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Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 52610 times) Print
poynt99

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It's not as complicated as it may seem...

« Reply #550 on: 2012-04-21, 16:53:30 »
I suppose you mean that the two signals are in opposition. I agree. I don't like the negative notation because we deal with ac currents.
Actually, it's pulsed DC, and it produces mainly a pulsed DC response on the loop. The meters are "polarity sensitive" as the good professor said, and therefore there is a positive and a negative response wrt the meters.

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For example, "no flux crossing between A-R900-D and the wires from CH1" means no flux through the green surface, "no flux crossing between A-R100-D and the wires from CH2" means no flux through the blue surface (see the image). It is implicit only if it is presumed that there is no flux outside the circle.
This doesn't make sense.

By this reasoning, if I understand it correctly, means if we take a measurement directly across R100, both from the left side and right side, the two readings should be the same, correct?

Accordingly, flux never crosses through the measurement wires as long as they are close to the loop. Also, are you assuming a non-infinite length solenoid?
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WaveWatcher

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« Reply #551 on: 2012-04-21, 19:17:10 »
Ex,

I wish you Good Luck. Perhaps your flare for details and formulae will clarify these concepts for us.  Afro
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"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
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exnihiloest
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« Reply #552 on: 2012-04-21, 21:34:16 »
Actually, it's pulsed DC, and it produces mainly a pulsed DC response on the loop. The meters are "polarity sensitive" as the good professor said, and therefore there is a positive and a negative response wrt the meters.
This doesn't make sense.

Induced emf from a pulsed DC is not a pulsed DC. The dc component is lost because only the time variation of the flux generates emf. For instance, a 1µs pulse from 0 to 10v, or from -5v to 5v or from -1000v to -990v and so on will induce exactly the same emf pulse. Only the signal transitions have an effect. It doesn't mean that the induced signal will have positive and negative alternance of equal amplitude (it's true only if the pulse has the same rise and fall time), but that the time integral of the induced pulse will give zero (the sum of the products time duration x voltage =0) although the time integral of the source signal is not null.

Quote
By this reasoning, if I understand it correctly, means if we take a measurement directly across R100, both from the left side and right side, the two readings should be the same, correct?
Accordingly, flux never crosses through the measurement wires as long as they are close to the loop.
Also, are you assuming a non-infinite length solenoid?

I'm not sure to understand what you mean.

a) It is right to assume that there is no flux through the blue area which is between R100 and the CH2 wires, even if CH2 is connected to R100 instead of A and D. So the scope really measures R100 voltage.
b) It is right to assume that there is no flux through the green area which is between R900 and the CH1 wires, if CH1 is connected to R900 instead of A and D. So the scope really measures R900 voltage.

But c) it is wrong to assume that there is no flux between R100 and the CH1 wires, CH1 being connected either to A/D or to R100. Between R100 and the CH1 wires, there is all the flux crossing the disk, which is also the surface of the measurement circuit. So CH1 measures R100 voltage minus 1v which is induced in the measurement circuit. Of course we retrieve the same result 0.9v or -0.9v depending on the sign convention, but it is easier to consider case b) to get the solution.

In other words, we can consider either that CH1 measures directly the voltage at R900 because there is no flux crossing the area between R900 and the scope wires, or that it measures R100 but with a correction, because we must take account of the induced emf in the circuit measurement which in this view, is now enclosing all the flux.

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poynt99

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It's not as complicated as it may seem...

« Reply #553 on: 2012-04-21, 22:25:55 »
Induced emf from a pulsed DC is not a pulsed DC. The dc component is lost because only the time variation of the flux generates emf. For instance, a 1µs pulse from 0 to 10v, or from -5v to 5v or from -1000v to -990v and so on will induce exactly the same emf pulse. Only the signal transitions have an effect. It doesn't mean that the induced signal will have positive and negative alternance of equal amplitude (it's true only if the pulse has the same rise and fall time), but that the time integral of the induced pulse will give zero (the sum of the products time duration x voltage =0) although the time integral of the source signal is not null.
You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.

