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



On Fri, 25 Jun 1999, Bob Sciamanda wrote:

billb wrote:
If a copper disk is spun on axis and immersed in a uniform b-field
perpendicular to the face of the copper disk, and if the leads to a
standard voltmeter are touched to the center of the spinning disk and also
slid along its rim, the meter will measure a real potential difference.
According to the voltmeter, the "radial e-field" is real. . . .

The EMF mechanism here is the QVxB "generator principle" force, not an E
field (I speak in YOUR lab frame). As in any circuit, surface charges will
build up and produce an E field to drive the current around the external
circuit, but the EMF force acting inside the armature is the QVxB force.
(The E field of the surface charges will of course also exist inside the
armature, but this is a "secondary" effect, and in fact will oppose the
current - all exactly as in a "standard" armature generator) There is no
"mystery" here (or there) [:)<

Aha, then I've found a big fuzzy spot in my understanding. It was my
understanding that VxB is itself an e-field: it is the same as the
relativistic e-field that an electron sees as it flys between the poles of
a cyclotron magnet. Wrong?

If we build a spinning disk-magnet device and sit it on the lab bench, and
the magnet is ceramic (nonconductor), then in the lab frame won't a nearby
test-charge "see" an e-field caused by the spinning magnet? At the
surface of the spinning disk-magnet, won't this e-field be radial?

If we place a metal plate close to the spinning magnet and parallel to its
face, won't this radial e-field cause the charges of the metal plate to
redistribute themselves until they produce a cancelling e-field and thus
cease their motion?

If we reverse the situation and spin the metal plate while keeping the
magnet still, won't the charges in the metal plate STILL redistribute
themselves as above? (Here neglecting any Tolman-force motions caused by
centripetal acceleration of mobile electrons in the metal.)


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