In an AC generator, a coil of 20 turns each with area \( 2\times10^{-2}\,\mathrm{m^{2}} \) is rotated at 50 revolutions per sec in a uniform magnetic field of 2.5 T. The maximum induced emf produced is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 314 V
In an AC generator the flux through the coil is \( NBA\cos\omega t \), so by Faraday’s law the peak emf is \( \varepsilon_{\max} = NBA\omega \), where \( \omega = 2\pi f \) with f in revolutions per second.
\( \omega = 2\pi\times50 = 100\pi\,\mathrm{rad\,s^{-1}} \)
\( \varepsilon_{\max} = 20\times2.5\times(2\times10^{-2})\times100\pi\,\mathrm{V} = 1\times100\pi\,\mathrm{V} \approx 314\,\mathrm{V} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
AC Generator – …
