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 10 π rad s -1 about an axis through its centre and perpendicular to the disc. What is the protential difference developed between the axis of the disc and the rim ? (π=3.14)

Physics · Electromagnetic Induction

JEE Main 2025 — 28 January, Evening Shift — Question 47

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 $10 \pi \mathrm{rad} \mathrm{s}^{-1}$ about an axis through its centre and perpendicular to the disc. What is the protential difference developed between the axis of the disc and the rim ? $(\pi=3.14)$

Option A:
0.0628 V
Option B:
0.5024 V
Option C:
0.2512 V
Correct
Option D:
0.1256 V

Answer: C

Step-by-step solution

$B=0.4 \mathrm{~T}$

$\mathrm{r}=20 \mathrm{~cm}$

$\omega=10 \pi \mathrm{rad} / \mathrm{s}$

$\mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^{2}$

$=0.2512 \mathrm{~V}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Motional EMF

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