According to my EM professor, an infinite line of current $I$ creates only a magnetostatic field (the regular $\vec B=(\mu_0 I\vec /2\pi r)\vec e_\varphi$ you get from applying Ampère's law), and not an electric field. I don't understand this statement since if I place myself in the inertial frame of reference of the moving charges, then I wouldn't register any current, just static charges, which would of course give place to an electrostatic field: $\vec E'=(\lambda'/2\pi\varepsilon_0 r')\vec{e}_r$. Therefore, if my professor's statement were true, then the invariant $E^2-c^2B^2$ would be violated when placing myself in the frame of reference of the moving charges. Also, the quasistatic approximation would not hold even for small intensities. All this does not make sense to me. Is it really true that a line of current does not give place to an electric field?
3 Answers
When we speak of a current-carrying wire, we usually mean a wire that is electrically neutral (as most wires are). There are as many positive charges as there are negative charges. The net charge is zero and there is no electric field. The electrons are flowing so there is a magnetic field. However, if you only have a single line charge flowing, then the net charge density is nonzero and both fields will be present.
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The phrasing of your professor's question is unclear, as it talks about an infinite line of current, which is, ofc, charged. So the implication in the question is:
There are two linear charge densities of $\pm \lambda$, with the positive one moving in the positive direction at $\beta$.
Note: please do not think of this thing as a real physical object, which in this case, would be a fixed lattice composed of antimatter copper ions and free positrons moving in the $+x$ direction...all that does is cause confusion. We are dealing with abstractions: co-located infinite linear charge densities.
So now we need to introduce observers and frames. They are Alice in the lab ($S$), and Bob, moving at $+\beta$ in $S$, which is his $S'$ frame.
(I am going to set $c=\epsilon_0=\mu_0=1$; feel free to add them--it's kinda cute how the mu's and epsilon's work out to be $c$).
In $S$, we have two charge distributions, which we can add:
$$ \lambda = \lambda_+ + \lambda_- = \lambda((+1) + (-1)) = 0 $$
And two currents, which we add.
$$ j = j_+ + j_- = v_+\lambda_+ + (0)\lambda_- = \beta\lambda $$
Note that the conservation equation:
$$ \frac{dj}{dx} - \frac{d\lambda}{dt} = 0 - 0 = 0$$
is trivially satisfied, making this both an electrostatic and magneto-static problem with:
$$ E = 0 $$ $$ B = j = \beta\lambda $$
where both fields are the so-called perpendicular components, with $E$ being radial and $B$, azimuthal. I'm not worrying about the $1/r$ parts. Note:
$$ (E^2-B^2) = -\beta^2\lambda^2 $$
Ok, now transform the charges and currents to Bob's frame.
$$ \lambda' = \lambda'_+ + \lambda'_- $$
The negative charge is contracted, while the positive charge is dilated (see Bell's Spaceship Paradox if that troubles you):
$$ \lambda' = \lambda_+/\gamma + \gamma\lambda_- $$
$$ \lambda' = \lambda/\gamma - \gamma\lambda$$
$$ \lambda' = (\frac 1 {\gamma} - \gamma)\lambda = -\gamma(1 - \frac 1 {\gamma^2})= -\gamma\beta^2\lambda$$
So Bob sees a negative charge density, giving an electrostatic field:
$$ E' = -\gamma\beta^2\lambda $$
For Bob, the current is:
$$ j' = v'_+\lambda'_+ + v'_-\lambda'_- $$
and by definition, $v'_+=0$, and by symmetry, $v'_- = -\beta$, so:
$$ j' = (-\beta)(\lambda'_-) = \beta\gamma\lambda $$
Bob sees a negative current going backwards, giving a positive magnetic field:
$$ B' = \beta\gamma\lambda $$
so now the invariant is:
$$ (E'^2 - B'^2) = \gamma^2\beta^4\lambda^2 - \beta^2\gamma^2\lambda^2$$
$$ (E'^2 - B'^2) = \gamma^2\beta^2\lambda^2(\beta^2-1)$$
$$ (E'^2 - B'^2) = -\beta^2\lambda^2 = (E^2-B^2)$$
QED.
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Yes, a single moving charge does produce an electric field as well as a magnetic field (although it isn’t really correct to describe that electric field as “electrostatic”). But a typical current-carrying wire is electrically neutral - there is just as much positive as negative charge per unit length in the lab frame - so the electric field produced by the negative charges exactly cancels the electric field produced by the positive charges, and there is no net electric field.
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