The coil of an A.C. generator having 500 turns and area of \(0.01\,\mathrm{m^{2}}\) rotates at a speed of 100 revolutions per minute in a uniform magnetic field \(3\times10^{-2}\,\mathrm{T}\). The peak emf will be:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(1.57\,\mathrm{V}\)
In an AC generator the flux through the coil is \( NBA\cos\omega t \), so the induced emf is \( NBA\omega\sin\omega t \) and the peak emf is \( NBA\omega \).
\( \displaystyle \omega = 2\pi\nu = 2\pi \times \frac{100}{60}\,\mathrm{s^{-1}} \approx 10.47\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle \varepsilon_0 = NBA\omega \)
\( \displaystyle \varepsilon_0 = 500 \times (3\times10^{-2}\,\mathrm{T}) \times (0.01\,\mathrm{m^{2}}) \times 10.47\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle \varepsilon_0 \approx 1.57\,\mathrm{V} \)
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 – …
