Question

A short bar magnet having magnetic moment $4A{m}^2$, placed in a vibrating magnetometer, vibrates with a time period of 8 seconds. Another short bar magnet having a magnetic moment $8A{m}^2$ vibrates with a time period of 6 seconds. If the moment of inertia of the second magnet is $9\times {10}^{-2}kg{m}^2$, the moment of inertia of the first magnet is :
(Assume that both magnets are kept in the same uniform magnetic induction field.)

• A. $9\times {10}^{-2}kg{m}^2$
• B. $8\times {10}^{-2}kg{m}^2$
• C. $5.33\times {10}^{-2}kg{m}^2$
• D. $12.2\times {10}^{-2}kg{m}^2$

Answer 1

Answer: (B)

$8\times {10}^{-2}kg{m}^2$

Torque on a dipole is given by:

$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$

$\tau =\mu B\sin \left(\theta \right)$

For small $\theta$, $\sin \left(\theta \right)\approx \theta$

$\therefore \tau =\mu B\theta ........\left(i\right)$

Hence, Magnet performs SHM.

$\tau =I{\omega }^2\theta ..........\left(ii\right)$

From (i) and (ii),

$\omega \propto \sqrt{\mu }$

Now Time period,

$T\propto \frac{1}{\omega }\propto \sqrt{\frac{I}{\mu }}$.

Hence, $\frac{8}{6}=\sqrt{\frac{I}{4}}\sqrt{\frac{8}{9\times {10}^{-2}}}$.

$I=8\times {10}^{-2}kg{m}^2$