Question

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

• A. $2a\alpha \beta$
• B. $a^2 \alpha \beta$
• C. $2a^2\alpha \beta$
• D. $a\alpha \beta$

Answer 1

Answer: (A)

At any time $t$, the side of the square $a=\left(a_{0}-\alpha t\right)$,
where $a_{0}=$ side at $t=0$.
At this instant, flux through the square:
$\phi= BA \cos 0^{\circ}= B \left( a _{0}-\alpha t \right)^{2}$
$\therefore $ emf induced $E =-\frac{ d \phi}{ dt }$
$\Rightarrow E =- B \cdot 2\left( a _{0}-\alpha t \right)(0-\alpha)=+2 \alpha a \beta$