JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2025
- Medium
- MCQ
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 ? \((\pi=3.14)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(0.2512 \mathrm{~V}\)
Step-by-step Solution
Detailed 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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