Question

A circular coil of $16$ turns and radius $10\text{cm}$ carrying a current of $0.75A$ rests with its plane normal to an external field of magnitude $5.0\times {10}^{-2}T$. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0s^{-1}$ . What is the moment of inertia of the coil about its axis of rotation?

Answer 1

Step 1: Calculate magnetic moment of the circular coil.

Given, Number of turns in the circular coil, $N=16$

Radius of the coil, $r=10cm=0.1m$

Current in the coil, $i=0.75A$

Magnetic moment,$m=NiA$

$m=Ni\pi r^2$
$m=16\times 0.75\times 3.14\times (0.1{)}^2=0.3768J/T$

Step 2: Find moment of inertia of the coil.

Given, Magnetic field strength, $B=5.0\times {10}^{-2}T$

Frequency of oscillations of the coil, $v=2s^{-1}$

As we know, Torque, $\tau \approx -mB\theta$ (for small angle $\sin \theta \approx \theta )$

$\tau =-mB\theta =I\alpha$

Here, $I$ = moment of inertia $\alpha =-\left(\frac{mB}{I}\right)\theta$

This is the equation of SHM with frequency, $v=\frac{1}{2\pi }\sqrt{\frac{mB}{I}}$

So, $I=\frac{mB}{4{\pi }^2{v}^2}$

$I=\frac{0.3768\times 5\times {10}^{-2}}{4\times (3.14{)}^2\times (2{)}^2}$

$I=1.19\times {10}^{-4}kg{m}^2\approx 1.2\times {10}^{-4}kg{m}^2$

Final answer: $1.2\times {10}^{-4}kg{m}^2$