I am trying to build a low power output generator with 12 pole pairs and 9 stator coils. The design will be similar to the one described in this link Basic Principles Of The Homebrew Axial Flux Alternator , but at a different scale/size. I am trying to find some way of estimating the number of turns I will need in the coils to produce an output voltage of 6.5V. The generator should produce this voltage at 100rpm. I am getting very confused about how to use Faraday's law to do my calculations.

$\varepsilon = N\frac{d(BA)}{dt}$

Faraday's Law as explained in this example where the area used is that of the magnet, while Faraday Law of Electromagnetic Induction states at the bottom that A is the area of the coil. Does this difference have to do with when the magnetic field is stationary while the coil is moving versus when the magnetic field is moving relative to a stationary coil? Also, should I be using $$\varepsilon = N\frac{d(BAcos(\theta))}{dt}$$ instead? Thanks very much in advance, and apologies if I am missing something very obvious here.


No expert, but the problem is interesting.


  • Generate $6.5V_{RMS}$ at 100 rpm with 9 stator coils.
    • $V_{MAX} = 9.2V$
    • 100rpm = 1.67rps
    • if radius to center of magnets/coils = 0.1m
    • $C = 2 \pi r = 2 \pi \times 0.1m = 0.628m$
    • at 1.67 rps, v = 1.05m/s
  • N48 Neodymium Bar Magnets 1 in x 1/2 in x 1/4 in
    • L = 0.0254m, W = 0.0127m, D = 0.00635m.
  • Separation z = 0.00635m.
    • Biggest problem is separation between magnets and windings. Closer they are the better.

From: How do you calculate the magnetic flux density? enter image description here

From: Magnetic Properties of Sintered NdFeB Magnets

N48 has a Remanence field of $B_r$ = 1.48T, which gives a flux density B = 0.164T, when you process the above equation.

$$V_{Ind} = N B l v$$

$$N = \frac {V_{Ind}} {B l v} = \frac {9.2V} {0.164T \times 0.0254m \times 1.05m/s} = 2108 turns$$

So that assumes you have one coil. Divide that by 3 (in series) gives 703 turns/coil (which is a lot). This coil has to be within length of magnet of 0.0254m. Whole area of coil has to fit within area of magnet for greatest effect.

The math is a bit distorted, but it should give you a ballpark to do your initial calculations. I'd guess, by the second or third prototype, you will have what you want. Not all three coils in series will generate maximum voltage at the same time, so some experimentation will have to come into play (i.e. more turns).

Realize that the coil will not experience a constant magnetic field. The further away windings are from the magnet the less the voltage induced. As already stated, biggest problem is air gap between windings and magnets. Air is a poor conductor of flux. I'd use a steel plate to attach your magnets.

  • $\begingroup$ Hi thanks for the reply! I thought about taking this approach, except I would assume the B field is constant, and then calculate dA/dt as you have. My point of contention though would be that vl gives you the area swept per second by the magnet. But is it not the rate of change of flux through the COIL's area? Currently I am using the length of the rectangular stator coil as the area, multiplied by the tangential velocity. Does this seem reasonable? I have a feeling this may depend on whether the magnets or coils are longer in length, but I'm not too sure. $\endgroup$ – masiewpao Mar 14 '17 at 14:41
  • 1
    $\begingroup$ As long as magnetic field changes, voltage will be induced. This gives you maximum voltage induced, which defines the sine wave. Ideally, full coil length fits within length of magnet because that is what produces the voltage. $\endgroup$ – StainlessSteelRat Mar 14 '17 at 15:38

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