A magnet of magnetic moment M oscillating fre | vedclass

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(C) The frequency of oscillation $n$ for a magnet in a horizontal magnetic field $B_H$ is given by the formula: $n = \frac{1}{2\pi} \sqrt{\frac{MB_H}{

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$2\sqrt{2} n$

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(C) The frequency of oscillation $n$ for a magnet in a horizontal magnetic field $B_H$ is given by the formula: $n = \frac{1}{2\pi} \sqrt{\frac{MB_H}{I}}$,where $M$ is the magnetic moment and $I$ is the moment of inertia.From this relation,we can see that $n \propto \sqrt{MB_H}$.Given that the magnetic moment becomes $M' = 4M$ and the magnetic field becomes $B_H' = 2B_H$.Substituting these values into the proportionality,we get the new frequency $n'$:$n' \propto \sqrt{M' B_H'} = \sqrt{(4M)(2B_H)} = \sqrt{8MB_H} = 2\sqrt{2} \sqrt{MB_H}$.Since $n \propto \sqrt{MB_H}$,it follows that $n' = 2\sqrt{2} n$.

$A$ magnet of magnetic moment $M$ oscillating freely in earth's horizontal magnetic field makes $n$ oscillations per minute. If the magnetic moment is quadrupled and the earth's field is doubled,the number of oscillations made per minute would be

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