Question

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

• A. $T=\sqrt{\frac{2m}{iB}}$
• B. $T=\sqrt{\frac{\pi m}{2iB}}$
• C. $T=\sqrt{\frac{2\pi m}{iB}}$
• D. $T=\sqrt{\frac{\pi m}{iB}}$

Answer 1

The correct option is C $T=\sqrt{\frac{2\pi m}{iB}}$

Torque on circular loop, $\tau =MB\sin \theta$
Where, $M=$ magnetic moment
$B=$ magnetic filed

Now, using $\tau =I\alpha$

$\therefore \tau =MB\sin \theta =I\alpha$

Here, $(M=iA=\pi R^{2}i$ and moment of inertia of circular loop is, $I=\frac{m{R}^2}{2})$

$\Rightarrow \pi R^{2}iB\theta =\frac{m{R}^2\alpha }{2}$

$\alpha =\frac{2\pi iB}{m}\theta$

Comparing the above equation with, $\alpha ={\omega }^2\theta$, we get,

$\omega =\sqrt{\frac{2\pi iB}{m}}$

Now, $T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{2\pi iB}{m}}}$
$\therefore T=\sqrt{\frac{2\pi m}{iB}}$

Hence, $\left(C\right)$ is the correct answer.