Question

A wire bent as a parabola $y=k x^{2}$ is located in a uniform magnetic field of induction $B$ , the vector $B$ being perpendicular to the plane $xy$ . At $t=0$ , the sliding wire starts sliding from the vertex $O$ with a constant acceleration a linearly as shown in Fig. Find the emf induced in the loop -

• A. $B y\sqrt{\frac{2 a}{k}}$
• B. $B y\sqrt{\frac{4 a}{k}}$
• C. $B y\sqrt{\frac{8 a}{k}}$
• D. $B y\sqrt{\frac{ a}{k}}$

Answer 1

Answer: (C)

$d\phi=B.dA=2B \, x \, dy$ and $y=k x^{2}$
$\therefore x=\sqrt{\frac{y}{k}}$
$\therefore \, Emf E=\frac{d \phi}{d t}=2B\sqrt{\frac{y}{k}}\frac{d y}{d t}$
using $v^{2}=2as$
$\frac{d y}{d t}=v=\sqrt{2 a y}$
or $\left|\epsilon \right|=\frac{d \phi}{d t}=2B\sqrt{\frac{y}{k}}\sqrt{2 a y}$
or $\left|\epsilon \right|=$ $B y \sqrt{\frac{8 a}{k}}$