Question

A short bar magnet placed with its axis at $30^\circ $ with a uniform external magnetic field of $0.25T$ experiences a torque of magnitude equal to $4.5 \times {10^{ - 2}}J$. What is the magnitude of the magnetic moment of the magnet ?

Answer 1

Hint: If a bar magnet is placed in uniform external magnetic field then it experiences a torque which is given by
$\sigma = mB\sin \theta $
Where
m $ = $ magnetic moment of the magnet
B $ = $ external uniform magnetic field
$\theta = $ angle between m and B.

Complete step by step answer:
Here given that a short bar magnet is placed with its axis at $30^\circ $ with a uniform external magnetic field then the bar magnet experience a torque
i.e., $\sigma = mB\sin \theta $
where
m $ = $ magnetic moment
B $ = $ magnetic field \[\left( {external} \right)\]
$\theta = $ angle between axis of bar magnet and external magnetic field
Given that
$B = 0.25T$
$\theta = 30^\circ $
$\sigma = 4.5 \times {10^{ - 2}}J$
$m = ?$
$\because \sigma = mB\sin \theta $
$\therefore m = \dfrac{\sigma }{{B\sin \theta }}$
So, $m = \dfrac{{4.5 \times {{10}^{ - 2}}}}{{0.25 \times \sin 30^\circ }}$
$\because \sin 30^\circ = \dfrac{1}{2}$
$\therefore m = \dfrac{{4.5 \times {{10}^{ - 2}}}}{{0.25 \times \dfrac{1}{2}}}$
$m = \dfrac{{4.5 \times 2 \times {{10}^{ - 2}} \times {{10}^2}}}{{25}}$
$m = \dfrac{9}{{25}} = 0.36\dfrac{J}{T}$
or $m = 0.36amp{m^2}$
Hence the magnitude of magnetic moment of bar magnet is $0.36\dfrac{J}{T}$ or $0.36amp{m^2}$

Note:
If angle $\theta = 0^\circ $ then torque $\sigma = mB\sin 0^\circ $
$\sigma = 0$
Then magnet is in stable equilibrium
If angle $\theta = 180^\circ $ then the bar magnet is in unstable equilibrium.