A small circular loop of conducting wire has radius $$a$$ and carries current $$I$$. It is placed in a uniform magnetic field $$B$$ perpendicular to its plane such that when rotated slightly about its diameter and released, it starts performing simple harmonic motion of time period $$T$$. The mass of the loop is $$m$$ then:

We have a thin circular loop of radius $$a$$, carrying a current $$I$$ and possessing mass $$m$$. In a uniform magnetic field $$B$$ directed perpendicular to the plane of the loop, the loop behaves like a magnetic dipole. Its magnetic dipole moment is, by definition,

$$\mu = I \,A = I\,(\pi a^{2}).$$

Next, we need the moment of inertia of the loop about the diameter about which it is allowed to oscillate. For a thin circular ring, the perpendicular-axis theorem gives

$$I_{x}+I_{y}=I_{z}.$$

Here $$I_{z}=m a^{2}$$ (axis through the centre and perpendicular to the plane). Because the two diameters in the plane are equivalent, $$I_{x}=I_{y}$$; hence

$$2I_{x}=m a^{2}\quad\Longrightarrow\quad I_{x}=I_{y}=\frac{1}{2}m a^{2}.$$

Thus the moment of inertia about the diameter of oscillation is

$$I_{\text{m.o.i.}}=\frac{1}{2}m a^{2}.$$

When the loop is given a small angular displacement $$\theta$$ from its equilibrium position, the restoring magnetic torque is

$$\tau = \mu B \sin\theta.$$

For small angles, $$\sin\theta \approx \theta$$, so

$$\tau \approx \mu B\,\theta.$$

The rotational form of Newton’s second law states

$$I_{\text{m.o.i.}}\;\frac{d^{2}\theta}{dt^{2}} = -\tau,$$

the negative sign indicating that the torque is restoring. Substituting the expressions for $$\tau$$ and $$I_{\text{m.o.i.}}$$, we obtain

$$\frac{1}{2}m a^{2}\;\frac{d^{2}\theta}{dt^{2}} \;=\; -\,\mu B\,\theta.$$

Simplifying,

$$\frac{d^{2}\theta}{dt^{2}} + \underbrace{\left(\frac{2\mu B}{m a^{2}}\right)}_{\omega^{2}}\theta = 0.$$

This is the standard differential equation of simple harmonic motion, $$\displaystyle \frac{d^{2}\theta}{dt^{2}}+\omega^{2}\theta = 0,$$ whose angular frequency is

$$\omega = \sqrt{\frac{2\mu B}{m a^{2}}}.$$

The time period is related to angular frequency by the formula

$$T = \frac{2\pi}{\omega}.$$

Substituting $$\omega$$,

$$T = 2\pi\sqrt{\frac{m a^{2}}{2\mu B}}.$$

Now we replace $$\mu$$ by $$I\pi a^{2}$$:

$$T = 2\pi\sqrt{\frac{m a^{2}}{2\,I\pi a^{2} B}} = 2\pi\sqrt{\frac{m}{2I\pi B}}.$$

Notice that $$a^{2}$$ cancels out. Extracting the common factors inside the square root,

$$T = 2\pi\;\frac{\sqrt{m}}{\sqrt{2I\pi B}} = \frac{2\sqrt{\pi}\,\sqrt{m}}{\sqrt{2I B}} = \sqrt{\frac{2\pi m}{I B}}.$$

Hence, the correct answer is Option C.