Electromagnetic Induction |JEE Main 2025 Question & Solution

JEE MAIN 2025

28-01-2025 SHIFT-2

Question

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 $10 \pi \mathrm{rads}^{-1}$ 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 ? ( $\pi=3.14$ )

Select the correct option:

0.1256 V

0.2512 V

0.0628 V

0.5024 V

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\quad \mathrm{B}=0.4 \mathrm{~T} ; \mathrm{r}=20 \mathrm{~cm} ; \omega=10 \pi \mathrm{rad} / \mathrm{s}$
$\mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^{2}=0.2512 \mathrm{~V}$

Question Tags

JEE Main

Physics

Medium

Start Preparing for JEE with Competishun

Video Solution

BY competishun

Watch Solution

YouTube

Report Issue