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

Length of a side of the square coil, 1 = 10 cm = 0.1 m

Current flowing in the coil, I = 12 A

Number of turns on the coil, n = 20

Angle made by the plane of the coil with magnetic field, 0 = 30°

Strength of magnetic field, B = 0.80 T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

τ = n BIA sin0

Where, A = Area of the square coil = 1×1 = 0.1 x 0.1 = 0.01m²

So,

τ = 20 x 0.8 x 12 x 0.01 x Sin30°

= 0.96 N m

Hence, the magnitude of the torque experienced by the coil is 0.96 N m.