Re: Faraday "magnetic" induction without a magnetic field.

On 14 May, 06:01, Benj <bjac...@xxxxxxxxxxx> wrote:
On May 13, 4:13 pm, blackhead <larryhar...@xxxxxxxxxxxx> wrote:

On 11 May, 06:54, Benj <bjac...@xxxxxxxxxxx> wrote:
The magnetic vector potential, A, is seen as more fundamental than B.
If Faraday's law used A instead of B, would that be better for you?

Faraday's law DOES "use" A and also "uses" B. What you don't get is
that a current density J creates BOTH a retarded magnetic field B and
a retarded A in space.
It ALSO creates a retarded electric field E
(electrokinetic) in space as well. Note that since all these things
are retarded: The curl of A EQUALS B but A does not CREATE B. Also
while the rate of change of A is equal to the negative of the Electric
field, A does NOT "create" E. The current source J creates A, B and E
all of which are retarded because of the speed of light. Hence the
true Faraday relationship is between J and E even though E is
"measured by" [to use Maxwell's words] the rate of change of A. It is
J that is the source that creates the E. The source of E is NOT B nor
A!

OK, but Faraday's Law establishes a relationship between the rate of
change of B in a closed loop and the induced EMF around that loop that
makes engineers' and physicists' lives easier. You could develop a law
using just d/dt(J(t -R/c)) to calculate E(R) but it would involve the
calculation of d/dt J at reatarded times which looks hopelessly
complicated except for simple problems.

You could argue the idea of the magnetic field is bogus as well
because it arises as a result of the retarded nature of E and
Galillean invariance.

What does this "electrokinetic E" field do which the Magnetic Vector
Potential doesn't?

Accelerates charges for one thing!  Just as any true E field would. We
call it "electrokinetic" however, because it's properties are quite
different from a static E field.

I suspect you mean it increases the KE of charge and is related to the
part of the retarded E that does work on charge, the orthogonal part
being B and just changing its direction.

If a wire of 0R carries a finite current, then the 2 points of the
wire are indeed at the the same potential.

This is obviously a totally false statement. In fact this is the whole
error that is the basis of the Lewin scam. Wrap a single turn about a
solenoid and cut it at one point. Measure the voltage across the cut.
It's 1 volt.
Now put a 1 megohm resistor across the gap. Measure the
voltage again. Duh. Still 1 volt and with 1 microamp flowing. You DO
agree that 1 microamp is a finite current, right?

I was wrong using the idea of potential at a point because it's not
uniquely defined. It is OK, however, to say what E_total.dl and hence
the voltage is along a path. The voltage "across" two points in a non-
conservative field is meaningless because you still have to define the
"across" path.

So in your example, the voltage "across" the cut using a path through
the resistor/cut will be 1v, taking a path through the loop it will be
0V, taking a path through the voltmeter loop it will be 1v, if the
measurement loop area is kept small to minimize the flux enclosed.

 What if we make the
loop superconducting? Does the current or voltage change? No they
don't. What if I make the resistor smaller? Well more current flows
but the voltage stays the same. Eventually I can get so much current I
can drop the loop out of superconductivity, but the voltage does not
change until I do because the wire has no resistance.

Yes, and this E consists of an internal E_int of the wire opposing the
induced E so that the current through it satisfies the boundary
condition. If there was no opposing E, then only the induced E would
exist and the current through it would be infinite in the ideal case,
very large in the practical case.

So you are saying a shorted turn on a transformer does NOT have a lot
of current flowing in it?  Duh.

No. I'm saying that if a section of conductor has an E_induced inside
that gives rise to say 100A flowing through it, while the rest of the
circuit requires 1A, then the charge distributes itself in the circuit
to create an E_oppose that forces 1A to flow through that conductor
and so maintaining charge continuity through out the circuit.

Look how much E does it take to get currents flowing in a material
(wire).

Formula is:  J = E/Rho  Rho is resistivity.

So what happens if we want to make resistance very very low (we won't
go quite to zero here)? Well Rho has to become very very small and so
does E because the ratio of E/Rho is a constant giving a certain
current. So thus with R = 0 a virtually non-existent E field can give
rise to enough current to drive the loop out of superconductivity.
Thus it is OBVIOUS that for R=0 there is the induced E field and
virtually NO "opposing E".

Eh? If R = 0, then to keep J finite, E = 0. So the opposing E has to
cancel the induced E.

Hence the full 1 volt will appear across
the "gap R" no matter how small we make it.

Be careful using "across". The voltage through the gap/resistor will
be 1v, yes.

Now if we make the
resistance of the wire higher and higher while drawing heavy current
then there will be voltage drop in the wire which will create your
"opposing E" and the voltage at the gap resistor will be lower than 1
volt.

It's the complete opposite. If the conductor is perfect, then E inside
is virtually 0, to drive a finite current through it, and this is
because E_oppose cancels the E_induced. Look at your working above. As
you increase the resistance of the wire, E_oppose falls because the
total E inside has to increase because more energy is needed to get
charge through the resistance.

(The way the output voltage drops on a heavily loaded
transformer secondary compared to the OCV)

9 Thus it follows that heavy copper wire in this case is NOT a "heavy
shorting bar". To think that it is creates the paradox.

This isn't true because of my answer to your point 8.

Is true because your answer to point 8 is dead wrong.

Even if there is an induced E in the leads, this will be cancelled by
the internal E to maintain a small I through it.

Just as wrong as your point above.

You seem to have a problem with understanding that E inside a
conductor drives charge through it at a certain rate.

No you do. I just gave you the equation for it. And it proves that you
are saying erroneous things.
You need to understand how conduction of current works before trying
to understand something complicated like this problem.

And you should try learning a little classical field theory before you
start accusing others of being ignorant.
Ahem, maybe reading a freshman text or two might help... :)

So do you agree that a conductor carrying 1A has the same total
internal E to drive the charge through it, regardless of the sourse of
E?

.

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