Re: "Faraday's Disk" which started it all

[William Beaty <billb@ESKIMO.COM>]

On Fri, 2 Jul 1999, Bob Sciamanda wrote:

> More from the hip:
> A magnet in translational motion with a velocity V will have an electric
> dipole polarization P=(1/c)VxM .   EG., if V is in the x direction and M is
> in the y direction,  P will be in the z direction - this is tantamount to a
> polarized dielectric with two bound surface charge layers - positive on its
> upper face and negative on its lower face (planes of constant z). This
> electrically polarized slab has no electric monopole moment (it is overall
> neutral) and no monopole E field.  But it does have a dipole moment (P times
> its volume), a vector in the z direction; and it will exhibit an external
> dipole E field.  (This much is NOT from the hip - this is "conventional
> wisdom".)

This matches my own thinking.


> Now (from the hip) spin a doughnut shaped disc magnet about its own axis.
> Extrapolating the above, I would expect an electrical polarization to exist,
> with bound cylindrical surface charge layers on its inner (say negative) and
> outer (positive) curved surfaces.  Now this electrically polarized slab has
> zero monopole moment and ZERO DIPOLE MOMENT.  It therefore produces zero
> external monopole field and zero external dipole field.  (In fact, it would
> seem that even all higher order fields are also zero, like a charged
> cylindrical capacitor).

OK!  This sounds like a sensible answer to my original question.


I'm still leery of the the rotating motion that each small bit of the
donut-magnet experiences, but perhaps this doesn't produce any EM effects.


If the donut-magnet is not a thin shell, then the tangential speed of the
inner rim will be smaller than the tangential speed on the outer rim.  As
a result, wouldn't the polarization give some bound layers of charge where
the charge on the inner rim is *less* than the charge on the outer rim?
Suddenly I see an obvious way out of this problem.  The mismatched charges
might simply mean that the charge on the outer rim is balanced by opposite
charges distributed throughout the rest of the magnet, rather than a layer
of opposite charge existing exclusively on the surface of the inner rim.



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