JEE Mains · Physics · STD 12 -6. Electromagnetic induction
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 _______.
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(5\)
Step-by-step Solution
Detailed explanation
\(\phi=\mathrm{NAB} \cos (\omega \mathrm{t})\) \(\varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{NAB} \omega \sin (\omega \mathrm{t})\) \(\varepsilon_{\max }=\mathrm{NAB} \omega\) \(=200 \times 0.2 \times 0.01 \times \pi\)…
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