Re: Test charge in a change magnetic field...

On Tue, 18 Nov 2008, delt0r wrote:

Here is my question. Perhaps followup on that page is a better idea.
But here is fine.

http://mathbin.net/2491

To sum up. I have a stationary test charge in a changing magnetic
field. What happen?

In my solution the answer is not unique even in the same coordinate
system. I suspect I need to use vector potential, but without forcing
some gauge on that I still have the same problem.

Perhaps you cannot consider a magnetic field independent from its
cause. But that causes other issues... ie a magnetic field has real
energy density.

Can you have a changing magnetic field Bz = at everywhere? No. Can we have
it over a finite volume? First, just consider a single point.

So, curl(E) = -dB/dt, so dEx/dy - dEy/dx = -a.

So, you have a solution Ex = by, Ey = cx, b - c = -a. As you note, there
appears to be no mistake. You can also add an arbitrary electrostatic
field (i.e., constant wrt time E, produced by a stationary charge
distribution), which would have curl = 0.

OK, note that the magnetic field energy is changing at the point. The
magnetic field energy will be u = mu0 B^2/2, so, ignoring the electric
field energy we can write

du/dt = mu0 a^2 t = - div(S) = -div(ExH) = -div(mu0 ExB)
= mu0 a(c-b)t^2 = mu0 a^2 t as required.

So, we see that your solution gives the required energy flow to account
for the change in magnetic energy. This will be the case even if E at the
point in question is zero.

However, note that S = mu0 at ( cx i - by j ); i,j are unit vectors. This
is in the xy plane, but in a direction which depends on the choice of c
and b.

What might this mean physically?

How do we move electromagnetic energy around? With electromagnetic waves.
What kind of electromagnetic wave will give us a Poynting vector in the xy
plane and a changing magnetic field in the z direction? Answer: a plane
wave with plane of polarisation (i.e., its E field) and direction of
propagation both in the xy plane. Your solution describes an arbitrary sum
of such waves. (Note that the linear time dependence means that they're
not time-harmonic, i.e., sinusoidal, waves.

Which direction the wave(s) will come from in a particular case will
depend on the source - in this sense, you can't consider the field
independently from the source of the field.

Or, as a mathematician might say, it isn't enough to have a solution to
your differential equation - your solution must also satisfy the boundary
conditions.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://espace.uq.edu.au/list/author_id/1189/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
.

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