Question

The magnetic field $B$ shown in the figure is directed into the plane of the paper. $\text{ACDA}$ is a semicircular conducting loop of radius $r$ with the centre at $\text{O}$. The loop is now made to rotate clockwise with a constant angular velocity $\omega$ about an axis passing through $O$ and perpendicular to the plane of the paper. The resistance of the loop is $R$. Obtain an expression for the magnitude of the induced current in the loop.

• A. $\frac{B\omega r^2}{4R}$
• B. $\frac{B\omega r^2}{R}$
• C. $\frac{B\omega r^2}{2R}$
• D. $\frac{2B\omega r^2}{R}$

Answer 1

The correct option is C $\frac{B\omega r^2}{2R}$

When the loop rotates through an angle $\theta \text{radian}$ in time $t$, which is less than $\pi$, the area inside the field region is,

$A=\frac{\theta }{\pi }\frac{\pi r^2}{2}=\frac{\theta r^2}{2}=\frac{\omega t{r}^2}{2}[\text{as}\theta =\omega t]$


The flux of the magnetic field at time $t$ is,

$\varphi =BA=B\frac{\omega t{r}^2}{2}$

The induced emf be,

$E=-\frac{d\varphi }{dt}=-\frac{B\omega r^2}{2}$

So, the magnitude of the induced current will be,

$i=\frac{E}{R}=\frac{B\omega r^2}{2R}$

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