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



On Sat, 3 Jul 1999, Bob Sciamanda wrote:

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.


I believe each magnetic bit m acquires an electrical polarization p.

Right, but by "rotation" I don't mean tangential translation. Each small
volume in the spinning disk moves tangentially and also accelerates
inwards. It also rotates. This rotation is about its own axis, not about
the axis of the entire disk. To calculate the e-field at any point near
the spinning magnet, we would integrate the fields contributed by each
small volume of the magnet. I assume that the *tangential* motion of each
small volume would produce an e-field. Would any additional e-fields be
produced by their inwards acceleration, or by the axial rotation of each
small volume? For a real-world disk at realistic velocities, I suspect
the answer is no. I assume that we need only look at the tangential
motion.


But, in the rotational situation the numerous little vectors p are so
arranged that the net P presented to the outside world is zero.

Are you certain about this? If the polarization of the rotating disk is
equivalent to a charged rim which surrounds an oppositely-charged center,
then the e-field must extend outside the volume of the disk. Yes, such a
charge distribution has no net dipole moment, and its e-field must drop
off fairly rapidly at a few radii from the disk. But for distances within
a few radii, it certainly would produce a significant external field. I'm
visualizing it as follows:




+--------------+ cross-section of spinning
charged disk




__ __
_/ \ / \_ disk with
/ _ \ / _ \ e-field lines
\ / \ \ / / \ /
------ +-------------+ ------
/ \_/ / \ \_/ \
\_ / \ _/
\__/ \__/



The field-lines radiate from the concentration of "charge" at the edge of
the disk, then they curve around and dive diagonally into the distributed
opposite "charge" on the flat faces of the disk.


This is due to an
ORDERED, cylindrical arrangement; but the external result might be compared
(and contrasted) to a ferromagnet above its Curie point, where the zero net
external effect (magnetic in this case) is due to a RANDOM arrangement.


At a distance from an unmagnetized magnet, the fields do sum to zero, but
if we examine the iron at a scale where the domains appear large, then
we'd detect external fields at the surface of the iron. This is similar
to the spinning disk: the fields are zero at a large distance, but if the
rim is charged one way and the rest of the disk is charged oppositely,
then the fields are non-zero within a few radii of the disk.

Another way to see it: the external fields around the disk would be zero
only if the internal electric dipoles were arranged end-to-end IN A
CIRCLE. An "electret" has a large external field because the lines of
polarization point outwards through the surface of the material. In the
spinning disk, the electric dipoles at the rim are pointing outwards
through the surface. In the central regions of the disk, there are no
lines of polarization which turn around backwards to connect with the rim.
Therefor we should find that an e-field flux extends out of the central
volume of the disk-magnet and loops around to connect with the outer rim.


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.

I'm not sure I follow these interesting speculations, but let me repeat the
substance of a discussion we had on an earlier thread. These "surface
layers of bound charge" are calculational artifices invented to ease the
evaluation of the field/potential produced by a volume distribution of
dipole polarization. They not only ease calculations, they also allow an
attractive EQUIVALENT conceptual model.

Not speculations. If we model the disk as a series of electrically
polarized cylindrical shells, then we make a mistake. We ignore something
important: that the direction of polarization is radial rather than
parallel. I'm looking at the situation in terms of flux. For each thin
cylinder, all of the e-field flux which exits from the inner surface of a
cylinder *cannot* enter the outer surface of the next smaller cylinder.
The polarization of the next cylinder is smaller: the next cylinder has a
smaller circumference and is also moving more slowly. As a consequence,
field lines must "leak out" of the disk's surface. Lines of polarization
which dive towards the center of the disk cannot simply vanish. They must
escape from the surface of the disk as flux. They reconnect with the
flux which exits from the disk's outer rim.

Any models upon which you may speculate to give the
same result may be equally useful - the actual volume distribution of
polarization in this rotational case may be equally served by a myriad of
"equivalent monopole charge distributions".

Yes, the monopole charge distribution is an artificial mathematical tool.
If we only want to calculate the external e-field of the disk, then it is
an appropriate tool. In addition, it is a very useful tool: it instantly
highlights something that we otherwise might miss. If we look exclusively
at the polarization in a tiny volume, then we might miss the fact that
flux *must* escape from the central regions of the disk and connect to the
flux which extends from the rim of the disk.


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