A coil of turns area rotated |AlternatingCurrentJEEMain2024

JEE MAIN 2024

01-02-2024 S2

Question

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 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 β is

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Solution

$\begin{aligned} & \phi=\mathrm{NAB} \cos (\omega \mathrm{t}) \\ & \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{NAB} \omega \sin (\omega \mathrm{t}) \\ & \varepsilon_{\max }=\mathrm{NAB} \omega \\ & =200 \times 0.2 \times 0.01 \times \pi \\ & =\frac{4 \pi}{10}=\frac{2 \pi}{5} \text { volt }\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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