A small circular loop of conducting wire has radius $a$ and carries current $I$. It is placed in a uniform magnetic field $\mathrm{B}$ perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period $T$. If the mass of the loop is $m$ then
JEE MAIN 2020, Medium
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$\overrightarrow{\mathrm{T}}=\overrightarrow{\mathrm{M}} \times \overrightarrow{\mathrm{B}}=-\mathrm{MB} \sin \theta$
$\mathrm{I} \alpha=-\mathrm{MB} \sin \theta$
for small $\theta$
$\alpha=-\frac{\mathrm{MB}}{\mathrm{I}} \theta$
$\omega=\sqrt{\frac{\mathrm{MB}}{\mathrm{I}}}=\sqrt{\frac{(\mathrm{I})\left(\pi \mathrm{R}^{2}\right) \mathrm{B}}{\left(\frac{\mathrm{mR}^{2}}{2}\right)}}$
$\omega=\sqrt{\frac{2 \mathrm{I} \pi \mathrm{B}}{\mathrm{m}}}$
$\therefore \mathrm{T}=\frac{2 \pi}{\omega}=\sqrt{\frac{2 \pi \mathrm{m}}{\mathrm{IB}}}$
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