Re: rotating magnetic field

On Apr 16, 12:44 pm, Benj <bjac...@xxxxxxxxxxx> wrote:
On Apr 16, 6:58 am, pmb <pm...@xxxxxxxxxxx> wrote:

v is the velocity of the charge WITH RESPECT TO the field.

That is incorrect. The v is the velocity of the particle with respect
to the observer's frame of reference.

Lets say I have an electron and it's moving in some magnetic field at
rest (lab frame). We observe a sideways thrust on it due to the qVxB.
OK?

Let’s be more precise. In the inertial frame of reference S let there
be a Cartesian set of spatial coordinates (x, y, x). In frame S let
the electric field be zero and let there be a constant and uniform
magnetic field pointing in the +y direction. Let the velocity be in
parallel to the x-axis and in the +x direction. Let the charge be
positive. Then the force will be given by qv x B and therefore it will
therefore point in the + z direction.

Now I sit on the particle and move with it.

Let S’ be the rest frame of the charged particle. S’ is moving in the
+x axis and let its spatial axes x’, y’, and z’ be parallel to x, y,
and z. S and S’ are then said to be in standard configuration with S.

Obviously since as
you say the field has not motion and has not relation to the V
producing the force, it must be that v is now zero as the particle is
not moving with respect to me. Hence no sideways force exists.

That is incorrect. You are neglecting to take into account the Lorentz
transformation of the electric field. Most texts on electrodynamics
explain this fact, usually in the section on Faraday’s law. In fact
you can get an approximate value (i.e. the value when c << c) by
demanding that the force has the same value and direction in S’ that
it does in S
(in Newtonian mechanics force is a Galilean invariant). In this case
the electric field, as measured in S, has the same value in S. The
fact that the electric and magnetic fields have different values in
different inertial frames is important to know in order to understand
what is going on with the homopolar generator.

Pete
.

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