Question

​A square loop of side 20 cm carrying current of 1A is kept near an infinite long straight wire carrying a current of 2A in the same plane as shown in the figure. Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor.​

Answer 1

Explanation:

A square loop of side 20 cm carrying current of 1 A is kept near an infinite long straight wire carrying a current of 2 A in the same plane as shown in the figure.

Calculate the magnitude and direction of the net force exerted on the loop due to the current carrying conductor.

Answer 2

The net force exerted on the loop due to the current carrying conductor is F = 5.33 x 10^-7 N

Explanation:

We are given:

I1 = 2A  , I2 = 1A

d1 = 10 cm , d2 = 30 cm

μo = 4 π x 10^-7 TmA^-1

We have

F = μo I1 I2 / 2πd x l

Now net force on the side will be;

F = μo 2 x 1 / 2π x 20 x 10^-2 [ 1/ 10 x 10^-2 - 1 / 30 x 10^-2 ]N

F = 4 x 10^-7 x 20 [ 20 / 30 x 10 ]N

F = 16 / 3 x 10^-7 N

F = 5.33 x 10^-7 N

This net force is directed towards the infinitely long straight wire.

Thus the net force exerted on the loop due to the current carrying conductor is F = 5.33 x 10^-7 N