JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2024
- Hard
- Numerical
A rod of length \(60 \mathrm{~cm}\) rotates with a uniform angular velocity \(20 \mathrm{rad} \mathrm{s}^{-1}\) about its perpendicular bisector, in a uniform magnetic field \(0.5 \mathrm{~T}\). The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is________\(V\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(0\)
Step-by-step Solution
Detailed explanation
\(\because \mathrm{V}_0-\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{B} \omega \ell^2}{2}\)
\(\mathrm{~V}_0-\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{B} \omega \ell^2}{2}\)
\(\therefore \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=0\)
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