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

[Bob Sciamanda <trebor@VELOCITY.NET>]

----- Original Message -----
From: William Beaty <billb@ESKIMO.COM>
To: <PHYS-L@LISTS.NAU.EDU>
Sent: Wednesday, June 30, 1999 5:42 PM
Subject: Re: "Faraday's Disk" which started it all


> . . .
> Does leaping between inertial frames cause us to begin seeing an e-field?
> To me it seems equivalent to say that the cause of the e-field is the
> velocity of the sources relative to the frame in which the measurement is
> made.  If each microscopic dipole in the above magnet is moving with
> respect to my frame, then I would discover that each dipole has a tiny
> e-field associated with it.  These small regions of e-field superpose, and
> the little e-fields add up to an extended region of fairly uniform e-field
> which exists in the space above the face of the moving magnet.
> . . .
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.
> . . .

> 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).  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. In the space between
the dees (twice a revolution), they are momentarily accelerated by a
horizontal E field; V increases and they spiral into a larger radius orbit.
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. I am sure that the device could not have been
easily conceived or designed if one thought only from the electron rest
frame.  (Something like trying to understand the solar system by viewing it
from earth, rather than from the sun, only much worse.) But it is true that
in the electron frame the magnet would produce an E field.

> My conclusion: if we move a large, flat magnet in a direction along the
> plane of its pole faces, then in the region above the pole face we will
> detect an e-field.  The direction of the e-field will be perpendicular to
> the magnet's direction of motion (this assumes that the b-field above the
> magnet is everywhere perpendicular to the pole face.)  If we dangled a
> stationary, very strong test-charge above this moving magnet, the
> test-charge would be visibly deflectd by this e-field.  Also, an observer
> who moves along with the magnet will not measure any e-field, but would
> explain the force on the test-charge as arising from the Lorentz force
> qVxB. All OK so far?

Yes

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

-Bob

Bob Sciamanda
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor