The time period of a freely suspended magnet | vedclass

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(B) The time period of a freely suspended magnet is given by $T = 2\pi \sqrt{\frac{I}{MB_H}} = 4 \ s$.When the magnet is cut into two equal halves alo

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$2$

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(B) The time period of a freely suspended magnet is given by $T = 2\pi \sqrt{\frac{I}{MB_H}} = 4 \ s$.When the magnet is cut into two equal halves along its length,the new magnetic moment becomes $M' = \frac{M}{2}$.The new moment of inertia $I'$ for one half is calculated as $I' = \frac{m'l'^2}{12}$,where $m' = \frac{m}{2}$ and $l' = \frac{l}{2}$.Thus,$I' = \frac{(m/2)(l/2)^2}{12} = \frac{1}{8} \cdot \frac{ml^2}{12} = \frac{I}{8}$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B_H}} = 2\pi \sqrt{\frac{I/8}{(M/2)B_H}}$.Simplifying this,$T' = 2\pi \sqrt{\frac{1}{4} \cdot \frac{I}{MB_H}} = \frac{1}{2} \cdot 2\pi \sqrt{\frac{I}{MB_H}} = \frac{T}{2}$.Substituting $T = 4 \ s$,we get $T' = \frac{4}{2} = 2 \ s$.

The time period of a freely suspended magnet is $4 \ s$. If it is broken in length into two equal parts and one part is suspended in the same way,then its time period will be: (in $s$)

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