The time period of a thin bar magnet in earth's magnetic field is $T$. If the magnet is cut into two equal parts perpendicular to its length, the time period of each part in the same field will be:
A. $\dfrac{T}{2}$
B. $T$
C. $\sqrt 2 T$
D. $2T$

Hint: Simple harmonic motion is a form of periodic motion in mechanics and physics in which the restoring force acting on a moving item is inversely proportional to the size of the object's displacement and drives the object toward its equilibrium point. If friction or other energy loss does not halt the oscillation, it can continue indefinitely.

Formula use:
The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
Here, $T$ is time period of oscillation of bar magnet, $I$ is moment of inertia of bar magnet, $M$ is magnetic moment and $B$ is magnetic field intensity of bar magnet.

Complete step by step solution:
In the question, we have given the time period of a thin bar magnet is $T$. In order for the model to be true, the net force exerted on the object at the pendulum's end must be commensurate to the displacement.

As we know, the motion of a simple pendulum may be described by the basic harmonic motion and simple harmonic motion can also be used to describe molecular vibration. The time period of bar magnet is given by,
$T = 2\pi \sqrt {\dfrac{I}{{MB}}} $
From the given information, the magnet is cut into two equal parts perpendicular to its length. So, the new inertia is:
$\dfrac{{(\dfrac{m}{4}){{(\dfrac{l}{2})}^2}}}{{12}} = \dfrac{{m{l^2}}}{{16 \times 12}}$

As we know that the moment of inertia is $I = \dfrac{{m{l^2}}}{{12}}$, then:
$\text{New inertia}\,I' = \dfrac{I}{{16}}$
The new magnetic dipole moment, $M' = \dfrac{q}{2} \times \dfrac{1}{2} = \dfrac{M}{4}$
Substitute the obtained values in the above formula, then we obtain:
$T' = 2\pi \sqrt {\dfrac{{\dfrac{I}{{16}}}}{{\dfrac{M}{4}B}}} \\$
$\Rightarrow T' = \dfrac{1}{2} \times 2\pi \sqrt {\dfrac{I}{{MB}}} \\$
$\therefore T' = \dfrac{T}{2} \\$
Therefore, the time period of each part in the same field will be $\dfrac{T}{2}$.

Thus, the correct option is A.

Note: Keep in mind that $M' = \dfrac{M}{2}$ is the proper substitute and not $M = \dfrac{{M'}}{2}$while trying to recall the time period formula because $T \propto \sqrt I \propto \sqrt m $. A simple harmonic motion is referred to the movement of a particle along the straight line with an acceleration that is proportional to the particle's distance from a fixed point on the line.