Question

A closely wound solenoid of $2000$ turns and area of cross-section $1.6 × 10^{-4}\ m^2$, 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 $30º$ with the axis of the solenoid?

Answer 1

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 = nA $I$
= 2000 × 1.6 × 10-4 × 4
= 1.28 Am2

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b. Magnetic field, B = 7.5 × 10-2 T
Angle between the magnetic field and the axis of the solenoid, $\theta$ = $30\degree$
Torque, τ = $MB\sin\theta$
=1.28 × 7.5 × 10-2 $\sin30\degree$
= 4.8 × 10-2 Nm
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 4.8 × 10-2 Nm.