Finding emf induced in loop

Let the equation of the parabola (which is not given in the problem) be $y=kx^2$ (I had to guess from the answer options)

Then the velocity of the connector at a position $(x, y)$ , assuming it to be zero when the wire is at the apex of the parabola (this is also not given), is $v=\sqrt {2ky}$

Area swept by the connector in time $dt$ at that instant is $dA=2xdy=\dfrac {2}{\sqrt k}y^{\frac 12}dy$ ,

so that the rate at which the area is swept out is

$\dfrac {dA}{dt}=\dfrac {2}{\sqrt k}y^{\frac 12}\dfrac {dy}{dt}$

$=\dfrac {2}{\sqrt k}y^{\frac 12}v$

$=\dfrac {2}{\sqrt k}\sqrt {2a}y$

$=\sqrt {\dfrac {8a}{k}}y$

So the e. m. f. Induced is

$E_{ind}=\sqrt {\dfrac {8a}{k}}By$

I think the ratio of $E$ and $y$ is asked in the problem, which is

$\boxed {\sqrt {\dfrac {8a}{k}}}$ .