Question

When a bar magnet is suspended in a uniform magnetic field, the torque acting on it will be :

a) maximum	e) $\theta ={45}^o$ with the field
b) half the maximum field	f) $\theta ={60}^o$ with the field
c) $\sqrt{3}/2$ times the maximum field	g) $\theta ={30}^o$ with the field
d) $1/\sqrt{2}$ times the maximum field	h) $\theta ={90}^o$ with the field

• A. a-g, b-h, c-d, d-e
• B. a-e, b-f, c-g, d-h
• C. a-f, b-e, c-g, d-h
• D. a-h, b-g, c-f, d-e

Answer 1

Answer: (D)

a-h, b-g, c-f, d-e

Torque acting on the bar magnet suspended in a uniform field B is given by,

$T=MBsin\theta$ where $M$ is the moment of magnet.

Torque is maximum when $sin\theta =1$ i.e. $\theta ={90}^o$

$\therefore$ maximum torque $=MB$

Torque is half when $sin\theta =1/2$ i.e. $\theta ={30}^o$

Torque is $\sqrt{3}/2$ the maximum when $sin\theta =\sqrt{3}/2$ i.e. $\theta ={60}^o$

Torque is $1/\sqrt{2}$ the maximum when $sin\theta =1/\sqrt{2}$ i.e. $\theta ={45}^o$