JEE MAIN - Physics (2025 - 28th January Evening Shift - No. 4)
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 $ rad 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? $(\pi=3.14)$
0.5024 V
0.0628 V
0.2512 V
0.1256 V
Explanation
$$\begin{aligned}
& \mathrm{B}=0.4 \mathrm{~T} \\
& \mathrm{r}=20 \mathrm{~cm} \\
& \omega=10 \pi \mathrm{rad} / \mathrm{s} \\
& \mathrm{E}=\frac{1}{2} \mathrm{~B} \omega \mathrm{R}^2 \\
& =0.2512 \mathrm{~V}
\end{aligned}$$
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