The coil of an ac generator rotates at 25 revolutions per second in a uniform magnetic field. At \(t=0\), the plane of the coil is perpendicular to the field. The emf induced in the coil first becomes maximum after a time of
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(10\ \mathrm{ms}\)
At \( t=0 \) the plane of the coil is normal to the field. So the area vector is along the field and \( \theta=0^\circ \). The emf \( \varepsilon=\varepsilon_0\sin\omega t \) is zero at this moment. It first becomes maximum when \( \omega t=90^\circ \), that is, after a quarter of a rotation.
\( \nu=25\ \mathrm{s^{-1}} \)
\( \begin{aligned} T &= \frac{1}{\nu} \\ &= \frac{1}{25}\ \mathrm{s} \\ &= 40\ \mathrm{ms} \end{aligned} \)
\( \begin{aligned} t &= \frac{T}{4} \\ &= \frac{40}{4}\ \mathrm{ms} \\ &= 10\ \mathrm{ms} \end{aligned} \)
NCERT (2026-27):
AC Generator – “When the coil is rotated with a constant angular speed ω, the angle θ between the magnetic field vector B and the area vector A of the coil at any instant t is θ = ωt (assuming θ = 0° at t = 0). …”
“Thus, the instantaneous value of the emf is”
\( \varepsilon=NBA\omega\sin\omega t \)\( \qquad (6.19) \)
“where NBAω is the maximum value of the emf, which occurs when sin ωt = ±1. …”