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

On 15 May, 07:04, " Don Kelly" <d...@xxxxxxx> wrote:
"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 0. In terms of circuit theory it looks like a
source i.e a battery in this case. While circuit theory is a quasistatic
approximation to field theory, it works very well in this situation.
---------------
 > 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 read this, and find nothing in this to contradict what I have been saying.
It does deal with the meters but assuming these  are ideal, consider what he
indicates for the meters, as applied to this case.
The key equations are 3 and 8 (8 essentially is KVL ) ,  and I have no
problem with these. Beyond this he uses Ohm's Law (which is derived from
Kirchoff's laws ) and notes that V1 =R1*I and V2=R2*I assuming a
counterclockwise direction.
so in the Lewin circuit V1=100*I and V2 =-900*I  and alpha is the  induced
voltage (1V)
V1-V2 =1000*I =alpha =1 Volt =E induced and V1 =0.1V , V2=-0.9V with a ccw
current of 1ma
or Einduced -(V1-V2) =1 -0.1-0.9 =0
Lewin and I assumed current in the clockwise direction  in the circuit model
so that V1=-0.1V and V2=+0.9V and a clockwise current of 1ma with Einduced
rising in the clockwise direction.
Any disagreement is simply the assumed positive direction of the flux and
the resulting direction of the induced voltage and current.

So- what does this paper do?- it simply reinforces what I have
said -considering the induced emf as a source (equivalent battery ) in the
circuit analysis is quite alright (as has been known in practice for more
years than you and I together have been around).

Romer's paper doesn't try to model the induced EMF with a battery. In
fact, he doesn't even introduce any equivalent circuit what so ever to
model his circuit in a non conservative field, which you are.

Neither Romer's paper nor I
attempt to pin down the  distribution of the total induced voltage. What I
have been saying is in agreement with Romer. He just dresses it up a bit
more.

Romer sticks to Faraday's law and the definition of a voltmeter
whereas you appeared to be dressing things up using conservative
circuits to model non conservative circuits.

---------------------------------
 > >This induced voltage is





completely independent of the resistance distribution. The IR voltages
are
dependent so we get:

Einduced= sum of IR drops. (and Lewin did this to get the 1ma -assuming
linear resistors.)

What you are saying is a distortion of KVL

--
bhead
Isn't Kirchoff's voltage law: The sum of the measured voltages around
a loop = 0?

--
DK
Not quite -drop the "measured" - sometimes the meter may measure a drop as
+ and other times the meter can be connected to register a rise as +
Lets see, in Lewin's battery circuit he measures +0.9V and +0.9V. Are you
saying that the sum is 1.8V?

I can sum voltage rises around the loop to zero or I can say sum of rises
=sum of drops (in the same direction as a drop is a negative rise) Same
thing.

a) There is a source of 1V in the circuit
b)there are two resistors in the circuit.
c) the circuit is drawn as a 4 terminal loop
A--------B
| |
D -------C

In both cases, Lewin assumes a 0 resistance from A to B and from C to D
In both cases, he looks at the voltage "measured" Vad and Vbc
Case 1:
Battery and resistor between A and D so Vad =1-100I =0.9 V as "measured"
because I =1/(100+900) =1ma (implicitly indicating that Lewin assumed
(correctly) a series circuit
Vbc=0.9V
Vab=0, Vcd=0 because of 0 resistance. Kirchoff is happy..and meters read
0.9 each

Case 1a -we move the battery between A and B, this doesn't change the
current or the IR drops so that
Vad=0-0.1 =-0.1V (or Vda =+0.1), Vab=-1V, Vbc=0.9V, Vcd=0 Around the
loop
sum =0
Kirchoff is happy but meters read -0.1 and +0.9

----
bhead
How can you have Vab = -1V in the second case when the current through
it is the same as in the first case where you have Vab = 0v? The E
causing charge to move at the same rate must be the same and so E.dl =
V through it must be the same.
------
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.
---------------
DK

 As to moving a battery  from one part of the circuit to another-

Since you didn't read what I said, here it's again:

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.

Where did I even mention a battery being moved from one part of a
circuit to another?

that is
perfectly valid and there is a circuit theorem (relatively obscure but easy
to develop logically) which deals with this- the model can be changed as
long as the KVL/KCL equations are not changed. In this case, with a single
loop with a battery E and resistances R1 and R2
E+(V1+V2)= 0 where E, V1=-R1*I and V2=-R2*I  are voltage rises in the
direction of the current. Topologically it is a ring with all elements in
series and which element is located at which part of the ring doesn't affect
the equation- so I can damn well put it where I want. Admittedly the meters
will measure different things depending on the battery location. I gave
case1 (Lewin's start case) and then  legitimately considered having the
battery  moved through the loop to each of the sides -all the while
retaining the KVL/KCL relationships - I did not consider putting the battery
in parallel with a shorted link and assumed that you would not be so stupid
as to think that.   You still seem to have missed the point that while the
circuit hasn't changed, the voltmeters will be measuring different things
depending on the location of the source- Go over the cases  that I gave you
(and you clipped) and think about it.

In any case, if you are using Romer as support, he doesn't give it to you..
If you are looking for some support from circuit analysis approaches- again
you are out of luck.

You haven't shown this to be true.


Don Kelly
d...@xxxxxxxxxxxx
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