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Author Topic: Professor Walter Lewin's Non-conservative Fields Experiment  (Read 251642 times)

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I think the only way to settle this is to see who understand how the system works.  

I posted a problem below.

The total induced emf is 1V
Scope reading position in-plane reads .2V
R1 = 1 Ohm
Find R2.  



 ;)  :)

I'll think about it. However, anyone offering an answer should think about this:

Electric potential (emf) only has meaning for electric fields produced by static charges (battery, capacitor stored energy, etc.). It has no meaning for electric fields produced by induction.

...

I started to build a set of graphics to show what is happening and had to break for dinner.


---------------------------
"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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...
The total induced emf is 1V
Scope reading position in-plane reads .2V
R1 = 1 Ohm
Find R2. 


Impossible to find, without knowing the surface topology between the wires connecting the voltmeter and R2. There is a part of the magnetic flux generated by the current in the circular circuit that loops outside around, in a reversed direction.  It is clear that the flux crossing this surface is not null and therefore generates an emf that disturbs the measurement.
If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm.


   

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If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm.

Yes, very possible.

I must question how the total emf induced on the loop is determined.
Measurement?
Using amper-turns-ratio and applied voltage to the solenoid?
Then, there is another problem that hasn't been accepted as fact; what are the measured potentials of the same two points at different secondary loop emf levels?
The ratio of IR1:IR2 will change with secondary loop emf and current levels and meter circuit topology and relative orientation.

The question cannot be answered without making several assumptions.



---------------------------
"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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Impossible to find, without knowing the surface topology between the wires connecting the voltmeter and R2. There is a part of the magnetic flux generated by the current in the circular circuit that loops outside around, in a reversed direction.  It is clear that the flux crossing this surface is not null and therefore generates an emf that disturbs the measurement.
If this surface is null because the wires would tightly follow the circle and then would be kept side by side, from R2 to the voltmeter, then R2=0.25 Ohm.




There is a flux outside in reverse direction, but it's minimal so we'll ignore those. 

R2 = 0.25 Ohms is correct.  You should show us how you derived your answer.

Here is another problem.  I think if you can solve these problems, you can solve any problem in decoupled and in plane measurement. 

R1 = 6 Ohms
R2 = 18 Ohms
In plane reading .2V
Find the loop's induced EMF

   

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It's not as complicated as it may seem...
"If it did not, induction would not work": this is the main point to verify. This objection is logical and I have already tried to verify it.

If the induced E field has an effect on a point charge, then it has also an effect on charges in a small circuit not encircling the varying flux.

You didn't measure anything because 1) you're making an incorrect assumption, and 2) you are going about "measuring" it incorrectly.

Any loop/circuit not including the solenoid will not produce an emf.

If you want to measure the induced electric field in its non-integral form, you should use two small metallic balls spaced about 3/4" apart. They have magnet wire leads twisted together, which connect to a scope. Position the balls so that one is closer to the solenoid axis, and the other is radial outward. You will measure a difference in potential across the two balls.
   

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There is a flux outside in reverse direction, but it's minimal so we'll ignore those. 

R2 = 0.25 Ohms is correct.  You should show us how you derived your answer.

Here is another problem.  I think if you can solve these problems, you can solve any problem in decoupled and in plane measurement. 

R1 = 6 Ohms
R2 = 18 Ohms
In plane reading .2V
Find the loop's induced EMF



On the run here and still wondering how you determine the emf induced but, I'll say .8V is what others would calculate. If things are as they should be, it will be closer to .9V and that still isn't what may actually be induced into the loop with no metering connected.



---------------------------
"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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If you want to measure the induced electric field in its non-integral form, you should use two small metallic balls spaced about 3/4" apart. They have magnet wire leads twisted together, which connect to a scope. Position the balls so that one is closer to the solenoid axis, and the other is radial outward. You will measure a difference in potential across the two balls.

I hadn't thought of that method. Nice  O0


---------------------------
"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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It's not as complicated as it may seem...
I hadn't thought of that method. Nice  O0

Here's a 50 pin chip that was designed to help detect E fields.  8)
   

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It's not as complicated as it may seem...
Here is another problem.  I think if you can solve these problems, you can solve any problem in decoupled and in plane measurement. 

