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
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.
(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 \(= 0.048\ Nm^2\)
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