Q. 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 centre 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
A) 3 × 10⁻³ N m
B) 1.5 × 10⁻³ N m
C) 1.5 × 10⁻² N m
D) 3 × 10⁻² N m
Explanation
[Solution from source] Magnetic moment m = NIA = 2000 × 2.0 × 1.5 × 10⁻⁴ = 0.6 A m². Torque τ = mB sin 30° = 0.6 × 5 × 10⁻² × 0.5 = 1.5 × 10⁻² N m. Using cos 30° or dropping the factor N leads to the other values. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.