The armature coil of a generator has 20 turns and its area is $0.127\, m^2$. How fast approximately should it be rotated in a magnetic field of $0.2\, Wbm^{-2}$, 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
Given:
$N = 20$, $A = 0.127 \, m^{2}$, $B = 0.2 \, Wb/m^{2}$, $E_{0} = 160 \, V$
Formula: $E_{0} = N B A \omega$
$\omega = \frac{E_{0}}{N B A}$
$\omega = \frac{160}{20 \times 0.2 \times 0.127}$
$\omega = \frac{160}{0.508}$
$\omega = 314.96 \, rad/s$
Angular speed, $\omega = 2 \pi n$
$n = \frac{314.96}{2 \pi} = 50 \, rps$
Answer: 50 revolutions per second.