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(B) The magnetic torque on the loop is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.For a small angular displacement $	heta$,the magnitude of the t

Vedclass · Accepted Answer

$T=\sqrt{\frac{2 \pi m}{IB}}$

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(B) The magnetic torque on the loop is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.For a small angular displacement $	heta$,the magnitude of the torque is $	au = MB \sin 	heta \approx MB 	heta$.The restoring torque is $	au = -MB 	heta$.Using Newton's second law for rotation,$	au = I_{moment} \alpha$,where $I_{moment}$ is the moment of inertia of the circular loop about its diameter,$I_{moment} = \frac{ma^2}{2}$.Thus,$\frac{ma^2}{2} \alpha = - (I \pi a^2) B 	heta$.$\alpha = - \frac{2 I \pi B}{m} 	heta$.Comparing this with the $SHM$ equation $\alpha = - \omega^2 	heta$,we get $\omega^2 = \frac{2 I \pi B}{m}$.Therefore,$\omega = \sqrt{\frac{2 I \pi B}{m}}$.The time period $T$ is given by $T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{m}{2 I \pi B}} = \sqrt{\frac{4 \pi^2 m}{2 I \pi B}} = \sqrt{\frac{2 \pi m}{IB}}$.

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