No. 002 · Section P
Interactive · Open Source
Magnetic flux through a surface is the integral $\Phi_B = \int \mathbf{B} \cdot d\mathbf{A}$ — the field, dotted with the area it passes through. The dot product is doing all the work. Hold a loop face-on to the field and you catch every field line. Turn it edge-on and you catch none: the field slides past the opening without ever going through it.
Close that surface up — wrap it all the way around a magnet, around the Earth, around a galaxy — and the integral collapses to zero, exactly:
$$\oint \mathbf{B} \cdot d\mathbf{A} = 0$$
Every field line that leaves comes back. There are no magnetic sources to find, which is why cutting a magnet in half gives you two magnets instead of a lone north pole. Flux is conserved and, on its own, does nothing at all.
The physics starts when you make it move. Faraday’s law says the EMF around a loop is the negative rate of change of the flux through it:
$$\varepsilon = -\frac{d\Phi_B}{dt}$$
So take a coil of $N$ turns and area $A$, drop it into a field $\mathbf{B}$, and spin it at angular velocity $\omega$ about an axis perpendicular to the field. The angle between the field and the coil’s normal is just $\theta = \omega t$, so the flux is a cosine — and its derivative is where the $\sin\theta$ you came for actually lives:
$$\Phi_B(t) = NBA\cos(\omega t) \quad \Longrightarrow \quad \varepsilon(t) = NBA\,\omega \sin(\omega t)$$
Read that carefully, because it is the counterintuitive part. Peak voltage happens when the flux is zero. Edge-on, the coil catches nothing — but it is sweeping through field lines faster than at any other moment, and the rate of change is what Faraday charges you for. Not the flux. The slope of the flux.
Earth drawn to scale with a tilted dipole (11.5° from the spin axis). Field magnitude from
$B(\theta) = B_0\sqrt{1 + 3\cos^2\theta}$ with $B_0 = 31\,\mu$T at the magnetic equator.
The Mathematics
Four Equations Running the Instrument
1 · The dipole field. Earth’s field, to good first order, is a dipole of moment $\mathbf{m}$. At radius $r$ and colatitude $\theta$ measured from the magnetic pole:
$$B_r = \frac{\mu_0}{4\pi}\frac{2m\cos\theta}{r^3}, \qquad B_\theta = \frac{\mu_0}{4\pi}\frac{m\sin\theta}{r^3}$$
There is your first $\sin\theta$ — it is the horizontal component of the Earth’s field. Combining the two gives the magnitude the instrument reports, $B = B_0\sqrt{1+3\cos^2\theta}$, which is exactly twice as strong at the poles as at the equator. The field lines themselves trace $r = C\sin^2\theta$.
2 · The dip angle. The angle the field makes with the ground follows from the ratio of those two components, $\tan I = 2\tan\lambda$ for magnetic latitude $\lambda$. At the equator the field is perfectly horizontal; at the pole a compass needle wants to point straight down.
3 · Flux through the spinning coil. The coil’s normal $\hat{\mathbf{n}}$ sweeps a circle as it turns, and only the projection onto the field survives the dot product:
$$\Phi_B(t) = N\,\mathbf{B}\cdot\mathbf{A} = NBA\cos(\omega t)$$
4 · Faraday, and the sine. Differentiate, flip the sign, and the cosine becomes a sine scaled by $\omega$:
$$\varepsilon = -\frac{d\Phi_B}{dt} = NBA\,\omega\sin(\omega t)$$
The $\omega$ out front is why spinning faster raises the voltage twice over — more cycles per second and a taller peak. The minus sign is Lenz’s law: the induced current always opposes the change that made it. Flip that sign and a stray nudge would amplify itself forever, so the minus sign is really just energy conservation in disguise.
What the numbers say. Run the defaults — 200 turns, 100 cm², 3600 rpm at the magnetic equator — and the peak EMF is about 23 millivolts. Not a typo. The Earth’s surface field is roughly 31 µT, some ten thousand times weaker than a fridge magnet, so a planet-sized magnet spinning your coil at 60 Hz will not light a bulb. Slide the latitude to the pole, where $\sqrt{1+3\cos^2\theta}$ doubles the field, and you get 47 mV. Still millivolts.
That is the honest lesson hiding in the toy. Real generators do not use a bigger planet; they use a stronger $B$, more turns, and more area, because those are the only three knobs the equation gives you. But the equation is the same equation. A hydroelectric dam and this coil differ by a few orders of magnitude, and by nothing else.