Electromagnetic Induction

JEE Main level, numerical answer, medium

8 of 153

Question 8 of 153

A rod of length 60 cm60 \mathrm{~cm} rotates with a uniform angular velocity 20rads−120 \mathrm{rad} \mathrm{s}^{-1} about its perpendicular bisector, in a uniform magnetic filed 0.5T0.5 T. The direction of magnetic field is parallel to the axis of rotation. The potential difference between the two ends of the rod is _____V.

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Answer: 0


∵V0−VA=Bωℓ22 V0−VB=Bωℓ22∴VA=VB∴VA−VB=0\begin{aligned} & \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\end{aligned}

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