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 30° with the axis of the solenoid?
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 = n.A.I = 2000 × 1.6 × 10-4 m2 x 4A
= 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°
τ = MBsinθ = 1.28 JT-1 × 7.5x10-2 T × sin30o
= 4.8 × 10-2 N-m
= 4.8x10-2 J
But the Force is zero as the Magnetic field is uniform.