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Re: ORIGINAL simple magnets question

On Fri, 25 Jun 1999, John S. Denker wrote:

First answer: When all else fails, look at the Maxwell equations.
Trying to do electrodynamics without Maxwell equations would be like
teaching a snake to walk.

But Maxwell's equations don't describe the forces on a conductor, or on a
charge moving through EM fields.


The magnetic field has a source term for changing electrical fields, but
by symmetry we don't have any of those in this situation. The magnetic
field has a source term from flowing charges, but by hypothesis (c) we
don't have any (unbalanced) flowing charges, so we are left with zilch.
It's that simple.

Because "Faraday's Disk" does produce real-world currents, I suspect an
error in the above. Ah: if we ride along with an electron as it flys into
a cyclotron's gap, we see no flowing charges anywhere. Or if we keep the
electron in the lab frame, but move the whole cyclotron, we see no flowing
charges anywhere, yet the electron is pushed sideways. The question still
is this: does the rim of a spinning disk-magnet behave anything like the
pole-pieces of a uniformly-moving cyclotron magnet?


Second answer: In order to make this result more intuitive, please
consider the particular case where the magnet is an electromagnet,
consisting of a 1-amp transistorized constant-current source feeding a
metal ring. We build two of the things. We rotate one of them, current
source and all, in the plane of the ring. In the (slightly
non-inertial) frame of the rotating current source, it continues to put
out one amp. In the lab frame, the rotating magnet's electrons are
circulating slightly faster, so the electron-current is larger. On the
other hand, the ions of that metal ring are circulating in the same
direction, creating a countervailing current. Therefore to first order
in v/c, the current is the same, and the magnetic field is the same
whether or not we spin the ring.

Correct. But if I use the above reasoning, I would end up saying that the
b-field in a moving cyclotron pole-piece has the same strength as the
b-field in a stationary pole-piece, and I'd totally miss the fact that a
cyclotron, in uniform motion relative to the lab frame, creates a
relativistic e-field. In your thought-experiment with the metal rings
above, if we rotate one of the rings, then *maybe* the b-field of that
rotating ring behaves like the b-field of linearly-moving conductor, or of
a linearly-moving cyclotron: it does remain constant, but a transverse
e-field still appears.


Let's change your thought-experiment a little. Suppose I have an infinite
uncharged straight wire with a 1-amp current. If I launch an electron
parallel to that wire, the electron sees a constant b-field, therefor it
won't be deflected. NO! The electron should actually see a radial
e-field because of the relative motion between itself and the wire. If
the electron should move in the same direction as the conventional current
in the wire, the electron will be repelled away from the wire, and if the
electron moves the other way, it will be attracted towards the wire. Or
in the frame of the electron, THE LINEAR MOTION OF THE WIRE IS IMPORTANT.
It is important even though the electron never sees the b-field change.
So, is the tangential motion of a circular wire also important to a nearby
electron? If the diameter of the circular wire is large, and if the
electron is close to the wire, then the electron should "see" it as being
like the infinite wire which is moving in a direction along the axis of
the wire. I conclude that, in the frame of an electron, a rotating
current-loop is different than a non-rotating current-loop. There are
e-fields involved.

To increase the generality of the argument, consider the following
arrangement of current loops, where each Q is a small loop; the tail on
the loop is meant to designate the constant-current source:


QQQQQ
QQQQQQQQQ
QQQQQQQQQQQ (1)
QQQQQQQQQ
QQQQQ


The foregoing argument applies to each of the little loops separately.


OK, how does this affect a single, moving electron? What if we have a
rectangular magnet composed of these current-loops:

QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ ----> moves
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ

If we hold an electron stationary above the pole-face of this magnet and
move the magnet as indicated, then the electron will experience a force.
Or, if we hold the magnet still and launch an electron so that it skims
across the face of the magnet (so it only sees the fairly uniform b-field
there), the electron will experience a force.

Next, what if the above magnet is part of a large, spinning, ring-magnet?


QQQQQQQQQQQQQ QQQQQQQQQQQQQ ___--->
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ ___---
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ spins
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQ
QQQQQQQQQQQQQQQQQQQQQQQQQQQQQ


It seems obvious that a stationary electron near the surface of such a
magnet would experience an e-field caused by the relative motion between
itself and the current loops. Since the b-field points out of the page,
and the motion is horizontal, then the e-field would be oriented
vertically in the above diagram. If the magnet was a disk rather than a
flat ring, would the e-field still be there? Faraday's Disk experiments
*seem* to suggest that it would: that disk magnets cause radial e-fields.



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