Question

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 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Answer 1

I = 12A
n = 20
B = 0.80 T
θ =30°
A = 10 x 10= 100cm²
= 100 x 10⁴m²
τ = n BIA sin θ
= 2d x 0.80 x l00 x l0^-4 x l2 x sin30°
= 0.96 Nm.