Quote
I'm not sure to understand what you mean.

a) It is right to assume that there is no flux through the blue area which is between R100 and the CH2 wires, even if CH2 is connected to R100 instead of A and D. So the scope really measures R100 voltage.
b) It is right to assume that there is no flux through the green area which is between R900 and the CH1 wires, if CH1 is connected to R900 instead of A and D. So the scope really measures R900 voltage.

But c) it is wrong to assume that there is no flux between R100 and the CH1 wires, CH1 being connected either to A/D or to R100. Between R100 and the CH1 wires, there is all the flux crossing the disk, which is also the surface of the measurement circuit. So CH1 measures R100 voltage minus 1v which is induced in the measurement circuit. Of course we retrieve the same result 0.9v or -0.9v depending on the sign convention, but it is easier to consider case b) to get the solution.

In other words, we can consider either that CH1 measures directly the voltage at R900 because there is no flux crossing the area between R900 and the scope wires, or that it measures R100 but with a correction, because we must take account of the induced emf in the circuit measurement which in this view, is now enclosing all the flux.
Interesting approach. It happens to work, but it is not the correct way to analyze this in-plane scenario.

For some odd reason, many here seem obsessed with the solenoid "flux". The flux outside the solenoid is zero, so according to you Ex, the measurement leads will never be influenced by the solenoid "flux" (and you would be correct). The "correction factor" you propose is nonsense. The scope indicates what is there across the two points on the loop. We never ask or ponder whether the measurement is relative to the R100 or R900 resistor  Huh?, the measurement simply IS what it IS. The scope certainly doesn't care.

I've stressed this a hundred times it seems, but it does not sink in; forget about the flux, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid. The flux has no direct influence on the loop or the measurement leads, ever.
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GibbsHelmholtz
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« Reply #554 on: 2012-04-22, 01:07:31 »
...

I've stressed this a hundred times it seems, but it does not sink in; forget about the flux, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid. The flux has no direct influence on the loop or the measurement leads, ever.

...

But the flux is what creating the E field (for this set up).  I don't think it just happens to work.  It would works with many more complex in-plane case.  I think the E field you're speaking of is the same as the flux method.  Can you use your method to solve this problem I posted a while ago?  Perhaps we can find a relationship.

 





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Professor Walter Lewin's Non-conservative Fields Experiment
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exnihiloest
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« Reply #555 on: 2012-04-22, 16:44:06 »
You're splitting hairs here Ex. Reference the Lewin experiment, a DC source is used to pulse the coil. Consequently, there is a corresponding unipolar response on the loop (polarity dependent on meter connection). i.e. the current does not change direction, it rises, then falls back to zero.

I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.
An induced current never rises and fall back to zero. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!

Quote
For some odd reason, many here seem obsessed with the solenoid "flux". The flux outside the solenoid is zero, so according to you Ex, the measurement leads will never be influenced by the solenoid "flux" (and you would be correct).

This idea is not according to me, but to your interpretation of what others say, including me, which is different.
The Faraday's law doesn't imply that the circuit be in the flux. The Faraday's law stipulates only that the surface of the circuit must be crossed somewhere by a varying flux. The flux can cross the surface even through only 1% of the surface and anywhere. If your loop is at 1km from the solenoid, still around, the same emf will be induced.

Quote
The "correction factor" you propose is nonsense. The scope indicates what is there across the two points on the loop. We never ask or ponder whether the measurement is relative to the R100 or R900 resistor  Huh?, the measurement simply IS what it IS. The scope certainly doesn't care.

