OverUnity Research

When dealing with B fields, your position of measurement matters.



Just noticed, you guys are on top of this:
http://www.overunityresearch.com/index.php?topic=739.0

I am rebooting this topic.
There is a larger one as listed by Matt,but that has changed course in the later posts.

As this topic has come up again in another thread,i will restart this one,so as to not over run the other thread with this topic.

This thread is about the following-->

https://www.youtube.com/watch?v=eqjl-qRy71w

https://www.youtube.com/watch?v=LzT_YZ0xCFY&t=934s

And ElectroBOOMs reply

https://www.youtube.com/watch?v=0TTEFF0D8SA

The circuit is below,although i have change the circuit i am using for testing,as will be seen in the next post i make that contains the video.

Brad

And here is the video of my test setup,and the results obtained from those tests.

https://www.youtube.com/watch?v=ldjIkYqV9xo

I made a comment on one of Lewins video's just yesterday,and he replied today-oddly enough.

His reply
Quote: You should first learn some very basic 1st year college Physics. *I demonstrated at the end of my lecture #16 of my E&M course at MIT that two identical voltmeters attached to the same 2 points in a circuit can show very different values. The reason is that in the case of an induced EMF (Faraday's Law) potentials are no longer determined and thus potential differences depend on the path. This is first year college Physics.* The same physics applies to the secondary windings of transformers as the EMF in the closed loop of secondary windings is induced (Faraday's Law). https://www.youtube.com/watch?v=b7i2uMx7gHo&list=PLyQSN7X0ro22zanLsOcvkaSY-IZqheFYM5&index=8 My demo was suggested by Romer in 1982. I first did it in 1984 in my MIT course (8.02) and I have done it every time that I lectured 8.02. *Electroboom insulted me* *by mentioning in one of his videos that: "The entire reason Dr. Lewin was reading two different voltages was due to bad probing".* It would have been fine if this comment had been made by someone who has little background in Physics. But it's quite worrisome if it is made by someone who has a masters degree in EE. This demo has now become a classic; it's done all over the world at many colleges and universities. In addition, Kirchhoff's loop rule (KVL) is older than Faraday's Law. My translation of the *original work by Kirchhoff from 1845 which was after the publication of Faraday's Law* "(...) it becomes easy to find the requirement, which [voltage] you must fulfill, so that the electrical condition on the [metal] disk can be a stationary one. When we look at a closed loop in [the disk], inside of which no electricity is being fed into [the disk], then the sum of all amounts of electricity in this loop must be 0." The original text can be found at google books in the following link, book pages 497-514: https://books.google.de/books?id=Ig8t8yIz20UC. *Clearly Kirchhoff was fully aware of the prerequisite for his "loop rule", i.e. that d_phi(t)/dt must be equal to zero! If d_phi(t)/dt is not zero, then Faraday's Law has to be used.* *Faraday's Law always holds regardless of the value of dphi(t)/dt that's why it is one of Maxwell's eqs and Kirchhoff's Loop Rule (KVL) is not.*

So it seems Lewin is still insistent that he is correct,and kirchhoff's law dose not apply to this circuit.


Brad

In the below video,i decided to try something a little different,where i use LEDs to prove a voltage difference across each resistor.

Once again,we find the total sum of voltages in the loop do not =0

https://www.youtube.com/watch?v=Lm0gGQQtcws

Kirchhoff's Loop Rule Formula
In any "loop" of a closed circuit, there can be any number of circuit elements, such as batteries and resistors. The sum of the voltage differences across all of these circuit elements must be zero. This is known as Kirchhoff's Loop Rule. Voltage differences are measured in Volts (V). When the current I in the loop is given in Amperes (A) and resistance of circuit elements is given in Ohms (Ω), the voltage difference across a resistor can be found using the formula V=IR .


