Q 12-06-047JEE MainJEE Main 2024 (1 Feb, Shift 2)Easy
A coil of 200 turns and area $0.20\ \text{m}^2$ is rotated at half a revolution per second and is placed in a uniform magnetic field of $0.01\ \text{T}$ perpendicular to the axis of rotation of the coil. The maximum voltage generated in the coil is $\dfrac{2\pi}{\beta}$ volt. The value of $\beta$ is ______.
Numerical value type. Enter your answer.
Answer: 5
$\omega = 2\pi\times0.5 = \pi\ \text{rad s}^{-1}$.
$$\varepsilon_0 = NBA\omega = 200\times0.01\times0.20\times\pi = 0.4\pi = \frac{2\pi}{5}\ \text{V} \Rightarrow \beta = 5$$
Solution by Sreeraj P, M.Sc Physics