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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