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

On 15 May, 07:52, " Don Kelly" <d...@xxxxxxx> wrote:
--
Don Kelly
d...@xxxxxxxxxxxx
remove the x to reply"blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

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On 14 May, 06:20, " Don Kelly" <d...@xxxxxxx> wrote:> "blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

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On 12 May, 08:18, " Don Kelly" <d...@xxxxxxx> wrote:> "blackhead"
<larryhar...@xxxxxxxxxxxx> wrote in message

{snipped text}





So- a con job with a purpose- trying to get people to think- that was
his
job.

I think you still see it as a con job because like others here, you
seem to think that E inside a resistance carrying 1A, say, depends
upon the nature of the external EMFs - whether they're generated by
batteries or changing magnetic fields. To push 1A through a
resistance, each electron has to be acted upon by a certain force and
it doesn't matter how this force per unit charge = E is generated. The
charge inside the resistance rearranges itself so creating an E_c to
oppose the external E_ext if boundary conditions require the total E_t
= E_c + E_ext inside to be of a certain value to give a certain
current through it.
---

So in Lewin's setup, we can safely say that if the induced EMF was 1v,
the induced current was 1mA and most of the measured V will be dropped
across the 100R and 900R. Internal Es inside the resistors/conductors
are generated to oppose the induced EMF to maintain the condition I =
1mA through the circuit.

Bull! You are trying to complicate a simple situation and also
attributing
things that I did not say or support. It appears that you have confused
the
reality and the model. The model doesn't deal with the geometry of the
situations which is the crux of the problem. In addition, there is no
such
thing as an induced current.

---
DK (my old messages are indicated with a > yours aren't and this is a
nuisance).
-------
bhead
The induced current is that current that flows in a circuit as a
result of an induced voltage around the circuit. In the example we
have been considering, there is an induced current. You created a
model with a conservative field to model a non-conservative field
which will end up confusing matters.
------
DK
Since voltage is induced and current is not, I consider the the use of
"induced current" to be misleading. I agree that an induced voltage will
cause a current to flow and this current depends on the loop impedance and
the induced voltage which itself is independent of the loop impedance.

I have not neither assumed nor created a model with a conservative field.
The integral around the closed loop of E.dl is not 0 when there is a
magnetic flux
---------->There is an induced voltage in the loop which
will produce a current such that KVL is satisfied.

For a non-conservative field, the sum of the voltages around the loop
<> 0, whereas in your model, they do. You've got a battery with its
ends located at different point modeling an EMF where the ends are at
the same point.
--------
DK
The induced voltage is not zero and is dependent upon the loop dimensions..
However, when  you have a closed loop with resistances, and use circuit
theory, the sum of all voltages, not just the induced voltages, is 0. The
circuit, as opposed to the field, is conservative -at least in terms of
energy.

Yes, *all* voltages.

------------------

bhead
The sum of the measured voltages around the loop won't equal zero,
whereas the usual form of Kirchoff's voltage law around a loop says it
does.
----
DK
Sorry, I disagree with you. They do sum to 0 KVL is happy.

It has been experimentally confirmed that this isn't the case by Romer
before Lewin, in his paper:

Romer, R.H. (1982), "What do ‘voltmeters’ measure? Faraday's law in a
multiply connected region", Am. J. Phys., Vol. 50 No.12, pp.1089-93.
-------------
DK
I have read this paper and I agree with it- and more importantly from the
point of this discussion, it agrees with what I have said.  Romer's paper
definitely doesn't confirm your contention.

The key equations are 3 and 8 and these are combined with trhe use of Ohm's
law to essentially come up with the KVL equation that I gave. Note that he
says V1-V2 =alpha which is the induced voltage for a ramp flux (alpha*t).
Lewin set this at 1V
For the  induced voltage acting counter clockwise, there will be a cccw
current as Romer indicates
What he says is alpha =V1-V2  or alpha-(V1-V2) =0 This is nothing but
Kirchoffs Voltage Law

The usual form of Kirchoff's voltage law is v1-v2 = 0. Sure, you can
fix it by adding in the induced EMF term, but Kirchoff's law is
usually used with *measured* voltages in the circuit that sum to zero.

Applying this to the Lewin circuit V1=0.1 , V2=-0.9, alpha =1
and 1 -(0.1-(-0.9)) =0   This is completely equivalent to what you would get
with the use of KVL and an equivalent 1V source in the circuit.
The only difference between what I said and this is that I took the induced
voltage and current in the opposite direction (as did Lewin)  -1+0.1+0.9=0
which can be represented by the battery case  that you don't like.






------
DK
SSheesh!!! I thought that you had enough experience with circuit analysis
to
deal with this.

*Read* what I have said- I moved the battery to between A and B. It has an
emf of 1V with B + with respect to A.
Where the battery is located in this series circuit doesn't change the
current or the IR drops. The sum of the voltages around the loop is still
the same. Kirchoff is happy.
Vda+Vab+Vbc+Vcd =0.1-1+0.9+0 =0

However it does change the AD meter reading. That is the point here.

Now who should read more?

So it's your battery trying to erroneously model between 2 different
points  the 1v EMF around a path starting and finishing at the same
point. Your model is just wrong.
---
This is a perfectly valid method in circuit analysis.  The basic KVL and KCL
equations are not changed. The meters will not measure the same as before
because they are now across different elements. The model is correct.
I assumed that you were not stupid enough to think that I put the battery
across the 0 impedance lead.

I didn't think you put it across the 0R impedance. I simply thought
that you had replaced an EMF starting and ending at the same point
with one that starts and ends at different points.

Thank you for the Romer reference as it vindicated my position and puts
"paid" to yours.

My position was that for the following:
[view in fixed font text]

+ ----(v2)------- -
| |
- --A-----------------B-- +
| | | |
| # # |
| # * * # |
(v1) R1 * L1 * R2 (v3)
| # # |
| # * * # |
| | * * | |
| | | |
+ --D-----------------C-- -
| |
- ----(v4)------- +

The sum of the voltmeter readings <> 0 where the solenoid is driven at
a constant rate, but the sum of all the voltages in the circuit does
equal zero. Romer's paper seems to confirm this position.

Larry

[snipped text]

.

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