A 100 turn coil of area 10 cm2 rotates at 2 revolutions per second in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. The maximum voltage generated in the coil is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 12.56 mV
A coil rotating in a uniform magnetic field produces a sinusoidal emf \( \varepsilon = NBA\omega\sin\omega t \), so the peak emf is \( NBA\omega \).
\( N = 100 \)
\( B = 0.01\,\mathrm{T} \)
\( A = 10\,\mathrm{cm^{2}} = 10\times10^{-4}\,\mathrm{m^{2}} = 10^{-3}\,\mathrm{m^{2}} \)
The coil makes 2 revolutions per second, so its angular speed is
\( \omega = 2\pi\nu = 2\pi \times 2 = 4\pi\,\mathrm{rad\,s^{-1}} \)
\( \varepsilon_{\max} = NBA\omega \)
\( \varepsilon_{\max} = 100 \times 0.01\,\mathrm{T} \times 10^{-3}\,\mathrm{m^{2}} \times 4\pi\,\mathrm{rad\,s^{-1}} \)
\( \varepsilon_{\max} = 4\pi\times10^{-3}\,\mathrm{V} \approx 12.56\,\mathrm{mV} \)
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 – …
