Question

A square shaped current carrying loop MNOP is placed near a straight long current carrying wire AB as shown in the fig. The wire and the loop lie in the same plane. If the loop experiences a net force F towards the wire, find the magnitude of the force on the side ‘NO’ of the loop.

Answer 1

The force acting on the section MN and the force on the section PO will cancel as the wires are located at equal distances from the infinite wire, but have current flowing in opposite directions.

The force acting on the whole loop,

F = `(mu_0 I_1 I_2 L)/(2 pi L) - (mu_0 I_1 I_2 L)/(2 pi (2 L))`

= `(mu_0 I_1 I_2 L)/(4 pi L)`

Towards the wire.

The force acting on the side ‘NO’ is given by:

`F_"NO" = (mu_0 I_1 I_2)/(2 pi(2 L))L = (mu_0 I_1 I_2)/(4 pi)`

=  F

i.e., Away from the wire.