Question

A wire bent as a parabola y = kx² 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 Fig. 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$\sqrt{\frac{8a}{k}}$

Answer 2

df = B.dA = 2B x dy and y = kx²

\ e =$\sqrt{\frac{y}{k}}$

\ =$\frac{d\varphi}{dt}$= 2B $\sqrt { \frac { \mathrm { y } } { \mathrm { k } } }$ $\frac{dy}{dt}$

using v² = 2as

$\frac{dy}{dt}$= v =$\sqrt{2ay}$

or | e | =$\frac{d\varphi}{dt}$= 2B $\sqrt { \frac { \mathrm { y } } { \mathrm { k } } }$ $\sqrt{2ay}$

or | e | = By$\sqrt{\frac{8a}{k}}$