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Question:

A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10⁻⁸ m² carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10⁻⁶ tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be?

1.5 × 10⁻⁷ Nm

3 × 10⁻⁶ Nm

3 × 10⁻⁷ Nm

1.5 × 10⁻⁶ Nm

Solution:

Magnetic moment of the loop:
M = NIA = 2000 × 2 × 1.5 × 10⁻⁸ = 0.6 J/T
torque τ = MBsin30° = 0.6 × 5 × 10⁻⁶ × (1/2) = 1.5 × 10⁻⁶ Nm