Brad

One of Lewins later video's about Kirchhoff's loop rule .

https://www.youtube.com/watch?v=wz_GqO-Urk4

Well, I think it is also the solution for Tariel Kapanadze device. The current in single loop can be different in various parts of circuit  IF there is a transformer which speed up electrons in coil by the use of polarized RADIO wave aka Tesla longitudinal wave aka radiant energy. The schematic I saw recently which may leaked from some source close to Kapanadze (or may not - I don't know) seem to confirm what I thought long time ago.
Would be interesting to know whatis Dr Lewin opinion on such hypothesis

Brad

By not using a solenoid to drive both chunks of wire you have deviated from the original circuit whereby both connecting wires are induced.

If you redraw your circuit such that the bottom loop is the secondary of the toroid transformer, then the 82 mV is the correct reading across that chunk of wire, as it threads the toroid.

However, by also threading the ground lead through the toroid, you have effectively nulled out the useful 82 mV for the measurement  so surely you measured zero mV.

The top loop is just a wire connecting the two resistors, there is no induction in it as it does not actually thread the toroid. It is"downstream" of the resistors.

Then the correctly measured 82mV equates to 74.5 uA in the loop. (82mV/1100 ohms=74.5uA)

Using the current then we can find the voltage on each resistor

For the 1k ohm we have 74.5 uA X 1000 ohms = 74.5 mV

For the 100 ohms we have 74.5 uA X 100 ohms = 7.45 mV

Adding those to values we arrive back at the 82 mV

Surely you see that if you removed the resistors and only left the bottom wire loop, you have a transformer with a single turn secondary of 82mV, so that reading is correct for that chunk of wire that threads the toroid and takes part in the induction, whereas the top wire does not.

Your idea to use the toroid does have merit in reducing measurement artifacts but it must be applied to both chunks of wire to simulate a solenoid induction method in that each chunk of connecting wire will act as the secondary of a transformer. You can do this by also looping the top wire through the toroid with the correct phasing or using an additional identically driven toroid on the top wire.

In this last case you would put the toroid primaries in series and then wind up with 41mV measured on each secondary chunk of wire, which if properly phased will add to give you 82mV of drive.

Hope this helps

edit: corrected mA to uA per Gyula who noticed the error

Quote from: ion on 2019.06.22, 18:16:30
Brad

By not using a solenoid to drive both chunks of wire you have deviated from the original circuit whereby both connecting wires are induced.

If you redraw your circuit such that the bottom loop is the secondary of the toroid transformer, then the 82 mV is the correct reading across that chunk of wire, as it threads the toroid.

However, by also threading the ground lead through the toroid, you have effectively nulled out the useful 82 mV for the measurement  so surely you measured zero mV.

The top loop is just a wire connecting the two resistors, there is no induction in it as it does not actually thread the toroid. It is"downstream" of the resistors.

Then the correctly measured 82mV equates to 74.5 mA in the loop. (82mV/1100 ohms=74.5mA)

Using the current then we can find the voltage on each resistor

For the 1k ohm we have 74.5 mA X 1000 ohms = 74.5 mV

For the 100 ohms we have 74.5 mA X 100 ohms = 7.45 mV

Adding those to values we arrive back at the 82 mV

Surely you see that if you removed the resistors and only left the bottom wire loop, you have a transformer with a single turn secondary of 82mV, so that reading is correct for that chunk of wire that threads the toroid and takes part in the induction, whereas the top wire does not.

Your idea to use the toroid does have merit in reducing measurement artifacts but it must be applied to both chunks of wire to simulate a solenoid induction method in that each chunk of connecting wire will act as the secondary of a transformer. You can do this by also looping the top wire through the toroid with the correct phasing or using an additional identically driven toroid on the top wire.

In this last case you would put the toroid primaries in series and then wind up with 41mV measured on each secondary chunk of wire, which if properly phased will add to give you 82mV of drive.

Hope this helps

I would agree. ;)  It is not the same where only one resistor connection is the source to the circuit.

Reminds me of when TK was working on the Rose Ainsley circuit where it was suggested that using 2 leds in parallel but opposite polarity in line with the circuit loop to see which way current was flowing more. But Tk inserted a 1:1 transformer primary with one led across that and the other led across the secondary.  To me, that is not the same as 2 leds alone.

