The figure shows a circular loop of radius `a` with two long parallel wires `(numbered 1 and 2)` all in the plane of the paper . The distance of each wire from the centre of the loop is `d`. The loop and the wire are carrying the same current `I`. The current in the loop is in the counterclockwise direction if seen from above.
(q) The magnetic fields(B) at `P` due to the currents in the wires are in opposite directions.
(r) There is no magnetic field at `P`.
(s) The wires repel each other.

(5) Consider `dgtgta`, and the loop is rotated about its diameter parallel to the wires by `30^(@)` from the position shown in the figure. If the currents in the wire are in the opposite directions, the torque on the loop at its new position will be ( assume that the net field due to the wires is constant over the loop).

Options

A. `(mu_0I^2a^2)/(d)`
B. `(mu_0I^2a^2)/(2d)`
C. `(sqrt3mu_0I^2a^2)/(d)`
D. `(sqrt3mu_0I^2a^2)/(2d)`

Answer

B

Explanation

When `d gt gt a`, then magnetic field at O due to currents in wires 1 and 2 will be
`B=2xx(mu_0)/(4pi)(2I)/(d)=(mu_0I)/(pid)`
Torque on the loop `=BIAsin30^@`
`=(mu_0I)/(pid)xxIxx(pia^2)xx1/2=(mu_0I^2a^2)/(2d)`