Re: Is B just that part of E that does no work?
- From: srp@xxxxxxxxxxxx
- Date: Thu, 26 Jul 2007 09:57:11 -0700
On 26 juil, 11:59, blackhead <larryhar...@xxxxxxxxxxxx> wrote:
On Jul 25, 11:25 pm, blackhead <larryhar...@xxxxxxxxxxxx> wrote:
B and E are two fields that act on charge, with the former doing no
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.
After reading the responses, some of my ignorance has been cleared up.
E and B are orthogonal components of the E-M field, and so E being a
part of B is nonsense. So my question should have been "Does the E and
B of the E-M field depend upon v of q used to measure it?"
Absolutely. All three are intimately interdependant
true only for particles moving in a straight lineFrom the Lorenz equation, you have this relation that is
E=vB
or if you prefer v=E/B
I could give you simple equations to calculate this
if you wish.
If you increase the velocity of an electron by any means
whatsoever, even mechanical, the combined E and B fields
of electrons and their carrying energy will increase
accordingly.
I refer you to the Barnett effect on this issue.
There is also the Einstein-de Haas effect showing
what happens when you increase the magnetic field
about a rod of ferromagnetic material
Take the case the E of an E-M field takes where there
is a changing magnetic field. E depends upon the velocity
of q, so that |E| is proportional to |v| with a direction along
the path taken by q and so E is zero when q is static.
If you have an externally produced E field causing q to move,
stopping q from moving by any means cannot reduce the
externally produced E field. Its "pressure" to induce motion
will still be present.
Is it not true that E depends also on the velocity of q
You have to distinguish between an externally produced E
field being used to cause charges to move, which does not
depend on the velocity of q and the intrinsic E field of q,
which does depend on the velocity of q and which is
made up of the combined E field of Q and of its velocity
related energy.
through the EM field of a static Q distribution?
If you talk of the static Q distribution that produces the
external field, that field does not depend on the velocity
of the test q moving in that field, but the reverse is true.
So whereas it may have a static E when a static q is
used, it takes on other Es depending upon the velocity
of q used to measure it.
Yes.
Thanks for your helpful comments in advance.
André Michaud
.
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- Is B just that part of E that does no work?
- From: blackhead
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