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The picture above shows, that the voltage of a phase is greatest, when the magnet alligns with the coil of the phase. Why is that?

To my knowledge of lenz's law the voltage and induced current should be 0 when the magnet aligns with a coil, since there is no opposing emf generated in the coil.

  • $\begingroup$ Makes sense. I've edited my first post. $\endgroup$
    – rogerg
    Aug 21, 2019 at 20:46
  • $\begingroup$ Hi Rogerg, I'm asking the same question to myself. If you run the applet from the link below and go to the generator tab: phet.colorado.edu/sims/cheerpj/faraday/latest/… you can see how the voltage changes it's direction when the magnet aligns with a coil. I'm so confused $\endgroup$
    – Ayrix
    Nov 7, 2022 at 22:47

1 Answer 1


The smaller is the gap between coil's core and the magnet, the greater is the flux change through the coil from the same magnet displacement. Thus the EMF is the highest at this phase (Faraday's law).

Now, if you add a load to the generator (in the most extreme case, short the winding of the coil), then the induced current in the coil will produce a counter-flux that will resist the magnet's motion. This is what Lenz' law is trying to tell you.

If, on the other hand, you leave the ends of the winding open, there will be no current, and no force resisting the magnet's movement. The apparent confusion in your question is that you must have a changing current in the winding, not just the EMF, to produce the opposing magnetic field.

You can also look at the whole thing from the conservation of energy standpoint. You have an ideal generator. If you give the rotor a rotating push, nothing is going to resist its rotation as long as the generator is unloaded. As soon as you connect a light bulb to the generator, the bulb will light up, at the same time eating up the kinetic energy of the rotor and slowing it down. Thus the Lenz' law can be qualitatively understood as a facet of the energy conservation law.

  • $\begingroup$ sorry for my late answer, was away for a few days. Im not sure if i understand it. The way i understand it is, that the counter flux/emf makes the electrons flow in a direction in the coil, therefore creating a difference in potential energy/a voltage. If there is no load connected and no current flowing, why is it, that the voltage is highest at this point? $\endgroup$
    – rogerg
    Aug 26, 2019 at 10:24
  • $\begingroup$ I think I cannot give a better answer that “this is how the Nature works.” This is captured by the Faraday's law. The law says that the higher is the rate of flux change ${dB}/{dt}$ through the area $A$, the higher is the EMF. In an open circuit, there is no current, and the voltage=EMF. If you close the loop, the voltage will be 0 (you are probing two points on an ideal conductor, so the Ohm law will give you $V=I R=I×0=0$), but the current will generate equal and opposing $B$. Scroll down that page and peruse links. HTH. $\endgroup$ Aug 27, 2019 at 1:52
  • $\begingroup$ Sometimes it helps to simplify the idea into an experiment. If you connect a wire between the pos/neg poles of a car alternator, to represent the load, then voltage will flow through the wire and you will notice it takes more energy to turn the alternator then when it has no load attached. The EMF is present in an open loop but it is immediately drawn to earth ground with little to no force. If there is any voltage, which is debatable, it is immediately released to Earth ground. The closed wire loop gives it a path to follow which is more attractive than Earth ground. $\endgroup$ Oct 19, 2022 at 14:50

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