The time period of a bar magnet suspended hor | vedclass

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(A) The time period $T$ of a bar magnet oscillating in a uniform magnetic field $B_H$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$.Here,$

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Is directly proportional to the square root of its mass

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(A) The time period $T$ of a bar magnet oscillating in a uniform magnetic field $B_H$ is given by the formula: $T = 2\pi \sqrt{\frac{I}{MB_H}}$.Here,$I$ is the moment of inertia of the bar magnet,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the earth's magnetic field.The moment of inertia $I$ for a rectangular bar magnet of mass $m$,length $l$,and breadth $b$ is given by $I = \frac{m(l^2 + b^2)}{12}$.Substituting $I$ into the time period formula,we get $T = 2\pi \sqrt{\frac{m(l^2 + b^2)}{12MB_H}}$.Since $T \propto \sqrt{m}$,the time period is directly proportional to the square root of its mass.

The time period of a bar magnet suspended horizontally in the earth's magnetic field and allowed to oscillate:

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