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Re: [Phys-l] induced electric field



On 11/23/2009 04:15 PM, LaMontagne, Bob wrote:
Instead of the differential form of Maxwell's equations, I find the
integral form a little easier to visualize here.

That's OK if you are willing to do a lot of work,
and carefully keep track of the details. Do not
try to integrate the field to find the potential.
In this problem, there is a perfectly good field,
but it is not the gradient of any potential. The
integrals in question are all path-dependent.

If I take your question as-is and don't turn it into a question that
may be easier to answer, I have to conclude that the question is not
answerable. If I form a circular loop in the x-y plane and integrate
around that circle, I can easily find a value for Integral(E dot ds)
= - d(B_flux)/dt. The result is dB/dt times the area of the circle.
The EMF around this loop has this same value. If a circular loop of
wire is placed in the same position as the loops we integrated over,
the current through the wire loop will be the EMF/R (where R is the
resistance of the loop). Since no boundaries are specified, I can
move the loop to any position that I want in the x-y plane and I will
get the same current in the same direction (clockwise or
counterclockwise). If I now take two loops of wire and place then so
their edges are almost just touching, the current in one is flowing
in the opposite direction as in the other at the point where they
almost touch. So if we are thinking in terms of electric fields
pushing an charge, the field obviously has two opposite directions at
the same point - depending on which loop I am looking at.

That last "obvious" conclusion is unobvious and
indeed untrue.

There is only one E-field at any point. Any stationary
test charge will get pushed in the direction of the E-field.

Knowing the total "EMF (total push) around a loop does not
mean that the push is evenly distributed around the loop.
In this situation it is verrrry unevenly distributed.

Draw some loops on top of this field diagram:
http://www.av8n.com/physics/non-grady.htm#fig-betatron
and observe that
a) At any particular location, the push is in a
definite direction
b) The total push around any loop is always clockwise.
The gears do not mesh.

This diagram is the answer to a slightly different
question, but it suffices to show that non-meshing
gears are not absurd.

Not every field is the gradient of some potential.

I take this
absurdity to indicate that the question cannot be answered exactly as
it is posed.

The original problem was slightly underspecified, but
it is not absurd. Mild assumptions suffice to allow
the question to be answered in quantitative detail.

Not every field is the gradient of some potential.