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 in motion is calculated as: \[ E = B \frac{dA}{dt} \] The area \(A\) contained within the rails at time \(t\) is expressed as: \[ A = \frac{1}{2} l^2 \] Given that the length \(l\) of the moving bar is directly proportional to time \(t\), it can be represented as: \[ l = vt \] Consequently, the area becomes: \[ A = \frac{1}{2} (vt)^2 = \frac{1}{2} v^2 t^2 \] Differentiating \(A\) with respect to \(t\) yields: \[ \frac{dA}{dt} = v^2 t \] Therefore, the induced EMF is: \[ E = B v^2 t \] When this is compared to the expression \( E \propto t^n \), it is determined that \( n = 2 \).
