Question

A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^–4 m², carrying a current of 4.0 A, is suspended through its centre allowing it to turn in a horizontal plane.

1. What is the magnetic moment associated with the solenoid?
2. What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10^–2 T is set up at an angle of 30° with the axis of the solenoid?

Answer 1

Given: Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 × 10⁻⁴ m²

Current in the solenoid, I = 4 A

(a) The magnetic moment along the axis of the solenoid is calculated as:

M = nAI

= 2000 × 1.6 × 10⁻⁴ × 4

= 1.28 Am²

(b) Magnetic field, B = 7.5 × 10⁻² T

Angle between the magnetic field and the axis of the solenoid, θ = 30°

Torque, τ = MB sin θ

= 1.28 × 7.5 × 10⁻² × sin 30°

= `1.28 xx 7.5 xx 10^(−2) xx 1/2`

= 1.28 × 7.5 × 10⁻² × 0.5

= 4.8 × 10⁻² Nm

Since the magnetic field is uniform, the force on the solenoid is zero.

The torque on the solenoid is 4.8 × 10⁻² Nm.