Direct calculation shows that the result
F_B/F_E = (v/c)^2 holds even in the relativistic case.
In sum, it can quite readily be shown from: F_E=qE, F_B=qvxB and B=(vxE)/c^2
(As before, I take the particles to be travelling directly across from each
other along parallel paths.)
So, unless other phenomena intervene, such a particle pair experience a net
electromagnetic repulsion.
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 10:40 PM
Subject: Re: electrodynamic stability?
| I don't know whether to trust this reference (www.awe.co . . .).
|
| Look at the bottom of page 2 of:
| http://web.mit.edu/6.014/www/handouts/lecture21_notes.pdf
|
| This makes no reference to any different result for the relativistic case,
| noting only that the plasma must be neutral for the pinch effect to win.
Given
| time I (you, too) will look into calculational details.
|
| My guess is that the pinch effect wins in the "awe.co diode" for reasons (of
| construction) other than just relativistic velocities.
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