Re: [Phys-l] induced electric field

[John Denker <jsd@av8n.com>]

On 11/17/2009 11:08 AM, Philip Keller wrote:
> Thank you for the reply -- I will need to read this all more slowly.
> But for now, I hope this is an easy question.  What happens to an
> isolated point charge in a region in space with a uniformly
> increasing uniform mag field?  Does it experience a force?  

Sure.

> If so, in what direction? 

In the same direction as the E field.  F = q E.
For ordinary not-too-weird-shaped regions, the
field will be azimuthal, i.e. in the dθ direction.

>  Does it matter where I release it?

Sure.  The E field is a strong function of position.

The B field is uniform, and dB/dt is uniform, but
that does  not  mean that the E field is uniform.
It means the curl of E is uniform.

Uniform curl E is not the same as uniform E.

  As usual, I can barely bear to express this in
  terms of the vector cross product.  The bivector
  wedge product makes things very much easier and
  nicer.

On 11/17/2009 11:59 AM, Carl Mungan wrote:
> > Off-list I was pointed to the following URL:
> >
> > http://home.minneapolis.edu/~carlsoro/note.htm
> >
> > Looking at the 2nd panel (Note 2), I venture to say that circled step 
> > 2 does not actually give E 

Looks OK to me.

> > because I think E isn't constant along a 
> > circle inside a square B-field region. 

Why not?

Note:  I assume that meant to talk about |E| (not the vector
E) being constant.

There is an implicit assumption in all these discussions
that there are no _other_ imposed fields.  This is a
nontrivial assumption, because the Maxwell equations are
linear, so you can always superpose solutions.  But it
is conventional and reasonable to assume that the E field
is zero at the center of symmetry.  Given that assumption
there is only one solution to the Maxwell equations, and
it gives constant |E| on the indicated circle.



> > ps: The question of how to produce a uniform B inside a square region 
> > (and zero outside) is a different issue, and also not particularly 
> > obvious to me. Just winding a solenoid around a square form 
> > presumably won't do the trick. (Because we then have a similar 
> > problem to the lack of circular symmetry for Faraday's law - except 
> > now for Ampere's law.) I could use a C-shaped magnet with square 
> > cross-section and a tiny gap and ignore (probably to my peril) the 
> > fringing field.

Physically and operationally there is no problem taking
a square form and winding wire around it to make a square
solenoid.  For a long solenoid, not too close to the ends,
it has a B field that is uniform inside and zero outside.
Use the same reasoning you use for non-square solenoids.
Hint: easy Gaussian pillbox argument.  Also there are other 
arguments that lead to the same conclusion.