A magnetic needle of negligible breadth and t | vedclass

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

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$T/n$

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(A) The time period of a magnetic needle oscillating in a magnetic field $B$ is given by $T = 2 \pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic dipole moment.When the magnet is broken into $n$ equal parts perpendicular to its length,the length of each part becomes $l' = l/n$.The new mass of each part is $m' = m/n$.The new moment of inertia of each part is $I' = \frac{m' (l')^2}{12} = \frac{(m/n) (l/n)^2}{12} = \frac{I}{n^3}$.The new magnetic moment of each part is $M' = M/n$.The new time period $T'$ is given by $T' = 2 \pi \sqrt{\frac{I'}{M' B}} = 2 \pi \sqrt{\frac{I/n^3}{(M/n) B}} = 2 \pi \sqrt{\frac{I}{n^2 MB}} = \frac{1}{n} \left( 2 \pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{n}$.

$A$ magnetic needle of negligible breadth and thickness compared to its length oscillates in a horizontal plane with a period $T$. What is the period of oscillation of each part obtained after breaking the magnet into $n$ equal parts perpendicular to its length?

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