Question

A conducting bar moves on two conducting rails as shown in the figure. A constant magnetic field \( B \) exists into the page. The bar starts to move from the vertex at time \( t = 0 \) with a constant velocity. If the induced EMF is \( E \propto t^n \), then the value of \( n \) is ________________________.

Hint:

For motional EMF in a varying area, always express the area as a function of time and use Faraday’s law.

Answer 1

Correct Answer: 2

The induced EMF in a moving conductor is given by: \[ E = B \frac{dA}{dt} \] The area enclosed by the rails at any time \( t \) is: \[ A = \frac{1}{2} l^2 \] Since the length of the moving bar is proportional to time \( t \), we assume: \[ l = vt \] Then: \[ A = \frac{1}{2} (vt)^2 = \frac{1}{2} v^2 t^2 \] Differentiating with respect to \( t \): \[ \frac{dA}{dt} = v^2 t \] Thus, the induced EMF is: \[ E = B v^2 t \] Comparing with \( E \propto t^n \), we get \( n = 2 \).