Exercise 5.3.3: The rotating coil generator

The normal of the coil makes the angle $\theta =\omega t$ with $B$ ($\omega =2\pi n/60$), so the flux through one turn is $\Phi =BS\cos \omega t$ and the flux linkage is $\Lambda =N\Phi =NBS\cos \omega t$. Then

$\omega =314.2rad/s$, $f=50\mathrm{Hz}$, $\hat{e}=100\times 0.5\times 0.02\times 314.2=314\mathrm{V}$ and $E_{\mathrm{rms}}=\hat{e}/\sqrt{2}=222\mathrm{V}$.

Both $\hat{e}$ and $f$ are proportional to $\omega$: at $1500\mathrm{rpm}$, $\hat{e}=157\mathrm{V}$ at $25\mathrm{Hz}$. The ratio $\hat{e}/f=2\pi NBS$ is fixed by the machine, the "volts per hertz" of a generator or of a motor run from an inverter.

Change the speed of the coil: the peak emf and the frequency grow together.

Units: $\mathrm{T}\cdot {\mathrm{m}}^2\cdot {\mathrm{s}}^{-1}=\mathrm{V}$. This is the alternator. With the ends of the coil connected to slip rings one obtains the sinusoid. With a split-ring commutator the sign is reversed every half turn and the output is $|\sin \omega t|$, the DC generator.