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

[William Beaty <billb@ESKIMO.COM>]

On Thu, 1 Jul 1999, Bob Sciamanda wrote:

> For details of how a current loop (a magnetic dipole) is also an electric
> dipole when viewed in motion, see pg 340 of Panofsky & Philips (1rst ed).
> Essentially they show that an upright, rectangular current loop carrying a
> ccw current, when viewed moving to the right develops a positive charge in
> the upper leg and a negative charge in the bottom leg - an electric dipole -
> because of the different carrier velocities in the two legs.

This agrees with my own understanding.  Because of Lorentz contraction,
the two parallel legs develop opposite charges, and an e-field appears
between them. (Even so, I can't clearly visualize what happens at the
corners of such a loop, where the altered charge-density of the
parallel-moving wire connects to the UN-altered charge density of the
perpendicularly-moving wire.)

As was true at the start of "simple magnet question", my main troubles
arise when the motion is rotational.  If the two parallel legs of the
square current-loop do not move at the same speed, does one leg become
less charged than the other?  Or a similar question: if we begin to rotate
a circular current-loop about its axis, does the charge density in the
loop appear to change because of Lorentz contraction?  Does an uncharged
current loop appear to become charged if we force it to rotate? Such a
thing is impossible.  Yet I cannot see any obvious way to avoid it.  If
there is a solution to this, perhaps it is outside of SR.


> > In other words, a cyclotron can deflect an electron because, in the
> > inertial frame where the electron is (momentarily) stationary, all the
> > little dipoles in the cyclotron magnet are moving, and together they are
> > creating an e-field which the electron can experience.  Is this an
> > incorrect model?
> The cyclotron is best understood in the lab frame (the rest frame of the
> cyclotron, its magnet, the operating personnel, etc - the frame which
> inspired its invention).

Then we have a disagreement in philosophy.  I want to obey SR and insist
that the "lab frame" can vary:  move a magnet past a test-charge, or move
a test-charge past a magnet.  In either case a force will arise.  I keep
bringing up the cyclotron because it is a clear example of a system where
a test-charge is deflected as a result of the *relative* motion between
a test-charge and a magnet, yet the b-field is uniform.  If the force
depends only on the *relative* motion between the two, then we must not
shy away from holding the test-charge stationary and moving the magnet
instead.


>  In this frame the electrons (while inside the dees) are subject only to
> a vertical magnetic field and so are forced into a horizontal, circular
> orbit by the QVxB magnetic force.

If I (now) understand things correctly, the qVxB force depends on the
frame of the observer: the force q(E + VxB) is "real" and unchanging, but
if the observer jumps to a different inertial reference frame, then E and
VxB will vary (although their sum will not.)

I was confused earlier because E does vary as the relative motion of the
magnet varies, but this is not specifically expressed in the equation above.

> Things are much more complicated when viewed from the non-inertial rest
> frame of the electron and no more insightful understanding of the apparatus
> is gained for your trouble.

True.  When discussing an electron which moves between the cyclotron's
poles, we must focus on the forces felt by an electron *at a particular
instant.*  Either that, or assume that we could restrain the electron so
it moves uniformly, and then we could measure the perpendicular force.


> > Finally, what happens when we dangle a test charge above the flat face of
> > a *rotating* disk-magnet, where the plane of the disk is horizontal, the
> > b-field points upwards, and the test-charge is not aligned with the axis
> > of the disk?  Won't this test-charge experience an e-field and therefor a
> > force? After all, that test-charge is close to billions of tiny moving
> > dipoles within the magnet, and in the frame of that stationary
> > test-charge, each of those dipoles creates an e-field.  If this is wrong,
> > where does my error lie?
>
> What you say is true.  The rotational case is not easily analyzed, eg. by a
> simple Lorentz transformation, and I confess that I do not have a detailed
> understanding here.

Since this apparantly is a controversial topic, then there is a small
chance that *nobody* has a detailed understanding.  Maybe not a "hole in
physics", but a foggy region which needs some illumination.


>  It seems that the external field will be very much
> weaker than in the translational case, as I proposed in an earlier post.

I probably misunderstood.  I thought you had that the e-fields of moving
magnets in general would be limited to the physical substance of the
magnet.  Yet this contradicts q(E + VxB) if the test charge is stationary
and the magnet is moving.  If the e-field outside of a linearly-moving
magnet is strong, then MAYBE the e-field outside of a rotating magnet is
also strong.  Or maybe it is weak as a consequence of the geometry of
rotating systems.

A thought... If we magnetize a spherical shell such that the direction of
magnetization is radial, we do not obtain a monopole!  Instead the
external fields sum to zero.  Perhaps something similar applies to the
e-field of a rotating disk-magnet.  (If so, then the concept-network by
which I understand Homopolar Generators is totally wrecked.)


>  A
> lucid, detailed analysis of this is hard to come by, and what is available
> is often questionable.  You have provoked me to make the time to seriously
> search for a better understanding here.

I hope those references in my other message end up shedding some light
rather than just deepening the controversy.


> Anyway, experimentally a stationary
> copper disc and a rotating disc magnet will NOT work as a generator.  The
> homopolar generator only works when the QVxB force is used by rotating a
> conductor in the magnetic field (as noted before, this conductor can be the
> magnet itself.) So, experimentally this external E field is zero or very
> weak in the rotational case.

I must disagree.  True, if we rotate only the magnet, then there can be no
"generator effect" and no current.  This doesn't prove that there is no
e-field around the rotating magnet.  If a cylindrical/radial e-field does
exist, then if we take the line-integral of the force it applies to each
charge in a nearby closed circuit, the force should sum to zero, and
therefor the e-field cannot create a direct current.  Current can only
flow in the generator circuit when there is relative motion between the
conductive sections:  motion as measured between and the wires of the
external circuit the conductive disk. If this is true regardless of the
strength of the e-field, then a Homopolar generator says nothing about the
existence of that e-field.


> I do recall someone (a very early author) calling the unipolar generator
> (simply brushes on a rotating magnet) the "electromagnetic version of the
> Foucault pendulum" since it shows that "absolute" rotation is measureable.

I think this person is wrong.  To generate a current, relative rotation
between the "conductor disk" and the "external circuit" is necessary.  If
we keep the "external circuit" stationary and ignore that we have done so,
then it will *appear* we can detect the absolute rotation of the
conductive disk.  If we hold the conductive disk stationary and rotate the
brushes, the meter, and the external circuit, we still obtain a current.
This shows that the Homopolar Generator does not rely on absolute
rotation.  The rotation is always a relative motion between a "rotor" and
a "stator" section.  The meter and its leads are the "stator".  Hold the
"rotor" still and spin the "stator" around it, and the generator still
functions.



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