Question

A wire as a parabola $y=4x^2$ is located in a uniform magnetic field of inductance B perpendicular to the XY plane. At t=0 a connection starts translation wise from the parabola apex with constant acceleration $\alpha$. The induced emf in the loop, thus formed, as a function of y is:

• A. $e=\sqrt{2\alpha }\cdot By$
• B. $e=By\sqrt{\frac{\alpha }{2}}$
• C. $e=\frac{By\sqrt{\alpha }}{2\sqrt{2}}$
• D. $e=\frac{By\sqrt{\alpha }}{4}$

Answer 1

The correct option is A $e=\sqrt{2\alpha }\cdot By$

$\overrightarrow{B}=B\uparrow$

Velocity of rod after travelling $dy$ distance $u$.

we know that,

$S=ut+\frac{1}{2}a{t}^2$

$y=0+\frac{1}{2}\alpha t^2$ ($t=\sqrt{\frac{2y}{\alpha }}$)

$\frac{dy}{dt}=\alpha t$

$V=\alpha t=\sqrt{2\alpha y}$

emf induced $\epsilon =VB\left(2x\right)$

$=\sqrt{2\alpha y}\times 2xB$

$x=\frac{\sqrt{y}}{2}$

$\epsilon =\sqrt{2\alpha }.By$