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Paul Colby

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- #1

Paul Colby

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$$\vec{j}=\rho \vec{v},$$

and from this you get the em. field as for any other charge-current distribution using the retarded potential or directly Jefimenko's eqs.

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Paul Colby

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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.

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$$\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{E}=\rho, \quad \vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0, \quad \vec{j}=\sigma \gamma (\vec{E}+\frac{1}{c} \vec{v} \times \vec{B}),$$

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

$$\vec{\nabla} \times \vec{E}=-\frac{1}{c} \partial_t \vec{B}.$$

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Paul Colby

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##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.

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Paul Colby

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

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