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



"blackhead" <larryharson@xxxxxxxxxxxx> wrote in message news:b6d371d4-cf87-4122-ab04-3448814568f2@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 12 May, 08:18, " Don Kelly" <d...@xxxxxxx> wrote:
"blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

news:0f41a387-cc6c-4335-b830-d9d457106268@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
On 10 May, 05:42, " Don Kelly" <d...@xxxxxxx> wrote:

> "blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

>news:9c568efc-2365-4ae8-a21c-7c88e19f3e36@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> On 7 May, 06:14, " Don Kelly" <d...@xxxxxxx> wrote:

> > "blackhead" <larryhar...@xxxxxxxxxxxx> wrote in message

> >news:46fc322f-33af-47c5-b7b1-87ae50e74ee8@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> > On 6 May, 05:43, " Don Kelly" <d...@xxxxxxx> wrote:

<snipped text>

> > I don't think you've understood the point of the video: Two resistors
> > are in parallel, yet havehave different voltages across them.

> > I understood the point, It was a con job and it got you.

> What was he hiding if it was a con job?
> ------------------
> Nothing but he wasn't telling you everything- and stated things
> deliberately
> in such a way that many fell for it. He actually wanted someone to think
> and
> point out the fallacies involved.

What did he leave out?





> > In one case the voltage across a resistor and a battery (1-0.1) was
> > measured
> > and compared to the voltage across the other resistor (0.9)= > > agreement.
> > In the second case the voltage across one resistor(0.1) and the second
> > resistor (0.9) were measured . The `distributed battery`` was not
> > included
> > in the measurement.

> He calculated the *measured* voltage across the resistors. He was
> comparing *measured* voltages across the resistors that you would
> observe in practise.
> ----------
> Right and then he took the "measured" voltage across the battery plus > the
> 100 ohm resistor (0.9V) which he compared to the 0.9V "measured" across
> the
> 900 ohm resistor. These agreed.
>Then with the battery out of circuit, he
> took as you indicate, the calculated voltages across the resistors (and
> leads) only and got -0.1V vs 0.9V -the previously calculated IR drops > and
> implied that that was what the meter if it existed would read.!!!!!

Within experimental error. Put it this way, for an achievable setup,
the voltage measured would be closer to -0.1V than 0.9V, so showing
the effect exists.

> These would not be the voltages that would be observed in practice.

Not 100% perhaps but enough to show the effect beyond reasonable
doubt.

> Implicit in this and not mentioned is that there was no induced voltage > in
> either the left or right side of the circuit -There is no reason to
> believe
> that to be true.

The solenoid isn't ideal so you will get a small changing flux inside
the loops containing the voltmeters, which you can minimise even
further by keeping these loops small. Also, the flux won't change at a
constant rate perfectly, so giving a constant EMF. It boils down to
being within or not experimental error.

>We actually have no information about the distribution of
> incremental voltages around the circuit- just that the total was 1V. The
> distribution in a real circuit (not a blackboard model as this was)
> depends
> on the geometry of the loop.

They distribution of the voltages around the circuit depends on the I-
V relationship of the elements that make up the circuit and the rate
of change of the flux inside it.
-----------
Not quite The distribution of the IR drops depend on the I-V relationships
of the elements but the distribution of the induced voltages in the loop has
nothing to do with this but depends on the rate of change of flux in a
particular loop, independent of the material or resistances in the path but
definitely dependent on the geometry of the path. Re-examine Faraday and
Maxwell.
-----------------------





> > Secondly, go back to the definition of series and parallel circuits. A
> > series has a single current common to all elements of the circuit and
> > voltages across elements differ. A parallel circuit has a common > > voltage
> > across all elements and currents in the elements differ (implying a
> > splitting of current between paths). In this primitive circuit we have > > a
> > common (single) current and different voltages across elements in the
> > circuit.- A single loop satisfies the criterion for a series circuit. > > It
> > doesn`t satisfy the criterion for a parallel circuit.

> > The resistors are NOT in parallel- either with a battery or with an
> > induced
> > voltage source. There is no common voltage across the resistors. In > > fact
> > there is a voltage source- either a battery or a distributed induced
> > voltage
> > in a single loop which contains resistors -that is series.
> > If the source produced a current which was split between two branches-
> > then
> > that is parallel- this isn`t the case here.

