Re: [Phys-l] induced electric field

[John Denker <jsd@av8n.com>]

On 11/23/2009 02:38 AM, John Mallinckrodt wrote:
> On Nov 22, 2009, at 8:36 PM, John Denker wrote:
>
> > I think the shoe is on the other foot.  Some folks are
> > assuming without proof -- indeed without evidence -- that
> > the squarish solution exists.
> >
> > ... proponents of squarish solutions are encouraged
> > to exhibit some such thing, in specific and formal terms,
> > and show that it solves the relevant Maxwell equation in
> > the given situation.  I don't think it exists:  too much
> > curl near the corners of the "square" and not enough
> > elsewhere.
>
> Well, in fact, I actually did do some numerical calculations based on  
> a Biot-Savart-like transformation of Faraday's law and they clearly  
> do support the "squarish solution."

OK.  Works for me.

I owe apologies to a lot of people. 

I was solving the wrong problem.  There *is* a problem 
for which the symmetrical solution works, but it isn't
the problem that was asked.  I can explain what that is,
if anybody is interested, but it would be a digression
and I don't want to hijack this thread.  I've caused
enough confusion already.

I now agree that the method and the answer at
   http://home.minneapolis.edu/~carlsoro/note.htm
are wrong for the square solenoid.  For one starters,
it is incomplete;  it doesn't say what happens when
r is greater than L/2 and less than sqrt(2) L/2.

A field of the form 
   E = x j + y i                   [1]
has the right amount of curl, but does not solve 
the given problem (square solenoid).  You can add 
to equation [1] any curl-free field and get another
expression, and in fact equation [1] differs from
the right answer by a complicated but curl-free
field.



> It shouldn't surprise anyone that the solution we seek is identical  
> in form to that for the B-field produced by a uniform current density  
> within a region having a square cross section.  For a quick and dirty  
> solution I set up a spreadsheet with a 20x20 array of 400 "wires"  
> carrying "current" in the +z direction and used it to calculate the  
> resulting field at arbitrary positions in the x-y plane.  

I assume the quote "wires" are really sub-regions within 
the solenoid and the quote "current" is really d(flux)/dt.

I'm OK with that. As Feynman liked to say, the same equations 
have the same solutions, and finding the E-field in the 
vicinity of some d(flux)/dt is isomorphic to finding the 
B-field in the vicinity of some current-carrying wires.

> Inside the  
> square, the solution is a little too too susceptible to the varying  
> distance to the nearest wire to be reliable, but outside the solution  
> is quite stable and smooth.

Hmmm, I ought to try this.  I would have thought the method
would work even inside the wire.

Suppose we divide the big solenoid into N^2 cells.  The
contribution from each cell falls off like 1/r, which is
nasty when r is small, but really we are integrating 
(1/r) [r d(r) d(theta)], which is is no problem at small
r.  In fact it is the slowly-converging contributions 
from large r that cause the most trouble, in the limit
of large N.

To say the same thing another way, the distance-dependence
of the nearest cell goes like 1/(L/N) which is nasty when N
is large, but the strength of the source goes like (1/N)^2
so all-in-all the method gets less sensitive to nearby
cells when N becomes large.

You can stabilize the numerics further as follows:  If the
rows (and columns) are labeled 1 though N inclusive, as
long as you are not in row (or column) 1 or N, you can
ignore contributions from "this" cell *and* from the 8
nearest cells, because it all cancels by symmetry.  As
you get deeper into the interior, you can ignore more,
so the solution becomes more and more stable.

That is to say, the interior field is the exterior field
of a smaller solenoid.  (This is true for round solenoids
too, not just square ones.)

==============

Another possibility to consider is two-dimensional Biot
Savart, so that the only source term is the current 
sheet on the boundary of the big solenoid.

I'm not going to have much time to work on this;  I've
spent most of the last week in the hospital.  I'm not
the patient, but still it eats up a lot of time.....