Physics · Electromagnetic Induction

JEE Main 2024 — 4 April, Shift 2 — Question 56

A rod of length 60 cm rotates with a uniform angular velocity $20 \mathrm{rad} \mathrm{s}^{-1}$ about its perpendicular bisector, in a uniform magnetic field 0.5 T . The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is $\qquad$ V.

Answer: 0

Numerical answer — enter this value.

Step-by-step solution

$\because \mathrm{V}_{0}-\mathrm{V}_{\mathrm{A}}=\frac{\mathrm{B} \omega \ell^{2}}{2}$ $\mathrm{V}_{0}-\mathrm{V}_{\mathrm{B}}=\frac{\mathrm{B} \omega \ell^{2}}{2}$ $\therefore \mathrm{V}_{\mathrm{A}}=\mathrm{V}_{\mathrm{B}} \therefore \mathrm{V}_{\mathrm{A}}-\mathrm{V}_{\mathrm{B}}=0$

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF