Question

A long horizontal wire P carries a current of $50$A. It is rigidly fixed. Another wire Q is placed directly above and parallel to P, as shown in Figure. The weight per unit length of the wire Q is $0.025$ $N{m}^{-1}$ and it carries a current of $25$A. Find the distance 'r' of the wire Q from the wire P so that the wire Q remains at rest.

Answer 1

Let the length of wire Q be $1m$ i.e. $l=1m$
So weight of wire Q $mg=0.025\times 1=0.025N$
Since current in both the wire is flowing in opposite directions, so both wire will repel each other.
Force acting on the wire Q due to P $F=\frac{{\mu }_o{I}_1{I}_{2}l}{2\pi r}$
$\therefore$ $F=\frac{4\pi \times {10}^{-7}\times 50\times 25\times 1}{2\pi \times r}=\frac{25\times {10}^{-5}}{r}$
For the wire Q to be in equilibrium, $F=mg$
$\therefore$ $\frac{25\times {10}^{-5}}{r}=0.025$
$⟹r=0.01m=1cm$