A uniform magnetic field of 0.4 T acts perpendicular to a circular copper disc 20 cm in radius. The disc is…

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 potential difference developed between the axis of the disc and the rim ? $(\pi=3.14)$
  1. $0.5024 \mathrm{~V}$
  2. $ \mathrm{~V}$
  3. $0.2512 \mathrm{~V}$
  4. $0.1256 \mathrm{~V}$

Solution

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

Asked in: JEE Main 2025 (28 Jan Shift 2)

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