A bar magnet is oscillating in the Earth's ma | vedclass

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(A) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$

Vedclass · Accepted Answer

Motion remains $S.H.M.$ with time period $= 2T$

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(A) The time period of a bar magnet oscillating in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B$ is the magnetic field.For a bar magnet of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$.Substituting this into the formula,we get $T = 2\pi \sqrt{\frac{ml^2}{12MB}}$.This shows that $T \propto \sqrt{m}$.If the mass $m$ is quadrupled $(m' = 4m)$,the new time period $T'$ becomes $T' = 2\pi \sqrt{\frac{4ml^2}{12MB}} = 2 	imes (2\pi \sqrt{\frac{ml^2}{12MB}}) = 2T$.The motion remains $S.H.M.$ because the restoring torque $	au = -MB \sin 	heta \approx -MB 	heta$ for small oscillations.

$A$ bar magnet is oscillating in the Earth's magnetic field with a period $T$. What happens to its period and motion if its mass is quadrupled?

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