R1 = 6 Ohms
R2 = 18 Ohms
In plane reading .2V
Find the loop's induced EMF

On the run here and still wondering how you determine the emf induced but, I'll say .8V is what others would calculate. If things are as they should be, it will be closer to .9V and that still isn't what may actually be induced into the loop with no metering connected.

Gibbs,

Are you planning on making your future tests somewhat challenging?  :P  :)

The answer to this last one is: loop's induced emf=0.2666V
   
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[/quote]
On the run here and still wondering how you determine the emf induced but, I'll say .8V is what others would calculate. If things are as they should be, it will be closer to .9V and that still isn't what may actually be induced into the loop with no metering connected.



WaveW,

There's really nothing big about this.  You just need to perform KVL to find the answer.

a) dB/dt + V(meter) + I(loop) x 6 Ohms = 0

The two unknown is dB/dt which is what we need to find and I(loop).  We need another equation to find I.  Do another loop.

b) dB/dt + V(meter) + I(loop) x 18 Ohms = 0

In this loop, dB/dt = 0 .  Everything else is known except for I(loop).  It turns out to be .0111 amps.  Solve for equation a).

dB/dt = .2 + .0666

It's lengthy but it's the correct format.

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

Whatever Poynt, I just think you got lucky.  :P

Try this one.  Now we got 2 identical loops, .2666V induced in each loop.  Find the meter reading V.  :P







   

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WaveW,

There's really nothing big about this.  You just need to perform KVL to find the answer.


I have the feeling you aren't going to tell me how you determine the induced emf.

So, you cranked your drive voltage down by 75% or you decreased the area of the loop by 75% or a mix of both?

Have you determined the induced emf by any other method than using KVL during this whole story?



---------------------------
"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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I have the feeling you aren't going to tell me how you determine the induced emf.

So, you cranked your drive voltage down by 75% or you decreased the area of the loop by 75% or a mix of both?

Have you determined the induced emf by any other method than using KVL during this whole story?



WaveW,

The induced emf is the unknown.  It is dB/dt of the loop. It's not 1V given like Lewin's . 

I think KVL is all we need.



   

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It's not as complicated as it may seem...
Whatever Poynt, I just think you got lucky.  :P

Try this one.  Now we got 2 identical loops, .2666V induced in each loop.  Find the meter reading V.
Gibbs, I think we're still going to see 0.2V on that meter.
   
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Gibbs, I think we're still going to see 0.2V on that meter.

Hm... looks like you did got lucky on the last one.

The answer is .466 V   :P



   

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Forgive me Gibbs.

KVL does not and never will apply to non-conservative fields. KVL may produce the correct results depending upon the view of the observer/meter. Potentials measured between points on a closed loop are meaningless.

The only correct measurement you can obtain is zero volts unless you break the loop in one place and measure potential difference between the open ends.

So, the 'now known' emf is taken from KVL?

I submit to you that the real emf is still unknown.



---------------------------
"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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It's not as complicated as it may seem...
Hm... looks like you did got lucky on the last one.

The answer is .466 V  :P

It's possible Gibbs, however I won't be fully convinced until I try it. Good one nonetheless ;)
   
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Forgive me Gibbs.

KVL does not and never will apply to non-conservative fields. KVL may produce the correct results depending upon the view of the observer/meter. Potentials measured between points on a closed loop are meaningless.

The only correct measurement you can obtain is zero volts unless you break the loop in one place and measure potential difference between the open ends.

So, the 'now known' emf is taken from KVL?

I submit to you that the real emf is still unknown.



Hi WaveW,

I've observed your thinking for a while.  I want to ask what do you consider the real value?  I see three value: the induced voltage, Ohm's law voltage, and meter voltage.  They are all different.  The induced voltage and meter voltage change, but the Ohm's law voltage can never change between two points of interest. 