"The measurement simply IS what it IS": yes, I agree. The measurement that you imagine by connecting CH1 to R100 while the wires are enclosing the disk surface is the measure of the potential difference at the resistance terminals + the emf induced in your measurement circuit, because the total flux of the solenoid is crossing your measurement circuit.
It is what it is: 0.9v, which is a real measurement that is what it is, but naturally not the potentiel difference felt by the R100 resistance.
Of course you will not get the same measurement by connecting CH2 to R100 by the other side, which gives the real potential difference at R100.
The nonsense is only to imagine that both measurements should give the same result.
My "correction factor" was only to obtain the potential difference at R100 terminal from the measurement of CH1 (in the same way that when you measure the potential difference at the terminals of two identical resistances in series, you can use a "correction factor" to obtain the potential difference at only one: by dividing by two).

Quote
I've stressed this a hundred times it seems, but it does not sink in; forget about the flux, it is simple a means to an end, and that END is the resulting induced electric field. The E-field is what counts. No E-field, no induced emf. The E-field is all around the solenoid, inside and outside, the flux is internal only to the solenoid.

You can get the same results by using the E-field, as by using the flux. But you can't oppose one method to the other, because the two methods are strictly equivalent due to the Stokes theorem. If one failed, the other too. I said it somewhere else but you missed the point:
"According to Stockes theorem, the surface integral of the curl of a vector field over a surface is linked to the line integral of the vector field over its boundary.".

Demonstration that they are the same:

Emf = -dΦ/dt           (Faraday's law = flux crossing method)
Φ = ∫∫B.n.dS           (definition of the flux, B and n are vectors, n is the unit vector, normal to the circuit surface S).
(1) Emf = -dΦ/dt = -∂(∫∫B.n.dS)/∂t =  -∫∫∂B/∂t .n.dS

Emf = ∮E.dl                 (integral of the electric field over the circuit = E-field method, l is refering to the circuit path).
∮E.dl =∫∫(∇xE).n.dS     (Stokes theorem applied to the E-field, S being the surface enclosed by l)
and ∇xE = -∂B/∂t         (Maxwell)
so ∮E.dl = -∫∫(∂B/∂t).n.dS =  equation (1) =  -dΦ/dt

QED

The only interest of the E-field, is that it gives the local field acting on the electrons, and therefore the force F=q.E which is the cause of their motion. On the contrary, if we supposed that the flux would be the cause of the current, the phenomenon would be a non local (therefore outside classical physics) because there is no flux at the position of the wire enclosing the solenoid. Nevertheless in our approximation of the quasi-stationary states (we suppose that there is no EM waves), the both methods are strictly equivalent to calculate the emf but the method by the flux is by far the simpler.

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The flux has no direct influence on the loop or the measurement leads, ever.

I agree, but the flux variation through the circuit surface is an event concomitant with the local E-field along the circuit, and so, there is no discrepancy in using one method or the other, as proved by the equations.

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exnihiloest
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« Reply #556 on: 2012-04-22, 17:15:38 »
But the flux is what creating the E field (for this set up). 
[...]
I think the E field you're speaking of is the same as the flux method. 

I confirm the second affirmation. For the first one I would rather say that it is the motion of charges in the solenoid coil that creates both the flux inside and the E-field outside. This is disputable. One can oppose that a flux variation can be obtained from a moving permanent magnet. But even in this case, the magnet field coming from electron spins and orbitals, there are still moving charges.

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exnihiloest
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« Reply #557 on: 2012-04-22, 17:34:24 »

Very smart proposition, Gibbs! I confess that I have been very surprised that we should get more than 1v, according to my previous method. We should have 1v + 0.9v = 1.9v!
And we will, the method works: it is finally easy to see it, by removing R900 and to follow a wire from the scope: the circuit has 2 turns before returning to the scope.



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GibbsHelmholtz
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« Reply #558 on: 2012-04-22, 18:03:15 »
Very smart proposition, Gibbs! I confess that I have been very surprised that we should get more than 1v, according to my previous method. We should have 1v + 0.9v = 1.9v!
And we will, the method works: it is finally easy to see it, by removing R900 and to follow a wire from the scope: the circuit has 2 turns before returning to the scope.



