The time period of a freely suspended magnet | vedclass

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(D) The time period of a freely suspended magnetic dipole is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia,$M$ is the mag

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(D) The time period of a freely suspended magnetic dipole 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 magnet of length $L$,mass $m$,and pole strength $q_m$,the magnetic moment is $M = q_m L$ and the moment of inertia is $I = \frac{mL^2}{12}$.When the magnet is broken into two equal parts along its length,the new length is $L' = L/2$,the new mass is $m' = m/2$,and the new pole strength remains $q_m$.The new magnetic moment is $M' = q_m (L/2) = M/2$.The new moment of inertia is $I' = \frac{(m/2)(L/2)^2}{12} = \frac{mL^2}{8 	imes 12} = I/8$.The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} T$.Given $T = 2 \ s$,the new time period is $T' = \frac{2}{2} = 1 \ s$.

The time period of a freely suspended magnet is $2 \ s$. 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.....$s$.

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