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Author Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 311247 times)

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It's not as complicated as it may seem...
0.5V is correct with the way the probes are.  O0 Good job.

Now, of course there will also be 0.5V across points 9' and 1' for a total of 1V, agreed? Is this 1V in-phase or out-of-phase with the combined resistor voltages of 1V?
   
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Now, of course there will also be 0.5V across points 9' and 1' for a total of 1V, agreed? Is this 1V in-phase or out-of-phase with the combined resistor voltages of 1V?



I just see it as a 1V battery connected to a 1k Ohm resistor.  It's in phase in my opinion. 


   

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It's not as complicated as it may seem...
I just see it as a 1V battery connected to a 1k Ohm resistor.  It's in phase in my opinion. 

Are you certain? What happened to Faraday and Lenz?
   
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Are you certain? What happened to Faraday and Lenz?




You just got ahead of me. lol  There is also parasitic inductance in the wire that will not effect by the B field.  Those will have some effect of Lenz.  Looks like I'll talk about it now anyway.


Faraday original equation is contain within an absolute value.  He doesn't care much about polarities.  When we take a close loop integral path dependence, we get two answers, one is positive and one is negative. 

dB/dt = 1V
dB/dt = -1V

If you apply KVL with this form, it won't work.  One will add up to zero and one is doubled.  What makes it works is we have to assign a sign value to dB/dt .  This is where Lenz come in.  The purpose of Lenz is to assign a polarity to dB/dt to make KVL works. 

Now Lenz is not like everyone think it is.  Lenz superimposed a current so we can know the direction of EMF.  The assumption of current does not reflect the reality situation.  It is just an assumption to obtain EMF polarities.  Current can flow against or with the EMF. 

   

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It's not as complicated as it may seem...
FOCUS Gibbs:  8)

Will the voltage indicated on the oscilloscope for the wires be in or out of phase with the voltage indicated for the resistors?
   
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FOCUS Gibbs:  8)

Will the voltage indicated on the oscilloscope for the wires be in or out of phase with the voltage indicated for the resistors?



 ;D At least I don't have to write the last post tomorrow. lol

Voltage against voltage?  Hm... I never thought of that.  Well, if a battery is connected to a resistor, the positive of the battery and resistor join at one junction.  Is this what you mean by out of phase? lol  Then yes.

   

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No, not what I mean.

Phase or relative polarity.

Using decoupled measurements and going around the loop and summing the 4 voltage drops (or PD's as WW calls them), will the total sum be 0V or 2V? Hint; it can't be both.
   
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No, not what I mean.

Phase or relative polarity.

Using decoupled measurements and going around the loop and summing the 4 voltage drops (or PD's as WW calls them), will the total sum be 0V or 2V? Hint; it can't be both.



0 V , but both probes has to move in 1 direction when measuring PD on 4 voltage drops. 

   
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Are you certain? What happened to Faraday and Lenz?

It's sure. The voltages are in phase, provided that you see each of them by turning clockwise or counterclockwise. There is nothing that could change the relative phase of the voltages because they are synchronous with the same dB/dt, which is the sum of the main field and of the opposing field due to Lenz's law.
What is measured is always the voltage induced by the magnetic flux crossing the surface delimited by the 2 points connecting the probe, and the rest of the circuit. We don't measure an independent circuit, we measure the emf of a circuit constituted by both the circuit to be tested and the circuit of the measurement apparatus.
When the probe is connected to two diametrically opposed points, there are 2 half disk surfaces each side, crossed by half the magnetic flux, and contributing equally to half the emf, i.e. 0.5v, provided that the total emf is 1v. It doesn't even depend on which side of the resistance we connect the probe.


   

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It's not as complicated as it may seem...
Gibbs, you are now correct. 0V.

And of course the probes are kept the same polarity orientation as we travel around the loop, that IS how it is always done in ALL cases, agreed?

@Ex: You will NOT measure an in-phase PD across the wires, relative to that across the resistors. You need to pay closer attention to the question and diagram being discussed. Your diagram is in reference to the in-plane measurements, while the last question and diagram reference the decoupled mode of measurement, which btw, is the only method of the two which provides the TRUE PD measurement of all the components in a dynamic circuit such as this.

The E field of the EMF/emf in all circuits, whether placed there physically (such as with a battery) or induced through induction, are always, and without exception, IN OPPOSITION  to the E fields of the loads, such as the resistors in our case.

Therefore, the TRUE PD sum of the loop is 0V at any instant of time, or total length of time.

While measuring the in-plane E field however, it is non-conservative, and is always 1V.

The notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was.  :P
« Last Edit: 2012-03-01, 13:40:01 by poynt99 »
   
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,,,
When the probe is connected to two diametrically opposed points, there are 2 half disk surfaces each side, crossed by half the magnetic flux, and contributing equally to half the emf, i.e. 0.5v, provided that the total emf is 1v. It doesn't even depend on which side of the resistance we connect the probe.



Even if it is decoupled, the reading would be .5V, .4V





The notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was.  :P


I think we can have FE generator even with 0V sum , but still trying to formulate an argument for it. 

   
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Wait.... you guys know how to calculate reading for any two points decoupled or in-plane, right?  Decoupled or in-plane are just the same and used the same method. 



   

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It's not as complicated as it may seem...
Why don't you show us your method Gibbs?
   
