JEE MAIN - Physics (2024 - 1st February Evening Shift - No. 28)
A coil of 200 turns and area $0.20 \mathrm{~m}^2$ is rotated at half a revolution per second and is placed in uniform magnetic field of $0.01 \mathrm{~T}$ perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is $\frac{2 \pi}{\beta}$ volt. The value of $\beta$ is _______.
Answer
5
Explanation
$\begin{aligned} & \phi=\mathrm{NAB} \cos (\omega \mathrm{t}) \\\\ & \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{NAB} \omega \sin (\omega \mathrm{t}) \\\\ & \varepsilon_{\max }=\mathrm{NAB} \omega \\\\ & =200 \times 0.2 \times 0.01 \times \pi \\\\ & =\frac{4 \pi}{10}=\frac{2 \pi}{5} \mathrm{volt}\end{aligned}$
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