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 30º 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 = nAI
= 2000 × 1.6 × 10−4 × 4
= 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°
Torque, $\tau=M B \sin \theta$
$=1.28 \times 7.5 \times 10^{-2} \sin 30^{\circ}$
$=4.8 \times 10^{-2} \mathrm{Nm}$
Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is $4.8 \times 10^{-2} \mathrm{Nm}$.