Question

A short bar magnet is placed in a uniform magnetic field of $2~\text{T}$ such that the axis of the magnet makes an angle of $45^\circ$ with the field. If the torque acting is $0.36~\text{Nm}$, the magnetic moment of the magnet is

• A. $0.54~\text{J/T}$
• B. $0.18~\text{J/T}$
• C. $0.72~\text{J/T}$
• D. $0.36~\text{J/T}$

Hint:

Torque on magnet in field: $\tau = MB \sin\theta$. Rearrange to get $M$ when torque is known.

Answer 1

The Correct Option is D

$\tau = MB \sin\theta \Rightarrow M = \dfrac{\tau}{B \sin\theta}$
$M = \dfrac{0.36}{2 \cdot \sin 45^\circ} = \dfrac{0.36}{2 \cdot \dfrac{1}{\sqrt{2}}} = \dfrac{0.36}{\dfrac{2}{\sqrt{2}}} = 0.36 \cdot \dfrac{\sqrt{2}}{2} \cdot \sqrt{2} = 0.36$