Thanks for the E field equivalent analysis, Exn.  

Yes, you can remove the R900 and see it has 2 turns, but that's not the only way.  You can pick the path through the R900 and it only has 1 turn.  I've come up with a rule using E field.

Pick a path
Count the number of turns it covers the flux
Multiply the Emf(1V) by the number of turns
Determine the direction of emf
Add or subtract resistance voltage according to the Emf polarity

Let's say you pick the path through R100, now you have 2 turns.  The total induce emf in those two turns is 2V.  Now you have to add or subtract the R100 .  If you follow the whole loop, you should see - + + - .  This tells you to subtract.  2V - .1V = 1.9V

Let's say you pick the path through R900, now you have 1 turn.  Total induced emf in the 1 turn is 1V .  Now you have to add or subtract the R900.  If you follow the whole loop, you see + - + - .  This tells you to add.  1V + .9V = 1.9V
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exnihiloest
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« Reply #559 on: 2012-04-23, 08:37:37 »
I agree 100%, gibbs. Now it becomes easy.  Afro

I would like now to go further with the question of whether or not it is possible to show an effect of an induced E-field on charges which are not in a circuit enclosing a varying flux.
I think that this thread is well intended for such a question.

We know for sure, provided that we accept Maxwell's equation ∇xE=-∂B/∂t, that there is no emf in a closed circuit whose the surface is not crossed by a varying flux, even if it lies in a not null induced E-field: near a toroid coil it is impossible to detect a current in a closed circuit without conductor passing through the toroid hole. This means that the contribution of the E-field at some places along the circuit cancels its contribution at some other places.

But we are not restricted to closed circuits. What with open circuits? I have tried in an already mentionned experiment with a piece of wire terminated by terminal capacitors at each end. But I couldn't detect any current going back and forth in my setup. I guess that the capacitors form a closed circuit due to the displacement currents between them. As the circuit is looped and there is no flux crossing, there is no emf.

This seems to be a problem but it opens other great prospects: could we engineer the path of displacement currents closing a circuit in order to modulate an enclosing flux and therefore to generate induced emf?

Therefore the question is double and both aspects being related to unknown domains (at my knowledge) could be promising.

Last Edit: 2012-04-23, 09:06:38 by exnihiloest
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GibbsHelmholtz
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« Reply #560 on: 2012-04-23, 14:24:53 »
I agree 100%, gibbs. Now it becomes easy.  Afro

I would like now to go further with the question of whether or not it is possible to show an effect of an induced E-field on charges which are not in a circuit enclosing a varying flux.
I think that this thread is well intended for such a question.

We know for sure, provided that we accept Maxwell's equation ∇xE=-∂B/∂t, that there is no emf in a closed circuit whose the surface is not crossed by a varying flux, even if it lies in a not null induced E-field: near a toroid coil it is impossible to detect a current in a closed circuit without conductor passing through the toroid hole. This means that the contribution of the E-field at some places along the circuit cancels its contribution at some other places.

But we are not restricted to closed circuits. What with open circuits? I have tried in an already mentionned experiment with a piece of wire terminated by terminal capacitors at each end. But I couldn't detect any current going back and forth in my setup. I guess that the capacitors form a closed circuit due to the displacement currents between them. As the circuit is looped and there is no flux crossing, there is no emf.

This seems to be a problem but it opens other great prospects: could we engineer the path of displacement currents closing a circuit in order to modulate an enclosing flux and therefore to generate induced emf?

Therefore the question is double and both aspects being related to unknown domains (at my knowledge) could be promising.



