Is it possible to 'twist' a magnetic field? I am not referring to the electrical field but the actual magnetic field.

In OAM (orbital angular momentum) we can twist light waves and sound waves however can this be done for a magnetic field?

I've seen research on curving antennas to twist electromagnetic waves and it works, however does this effect translate the same for stationary magnetic fields?

If I twist a wire or curve a wire around a cylinder and apply a current will the resultant magnetic field have an orbital angular momentum? Or is the field only ever perpendicular to the wire?

If not how could you achieve a twisted magnetic field? Or does orbital angular momentum only apply to waves and not fields, which doesn't make sense to me as you can twist an EM wave which has a magnetic field?

Imagine a magnetic field twisted like the below image:


  • $\begingroup$ Not quite certain what it is exactly that you mean. Do you mean a magnetic field whose curl is non-zero ? In that case the current through a wire can produce such a thing. Or maybe you mean something like a rotating bar magnet, which produces an electric dipole, analogous to quantum mechanical spin, but the roles of electricity and magnetism are reversed in a certain sense? $\endgroup$
    – Stratiev
    Commented Jun 29, 2020 at 15:46
  • $\begingroup$ Roger Penrose came up with an idea that splitting an object in an ergosohere of a black hole would measure a loss of negative energy and provide it to the other half, recently this was proven in a Nature publication using sound waves. EM and Light waves move too fast for a feasible experiment to be carried out but twisted magnetic fields might be possible. My question is how would it possible to twist a magnetic field, I am unsure which bracket of the two propositions you supposed that this would fall into. $\endgroup$
    – G Gr
    Commented Jun 29, 2020 at 16:04
  • $\begingroup$ As a thought experiment imagine a twisted magnetic field moving through a rotating magnetic field, I want to measure if any gain is achieved the same way the sound experiment was conducted. $\endgroup$
    – G Gr
    Commented Jun 29, 2020 at 16:11
  • $\begingroup$ Are you talking about magnetic helicity? $\endgroup$
    – DavidH
    Commented Jun 29, 2020 at 16:27
  • $\begingroup$ Helicity is more to do with self linkaged rather than twist. I will upload a picture of what I want to achieve. $\endgroup$
    – G Gr
    Commented Jun 29, 2020 at 16:38

1 Answer 1


A solenoidal coil produces a magnetic field with no component that corresponds to "twist". However, a straight wire carrying a current produces a magnetic field that it all "twist". So, a solenoidal coil with a straight wire running along its axis will produce a twisted field that winds around the axis more or less as you describe.

This article has a layman's description of the tokamak fusion reactor which obtains a twisted field using a current along its axis. The article also describes a stellarator reactor, which obtains a twisted field by distorting its toroidal coil.

However, this twist is not "orbital angular momentum". Because there is no electric field in this configuration, there is no Poynting vector; and without the Poynting vector there is no EM field momentum and hence no angular momentum.

  • $\begingroup$ So you technically can't test the theory in this way? Re: My Comments. $\endgroup$
    – G Gr
    Commented Jun 29, 2020 at 17:15
  • $\begingroup$ The image your link leads to represents a propagating wave, not a static field. $\endgroup$
    – S. McGrew
    Commented Jun 29, 2020 at 17:50
  • $\begingroup$ Yes it was to illustrate what I wanted to achieve with a static field. I understand that you can't have angular momentum without the momentum (propagating wave) so in part that answers the question really, just making sure there isn't another way. $\endgroup$
    – G Gr
    Commented Jun 29, 2020 at 18:17
  • $\begingroup$ A propagating wave is not necessary for there to be field momentum. The Poynting vector is present in a stationary field if there are both E and H fields present, and if they are not parallel. $\endgroup$
    – S. McGrew
    Commented Jun 29, 2020 at 19:53
  • $\begingroup$ But the Poynting vector is just the cross product of the E and B fields? It's still going to be a flat (not twisted) magnetic field, have a look at the the gif in this link as an example brilliant.org/wiki/poynting-vector $\endgroup$
    – G Gr
    Commented Jun 29, 2020 at 23:30

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