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.
(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 x 10^-2 T is set up at an angle of 30° with the axis of the solenoid?

Answer 1

Number of turns of the solenoid = 2,000
I = 4A
A = 1.6 x 10^-4m²

(a) The magnetic moment of the solenoid,
M = (IA) x number of turns
= 4 x 1.6 x 10^-4 x 2000
= 1.28 Am²

(b) τ - MB sin θ
B = 7.5 x 10^-2T
θ = 30°
τ - 1.28 x 7.5 x 10^-2 x sin30°
= 0.048 Nm
The net force on the solenoid is zero.