Question

The time period of oscillation of a freely suspended bar magnet
A. $T = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}} \\ $
B. $T = 2\pi \sqrt {\dfrac{{M{B_H}}}{I}} \\ $
C. $T = \sqrt {\dfrac{I}{{M{B_H}}}} \\ $
D. $T = 2\pi \sqrt {\dfrac{{IM}}{{{B_H}}}} $

Answer 1

Hint: By deriving the relevant formulas and verifying the inclusion of terms in those formulas, it is possible to determine the dependence of any physical quantity on another physical quantity. A magnet feels a torque when it is hanging freely in a magnetic field because of the interaction of the magnetic pole with the magnetic field. The magnetic field horizontal component produces the torque for a freely suspended magnet.

Formula used:
The expression of torque is,
$\tau = - mB\sin \theta $
Where, $m$ is the magnetic moment, $B$ is the magnetic field and $\theta$ is the angle between $m$ and $B$.

Complete step by step solution:
We are aware that a magnet receives a force equal to $m \times B$ if its pole strength is m and a magnetic field of B is present. Now imagine that a magnet is initially positioned with its length parallel to the direction of the magnetic field and hangs freely halfway in the earth's atmosphere.

Now, if the magnet is moved at an angle of $\theta $, the restoring torque on it will be (notice that the oscillations are caused solely by the earth's magnetic field's horizontal component; the vertical component plays no part in the oscillation in the horizontal plane).
$\tau = - mB\sin \theta \\ $
Also, we know that, $\tau = I\alpha \\ $
Where $I = $ moment of inertia about centre
Hence, $I\alpha = - mB\sin \theta \\ $
For small angles $\sin \theta = \theta \\ $
So, $I\alpha = - mB\theta \\ $
$\alpha = \dfrac{{ - mB\theta }}{I}$

Now its differential form is,
$\alpha = \dfrac{{{d^2}\theta }}{{d{t^2}}} \\ $
$\Rightarrow \dfrac{{{d^2}\theta }}{{d{t^2}}} = - \dfrac{{mB\theta }}{I}$
Now comparing above equation with SHM general form,
$\dfrac{{{d^2}\theta }}{{d{t^2}}} = - {\omega ^2}\theta \\ $
We get,
${\omega ^2} = \dfrac{{mB}}{I} \\ $
$\Rightarrow \omega = \sqrt {\dfrac{{mB}}{I}} $
And the time period is given by,
$T = \dfrac{{2\pi }}{\omega }$
Hence the time period of the suspended bar magnet is given by $T = 2\pi \sqrt {\dfrac{I}{{M{B_H}}}}$.

Hence, option A is correct.

Note: The length of the magnet affects the moment of inertia (I). Time period also depends on pole strength. The "Horizontal component of the earth's magnetic field" is the word for the magnetic field that appears as letter "B" in the formula. The magnet can only oscillate in the horizontal plane since it is freely hanging; it has nothing to do with the vertical component.