A short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3…
A short bar magnet placed with its axis at $45^{\circ}$ with a uniform external magnetic field of $28.3 \times 10^{-3} \mathrm{~T}$ experiences a torque of magnitude equal to $3.6 \times 10^{-5} \mathrm{~J}$. The magnitude of magnetic moment of the magnet is nearly
$1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
$1.2 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
$2.4 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
$1.6 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
Solution
We know that
$\tau=\mathrm{MB} \sin \theta$
$\Rightarrow \mathrm{M}=\frac{\tau}{\mathrm{B} \sin 45^{\circ}}=\frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times \frac{1}{\sqrt{2}}}=1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$