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Re: electrodynamic stability?



Take the simple case of two charges travelling along parallel paths (externally
constrained), a distance r apart, at a common non relativistic velocity v:

The B field of one at the location of the other is: B=(mu/4PI)qv/r^2
This exerts an attractive magnetic force (qVxB): F_B = (mu/4PI)(qv/r)^2

The repulsive electric force is: F_E=(1/4PI*epsilon)q/r)^2

The ratio is: F_B/F_E = Mu*epsilon* v^2 = 111*10^-19 * v^2 in favor of
repulsion (independent of r!).

Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net
http://www.velocity.net/~trebor
----- Original Message -----
From: "Ludwik Kowalski" <kowalskil@MAIL.MONTCLAIR.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Saturday, March 08, 2003 2:19 PM
Subject: electrodynamic stability?


| Does it make any sense to think that a very large number of electrons
| (billions) can stay together (despite the mutual repulsion) when they
| circulate rapidly forming a small "toroidal vortex?" The nonuniform
| magnetic field, created by them, is said to produce confining forces.
| It would be like a large "classical" atom without a nucleus (a cluster
| of charges, if you prefer).
| Ludwik Kowalski


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