Yes, this is a hot topic.  I think part of the reason you cannot have good detection is because of the weak signal.   I've done this experiment in a different way.  I have a plasma ball and tried to detect an E field using my volt meter.  The set up is as below.  As I varies the distance between them, the voltage drop.  Reverse the lead the voltage reverse polarity.   There is also a video that could be similar to this experiment.  He used LED to detect this E field.  I believe this E field also a displacement current, but 90 degrees to the displacement current of a closed loop containing magnetic flux. 

http://www.youtube.com/watch?v=7TlyO9o-fSo

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Professor Walter Lewin's Non-conservative Fields Experiment
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exnihiloest
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« Reply #561 on: 2012-04-23, 15:59:00 »

I'm not sure that this kind of setup detects an induced E-field. A double-diode-led detector circuit lights strongly when approaching the hot end of a coil and switches off on the other side. We should not have this effect by detecting a real induced E-field, which is uniformly distributed around a coil and symmetrically between the two ends.
For me the detector is a capacitively coupled to the hot part of the coil which is usually resonant thus building a high voltage. So only an ordinary electric field is detected, although there are real displacements currents from the coil to the dectector.

If we make a too sensitive device, it will always detect something. However the induced E-field is strong. If we have 1v/turn in a long solenoid coil, and we consider a circuit around of circumference 10 cm, this means that the E-field is 10v/m, which should be large enough to be easily detected.

In an old experiment intended for another goal, I had realized a coaxial coil and powered it with an HF current. The coaxial cable was to shield the capacitive coupling. I could light a low energy lamp bulbs at a distance:
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%201.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%20Ensemble.jpg

But
1) only one end of the shield of the coaxial cable was connecting to the ground, otherwise the induced emf in the shield would have been shortcut, and consequently it would have acted as a shorcut transformer secundary. But by connecting only one end, the induced E-field in the shield makes it acquiring the same voltage as the inner conductor. So this way I had not eliminated the capacitive coupling.
2) I don't know the physical principle that lights the lamp. At the beginning the gas is not ionized, so a magnetic field has no action. Could it be the induced E-field that excites the electrons? Or simply the electrical field due to the HV section of my resonant coi,l capacitively coupled to the lamp?
3) The lamp could be lit whatever its position around the coil. The frequency was high (some Mhz). So the last problem is that there are likely electromagnetic waves. A power is radiated, so I don't know exactly what this experiment can really show, expect a nice effect.

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GibbsHelmholtz
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« Reply #562 on: 2012-04-24, 01:28:47 »
I'm not sure that this kind of setup detects an induced E-field. A double-diode-led detector circuit lights strongly when approaching the hot end of a coil and switches off on the other side. We should not have this effect by detecting a real induced E-field, which is uniformly distributed around a coil and symmetrically between the two ends.
For me the detector is a capacitively coupled to the hot part of the coil which is usually resonant thus building a high voltage. So only an ordinary electric field is detected, although there are real displacements currents from the coil to the dectector.

If we make a too sensitive device, it will always detect something. However the induced E-field is strong. If we have 1v/turn in a long solenoid coil, and we consider a circuit around of circumference 10 cm, this means that the E-field is 10v/m, which should be large enough to be easily detected.

In an old experiment intended for another goal, I had realized a coaxial coil and powered it with an HF current. The coaxial cable was to shield the capacitive coupling. I could light a low energy lamp bulbs at a distance:
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%201.jpg
http://exvacuo.free.fr/div/Sciences/Experiences/Induction/Lampe%20Ensemble.jpg

But
1) only one end of the shield of the coaxial cable was connecting to the ground, otherwise the induced emf in the shield would have been shortcut, and consequently it would have acted as a shorcut transformer secundary. But by connecting only one end, the induced E-field in the shield makes it acquiring the same voltage as the inner conductor. So this way I had not eliminated the capacitive coupling.
2) I don't know the physical principle that lights the lamp. At the beginning the gas is not ionized, so a magnetic field has no action. Could it be the induced E-field that excites the electrons? Or simply the electrical field due to the HV section of my resonant coi,l capacitively coupled to the lamp?
3) The lamp could be lit whatever its position around the coil. The frequency was high (some Mhz). So the last problem is that there are likely electromagnetic waves. A power is radiated, so I don't know exactly what this experiment can really show, expect a nice effect.



