Question

The time period of a vibration magnetometer is ${T_0}$. Its magnet is replaced by another magnet whose moment of inertia is 3 times and magnetic moment is $\dfrac{1}{3}$ of the initial magnet. The time period now will be:
A. $3{T_0}$
B. ${T_0}$
C. $\dfrac{{{T_0}}}{{\sqrt 3 }}$
D. $\dfrac{{{T_0}}}{3}$

Answer 1

Hint: The time period of the vibration magnetometer is the square root of the ratio of moment of inertia of the magnet to the product of magnetic moment and the background magnetic field. Magnetic field for the setups using magnets with different moments will remain the same. Use this to find the ratio of time periods.

Complete answer:
If a magnetometer is suspended in a magnetic field $B$ then for a small angular displacement the magnet will start oscillating in simple harmonic motion with a time period $T = 2\pi \sqrt {\dfrac{I}{{MB}}}$. Here $I$ is the moment of inertia of the bar magnet and $M$ is the magnetic moment of the magnet.
Let the moment of inertia of the first magnet be ${I_i}$ with magnetic moment ${M_i}$ and the magnetic field for both set ups will be the same as $B$. Then the time period for the first magnet ${T_0} = 2\pi \sqrt {\dfrac{{{I_i}}}{{{M_i}B}}}$.
For the second magnet, moment of inertia is ${I_f} = 3{I_i}$ and magnetic moment, ${M_f} = \dfrac{{{M_i}}}{3}$. Therefore using the relation for time period we get, ${T_f} = 2\pi \sqrt {\dfrac{{{I_f}}}{{{M_f}B}}} = 2\pi \sqrt {\dfrac{3I_i}{\dfrac{{M_i}B}{3}}} = 3 \times 2\pi \sqrt {\dfrac{{{I_i}}}{{{M_i}B}}} = 3{T_0}$.
Comparing this result with the above mentioned choices, we can say that option A is the correct one.

Note: You can derive the time period of oscillation by balancing the moment (rotating acceleration) created by the displacement and the moment due to the magnetic field. This will set up a harmonic oscillation and thus a relation for time period.