What happens if the electric field is not perpendicular to the magnetic field for an electromagnetic wave? What could be the possible consequences if such a case was found? Are there examples or demonstrations where the two fields are not perpendicular?

  • $\begingroup$ Are you talking about electromagnetic waves? Electric fields and magnetic fields needn't be perpendicular. $\endgroup$ – Yashas Feb 13 '17 at 13:12
  • $\begingroup$ Yes, electromagnetic waves. $\endgroup$ – salvo9415 Feb 13 '17 at 17:42
  • $\begingroup$ Just put a plane wave in which they are perpendicular in a background electric or magnetic field and then the total field won't be perpendicular everywhere. $\endgroup$ – octonion Feb 13 '17 at 19:20

In a word: nothing.

It's perfectly easy to set up an electric field and a magnetic field at any arbitrary angle you might come up with - just set up a pair of electrodes and some Helmholtz coils at that angle and you're done. Similarly, if you want oscillating fields, you can set up monochromatic electric and magnetic fields with independent arbitrary polarization ellipses with only a modicum of work.

If you've got a single, monochromatic, linearly-polarized, plane-wave field, then yes, there is a standard argument that says that the electric field and magnetic field polarizations need to be orthogonal. However, this fails if you break any of those assumptions, particularly when you have non-plane-wave configurations or more than one plane wave.

The only actual constraints, in a vacuum, are the Maxwell equations $\nabla \times\mathbf E = -\frac{\partial\mathbf B}{\partial t}$ and $\nabla \times\mathbf B = \mu_0\varepsilon_0 \frac{\partial\mathbf E}{\partial t}$ (plus, obviously, the two divergence laws). Everything else is secondary.

  • $\begingroup$ I understand what you're saying, but what do you mean in the last sentence? Why the third and the forth law are the constraints? $\endgroup$ – salvo9415 Feb 13 '17 at 17:49
  • $\begingroup$ @salvo9415 I don't understand your question. Why do EM fields need to satisfy the Maxwell equations, you mean? $\endgroup$ – Emilio Pisanty Feb 13 '17 at 18:18

Not the answer you're looking for? Browse other questions tagged or ask your own question.