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A closely wound solenoid of 2000 turns and area of cross-section 1.5 × 10⁻⁴ m² carries a current of 2.0 A. It is suspended through its center and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field 5 × 10⁻² tesla making an angle of 30° with the axis of the solenoid. The torque on the solenoid will be

1. 3 × 10⁻² N-m
2. 3 × 10⁻³ N-m
3. 1.5 × 10⁻³ N-m
4. 1.5 × 10⁻² N-m

Correct answer: 3 × 10⁻² N-m

Solution

The torque on a current-carrying solenoid in a magnetic field is given by τ = nIABsinθ, where n is the number of turns, I is the current, A is the area, B is the magnetic field, and θ is the angle between the solenoid axis and the field. Substituting the values: τ = 2000 × 2 × 1.5 × 10⁻⁴ × 5 × 10⁻² × sin30° = 3 × 10⁻² N-m.

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