Like why have the transformer at all? If one led across the primary is biased alone, then why couldnt the other led also just be in parallel with the other but opposite polarity? ;)

Mags

Quote from: ion on 2019.06.22, 18:16:30
...
Then the correctly measured 82mV equates to 74.5 mA in the loop. (82mV/1100 ohms=74.5mA)

...

Dear Ion,

Please check : 82 mV / 1100 Ohm = 74.5 uA  (the mA is surely a typo).

Gyula

Quote from: gyula on 2019.06.22, 23:14:49
Dear Ion,

Please check : 82 mV / 1100 Ohm = 74.5 uA  (the mA is surely a typo).

Gyula0

Surely for some people the danger of being old and doing quick math in your head, the error applies to the other current transforms also.

But is the concept I am implying correct, can you comment on that?

Thanks

 author=ion link=topic=2369.msg75449#msg75449 date=1561227390]
Brad

QuoteIf you redraw your circuit such that the bottom loop is the secondary of the toroid transformer, then the 82 mV is the correct reading across that chunk of wire, as it threads the toroid.

I would disagree with that ION (but i could be wrong here),and here is why.
As the current flowing through the loop is the same at every point,then the bottom circuit link having the same resistance value as the top circuit link must also have the same voltage value across it !ohms law!.

QuoteBy not using a solenoid to drive both chunks of wire you have deviated from the original circuit whereby both connecting wires are induced.

Both connecting wires are induced by the electric field of the toroid transformer.
The induction is not limited to the small portion of wire threaded through the toroids center.

QuoteThe top loop is just a wire connecting the two resistors, there is no induction in it as it does not actually thread the toroid. It is"downstream" of the resistors.

As above.
Tests i have carried out before show that the secondary loop can be induced at any point throughout the loop,and is not limited to the small portion threading the toroid.

QuoteThen the correctly measured 82mV equates to 74.5 mA in the loop. (82mV/1100 ohms=74.5mA)

Using the current then we can find the voltage on each resistor

For the 1k ohm we have 74.5 mA X 1000 ohms = 74.5 mV

For the 100 ohms we have 74.5 mA X 100 ohms = 7.45 mV

Adding those to values we arrive back at the 82 mV

I know it is only a typo,but it is uA not mA
The 82mV is the maximum voltage obtainable from the single loop with the turn ratio and frequency used regardless of the total resistance value.

QuoteSurely you see that if you removed the resistors and only left the bottom wire loop, you have a transformer with a single turn secondary of 82mV, so that reading is correct for that chunk of wire that threads the toroid and takes part in the induction, whereas the top wire does not.

Ohms law states that if the current flowing through the loop is the same at all points of the loop(which it is),then the voltage across any two resistances with the same value within that loop must have the same voltage across them. As the bottom loop has the same resistance value as the top loop,and has the same value of current running through it,then the voltage across that bottom loop must be the same as the voltage across the top loop.

As stated above,the secondary loop is induced by the electric field,and not the magnetic field. The magnetic field is confined to the core,while the electric field is not. All parts of the loop are induced by the electric field,and induction is not limited to just the small portion of the loop that threads the toroid.

By placing the scope across the bottom loop,where it is around the outside of the toroid,you now have created a second circuit-which is the scope it self,where the resistance value of that circuit is now the impedance value of the scope.

As i said,i could be wrong here,but there is a quick test i can do to see if i am correct,which i will do shortly.


Brad

QuoteI would disagree with that ION (but i could be wrong here),and here is why.
As the current flowing through the loop is the same at every point,then the bottom circuit link having the same resistance value as the top circuit link must also have the same voltage value across it !ohms law!.

Note that the bottom wire is an active voltage source as it is the single turn secondary of your toroid. This voltage feeds two resistors and a piece of wire which is also a resistor although very low in value, so we can expect it will have close to zero volts across it. It is a load, not a voltage source.

QuoteBoth connecting wires are induced by the electric field of the toroid transformer.
The induction is not limited to the small portion of wire threaded through the toroids center.

