Using a bar magnet P,a vibration magnetometer | vedclass

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(B) The time period of a vibration magnetometer is given by $T = 2\pi \sqrt{\frac{I}{MB_H}}$,where $I$ is the moment of inertia,$M$ is the magnetic mo

Vedclass · Accepted Answer

$Q$ is a bar magnet identical to $P$,and its north pole is placed on top of $P$'s north pole.

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(B) The time period of a vibration magnetometer 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.When bar $Q$ is placed on top of $P$,the new moment of inertia becomes $I' = I + I = 2I$.For the time period to remain unchanged $(T' = T)$,the new magnetic moment $M'$ must satisfy the relation $T' = 2\pi \sqrt{\frac{2I}{M'B_H}} = 2\pi \sqrt{\frac{I}{MB_H}}$.Squaring both sides,we get $\frac{2I}{M'} = \frac{I}{M}$,which implies $M' = 2M$.This indicates that the magnetic moment of the system has doubled,which happens if $Q$ is an identical magnet placed with its poles aligned (north on north,south on south) with $P$.

Using a bar magnet $P$,a vibration magnetometer has a time period of $2 \ s$. When a bar $Q$ (identical to $P$ in mass and size) is placed on top of $P$,the time period remains unchanged. Which of the following statements is true?

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