An armature coil consists of 20 turns of wire, each of area \( A = 0.09\,\mathrm{m^{2}} \) and total resistance 15 Ω. It rotates in a magnetic field of 0.5 T perpendicular to the axis of rotation at a constant frequency \( \displaystyle\frac{150}{\pi}\,\mathrm{Hz} \). The value of maximum emf produced in the coil is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 270 V
A coil of N turns rotating in a uniform field B produces an emf \( \varepsilon = NBA\omega\sin\omega t \), whose maximum value is \( \varepsilon_{\max} = NBA\omega \), with \( \omega = 2\pi f \).
\( \displaystyle \omega = 2\pi f = 2\pi\times\frac{150}{\pi} = 300\,\mathrm{rad\,s^{-1}} \)
\( \varepsilon_{\max} = 20\times0.5\,\mathrm{T}\times0.09\,\mathrm{m^{2}}\times300\,\mathrm{rad\,s^{-1}} = 270\,\mathrm{V} \)
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 – …
