Question

A bar magnet of moment $M$ gives a time period ${}^{\prime }{T}^{\prime }$ at a place in a vibration magnetometer. Four such similar bar magnets are placed in the frame one over the other out of which one magnet is placed with opposite polarity. The new time period is:

• A. $\frac{T}{\sqrt{2}}$
• B. $\sqrt{2}T$
• C. $T$
• D. $2T$

Answer 1

The correct option is B $\sqrt{2}T$

Magnetometers are devices that measure magnetic fields.

The torque on a magnet is proportional to its magnetic moment. Hence,

$\tau =-BMsin\left(\theta \right)$

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

$\tau =-BM\theta ..........\left(i\right)$

This is similar to the equation of SHM.

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

Also, $\omega =\frac{2\pi }{T}...........\left(iii\right)$

From (i), (ii) and (iii),

$T\propto \frac{\sqrt{I}}{\sqrt{M}}$

In the second case, net magnetic moment will be $M_2=3M-M=2M$ and net moment of inertia will be $4I$.

$\frac{T_2}{T_1}=\sqrt{\frac{I_2{M}_1}{M_2{I}_1}}=\sqrt{2}$

$T_2=\sqrt{2}T$