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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2024
  • Hard
  • Numerical

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 _______.

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Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\(\phi=\mathrm{NAB} \cos (\omega \mathrm{t})\)

\(\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{NAB} \omega \sin (\omega \mathrm{t})\)

\(\varepsilon_{\max }=\mathrm{NAB} \omega\)

\(=200 \times 0.2 \times 0.01 \times \pi\)

\(=\frac{4 \pi}{10}=\frac{2 \pi}{5} \mathrm{volt}\)

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