In an AC generator, a 100 turns coil of area \( 0.10\,\mathrm{m^2} \) rotates at 50 Hz in a uniform magnetic field of 0.01 T. The maximum voltage generated in the coil is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 31.4 V
In an ac generator the flux through the rotating coil is \( NBA\cos\omega t \), so by Faraday's law the emf is \( NBA\omega\sin\omega t \). Its maximum value is
\( \displaystyle \varepsilon_0 = N B A \omega = N B A\,(2\pi\nu) \)
\( N = 100 \)
\( B = 0.01\,\mathrm{T} \)
\( A = 0.10\,\mathrm{m^2} \)
\( \omega = 2\pi \times 50 = 314\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle \varepsilon_0 = 100 \times 0.01 \times 0.10 \times 314 = 0.1 \times 314 = 31.4\,\mathrm{V} \)
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 – …
