In vacuum the two field components of a radio wave are directed perpendicular on each over. $\mathbf{E}$, $\mathbf{B}$ and $\mathbf{k}$ form a right hand system ($\mathbf{k}$ is the direction of propagation).

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This matches the right hand rule for a current carrying wire and the induces magnetic field around the wire.

Now, for the reasons of symmetry, a radio wave also could be drawn with $\mathbf{E}$, $\mathbf{-B}$ and $\mathbf{k}$. Aplicated to the image above, the green fieldlines (which showing the direction of the magnetic field) simply would show in the opposite direction. How induce a left hand radio wave?

  • $\begingroup$ left-handed meta materials do just that. They are the topic of much current research. $\endgroup$ – garyp Feb 16 at 14:30
  • $\begingroup$ Closely related question by the same user: physics.stackexchange.com/q/461167/44126 $\endgroup$ – rob Feb 16 at 15:25

There is no such symmetry argument ?

If you change $\overrightarrow{B}\to -\overrightarrow{B}$ and $\overrightarrow{E}\to \overrightarrow{E}$ the Maxwell Faraday equation $\overrightarrow{\nabla }\wedge \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ would not be true anymore.

  • $\begingroup$ Vincent, that seems to be another formulaton of my question. Furthermore, the current is from electrons. What about a current of protons (in a vacuum tube for example) or a current from positrons? $\endgroup$ – HolgerFiedler Feb 16 at 14:23
  • $\begingroup$ The charge does not interfere with Maxwell Faraday's equation. I think the wave you propose can not be a solution of Maxwell's equations (and therefore does not exist) because it cannot check Maxwell Faraday's equation. $\endgroup$ – Vincent Fraticelli Feb 16 at 14:27

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