A square coils of side 10 cm consists of 20 turns and carries a current of 12A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of `30^(@)` with the direction of a uniform horizontal magnetic field of magnitude `0.80T`. What is the magnitude to torque experienced by the coil ?

Explanation

Torque,
`tau= NIBA sin theta`
`=20xx12xx(10xx10^(-2))xx0.8 sin 30^(@)`.
`0.96N*m` [ Here, N= number of turns, I= currents, A= ara of the coil, B = magnetic field]