Question

A wire loop enclosing a semicircle of radius $R$ is located on the boundary of a uniform magnetic field $B$ . At the moment $t=0$ , the loop is set into rotation with constant angular acceleration α about an axis $O$ . The clockwise emf direction is taken to be positive and the magnitude of variation of emf as a function of time is denoted by E . If $t_1$ and $t_2$ are the time required to complete first half and second half circle respectively then, which of of the following options are correct?

($E_n$ is the inducede emf produced in nth half rotation.)

• A. $E_1=-\frac{1}{2}B{R}^2\alpha t$
• B. $t_1<t_2$
• C. $E_1=\frac{3}{2}B{R}^2\alpha t$
• D. $t_1>t_2$

Answer 1

Answer: (A), (D)

The correct options are
A $E_1=-\frac{1}{2}B{R}^2\alpha t$
D $t_1>t_2$

The emf is induced in the loop because area inside the magnetic field is continually changing.

From $\theta =0$ to $\pi ,2\pi$ to $3\pi ,4\pi$ to 5$\pi$ , the loop begins to enter the magnetic field. Thus the magnetic field passing through the loop is increasing. Hence, current in the loop is anticlockwise, and for $\theta =\pi$ to $2\pi ,3\pi$ to $4\pi ,5\pi$ to 6$\pi$ , etc. magnetic field passing through the loop is decreasing. Hence current in the loop is clockwise.
Let at any time, angle rotated is $\theta ,$ then $\theta =\frac{1}{2}\alpha t^2$

Area inside magnetic field. $A=\frac{1}{2}{R}^2\theta =\frac{1}{2}{R}^2\left(\frac{1}{2}\alpha t^2\right)=\frac{1}{4}{R}^2\alpha t^2$

Flux in the loop: $\varphi =BA=\frac{B}{4}{R}^2\alpha t^2$
emf: $e=-\frac{d\varphi }{dt}=-\frac{B}{2}{R}^2\alpha t\Rightarrow e\propto t$

Time taken to complete first half circle: $t_1=\sqrt{\frac{2\pi }{\alpha }}$
When the loop starts coming out: $\theta =\frac{1}{2}\alpha t^2$
$\beta =\theta -\pi ,\gamma =\pi -\beta =\pi -\theta +\pi =2\pi -\theta$Area within magnetic field:
$A=\frac{1}{2}{R}^{2}y=\frac{1}{2}{R}^2(2\pi -\theta )=\pi R^2-\frac{R^2\theta }{2}$$\begin{array}{cc}\text{Flux}& \varphi =BA=B(\pi R^2-\frac{R^2\theta }{2})\\ \text{emf}& e=-\frac{d\varphi }{dt}=\frac{B}{2}{R}^2\alpha t\\ & e=\frac{B{R}^2}{2}\alpha t\Rightarrow e\propto t\end{array}$

Time taken to complete second half revolution
$t_2=\sqrt{\frac{4\pi }{\alpha }}-\sqrt{\frac{2\pi }{\alpha }}$
We see that $t_2<t_1$

We can write induced emf as $e=(-1{)}^n\left[\frac{1}{2}B{R}^2\alpha t\right]$

where $n=1,2,3,\dots \dots$ is the number of half revolutions completed by loop. Smaller time will be taken to complete the second half revolution as compared to the previous half revolution.