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On Tue, 29 Jun 1999, Bob Sciamanda wrote:((((((((((((((((((((( ( ( ( ( (O) ) ) ) ) )))))))))))))))))))))
Hi Bill,
In order to clarify both of our understandings and clearly identify any
differences, let us begin by comparing our appreciations of the physics
of this ultra-simple situation: A disc magnet sits at rest on a table,
its static B field is vertical. A copper disc sits at rest on top of
the disc magnet. This is the situation as seen from the LAB frame. In
this LAB frame there is a static B field and no E field. There is no
electric force on the free charges in the copper disc (E=0), and because
the disc is stationary in this frame, there is no QVxB magnetic force on
them. There are thus no currents or separated charges in the stationary
copper disc.
There happens to be a second observer going by on a train moving with a
moderate horizontal velocity V (as measured in the LAB frame). In HIS
frame both the copper disc and the magnet are moving with a common
velocity -V . In HIS frame the free charges in the copper disc are
moving in a magnetic field and must experience the force Q(-V)xB'; where
B'=B. But in HIS frame there is also an electric field: E'=VxB. Thus in
HIS frame the free charges in the copper disc are in equilibrium under
the opposing electric (QE') and magnetic (-QVxB') forces, so that he
also observes no currents or separated charges in the copper disc.
Comments?
Aha, I see what I've been missing. qVxB affects charges only when they
are moving relative to an observer, and this "B" is the field measured by
the observer, not the field which the moving charges might measure.
Suppose we hang a test-charge over a large, flat magnet on a table (the
field is vertical.) If we move this magnet constantly in a straight line
under the test charge, the charge responds to an e-field, and we would
also measure the same e-field near that uniformly-moving magnet. However,
since the charge is not moving, there is no qVxB force in this situation.
And conversely, if we anchor our magnet to the table, and then move a
test-charge in a straight line so it passes above the magnet, the charge
responds to the qVxB force because the charge is moving through a b-field.
However, since the magnet is not moving with respect to the table and the
lab, we would measure no electric field in this situation.
I was screwed up about VxB earlier. Are we now on the same page?
William J. Beaty SCIENCE HOBBYIST website
billb@eskimo.com http://www.amasci.com
EE/programmer/sci-exhibits science projects, tesla, weird science
Seattle, WA 206-781-3320 freenrg-L taoshum-L vortex-L webhead-L