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A vibration magnetometer consists of two identical bar magnets placed one over the other such that they are mutually perpendicular and bisect each other. The time period of oscillation in a horizontal magnetic field is $4\, sec$. If one of the magnets is taken away, find the period of oscillation of the other in the same field (in $sec.$) :-
${2^{3/4}}$
${2^{7/4}}$
${2^{3/2}}$
None
Solution
Given that the time period of oscillation of the combination in horizontal magnetic field is $4 \mathrm{s}$
Time period of oscillation in vibration magnetometer is proportional to moment of inertia and inversely proportional to effective magnetic moment of the magnet. $T \propto \sqrt{\frac{I}{L}}$
Let Initial Moment of Inertia and Magnetic Moment of the combination are $I$ and $M$ respectively.
If one of the magnets is removed, the resultant magnetic moment will be $\frac{M}{\sqrt{2}}$
Similarly, Moment of Inertia after removing one magnet is $\frac{I}{2}$ (as moment of inertia passing through midpoint of magnets along perpendicular plane, is directly added for two magnets)
Hence,
$\frac{T_{1}}{T_{2}}=\sqrt{\frac{I_{1} M_{2}}{I_{2} M_{1}}}$
$=\sqrt{\frac{I \frac{M}{\sqrt{2}}}{M \frac{I}{2}}}=2^{1 / 4}$
$\text { Hence, } T_{2}=\frac{T_{1}}{2^{1 / 4}} \mathrm{sec}$
$T_{2}=\frac{4}{2^{1} / 4} \mathrm{sec}$
$T_{2}=2^{7 / 4} \mathrm{sec}$