Actually, this is also one of my concern about this effect.  Is it an induced E field or an E field from HV hot end.  The E field according to our thinking would be uniformly distributed as you mentioned.  I think I'll agree that it is HV hot end.  If so, the capacitive coupling would link back to ground end.  The experiment you performed as well as Tesla coil wireless lighting would also share the same concept.  We can consider this as  a closed circuit except in the form of extremely low current high voltage. 

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WaveWatcher

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« Reply #563 on: 2012-04-24, 03:17:16 »
Is it an induced E field or an E field from HV hot end.

I must say, this is a very good way to describe the reality of the experiments offered as 'not being from RF energy'.

We all know that charges accumulate best on physical points. While the battery and other larger surface objects provided little area for charge accumulation, either side may be a small point for accumulation. Both sides pointed is not required.

The 'AV plug' provides the pointed side of unbalanced charges so highest potential accumulates there. The circuit is closed due to capacitive coupling. I watched the whole series hoping a new proof would emerge. The only proof I saw was that the experimenter still doesn't have support for his claims.



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"As far as the laws of mathematics refer to reality, they are not certain; as far as they are certain, they do not refer to reality." - Einstein

"What we observe is not nature itself, but nature exposed to our method of questioning." - Werner Heisenberg
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GibbsHelmholtz
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« Reply #564 on: 2012-04-24, 04:26:29 »

The question I have left is:

If we induced 10kV into a uniform loop of wire... Where is the hot end? lol

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exnihiloest
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« Reply #565 on: 2012-04-24, 07:54:56 »

In practice, it is not so difficult than it looks. It depends on which end is the most coupled to the ground and which is the most coupled to the detector. For example the cold point of a big vertical Tesla coil would be the side near the ground, even if this coil end was not connected to the ground.
Generally the primary coil that induces the current determines the cold side because it is capacitively coupled to the secondary coil and it is at a low potential. If it is wound at the middle of an HV winding, then the cold point is the middle and there are 2 hot ends with a voltage half the HV.

It must be concluded that a Faraday cage is needed if we want perform correct measurements of induced E-field.
Another flaw to consider is the magnetic field which could disturb a pure induced E-field measurement. As a Faraday cage doesn't shield magnetic fields, only a toroid coil with a high permeability core, not saturated, can guarantee the lowest leakage of magnetic field.
 
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GibbsHelmholtz
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« Reply #566 on: 2012-04-24, 13:49:14 »

I'm just thinking if we have a loop with current flowing, is there a hot end since hot and cold tailing each other.  What if the current increasing or decreasing?  This is how induction manifest and it seems like a hot and cold end is needed for induction.

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exnihiloest
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« Reply #567 on: 2012-04-25, 09:32:31 »

There is only a potential difference between the two ends. The coupling with the vicinity determines the potential along the line, relative to the ground, the section of highest potential becoming the hot spot.
A potential is always relative.

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poynt99

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It's not as complicated as it may seem...

« Reply #568 on: 2012-04-25, 23:51:55 »
But the flux is what creating the E field (for this set up).
So what? Irrelevant. The flux has about the same relevance as the burning coal does in a steam engine. Neither of them ultimately do the work. They are simply a step in the energy conversion process. Ultimately, it is the E-field that does the work on the loop charges, and it is only the E-field that is far-reaching outside the solenoid.

Quote
I don't think it just happens to work.  It would works with many more complex in-plane case.  I think the E field you're speaking of is the same as the flux method.  Can you use your method to solve this problem I posted a while ago?  Perhaps we can find a relationship.
If your flux method works and you are happy with it, good for you guys Wink I'm going to continue applying the E-field for any analysis, as both the loop and the measurement leads measure the line integral of the E-field alone.
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poynt99

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It's not as complicated as it may seem...

