Question:
A closely wound solenoid of 2000 turns and area of cross-section 1.6 x 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 30degree with the axis of the solenoid?
Answer:
Number of turns on the solenoid, n = 2000
Area of cross-section of the solenoid, A= 
Current in the solenoid, I= 4 A
(a)The magnetic moment along the axis of the solenoid is calculated as:
M= nAI
= 
= 1.28 A
(b)Magnetic field, B =
Angle between the magnetic field and the axis of the solenoid, θ= 30°
Torque, 
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 
s
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