The time period of oscillation of a bar magne | vedclass

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(C) The time period of oscillation of a bar magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia

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$\sqrt{2} T_0$

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(C) The time period of oscillation of a bar magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic moment,and $B_H$ is the horizontal component of the Earth's magnetic field.For a bar magnet of mass $m$ and length $l$,the moment of inertia is $I = \frac{ml^2}{12}$.Since $M$ and $B_H$ remain constant,we have $T \propto \sqrt{I}$.Since $I \propto m$,we get $T \propto \sqrt{m}$.Let $T_0$ be the initial time period with mass $m$,and $T'$ be the new time period with mass $2m$.$\frac{T'}{T_0} = \sqrt{\frac{2m}{m}} = \sqrt{2}$.Therefore,$T' = \sqrt{2} T_0$.

The time period of oscillation of a bar magnet suspended horizontally along the magnetic meridian is $T_0$. If this magnet is replaced by another magnet of the same size and pole strength but with double the mass,the new time period will be

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