> OK, let's call them in series then.
> ---
> According to the definition of a series circuit-
> --

> A voltmeter across the 100Ohms will measure 0.1V, across the 900Ohms
> will measure 0.9V. So even though the voltmeters are physically in
> parallel with one another, they measure different voltages! I don't
> think you've grasped this point. Put another way, your voltmeter will
> read different voltages between A and B depending on where it's
> located in a circuit, even though there is no changing magentic field
> in the loop the voltmeter is a part of. And that's surprising, at
> least for me.

> ---------
> The voltmeter was across the resistance *and the leads* between the
> terminals where the meter was connected- and assumes (assumption not
> explicit but obvious as the voltages given were the calculated IR
> voltages)
> no induced voltage in these parts , then all the induced voltage would
> have
> to be in the top and bottom "shorted" bars (and the induced voltage in
> these bars depends on the flux and the geometry of the situation, not > the
> fact that the bars are assumed to have 0 resistance). If you wish you > can
> lump the induced voltage anywhere in the loop (or split it into several
> lumps in different places).
> If on the left side- then the meters as connected would read 0.9 on both
> left and right sides. If on the right side, the meters would read -0.1 V
> on
> both sides. In these cases, the voltmeters would be in parallel with > each
> other (this doesn't make the main circuit a parallel circuit). The
> "results"
> obtained correspond to the effective induced voltage being in the
> "shorted"
> bars and in this case the voltmeters would be measuring different legs > of
> the circuit as they would not be in parallel. Work it out. Consider > Vab=0,
> Vbc=0.9v, Vcd=-1V (source), Vba=100I (or Vab as measured would be -0.1V.
> Try
> it with the source moved to some other part of the circuit or split
> equally
> between the 4 legs- the IR drops, the source voltage and the current > will
> not change but the "meters will have different readings.

> No problems exist in this case. If, as would be the situation in a real
> circuit- thee induced voltage is distributed somehow, the voltmeters > would
> still read differently because they would be measuring the IR drop + > some
> of
> the induced voltage. Note that he said nothing about the voltage across
> the
> "shorting bars" leaving one to think it is 0 because the resistance is > 0.
> The loop resistance or variations in resistance has no effect on the
> induced
> voltage or its distribution in the loop- that is dependent on geometry
> which is not taken into account in lumped circuit analysis.

> 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.
----
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.
------


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?

As to the distribution of the induced voltage -again that depends on loop geometry. assume the resistors are perfectly bifilar and stuck out from the loop at right angles presenting a negligable distance between the terminals so that no induced voltage exists in the resistors per se but the total loop induced voltage is still 1 V (geometry of the loop is essentially unchanged) and the loop resistance is unchanged. we still get a current of 1ma and the same resulting IR drops.
I can think of more practical situations

In another response you wrote:

"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."

If the total loop had 0 resistance, then the current would be infinite or very large- that is true. It's called a short circuit.

However, the circuit has resistance in some parts the opposing E that you are looking for appears across this resistance limiting the current. There is no need for the opposing E to be at the same place as the induced E as you seem to want to place it. Only if the E.dl was uniform in magnitude around the circuit and the resistance was also uniform per unit length- would that be true. It would also mean that you would not be able to measure a voltage between any two points in this loop because at any section the induced voltage rise would equal the IR drop (negative rise) so the sum would be 0 and that would make the whole thing moot.
You say that the voltage between any two points of a 0 resistance conductor carrying current would be 0 (in response to Benj's point 8) is simply not true. It means that IR is true but no more than that.
The corollary that you seem to be implying in some of your statements is that the induced voltages only exist across the non-zero resistances. In that case, the induced voltage must produce a rise in the direction of the current and the IR voltage is a drop in the same direction- the sum is 0.

On this basis there seems to be an implication that no measurable voltages exist anywhere in the circuit.
Maybe I am misreading you- so clarify your situation recognising that the total voltage induced is equal to the total voltage drop. Would you care to address this minor problem?

Possibly you should now read the text that you omitted.
--
Don Kelly
dhky@xxxxxxxxxxxx
remove the x to reply

.

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