I saw this and thought it was interesting. If one can get beyond the pretentious music, super annoying robotic narration and ignore the completely off the wall theory discussion there remains a reasonably well documented experiment. The measurement shows the B field about a conducting cylinder due to an axial current is greatly increased by rotating the cylinder. I would think this additional B is due to a helical component in the drift velocity of the rotating electrons. Anyone familiar with how to compute this addition B?

The convection current is given by
$$\vec{j}=\rho \vec{v},$$
and from this you get the em. field as for any other charge-current distribution using the retarded potential or directly Jefimenko's eqs.

Yes, that's the trivial answer; given the currents one can calculate the fields. This much I know. The issue is what are the currents? For the stationary rod the currents are given by Ohms law as ##J=\sigma E## where ##\sigma## is the conductivity. The internal field, ##E## taken as the voltage drop across the rod divided by it's length. This gives an axial current and hence the B-field exterior to the rod. The question is how is the conduction current modified by the rotation of the rod?

BTW I'm starting to question the validity of the data presented in the video. It's much larger than I would expect based on a simple Drude type picture of the current flow.

I do not understand what's the statement of this movie. I've also found the same as a text, and it doesn't help any further for me too. I'm pretty sure that it should be explainable in terms of classical electrodynamics. I'd start with a calculation for DC. What one has to do is to solve the magnetostatic equations. Since we have a moving body, one should use the full relativistic equations to be sure to get all effects right. You can make approximations that the velocities involved are small compared to ##c## anyway, if it helps to solve the equations. So we have
$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{j}=\sigma \gamma (\vec{E}+\frac{1}{c} \vec{v} \times \vec{B}),$$
where ##\vec{v}## is the velocity of the conduction electrons and ##\gamma=(1-\vec{v}^2/c^2)^{-1/2}##. The latter you can approximate by ##1## I guess. Most likely it's even justified to use the non-relativistic constituent equation (aka Ohm's Law)
$$\vec{j} \simeq \sigma \vec{E}.$$
This neglects the Hall effect due to the magnetic field, which is negligable with high accuracy for usual "household currents" as here. Then one can also make the approximation ##\rho=0##, i.e., neglect the charge density built up due to the Hall effect, which is also usual in treating magnetostatic problems with conductors carrying "household currents".

I've not done the calculation (perhaps I find the time over the weekend), but I'd start with an ansatz
$$\vec{j}=j_{\parallel} \vec{e}_z+\rho \omega \vec{e}_z \times \vec{x}.$$
The first term is due to the applied electric field to get the usual conduction current, the second is the term from the rotation of the conductor around its axis. The magnetic field should be ##\vec{B}=B_\phi \vec{e}_{\phi}+B_z \vec{e}_z## (in usual cylindrical coordinates ##(r,\vartheta,z)##). That's a guess. As I said, I've to carefully check, whether this idea works out.

For the AC case, it should be safe to use the quasistatic approximation, i.e., to neglect the displacement current, but one has to use Faraday's Law
$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$

Hey, thanks. I think if there is an explanation it's classical certainly. For what it's worth I looked at a Drude model for the conduction electrons. I'm pretty convinced that there is a tiny centripetal radial E-field generated inside the metal. This field is needed to stop the radial drift of conduction electrons which must come to rest relative to the rotating atomic background. For a rotating solid conducting rod I get the potential drop between the axis of rotation and the outer radius as,

##V = \frac{m \omega^2 R^2}{2 e}##

where ##R## is the rod radius, ##\omega## the angular velocity, ##m## the electron mass, and ##e## the unit charge. Not that this is particularly relevant to the problem posed but for a 1 cm rod rotating at 1000 rpm I get, ##V= 3 \times 10^{-12}## volts which isn't big.

I think your ansatz is a good starting point. Also the ##v/c## terms are important. Most B-fields are the result of less than perfect cancellation between enormous forces at play inside material bodies. As far as AC is concerned I would like to see the DC case measured using a 3-axis hall probe. I actually have a toy one (phidget 333) which I could use.

Yes, this ##\vec{E}##-field is what I meant when I talked about the "Hall effect". In some sense it's a Hall effect but not due to an external magnetic field but by the magnetic field due to the current under consideration itself. I guess, one has to solve the equations to see, how much of this relativistic effects are needed or not. Do you have a better source about the outcome of these experiments. It's still not clear to me what's precisely measured.

Sorry, I haven't seen any other references other than the video. The more I look at this from a theoretical side the less plausible the "measured effect" becomes. In the DC limit following your ansatz, (kind of), I assume a current (density) induced in the conductor of the form,

##J = \sigma E + \alpha \omega \times r |E|##

where ##\alpha## is a parameter. This lead to an azimuthal current and therefore an axial B-field. A couple well spent minutes with ##\nabla \times B = J## and Stokes law convinces one that the B-field is confined to the interior of the rod which is where the sensor is not. This combined with a sinking feeling the current is for all intents and purposes axial lead me to doubt the results presented in the video. There may be a dynamic effect but I'm sincerely doubtful.