Also note the boo-boo in the equation for F_E;
the relevant line should read:
The repulsive electric force is: F_E=(1/4PI*epsilon)(q/r)^2
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Bob Sciamanda" <trebor@VELOCITY.NET>
To: <PHYS-L@lists.nau.edu>
Sent: Saturday, March 08, 2003 3:35 PM
Subject: 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
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