Question

A circular coil of wire of 200 turns and diameter 2.0 cm carries a current of 4.0 A. It is placed in a magnetic field of 0.70 T with the plane of the coil making an angle of 30° with the magnetic field. What is the magnetic torque on the coil?

Answer 1

Answer:

0.087976 Nm

Explanation:

The magnetic torque (τ) on a current-carrying loop in a magnetic field is given by;

τ = NIAB sinθ     --------- (i)

Where;

N = number of turns of the loop

I = current in the loop

A = area of each of the turns

B = magnetic field

θ = angle the loop makes with the magnetic field

From the question;

N = 200

I = 4.0A

B = 0.70T

θ = 30°

A = π d² / 4        [d = diameter of the coil = 2.0cm = 0.02m]

A = π x 0.02² / 4 = 0.0003142m²         [taking π = 3.142]

Substitute these values into equation (i) as follows;

τ = 200 x 4.0 x 0.0003142 x 0.70 sin30°

τ = 200 x 4.0 x 0.0003142 x 0.70 x 0.5

τ = 200 x 4.0 x 0.0003142 x 0.70      

τ = 0.087976 Nm

Therefore, the torque on the coil is 0.087976 Nm