Acircular coil of wire of 200 turns and diameter 6 cm carries a current of 7 a. it is placed in a magnetic field of 0.90 t with the plane of the coil making an angle of 30° with the magnetic field. what is the torque on the coil?

3.08 Nm

Explanation:

N = 200, diameter = 6 cm, radius = 3 cm, I = 7 A, B = 0.90 T, Angle = 30 degree

The angle made with the normal of the coil, theta = 90 - 30 = 60 degree

Torque = N I A B Sin Theta

Torque = 200 x 7 x 3.14 x 0.03 x 0.03 x 0.90 x Sin 60

Torque = 3.08 Nm

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