In a linear, stationary, isotropic, homogeneous, only time-dispersive medium one usually writes Ohm's law as:

$$\underline{\mathcal{J}}(\underline{r},\omega)=\sigma(\omega) \underline{\mathcal{E}}(\underline{r},\omega)$$ where $\mathcal{J}(\underline{r},\omega)$ is the current density (and so, movement of charges) that is function of the point $\underline{r}$ and the frequency of the EM field $\omega$, $\sigma(\omega)$ is the conducibility (so, in other words, the limit velocity acquired by charges when they feel a field at frequency $\omega$) and $\underline{\mathcal{E}}(\underline{r},\omega)$ is obviously the electric field.

My question: why does charges movement (read $\underline{\mathcal{J}}(\underline{r},\omega)$) is indipendent of magnetic field $\underline{\mathcal{B}}(\underline{r},\omega)$?

I think that when $\omega \neq 0$, the charges movement is not parallel to the electric field, but it makes alternating rotations imposed by magnetic field.

But Ohm's law works so well, so it's me that have a problem. Can you help me?

  • $\begingroup$ Are you aware of the Hall effect? Are you thinking about this in terms of light-metal interaction, so the magnetic field in question is the light’s own field? $\endgroup$ – Gilbert Dec 6 '18 at 21:28
  • $\begingroup$ No, see the comment under Philip Wood answer :) $\endgroup$ – Nameless Dec 7 '18 at 6:02

@PhilipWood gives a good answer, certainly applicable to the DC Hall effect. In the case of an electromagnetic waveguide, things are likely more complicated since there are time-varying fields. Also, due to the directions of the fields, the magnetic component of the Lorentz force will tend to move the charges along the length of the waveguide rather than across the cross-section. If you want to include all the physics, then yes this would introduce a deviation from first-order behavior.

And this is the point: The magnetic Lorentz force term, $J\times B$ is a second-order effect, with a prefactor proportional to $v/c$, where $v$ is the velocity of the electrons, and $c$ is the speed of light. You can see this because from Ohm, $J=\sigma E=Nq\mu E=Nqv$ ($\mu$ is mobility, $N$ is free carrier density, $q$ is electric charge), and from Faraday’s law, $|B|=|E|/c$. $$ |J\times B| = N q \left(\frac{v}{c}\right) |E|<<Nq|E|.$$

So the magnetic force is essentially a relativistic correction, which is practically irrelevant for all but very special circumstances.

The broader point is that whether or not you include the magnetic force has no bearing on whether Ohm’s Law is applicable. Ohm’s Law determines what current is formed when you apply a force to charges. This results in $J=\sigma E$ for Coulomb forces. But you are free to specify any force (electric, magnetic, etc.). Then the same procedure leading to $J=\sigma E$ applies, just instead of only using $F=NqE$, you include the other terms.


You say that Ohm's law works well, and that suggests to me that you're considering the current in a wire. In that case, the magnetic field will, initially, deflect moving electrons towards a (curved) surface of the wire, which will therefore acquire a negative charge, while the diametrically opposite surface of the wire, depleted of electrons, will acquire a positive charge. Thus a transverse electric field will arise that will in a very short time, cancel the transverse force due to the magnetic field, allowing electrons to flow as if there were no magnetic field.

Is this too naïve?

  • $\begingroup$ I’m thinking about an EM wave in a rectangular or circular waveguide, for example. That equation is the one used to know currents on the contour metal. In the metal, I think there is no generated eletric field that balance magnetic effects on charges. No? $\endgroup$ – Nameless Dec 7 '18 at 5:58
  • $\begingroup$ I don't see why the reasoning in my answer shouldn't apply to the metal of the waveguide, but it's a long time since I studied waveguides! $\endgroup$ – Philip Wood Dec 7 '18 at 10:46

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