The magnetic moments of two bar magnets of sa | vedclass

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(B) The period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is

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$3 \sqrt{3}$

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(B) The period of oscillation of a magnet in a magnetic field is given by $T = 2\pi \sqrt{\frac{I}{MB}}$,where $I$ is the moment of inertia and $M$ is the magnetic moment.Since the magnets are of the same size,their moments of inertia $I$ are the same. Let the moments of inertia be $I_1 = I_2 = I$. The total moment of inertia is $I_{total} = I + I = 2I$.Case $1$: Similar poles together. The net magnetic moment is $M_{net} = M_1 + M_2$. Given $M_1 : M_2 = 1 : 2$,let $M_1 = M$ and $M_2 = 2M$. So,$M_{net} = M + 2M = 3M$.The period $T_1 = 3 \, s = 2\pi \sqrt{\frac{2I}{3MB}}$.Case $2$: One magnet is reversed. The net magnetic moment is $M'_{net} = |M_2 - M_1| = |2M - M| = M$.The new period $T_2 = 2\pi \sqrt{\frac{2I}{MB}}$.Dividing the two equations: $\frac{T_2}{T_1} = \sqrt{\frac{2I/MB}{2I/3MB}} = \sqrt{3}$.Therefore,$T_2 = 3 	imes \sqrt{3} = 3\sqrt{3} \, s$.

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