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_(0)I^(2)(a^(2)))/d`
B. `(mu_(0)I^(2)alpha^(2))/(2d)`
C. `(sqrt(3)mu_(0)I^(2)a^(2))/d`
D. `(sqrt(3)mu_(0)I^(2)a^(2))/(2d)`

Answer

B

Explanation

Magnetic field at mid-point of two wires = 2 (magnetic field due to one wire) `=2[(mu_(0))/(2pi) I/d]=(mu_(0)I)/(pid)ox`
Magnetic moment of loop `M=IA =I pia^(2)` Torque on loop `=MB sin 150^(@)=(mu_(0)I^(2)a^(2))/(2d)`