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Question
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.
- What is the magnetic moment associated with the solenoid?
- 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 30° with the axis of the solenoid?
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Solution
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
- The magnetic moment along the axis of the solenoid is calculated as:
M = nAI
= 2000 × 1.6 × 10−4 × 4
= 1.28 Am2 - Magnetic field, B = 7.5 × 10−2 T
Angle between the magnetic field and the axis of the solenoid, θ = 30°
Torque, τ = MB sin θ
= 1.28 × 7.5 × 10−2 × sin 30°
= 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.
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