Question

A bar magnet is suspended by a thin wire in uniform magnetic field. When upper end of the wire is twisted by 120° , then the magnet gets deflected by 30° from its position. How much upper end of the wire be twisted so that magnet may rotate by 90° from its initial position

(a) 270°

(b) 360°

(c) 180°

(d) 45°

Answer 1

Correct option (a) 270° 

Explanation:

Torque = MB Sinϕ = Cθ;

We know that T = Cθ 

θ is net deflection.

MB Sin ϕ = Cθ 

 ϕ = angle by which magnet is deflected.