Question

In the adjoining diagram, a current-carrying loop pqrs placed with its sides parallel to a long current-carrying wire. The currents $i_1$ and $i_2$ in the wire and loop are 20A and 16 A respectively. If a= 15 cm, b= 6 cm, and d= 4 cm, what will be the force on current $i_2$ in the loop is clockwise instead of anticlockwise?

Answer 1

The repulsive force on the side ps of the current-carrying loop, due to current $i_1$ is
$F_1=\frac{{\mu }_0}{2\pi }\frac{i_1{i}_{2}a}{d}=(2\times {10}^{-7})\times \frac{20\times 16\times 0.15}{0.04}=2.4\times {10}^{-4}N$
This force will be towards RHS and $\perp$ to the current-carrying wire ps. Similarlly, the force acting on the side qr of the loop, due to current $I_1$ is
(Here R= d +b = 10 cm = 0.1 meter)
$\therefore F_2=(2\times {10}^{-7})\times \frac{20\times 16\times 0.15}{0.10}=0.96\times {10}^{-4}N.$
Direction of this force will be towards LHS and $\perp$ to current-carrying wire qr.
The force acting on the sides pq and rs of the loop will be equal and opposite.
Thus net force on the loop $=F_1-F_2=(2.4-0.96)\times {10}^{-4}=1.44\times {10}^{-4}N$
(Acting away from the current-carrying wire)
When the direction of current in the loop becomes clockwise, the net force on the loop remain same, but its direction now becomes towards the current-carrying wire.