Calculating the Maximum Emf of a Generator

Once $ω$ , the angular velocity, is determined, ${emf}_{0} = NAB ω$ can be used to find ${emf}_{0}$ . All other quantities are known.

Angular velocity is defined to be the change in angle per unit time:

ω = \frac{Δ θ}{Δ t} .

23.21

One-fourth of a revolution is $π/2$ radians, and the time is 0.0150 s; thus,

\begin{array}{l} ω & = & \frac{π / 2 rad}{0.0150 s} \\ = & 104 . 7 rad/s . \end{array}

23.22

104.7 rad/s is exactly 1000 rpm. We substitute this value for $ω$ and the information from the previous example into ${emf}_{0} = NAB ω$ , yielding

\begin{array}{l} {emf}_{0} & = & NAB ω \\ = & 200 (7 . 85 \times 10^{- 3} m^{2}) (1.25 T) (104.7 rad/s) \\ = & 206 V \end{array} .

23.23

The maximum emf is greater than the average emf of 131 V found in the previous example, as it should be.