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I posted a picture below.  Both of these are in-plane, but the one on the right can also be recognized as decoupled.  The first step is to convert it into in-plane.  If we look at it from the top (bird's view), they are all in-plane.  With this view we know exactly how much magnetic flux passing through the loop we choose. 

For the picture on the right, we go around the loop contain the scope to determine the reading.  This loop contain only half of the flux so we automatically know the total voltage is .5V.  Then we compensate that with the resistor also in the loop.  For right loop, compensate with .9V, for left loop, compensate with .1V.  Either will give .4V .  This is just taking a close loop integral.

However you set up your probes geometry, by looking it from bird's view, you can attain the answer by doing a loop integral. 

   
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@exnihiloest
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It's sure. The voltages are in phase, provided that you see each of them by turning clockwise or counterclockwise. There is nothing that could change the relative phase of the voltages because they are synchronous with the same dB/dt, which is the sum of the main field and of the opposing field due to Lenz's law.
What is measured is always the voltage induced by the magnetic flux crossing the surface delimited by the 2 points connecting the probe, and the rest of the circuit. We don't measure an independent circuit, we measure the emf of a circuit constituted by both the circuit to be tested and the circuit of the measurement apparatus.
When the probe is connected to two diametrically opposed points, there are 2 half disk surfaces each side, crossed by half the magnetic flux, and contributing equally to half the emf, i.e. 0.5v, provided that the total emf is 1v. It doesn't even depend on which side of the resistance we connect the probe.

Well put, fundamentally we are dealing with external forces and charge density at any given point which dictates the phenomena, equations do not dictate reality it is vice versa.

Regards
AC



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Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
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@Poynt99
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The notion that the true PD sum can be 2V (vs. 0V) is ludicrous. You would have a FE generator at your fingertips, if it was
I guess that depends on ones perspective, does your getting a measurement you do not expect mean the conservation of energy has been violated or that you simply do not understand what is happening in reality?. Personally I have measured many things I did not expect nor have many others and a violation of the conservation of energy never crossed my mind --- not once. As such I think using the COE as a mental crutch to justify our personal opinions is and has always been a losing proposition. Every phenomena must stand on it's own two feet without the COE otherwise the concept is completely biased which is anything but scientific, it must prove the COE not vice versa.

Regards
AC


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Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
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If anyone can give the reading on this one, then I think you understand the idea.  Total induced voltage still 1V.







   

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It's not as complicated as it may seem...
If anyone can give the reading on this one, then I think you understand the idea.  Total induced voltage still 1V.

From a bird's eye view, and with the leads in an in-plane configuration, the voltage will be 0.9V. Keep in mind that your right lead must travel outside the loop, because the solenoid is roughly the same diameter. You are showing that lead cutting across, which is not possible with an in-plane configuration.
   
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From a bird's eye view, and with the leads in an in-plane configuration, the voltage will be 0.9V. Keep in mind that your right lead must travel outside the loop, because the solenoid is roughly the same diameter. You are showing that lead cutting across, which is not possible with an in-plane configuration.


But my lead is not outside the loop.  I cut the loop like that on purpose. lol You're right, if it's outside the loop, it will be .9V .  I'm sure if you configured like the picture in a real experiment, you'll get a value, what is that value?  Let's round it to the thousandth placed, so you can't cheat with scope. lol You can use the scope to guide.


BTW, the two measured point are top and bottom, the right intersection does not contact each other.  


You can cut through the loop by wire it up and over the top of the solenoid.


   
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I'll give a hint.  This one reads .4V .





   
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@GibbsHelmholtz
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But my lead is not outside the loop.  I cut the loop like that on purpose. lol You're right, if it's outside the loop, it will be .9V .  I'm sure if you configured like the picture in a real experiment, you'll get a value, what is that value?  Let's round it to the thousandth placed, so you can't cheat with scope. lol You can use the scope to guide.


I find this absolutely remarkable, page after page of in plane/out plane probe lead nonsense and nobody has the common sense to simply stick their probes inside a couple pieces of iron pipe to shield from the influence of an external magnetic field and induced electric fields because the probe leads are effectively inside a faraday cage. Not to mention a little fact called the inverse square law as well as a failure to understand exactly what occurs in a simple loop of conductor with a couple resistors. 11 pages and counting and nobody can seem to explain this phenomena in anything resembling simple terms, Really?, Im really not feeling that expert vibe in fact it seems more like amature night.
Regards
AC


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Comprehend and Copy Nature... Viktor Schauberger

“The first principle is that you must not fool yourself and you are the easiest person to fool.”― Richard P. Feynman
   
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.675V - roughly.

Second thought... .425V

AC,

I've been watching and hoping you would offer your explanation  :)
   
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.675V - roughly.



Yep, but that's too rough. lol A smoother one is .8... V lol


AC, I suspect that is a complement but I have no idea and will try to understand the content. Thanks. lol

   

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It's not as complicated as it may seem...
In your examples Gibbs, are your leads in-plane or decoupled, or a mixture of both? I would appreciate the designation of each lead.
   

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It's not as complicated as it may seem...
11 pages and counting and nobody can seem to explain this phenomena in anything resembling simple terms, Really?, Im really not feeling that expert vibe in fact it seems more like amature night.
Regards
AC

What "phenomena" specifically are you referring to?
   
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