In the adjoining diagram a current carrying loop pqrs is placed with its sides parallel to a long current -carrying wire. The currents `I_(1)` and `I_(2)` in the wire and loop are 20 A and 16 A respectively . If a = 15 cm, b = 6 cm and d = 4 cm, what will be the force on current loop pqrs?
What will be the difference in the force, if the current `I_(2)` in the loop is clockwise instead of anticlockwise?

Explanation

The repulsive force on the side ps of the current carrying loop, due to to current `i_(1)` is
`F_(1) (mu_(0) i_(1) i_(2) a)/(2 pi d) = (2 xx 10^(-7)) xx (20 xx 16 xx 0.15)/(0.04) = 2.4 xx 10^(-4) N`
This force will be towards RHS and `_|_` to the current carrying wire ps. Similarly, the alternative force acting on the side qr of the loop, due to current `I_(1)` is
Here R = d + b = 10 cm = 0.1 meter)
`F_(2) = (2 xx 10^(-7)) xx (20 xx 16 xx 0.15)/(0.10) = 0.96 xx 10^(-4)N`
Direction of this force will be towards LHS and `_|_` to current carrying wire qr. The forces acting on the sides pq and rs of the loop will be equal and opposite . The net force on the loop `= F_(1) - f_(@) = (2.4 - 0.96) xx 10^(-4) = 1.44 xx 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 remian same but its direction now becomes towards the current carrying wire.