JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2025
- Medium
- Numerical

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 \(\mathrm{E} \propto \mathrm{t}^{\mathrm{n}}\), then value of n is ________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) 1
Step-by-step Solution
Detailed explanation
As the bar moves without change in orientation, the length of bar will be proportional to its distance from the vertex.
i.e. \(I=c(v t)\)
\(\begin{aligned}
& \text { induced emf } E=B / v \\
& =c B v^2 t
\end{aligned}\)
\(\Rightarrow n=1\)
i.e. \(I=c(v t)\)
\(\begin{aligned}
& \text { induced emf } E=B / v \\
& =c B v^2 t
\end{aligned}\)
\(\Rightarrow n=1\)
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