Question

A long horizontally fixed wire carries a current of $100$ ampere. Directly above and parallel to it is a fine wire that carries a current of $20$ ampere and weights $0.04$ newton per meter. The distance between the two wires for which the upper wire is just supported by magnetic repulsion is:

• A. ${10}^{-2}mm$
• B. ${10}^{-2}cm$
• C. ${10}^{-2}m$
• D. ${10}^{-2}km$

Answer 1

The correct option is C ${10}^{-2}m$

Force on the wire element per unit length is given by:

$F=\frac{{\mu }_o{I}_1{I}_2}{2\pi r}$

This balances the weight of the wire per unit length

$0.04=\frac{2\times {10}^{-7}\times 100\times 20}{r}$

$r={10}^{-2}m$