.

I would disagree here.

QuoteAs above.
Tests i have carried out before show that the secondary loop can be induced at any point throughout the loop,and is not limited to the small portion threading the toroid.

But you measure zero volts on the top wire, thus it is not induced the same as the bottom wire when measured normally (as you would normally measure the secondary of a transformer) You assume the 82mV is an error that you must correct by nulling it with the ground lead of the scope passing through the core. I say 82mV  is the correct voltage and the nulling trick is not needed and confuses the issue.

QuoteI know it is only a typo,but it is uA not mA
The 82mV is the maximum voltage obtainable from the single loop with the turn ratio and frequency used regardless of the total resistance value.

We agree here.

QuoteOhms law states that if the current flowing through the loop is the same at all points of the loop(which it is),then the voltage across any two resistances with the same value within that loop must have the same voltage across them. As the bottom loop has the same resistance value as the top loop,and has the same value of current running through it,then the voltage across that bottom loop must be the same as the voltage across the top loop.

True if all resistances are passive, but you have one as an active voltage source so it is not a passive loop.

QuoteAs stated above, the secondary loop is induced by the electric field, and not the magnetic field. The magnetic field is confined to the core,while the electric field is not. All parts of the loop are induced by the electric field,and induction is not limited to just the small portion of the loop that threads the toroid.

I believe this is true for solenoid induction, not so for a toroid singly threaded.

QuoteBy placing the scope across the bottom loop, where it is around the outside of the toroid, you now have created a second circuit-which is the scope it self, where the resistance value of that circuit is now the impedance value of the scope.

Then I must ask how you would normally use your scope to measure the secondary of a transformer? Do you always try to null the voltage to zero?

QuoteAs i said, i could be wrong here,but there is a quick test i can do to see if i am correct,which i will do shortly.

Brad


We shall see. I maintain that with the toroid feeding the loop, you have a simple transformer secondary feeding three resistors, 1000, 100 and one is a fraction of an ohm (the connecting wire). Trying  not to confuse the issue, as this is not the same circuit as solenoid induction, whereby the upper wire would have induction thus becoming an active voltage source like the bottom wire.

All for now, good luck with further test and discussion,  I may not be back as I entered the Lewin fray reluctantly.

Regards

Quote from: ion on 2019.06.23, 01:57:57


All for now, good luck with further test and discussion,  I may not be back as I entered the Lewin fray reluctantly.

Regards

I would encourage you to stay,as if i am incorrect,i would like you to correct me.
If i am correct,then would this not also help you further your understandings?,even though it is more likely that i am wrong.

Below is your picture i have added a question to.
That question is-->if there is 87mV across R2,then what would be the voltage across R1 ?
Dose not ohms law state that it should be 870mV,as the current flowing through the loop must be the same at all points of that loop?.


Brad

With 87 mV across the 100 ohm R2, the loop current must be  0.87mA

Thus the voltage across 1000 ohm R1 must be 870mV (naturally as it is  10X higher in resistance it will have 10X higher voltage)

The total applied voltage must be 957mV (not 82mV as shown on my drawing)

This 957mV being the sum of 87mV plus 870mV (this is the new transformer output)

Hope I haven't created another slipped decimal.

Regards

I beat this topic to death years ago, and I think proved beyond a doubt that KVL holds in all cases.

ElectroBOOM is absolutely correct, and Lewin is simply being short-sighted and stubborn.

ElectroBOOM went to great lengths to explain things, but another easier way in my opinion is to simply remove the measurement probes from the experiment. How do we do this? Easy, just bring your probe leads down from a hook situated above the ring so that the leads are placed across the resistors and wire segments perpendicular to the plane the ring is situated. I would suggest at least 1m above the apparatus.

This effectively removes the probes and leads from the experiment apparatus and will allow you to not only measure the actual voltage across the 2 resistors, but across the 2 wire segments as well. If one does this and then adds all 4 voltages (going the same direction) in the loop, the result will be close to zero.

