Question

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, its starts performing simple harmonic motion of time period $T$. If the mass of the loop is $m$then

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

Answer 1

The correct option is B

$T=\sqrt{\frac{2\pi M}{iB}}$

Step 1. Given data:

Current $i$ through circular loop having radius $a$

Uniform magnetic field $B$ perpendicular to loop

Step 2. Finding the time period

The torque on the loop is $\tau =MB\sin \theta =-I\alpha$

Where$I$ is moment of inertia $\alpha$ is angular acceleration $M$ is magnetisation,

Magnetisation is $M=iA=i\pi a^2$

And when$\theta$ is small $\sin \theta \approx \theta$, so we have

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

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

$\Rightarrow \alpha =-{\omega }^2\theta$, where $\omega =\sqrt{\frac{2\pi iB}{m}}$

where $\omega$ is the angular velocity

Also $\omega =\frac{2\pi }{T}\Rightarrow T=\frac{2\pi }{\omega }$

where $T$is the time period

$\Rightarrow T=\frac{2\pi }{\sqrt{\frac{2\pi iB}{m}}}=\sqrt{\frac{2\pi m}{iB}}$

Hence, option B is correct