About KVL and non conservative field.  I think Lewin's argument is based on philosophy and not the math.  Lewin's argument is this:

V1+V2+V... = dB/dt

KVL is this:

V1+V2+V... -dB/dt = 0

I  mean they are different only on philosophy.  I think we should come to an agreement with this.    Maybe there is higher philosophy about free energy with this but what important to me is how can we determine parameters based on known parameters.   Yes, even Poynt have a hard time with this. lol

Once again:

The closed loop voltage = dB/dt   (Faraday)
The closed loop voltage = 0 (KVL by bring dB/dt to the other side)


JUST FOLLOW THESE RULES.
   

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Hi WaveW,

I've observed your thinking for a while.  I want to ask what do you consider the real value?  I see three value: the induced voltage, Ohm's law voltage, and meter voltage.  They are all different.  The induced voltage and meter voltage change, but the Ohm's law voltage can never change between two points of interest. 

About KVL and non conservative field.  I think Lewin's argument is based on philosophy and not the math.  Lewin's argument is this:

V1+V2+V... = dB/dt

KVL is this:

V1+V2+V... -dB/dt = 0

I  mean they are different only on philosophy.  I think we should come to an agreement with this.    Maybe there is higher philosophy about free energy with this but what important to me is how can we determine parameters based on known parameters.   Yes, even Poynt have a hard time with this. lol

The induced voltage is always correct and fixed. Ohm's law voltage is always correct but the path must be considered. If the path includes more than one possibility your meter position will convince you the current is going either way and Ohm's law is broken. There are only two cases where the meter voltage is correct:

1. the loop is broken to insert the meter probes (Voltage)
2. the meter and the probes are made up like a single, solid and straight line. That is placed across the diameter of the loop. The reading should be zero.

Using the diagram where I 'failed', I calculate the induced emf to be .799V not considering the change from the induced emf of the previous exam. I don't truly know what that value was but assume that you have changed little and the actual induced emf is almost exactly the same as before.

KVL and Faraday are not differing philosophies. KVL is a round peg and Faraday a square hole. There are ways to make KVL fit. That doesn't mean it is the correct method.

KVL has never included dB/dt. Only the later adaptations of KVL include that. From that point on it is no longer KVL but Maxwell's flavor of Faraday.

Quote
Once again:

The closed loop voltage = dB/dt   (Faraday)
The closed loop voltage = 0 (KVL by bring dB/dt to the other side)


JUST FOLLOW THESE RULES.

Thanks, but if I followed those rules I would not be able to understand much of the machinery I design and diagnose on a daily basis.
All resolvers would cease to function. Older navigational gyroscope systems would instantly fail. Certain types of adjustable transformers and inductive couplers would fail.

This all has little to do with Lewin.

once again:

He was presenting a very old script from as far back as the 70's and WAS taught in all transformer theory classes, as far as I assumed.


I am dismayed with people I thought I knew and have worked side by side with for years. When I mentioned the subject of this discussion his ears perked. Within two minutes of explaining the discussion he shook his head and said he didn't know I was such an idiot.

I grabbed an old resolver from inventory and placed it on his desk. He ask, why? I said I'll answer that question when you can tell me how it works.

His answer was, "Oh, sorry 'bout that".... DUH!







---------------------------
"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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It's not as complicated as it may seem...
First this: Much of what we've been discussing lately doesn't directly relate to the Lewin experiment, but it most certainly is related and helpful to investigate. I have a pretty good grip on the dynamics involved in this experiment, but I am learning some new things here as well.

The induced voltage is always correct and fixed. Ohm's law voltage is always correct but the path must be considered. If the path includes more than one possibility your meter position will convince you the current is going either way and Ohm's law is broken.
Fellows: The PATH is important only when our measurement leads are in-plane with the loop. Ohm's law is NEVER broken when the measurement is taken decoupled.

The loop is a translator; it performs the integral of the E-field at radius r. That integrated E-field results in a loop emf. No emf can be induced if there is no conductive loop around the solenoid.

Now, once the loop is present, it is a real-time translator/integrator, and we must then treat the loop as any "normal" circuit, with an emf and loads. The only difference now between this circuit and one with a DC battery, is the emf source is localized with the battery. With an induced emf, the emf will be spread over any length (or lengths) of wire in the loop, and it will vary in amplitude with time.

