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. (a) What is the magnetic moment associated with the solenoid? (b) 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 30º with the axis of the solenoid?

Number of turns on the solenoid, n = 2000

Area of cross-section of the solenoid, A = 1.6 x 10^-4m²

Current in the solenoid, I = 4 A

Ans (a).

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

M = nAI = 2000 x 1.6 x 10⁻⁴ x 4 = 1.28 Am²

Ans (b).

Magnetic field, B = 7.5 x 10^-2 T

Angle between the magnetic field and the axis of the solenoid, 0 = 30°

Torque, τ = MBsinB

= 1.28 x 7.5 x10⁻² sin 30°

= 4.8 x 10⁻² Nm

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 x 10⁻² Nm.