Electromagnetic Induction

JEE Main level, single correct, medium

77 of 153

Question 77 of 153

A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is having a uniform angular velocity of 10πrads−110 \pi \mathrm{rad} \mathrm{s}^{-1} about an axis through its centre and perpendicular to the disc. What is the potential difference developed between the axis of the disc and the rim ? (π=3.14)(\pi=3.14)

Show answer and solution

Answer: (C) 0.2512 V0.2512 \mathrm{~V}

B=0.4 Tr=20 cmω=10πrad/sE=12 BωR2=0.2512 V\begin{aligned} & \mathrm{B}=0.4 \mathrm{~T} \\ & \mathrm{r}=20 \mathrm{~cm} \\ & \omega=10 \pi \mathrm{rad} / \mathrm{s} \\ & \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^2 \\ & =0.2512 \mathrm{~V}\end{aligned}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Electromagnetic induction simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Electromagnetic Induction (Motional and Rotational EMF), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.