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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
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Verified Answer
The correct answer is:
$1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
We know that
$\tau=\mathrm{MB} \sin \theta$
$\Rightarrow \mathrm{M}=\frac{\tau}{\mathrm{B} \sin 45^{\circ}}=\frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times \frac{1}{\sqrt{2}}}=1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
$\tau=\mathrm{MB} \sin \theta$
$\Rightarrow \mathrm{M}=\frac{\tau}{\mathrm{B} \sin 45^{\circ}}=\frac{3.6 \times 10^{-5}}{28.3 \times 10^{-3} \times \frac{1}{\sqrt{2}}}=1.8 \times 10^{-3} \mathrm{~J} \mathrm{~T}^{-1}$
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