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

Options

(a) 3 x 10⁻² N-m
(b) 3 x 10⁻³ N-m
(c) 1.5 x 10⁻³ N-m
(d) 1.5 x 10⁻² N-m

Correct Answer:

1.5 x 10⁻² N-m

Explanation:

Torque on the solenoid is given by ? = MB sin θ
where θ is the angle between the magnetic field and the axis of solenoid. M = niA
? = niA B sin 30⁰
= 2000 x 2 x 1.5 x 10⁻⁴ x 5 x 10⁻² x 1/2
= 1.5 x 10⁻² N – m