Q98 / 125 · BITSAT 2015 · Physics
A conducting square loop is placed in a magnetic field $B$ with its plane perpendicular to the field. The sides of the loop start shrinking at a constant rate $\alpha$. The induced emf in the loop at an instant when its side is ' $a$ ' is
- Correct answerA$2 a \alpha B$
- B$a^{2} \alpha B$
- C$2 a^{2} \alpha B$
- D$a \alpha B$
Solution
At any time $t$, the side of the square $a=\left(a_{0}\right.$ $-\alpha t),$ where $a_{0}=$ side at $t=0$
At this instant, flux through the square :
$$
\begin{array}{l}
\phi=B A \cos 0^{\circ}=B\left(a_{0}-\alpha t\right)^{2} \\
\therefore \text { emf induced } E=-\frac{d \phi}{d t} \\
\Rightarrow E=-B .2\left(a_{0}-\alpha t\right)(0-\alpha)=+2 \alpha a B
\end{array}
$$
Hence, the correct answer is option: 'A'.