Sorry that I don't have pictures.

Imagine I have a steel circle, like a hula hoop. On opposite sides of that circle I attach two springs, then stretch them. They meet in center where I hook them together. The springs keep pulling on each other, and the circle, but they are in equilibrium and the system stays static. Unless it rusts to pieces, it will stay that way for millennia. No energy is being used here.

Now let's replace one spring with a not very elastic string. One end is still attached to the circle. On the other end I add an electromagnet. The spring which remains also gets modified slightly - instead of a hook, I add a little metallic plate which the electromagnet can attract.

Now, when I turn on electricity, the magnet attracts the plate and keeps the spring stretched, just like in the first case. But in this case, I need to keep spending electrical energy in order to keep the spring taut. Work is being done, heat is being generated. The system is still static, unmoving - but as soon as the battery runs out, the magnet will release the spring.

Why is this? Why is there work being done even if nothing is moving, just like in the first case?


1 Answer 1


You are right in saying that it's weird that this static system does work. The reason it does work is because it is not a static system, but rather electrons are moving through a resistive medium (the electromagnet). Though a normal electromagnet would in fact consume power, this is not necessarily the case. Because the system is not moving, and there are no time-varying electric or magnetic fields, keeping the magnet in place requires no power, it is only because of the resistance of the wire that power is being consumed. If instead you were to replace the simple electromagnet with a superconducting electromagnet (or let's say a ballistic conductor, as the Meissner effect rather complicates things), and were to put the electricity in and then connect the two ends together, no power would be consumed and it would still hold the magnet! The only reason that the system consumes energy is because you are turning electrical energy into heat through resistance in a normal electromagnet. None of the systems do work in the traditional sense.

If you were to consider similar case where you replaced the second spring with your arm, and had to hold it, you would have to expend energy keeping it there. This is not because of some strange thermodynamic property, but simply because your arm is built like that.

  • $\begingroup$ Ahh, I forgot an intermediate stage with a static magnet. But, yes, this explains them both. $\endgroup$
    – Vilx-
    Commented Jul 15, 2019 at 19:49
  • $\begingroup$ I think there is an important misconception in this answer (and the original question). Just because energy is being expended does not mean work is being done! Indeed, I believe it would be more appropriate in each of these cases (the electromagnet, and your arm) to say that useful energy is being converted purely into heat, due to the nature of the system, without any resulting work. $\endgroup$
    – Rococo
    Commented Jul 15, 2019 at 20:23
  • 1
    $\begingroup$ @Rococo A fair point, as none of the systems do traditional macroscopic thermodynamic work. I've updated my answer to reflect this. Whether the initial electron-phonon interactions of joule heating and actin-myosin activation could count as work in some sense is an interesting question though. $\endgroup$
    – Ash Pera
    Commented Jul 15, 2019 at 20:56
  • $\begingroup$ Well, I was thinking of "work" in a more mathematical sense of "energy multiplied by time". I have not done the math but I suspect that this is what the "force times distance" also reduces to, no? Anyway, the resistance argument is pretty convincing. With a superconducting magnet it essentially turns into a static magnet (of sorts) and that is obviously not using energy. $\endgroup$
    – Vilx-
    Commented Jul 15, 2019 at 22:37
  • $\begingroup$ Hi @Vilx- Work has the same units as energy, and in a thermodynamic context can be thought of as the energy used in a process of interest, to be contrasted with heat. If you search for "work-energy theorem" on this site you can find much more information about their relation. $\endgroup$
    – Rococo
    Commented Jul 16, 2019 at 14:47

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