Question

A long horizontal rigidly supported wire carries a current $i_a=96A$. Directly above it and parallel to it at a distance, another wire of $0.144N$ weight per metre is carrying a current $i_b=24A$, in a direction same as the lower wire. If the weight of the second wire is balanced by the force due to magnetic repulsion, then its distance (in mm) from the lower wire is :

• A. $9.6$
• B. $4.8$
• C. $3.2$
• D. $1.6$

Answer 1

The correct option is C $3.2$

Given : $i_a=96A$ $i_b=24A$
Let the separation between the wires be $d$.
Thus magnetic force between them $F_m=\frac{{\mu }_o{i}_a{i}_b}{2\pi d}$
Weight of wire $W=0.144N$
$\therefore$ $W=\frac{{\mu }_o{i}_a{i}_b}{2\pi d}$
$⟹d=\frac{{\mu }_o{i}_a{i}_b}{2\pi W}$
$d=\frac{4\pi \times {10}^{-7}\times 24\times 96}{2\pi \times 0.144}=3.2\times {10}^{-3}m=3.2mm$