The only way to measure all the points on a loop with an induced emf, is to decoupled the leads, otherwise the leads become influenced by the induced E-field. Once decoupled, all measurements and calculations we normally use on DC circuits apply to this one as well. Ohm's Law, and KVL are completely valid.

Where KVL breaks down, is when we are observing or measuring the E-field, or its integral. We inadvertently do this when our measurement leads are in-plane with the induction loop. When the measurement leads are directly in-plane with the induced E-field, they are at its mercy. What ever the E-field is doing, the meter will reflect. The meter can not measure the induced emf in the wire segments this way.

As emf is induced in the wire segments, there is an excess and shortage of electrons created at each end, due to restrictions caused by the resistors. This bunching of charges at each end of the wire segments and resistors "distorts" the induced E-field so that it too is concentrated at the junctions of the resistor elements and the wire segments. This is why moving the in-plane measurement lead probes anywhere in-between the wire segments in the Lewin experiment, makes no discernible difference in what is indicated on the scope.

Now, decouple the measurement leads (raise them normal to the loop) and the induced E-field no longer has any influence on them. You are now able to measure the true emf and potential drops anywhere on the loop, without interfering with the induced emf.
   

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Gibbs,

Did you decide why one of your earlier measurements was zero when the meter circuit was very close to the plane of the loop?

What reason did you choose?


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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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Fellows: The PATH is important only when our measurement leads are in-plane with the loop. Ohm's law is NEVER broken when the measurement is taken decoupled.

Fellows: The PATH is always important when working with non-conservative fields. Ohm's law is NEVER broken.
I think everyone reading this thread must make there own decision.

This is no theory of mine so I cannot concede.

The decoupling methods depicted simply work. There is no argument that the methods fail to prevent unwanted errors by induction.

When you must modify a circuit law to suite your findings you are no longer using that circuit law.



---------------------------
"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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Gibbs,

Did you decide why one of your earlier measurements was zero when the meter circuit was very close to the plane of the loop?

What reason did you choose?

WaveW,

I still use only 1 method.

dB/dt + V(meter) + I x R segment = 0

Since the meter is close, there is no flux going through the meter loop.  Meter reading has to be the same as I x R segment.  It's not zero, it is just extremely small.

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

Hey, I got some more setup for you guys.  Determine the meter reading on this one.  If possible, show some KVL. lol

   

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It's not as complicated as it may seem...
Fellows: The PATH is always important when working with non-conservative fields. Ohm's law is NEVER broken.
I think everyone reading this thread must make there own decision.

WW, when your measurement leads are in-plane, what are you measuring, and what are you "working with"?

And yes, Ohm's Law CAN APPEAR to be broken when working with non-conservative fields, let's make that distinction. I have shown that, and I can prove it. Have you shown and proved the contrary?

My challenge to you is to prove/show that it CANNOT APPEAR to be broken.
   

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

What is the voltage indicated on CH2 of the scope? Assume 1V emf.
   
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Yes, very possible.

I must question how the total emf induced on the loop is determined.
Measurement?

You just need to open the loop and connect its two ends to a voltmeter. Or easier: connect the two voltmeter probes together, each one passing each side of the solenoid.

Quote
Using amper-turns-ratio and applied voltage to the solenoid?

It works also.

Quote
Then, there is another problem that hasn't been accepted as fact; what are the measured potentials of the same two points at different secondary loop emf levels?
The ratio of IR1:IR2 will change with secondary loop emf and current levels and meter circuit topology and relative orientation.

The question cannot be answered without making several assumptions.


There is no potentials. The induced emf doesn't derive from a potential.
At the position of the circuit, the circuit being removed, there is a circular electric field but no potential differences. The potential differences are created when you place a circuit with "obstacles" like an open circuit or resistances. The electric field move the charges which are prevented to move freely and so, positive or negative charges tend to accumulate on each side of the "obstacle" ("positive charges" meaning lack of electrons), creating the potential difference.

We logically assume that we have the maximum potential difference when the "obstacle" is total, i.e. when the circuit is open and we measure the voltage at its two ends. And this gives us the circular electric field (E=U/circuit length). Potential differences can be simply measured with a voltmeter, provided that there is no flux crossing the surface of the measurement circuit.

   
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