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)a^(2))/(2d)`
C. `(sqrt3 mu_(0)I^(2)a^(2))/d`
D. `(sqrt3 mu_(0) I^(2) a^(2))/(2 d)`

Answer

b

Explanation

Net magnetic field due to both the wires can be written as follows:
` B = 2 ((mu_(0)i)/(2 pi d)) = (mu_(0)i)/(pi d)`
According to given orientation angle between area vector and magnetic field is 30°. Hence torque on loop can be written as follows:
` tau = i pi a^(2) (mu_(0)i)/(pi d) sin (30^(@))`
` rArr tau = (mu_(0) i^(2) a^(2))/(2 d) `
Hence option (b) is correct.