A long current carrying wire, carrying current `I_(1)` such that `I_(1)` is flowing out from the plants of paper is placed at O. A steady state current `I_(2)` is flowing in the loop ABCD.

Options

A. The net force is zero
B. The net torque is zero
C. As seen from O, the loop will rotate in clockwise along OO' axis
D. As seen from O, the loop will rotate in anticlockwise direction along OO' axis.

Answer

A, C

Explanation

The current carrying wire is placed at O. it carries a current `I_(1)` which flows out from the plane of paper. Field lines are circular around the wire, due to `I_(1)`. Force on AB= zero, as the magnetic lines are parllel, to the arm AB of the loop. Therefore, `dF= IdlB sin theta or dF= IdlB sin 0^(@)=` zero. Force on CD= zero, as the magnetic lines of `I_(1)` are antiparallel to the arm CD of the loop. Therefore, `dF= IdlB sin theta= IdlB sin 180^(@)=` zero. Force on BC is perpendicular to plane of paper outwards, according to Fleming.s left hand rule. Force on CD is perpendicular to plane of paper inwards, according to Fleming.s left hand rule.

(a) The forces on BC and CD are equal and opposite.
`:.` The net force on ABCD is zero. Torque on loop ABCD: The equal and opposite forces, on BC and AD, constitute a torque which, as seen from O, will rotate the loop in clockwise direction, along OO. as axis.