire must be directed away from the lower wire. Then the magnetic interaction between the two wires must be repulsion i.e. the current in the upper one must be directed opposite to that in the lower one.