The time period of oscillation of a freely su | vedclass

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Question

(A) freely suspended bar magnet in a uniform magnetic field $B_H$ experiences a restoring torque $	au = -M B_H \sin 	heta$. For small oscillations,$

Vedclass · Accepted Answer

$T = 2\pi \sqrt{\frac{I}{M B_H}}$

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(A) freely suspended bar magnet in a uniform magnetic field $B_H$ experiences a restoring torque $	au = -M B_H \sin 	heta$. For small oscillations,$\sin 	heta \approx 	heta$,so $	au = -M B_H 	heta$. The equation of motion is $I \alpha = 	au$,where $I$ is the moment of inertia and $\alpha$ is the angular acceleration. $I \frac{d^2 	heta}{dt^2} = -M B_H 	heta$. This is the equation of simple harmonic motion $\frac{d^2 	heta}{dt^2} + \omega^2 	heta = 0$,where $\omega^2 = \frac{M B_H}{I}$. The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{I}{M B_H}}$.

The time period of oscillation of a freely suspended bar magnet with usual notations is given by

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