Quote from: poynt99 on 2019.06.23, 03:53:22
I beat this topic to death years ago, and I think proved beyond a doubt that KVL holds in all cases.

ElectroBOOM is absolutely correct, and Lewin is simply being short-sighted and stubborn.

ElectroBOOM went to great lengths to explain things, but another easier way in my opinion is to simply remove the measurement probes from the experiment. How do we do this? Easy, just bring your probe leads down from a vertical hook so that the leads are placed across the resistors and wire segments perpendicular to the plane the ring is situated.

This effectively removes the probes and leads form the measurement and will allow you to not only measure the actual voltage across the 2 resistors, but across the 2 wire segments as well. If one does this and then adds all 4 voltages (going the same direction) in the loop, the result will be close to zero.

That's why I reluctantly got into the fray as this has all been worked out by Poynt years ago.

I only jumped in when Brad added a new twist by using a toroid to inject the current into the loop and this is not the same as Lewins solenoid induction method that affects all wires in the loop.

I was trying to show that Brad's circuit is a simple transformer circuit, but as I have now exceeded my mental capacity, I will exit left.

Good to see you here Darren.

Regards

Hey Ernie, if we never push the envelope, then... :)

Keep pushing my friend.

Quote from: ion on 2019.06.23, 01:57:57
Note that the bottom wire is an active voltage source as it is the single turn secondary of your toroid. This voltage feeds two resistors and a piece of wire which is also a resistor although very low in value, so we can expect it will have close to zero volts across it. It is a load, not a voltage source.
.

I would disagree here.

But you measure zero volts on the top wire, thus it is not induced the same as the bottom wire when measured normally (as you would normally measure the secondary of a transformer) You assume the 82mV is an error that you must correct by nulling it with the ground lead of the scope passing through the core. I say 82mV  is the correct voltage and the nulling trick is not needed and confuses the issue.

We agree here.


True if all resistances are passive, but you have one as an active voltage source so it is not a passive loop.

I believe this is true for solenoid induction, not so for a toroid singly threaded.

Then I must ask how you would normally use your scope to measure the secondary of a transformer? Do you always try to null the voltage to zero?
Brad


We shall see. I maintain that with the toroid feeding the loop, you have a simple transformer secondary feeding three resistors, 1000, 100 and one is a fraction of an ohm (the connecting wire). Trying  not to confuse the issue, as this is not the same circuit as solenoid induction, whereby the upper wire would have induction thus becoming an active voltage source like the bottom wire.

All for now, good luck with further test and discussion,  I may not be back as I entered the Lewin fray reluctantly.

Regards

Here is the first video in relation to the above.

https://www.youtube.com/watch?v=HKoC3yOEtkw

The next one is uploading now.
I have also added a diagram showing the electric field around a toroid transformer.
As you can see,the whole loop of my secondary would be within that electric field,thus allowing induction via that field to take place anywhere around that loop,and not confined to the portion of the loop that is through the center hole of the toroid.

Brad

Quote from: poynt99 on 2019.06.23, 03:53:22
I beat this topic to death years ago, and I think proved beyond a doubt that KVL holds in all cases.

ElectroBOOM is absolutely correct, and Lewin is simply being short-sighted and stubborn.

ElectroBOOM went to great lengths to explain things, but another easier way in my opinion is to simply remove the measurement probes from the experiment. How do we do this? Easy, just bring your probe leads down from a hook situated above the ring so that the leads are placed across the resistors and wire segments perpendicular to the plane the ring is situated. I would suggest at least 1m above the apparatus.

This effectively removes the probes and leads from the experiment apparatus and will allow you to not only measure the actual voltage across the 2 resistors, but across the 2 wire segments as well. If one does this and then adds all 4 voltages (going the same direction) in the loop, the result will be close to zero.

As you know Poynt,i did all these tests along side you,and with your guidance,and we did end up with a sum of 0v.
But something was wrong there,and it would seem that we can place our measurement equipment in any position to gain the results we want to see.

In my test setup,i have removed the magnetic field problem,and now use only the electric field to induce the secondary loop. Now we can have our scope leads where ever we want,and there is no adverse effect due to the magnetic field.

