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 _______.
Pick an option to check your answer.
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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