A long, horizontal wire AB rests on the surface of a table and carries a current I. Horizontal wire CD is vertically above wire AB, and is free to slide up and down on the two vetical mental guides C and D (as shown in Fig) Wire CD is connected through the sliding contacts to another wire that also carries a current I, opposite in direction to the current in wire AB. The mass per unit length of the wire CD is `lambda`. To what euqilibrium heigth h will the wire CD rise, assuming that magnetic force on it is wholly due to the current in wire AB?

Answer

`(mu_0I^2)/(2pilambdag)`

Explanation

We need the magnetic and gravitational forces to cancel.
`lambdaLg=(mu_0I^2L)/(2pih) implies h=(mu_0I^2)/(2pilambdag)`