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 $t=0$, sliding wire starts sliding from the vertex O with a constant acceleration a linearly as shown in figure. Find the emf induced in the loop.

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

Answer 1

Answer: (C)

$By\sqrt{\frac{8a}{k}}$

$d\varphi =B.dA=2Bx.dy$ and $y=k{x}^2$
$\therefore x=\sqrt{\frac{y}{k}}$
$\therefore ϵ=\frac{d\varphi }{dt}=2B\sqrt{\frac{y}{k}}\frac{dy}{dt}$
using $v^2=2as$
$\frac{dy}{dt}=v=\sqrt{2ay}$
In y direction
or $|\epsilon |=\frac{d\varphi }{dt}=2B\sqrt{\frac{y}{k}}\sqrt{2ay}$
or $|\epsilon |=By\sqrt{\frac{8a}{k}}$