Question

A uniform circular loop of radius $a$ and resistance $R$ is placed perpendicular to a uniform magnetic field $B$. One half of the loop is rotated about the diameter with angular velocity $\omega$ as shown in figure. Then, the current in the loop is:

• A. zero, when $\theta$ is zero
• B. $\frac{\pi a^{2}B\omega }{2R}$, when $\theta$ is zero
• C. Zero, when $\theta =\pi /2$
• D. $\frac{\pi a^{2}B\omega }{2R}$, when $\theta =\frac{\pi }{2}$

Answer 1

Answer: (A), (D)

The correct options are
A zero, when $\theta$ is zero
D $\frac{\pi a^{2}B\omega }{2R}$, when $\theta =\frac{\pi }{2}$

Angle by which the plane of the loop is rotated in time $t$, $\theta =wt$

Flux passing through the rotating loop $\varphi =\overrightarrow{B}.\overrightarrow{A}$

$\therefore$ $\varphi \left(\theta \right)=BAcos\theta$ where $A=\frac{\pi a^2}{2}$

$⟹\varphi \left(\theta \right)=\frac{\pi a^{2}B}{2}cos\theta$ OR

$\varphi \left(t\right)=\frac{\pi a^{2}B}{2}coswt$

EMF induced $|E|=\frac{d\varphi }{dt}=\frac{w\pi a^{2}B}{2}sinwt$

Current in the loop is given by : $I=\frac{|E|}{R}$ where $R$ is the resistance of the loop

$\therefore I\left(\theta \right)=\frac{w\pi a^{2}B}{2R}sin\theta$

Thus, $\theta =0$ $⟹I=0$

Also, $\theta =\frac{\pi }{2}$ $⟹I=\frac{\pi a^{2}Bw}{2R}$