Question 8

A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 10-4m2, 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 10-2T is set up at an angle of 30º with the axis of the solenoid?

Answer

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 ×10-2T

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

Torque, T = MB sin θ

= 1.28 x 7.5 x 10-2 sin30°
= 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 

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