The armature coil of a generator has 20 turns and its area is \( 0.127\,\mathrm{m^2} \). How fast approximately should it be rotated in a magnetic field of \( 0.2\,\mathrm{Wb\,m^{-2}} \), so that the peak value of induced emf is 160 V?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 50 rps
For a coil rotating with angular speed \( \omega \) in a uniform field, the peak induced emf of the ac generator is
\( \displaystyle \varepsilon_0 = N A B\,\omega \;\Rightarrow\; \omega = \frac{\varepsilon_0}{NAB} \)
\( \displaystyle \omega = \frac{160}{20 \times 0.127 \times 0.2} = \frac{160}{0.508} = 315\,\mathrm{rad\,s^{-1}} \)
The options are in revolutions per second, so divide by \( 2\pi \):
\( \displaystyle \nu = \frac{\omega}{2\pi} = \frac{315}{2 \times 3.14} \approx 50\,\mathrm{rps} \)