« Reply #569 on: 2012-04-25, 23:55:20 »
I'm not splitting hairs, I justified what I said. I say that "+" or "-" are not relevant for signals with a null dc component and I explain why it is the case here.
An induced current never rises and fall back to zero. It rises and fall back below zero at some time, otherwise there would be a dc component. If transformers could pass DC, it would be known for a long time!
Yeah, you were, and you're doing it again.

I think you know very well what I mean, and there's nothing incorrect about what I said.  Roll Eyes
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exnihiloest
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« Reply #570 on: 2012-04-26, 07:55:07 »
...
I think you know very well what I mean,

Certainly not. I'm intellectually honest. What you say here is to judge someone on presumed intent.

Quote
and there's nothing incorrect about what I said.  Roll Eyes

Maybe what you mean is correct, but I know only what you say and what you say is not correct. In physics things must be clearly said, without ambiguity. It's not a game where everyone would have to guess what others really mean.
When you assert: "the current does not change direction, it rises, then falls back to zero", this is false for an induced current. And I say it, question of truth. An induced emf always rises above zero and then decreases under zero, perhaps several times if it oscillates, and the integral of emf over time is always zero, meaning that there is no DC component.
If you meant other thing that what you said and that I have reformulated here, then you should say it more clearly. If you don't agree with what I say, I would appreciate that you say why, in terms of electromagnetism.

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exnihiloest
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« Reply #571 on: 2012-04-26, 08:28:24 »
...
it is the E-field that does the work on the loop charges, and it is only the E-field that is far-reaching outside the solenoid.
If your flux method works and you are happy with it, good for you guys Wink I'm going to continue applying the E-field for any analysis, as both the loop and the measurement leads measure the line integral of the E-field alone.

I agree with your first point: the local E-field is the causal explanation for the motion of charges, so it's the best.
But you must also admit that the flux variation is a concomitant event and that the knowledge of the flux is an easier way to calculate the emf. The most important: the equality between the integration of the E-field along the circuit and that of the flux through the circuit surface, both giving the emf, is not a physical principle, it is a direct implication based only on logical mathematical arguments applied to Maxwell equations (Stokes theorem). Accepting Maxwell's equations implies to accept both methods to calculate the emf: one of them can't be true while the other would be false, without logical fallacy.


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GibbsHelmholtz
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« Reply #572 on: 2012-04-26, 12:50:47 »

If your flux method works and you are happy with it, good for you guys Wink I'm going to continue applying the E-field for any analysis, as both the loop and the measurement leads measure the line integral of the E-field alone.

Ya know what... I like that attitude. lol  Someone has to go against the flow. lol  As long as you understand both sides of the argument, you are free to choose any path.  I don't recommend everyone think the same thing, that's how things come to an end.  Wink

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poynt99

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It's not as complicated as it may seem...

« Reply #573 on: 2012-04-26, 14:09:54 »
When you assert: "the current does not change direction, it rises, then falls back to zero", this is false for an induced current. And I say it, question of truth. An induced emf always rises above zero and then decreases under zero, perhaps several times if it oscillates, and the integral of emf over time is always zero, meaning that there is no DC component.

Ex,

Draw out what V(D-A) looks like vs. time on a graph and post it. Use the actual voltage across the two points, i.e. what would be measured with decoupled leads. Assume same conditions as Lewin's experiment; step function on input to solenoid (as I have previously prescribed), and 1V induced emf. Actual time is not important, just draw out the basic shape of V(D-A) until it reaches equilibrium.

Now, also plot out the loop current. Assume induced i(t) is 1mA peak.

We may be talking about two different things, and this exercise will sort that out.
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poynt99

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It's not as complicated as it may seem...

« Reply #574 on: 2012-05-01, 13:27:56 »
How about a response Ex?  Wink
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