A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic field of 0.050 T oriented at 30° to the plane of the loop. What torque acts on the loop?

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm