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 _____. 

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

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

Answer 1

The Correct Option is B

The induced electromotive force (EMF) in a conductor undergoing motion is expressed as: \[ E = B \frac{dA}{dt} \] The area \( A \) enclosed by the conducting rails at time \( t \) is given by: \[ A = \frac{1}{2} l^2 \] Given that the length \( l \) of the moving bar is directly proportional to time \( t \), we establish the relationship: \[ l = vt \] Substituting this into the area equation yields: \[ A = \frac{1}{2} (vt)^2 = \frac{1}{2} v^2 t^2 \] Differentiating the area \( A \) with respect to time \( t \) gives: \[ \frac{dA}{dt} = v^2 t \] Consequently, the induced EMF \( E \) is determined to be: \[ E = B v^2 t \] By comparing this result with the proportionality \( E \propto t^n \), we deduce that \( n = 2 \).