Question

A circular coil has moment of inertia $0.8{\text{kg m}}^2$ around any diameter and is carrying current to produce a magnetic moment of $20{\text{Am}}^2$. The coil is kept initially in a vertical position, and it can rotate freely around a horizontal diameter. When a uniform magnetic field of $4\text{T}$ is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil, acquired after rotating by ${60}^{\circ }$, will be :

• A. $10{\text{rad s}}^{–1}$
• B. $10\pi {\text{rad s}}^{–1}$
• C. $20\pi {\text{rad s}}^{–1}$
• D. $20{\text{rad s}}^{–1}$

Answer 1

The correct option is A $10{\text{rad s}}^{–1}$

Given, Moment of inertia of circular coil, $I=0.8{\text{kg m}}^2$
Magnetic moment of circular coil, $M=20{\text{Am}}^2$

Rotational kinetic energy of circular coil is,
$KE=\frac{1}{2}I{\omega }^2$

Here, $\omega =$ angular speed of coil.

Potential energy of coil, when considered equivalent to a bar magnet is,

$U=–MB\cos \varphi$

From energy conservation,

$\frac{1}{2}I{\omega }^2=U_i-U_f$

$\Rightarrow \frac{1}{2}I{\omega }^2=-MB\cos {60}^{\circ }-(-MB)$

$\Rightarrow \frac{MB}{2}=\frac{1}{2}I{\omega }^2$

$\Rightarrow \frac{20\times 4}{2}=\frac{1}{2}\left(0.8\right){\omega }^2$

$\Rightarrow \omega =10\text{rad/s}$.

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