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.

Answer 2

(a) Associated magnetic moment,
\(μ_m = niA\)
\(μ_m = 2000 × 4 ×1.6 × 10^{-4}\) \(Am^2\)
\(μ_m = 1.28\  Am^2\)

(b) Toeque \(= μ_m B sin \theta\)
Toeque \(= 1.28 ×7.5 ×10^{-2} × sin 30^o\)
Toeque \(= 0.048\ Nm^2\)