Question

12. A flat circular coil of wire having 200 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?

Answer 1

Answer:

The magnitude of the magnetic torque on the coil is 1.98 A.m²

Explanation:

Magnitude of magnetic torque in a flat circular coil is given as;

τ = NIASinθ

where;

N is the number of turns of the coil

I is the current in the coil

A is the area of the coil

θ is the angle of inclination of the coil and magnetic field

Given'

Number of turns, N = 200

Current, I = 7.0 A

Angle of inclination, θ = 30°

Diameter, d = 6 cm = 0.06 m

A = πd²/4 = π(0.06)²/4  = 0.002828 m²

τ = NIASinθ

τ = 200 x 7 x 0.002828 x Sin30

τ = 1.98 A.m²

Therefore, the magnitude of the magnetic torque on the coil is 1.98 A.m²