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 _____.
Show Hint
For motional EMF in a varying area, always express the area as a function of time and use Faraday’s law.
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 \( n = 2 \).
