Question

The period of oscillation of a magnet of a vibration magnetometer is $2.45s$ at one place and $4.9s$ at the other. The ratio of the horizontal component of the earth’s magnetic field at the two places is

• A. 1 : 4
• B. 1 : 2
• C. 2 : 1
• D. 4 : 1

Answer 1

The correct option is D 4 : 1

Magnet kept with its axis at an angle $\theta$ with a magnetic field will undergo angular SHM.



Equation of angular SHM is given by
$\alpha =-{\omega }^2\theta --\left(1\right)$

Torque acting on the magnet
$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$
$=MB\sin \theta$
For very small $\theta$, $\tau =-MB\theta$
(because torque is in the opposite direction of $\theta$)

$\tau =-MB\theta$
$\Rightarrow I\alpha =-MB\theta$
$\Rightarrow \alpha =-\left(\frac{MB}{I}\right)\theta$
where $I$ is the moment of inertia.

Comparing with (1),
$\omega =\sqrt{\frac{MB}{I}}$
$\therefore$ Time period $T=\frac{2\pi }{\omega }$
$=2\pi \sqrt{\frac{I}{MB}}$
i.e $T\propto \frac{1}{\sqrt{B}}$
$\therefore \frac{B_1}{B_2}={\left(\frac{T_2}{T_1}\right)}^2={\left(\frac{4.9}{2.45}\right)}^2$
$=4:1$