JEE Physics 2025 Question with Solution

Given: A conducting bar moves with constant velocity on two conducting rails in a uniform magnetic field $B$ into the page.

Find: The power $n$ in $E \propto t^n$.

Use Faraday's law for motional emf:

$$E = B\frac{dA}{dt}$$

If the distance of the bar from the vertex at time $t$ is $x = vt$, then for the V-shaped rails the enclosed area is proportional to $x^2$. In the solution, this is written as

$$A = x^2 \tan\theta$$

So, substituting $x = vt$,

$$A = v^2 t^2 \tan\theta$$

Differentiating with respect to $t$,

$$\frac{dA}{dt} = 2v^2 t \tan\theta$$

Therefore,

$$E = B\frac{dA}{dt} = 2Bv^2 t \tan\theta$$

Hence,

$$E \propto t$$

Comparing with $E \propto t^n$, we get $n = 1$.

The solution lists $n = 2$, but the working shown there gives $E \propto t^1$. Therefore, from the extracted working, the numerical value is $1$.