A 200 turn circular coil of area \( 10^{3}\,\mathrm{cm^2} \) rotates at 60 revolutions per minute in a uniform magnetic field of 0.02 T perpendicular to the axis of rotation of the coil. The maximum voltage induced in the coil is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{4}{5}\pi\,\mathrm{V} \)
For a coil rotating in a uniform field, the flux is \( NBA\cos\omega t \), so by Faraday's law the induced emf is \( NBA\omega\sin\omega t \). Its maximum value is
\( \displaystyle \varepsilon_0 = N B A \omega \)
\( N = 200 \)
\( B = 0.02\,\mathrm{T} \)
\( A = 10^{3}\,\mathrm{cm^2} = 10^{3}\times10^{-4}\,\mathrm{m^2} = 0.1\,\mathrm{m^2} \)
\( \displaystyle \omega = 2\pi\times\frac{60}{60}\,\mathrm{rev\,s^{-1}} = 2\pi\,\mathrm{rad\,s^{-1}} \)
\( \displaystyle \varepsilon_0 = 200\times0.02\times0.1\times2\pi = 0.8\pi = \frac{4}{5}\pi\,\mathrm{V} \)
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 – …
