The time period of a thin bar magnet in Earth | vedclass

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Question

(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

$T/2$

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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.When the magnet is cut into two equal parts perpendicular to its length,the new magnetic moment $M' = M/2$ and the new moment of inertia $I' = I/8$ (since $I = \frac{1}{12}mL^2$ and mass $m$ becomes $m/2$,length $L$ becomes $L/2$).The new time period $T'$ is $T' = 2\pi \sqrt{\frac{I'}{M'B}} = 2\pi \sqrt{\frac{I/8}{(M/2)B}} = 2\pi \sqrt{\frac{I}{4MB}} = \frac{1}{2} \left( 2\pi \sqrt{\frac{I}{MB}} \right) = \frac{T}{2}$.

The time period of a thin bar magnet in Earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length,the time period of each part in the same field will be

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