Re: Is B just that part of E that does no work?
- From: John C. Polasek <jpolasek@xxxxxxxxxx>
- Date: Thu, 26 Jul 2007 21:30:50 -0400
On Wed, 25 Jul 2007 14:51:20 -0700, blackhead
<larryharson@xxxxxxxxxxxx> wrote:
B and E are two fields that act on charge, with the former doing no(??"B is an E"??).
work. So isn't B just that part of E which does no work? i.e. that
part of E which is normal to the velocity, v, of a charge and so could
be defined as E x v? So every E will have a B that is dependent upon
the path taken by a charge through it.
I want to calculate B for a straight wire using the above idea. I will
calculate the delayed E at a point due to the moving electrons, and
subtract that from E due to the stationary ions which should give zero
since a stationary charge experiences no force in practise. The
problem is why a moving charge, Q, experiences a force if E is zero
there? I suspect it's because it's E will have an effect on the
electrons in the wire because of the finite time it takes to effect
them, which in turn generate an E that effects Q so that an E normal
to v of Q is the net result.
Is this going to be a waste of time, so that I will end up with
something that conflicts with Maxwell's equations?
Thanks in advance.
It would help your case if you attach units to B & E.
That's what distinguishes physics from mathematics. Mistakes become
glaringly obvious. Unless of course you use cgs units then write
anything you want.
.
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- From: blackhead
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