I have already tried having the scope probes vertical as we did in the last test,and it makes no difference what so ever,as it shouldn't in this case.

The video's i am uploading show that my measurements are correct,and if you see a problem with them,then let me know.


Brad

Quote from: ion on 2019.06.23, 04:00:44
That's why I reluctantly got into the fray as this has all been worked out by Poynt years ago.

I only jumped in when Brad added a new twist by using a toroid to inject the current into the loop and this is not the same as Lewins solenoid induction method that affects all wires in the loop.


Regards

No it is not the same as Lewins tests.
My setup still induces a current through the closed loop,but eliminates the magnetic field problems associated with Lewins setup.

However,the results still show the same non 0 voltage around the loop.
Ohms law also states the voltages cannot sum to 0 in either case as we found when i did the tests with Poynt,so something is wrong here,and i want to find out what.

Looking at lewins circuit below,we can see that both the top link and bottom link will cancel each other out,and there sumed voltage will be 0.
This leaves the two different value resistors,and the voltage across them.
If the current through the circuit is the same at all points in time,then the voltage across each resistor cannot be the same value to cancel each other out,and result in 0 volts-->ohms law says no way.


Brad

Quote from: poynt99 on 2019.06.23, 03:53:22
I beat this topic to death years ago, and I think proved beyond a doubt that KVL holds in all cases.

ElectroBOOM is absolutely correct, and Lewin is simply being short-sighted and stubborn.

Message received via email today from Walter.

Quote:  *By teaching students that KVL always works, makes many believe that the closed loop integral of E dot dL is always zero. Which is not true.* Even ElectroBOOM who has a masters in EE believed that, and he therefore concluded (he was not the only one) that two identical voltmeters attached to the same 2 points in a circuit MUST always show the same values. In other words he does not even know that in the case of an induced EMF potential differences are no longer determined. None of my thousands of MIT students who take my 8.02 would make this mistake which is a HUGE BLUNDER! They have seen my Lecture #16 and the great demo with the 2 voltmeters which I do at the end! I believe that I teach Faraday's Law in the way it should be taught. All my students know that KVL is part of Faraday's Law and they have no problems with the concept that potential differences in a circuit can be path dependent. *In conclusion I believe that teaching students that KVL always holds is not only wrong Physics but you set them up for making terrible mistakes/blunders as ElectroBOOM did.*


Brad

I have been reading the other thread listed in Matts first post here--

https://www.overunityresearch.com/index.php?topic=739.25

It would seem to me that more attention should have been paid to what member Harvey had to say,but it also seems that !some! had no interest in what he had to say,and just dismissed him like he was nothing at all--very sad,as he seemed to be very switched on in the art of EE.


Brad

Here is another video regarding the placement of the scope probe leads across the link that passes through the toroid.

As you will see,things do not add up,and having the scope hooked up around the toroid is a no go.

The video also eliminates the thought that the connecting link wire passing through the toroid is the only point of induction in the circuit.
Along with it,2 other videos i made regarding the same thing,only on a larger scale.

https://www.youtube.com/watch?v=UkUNZ5ik4go

https://www.youtube.com/watch?v=n84bPUWLLLY

https://www.youtube.com/watch?v=cqXC7Qjrh1Y



Brad

Quote from: poynt99 on 2019.06.23, 03:53:22
I beat this topic to death years ago, and I think proved beyond a doubt that KVL holds in all cases.

ElectroBOOM is absolutely correct, and Lewin is simply being short-sighted and stubborn.

ElectroBOOM went to great lengths to explain things, but another easier way in my opinion is to simply remove the measurement probes from the experiment. How do we do this? Easy, just bring your probe leads down from a hook situated above the ring so that the leads are placed across the resistors and wire segments perpendicular to the plane the ring is situated. I would suggest at least 1m above the apparatus.

This effectively removes the probes and leads from the experiment apparatus and will allow you to not only measure the actual voltage across the 2 resistors, but across the 2 wire segments as well. If one does this and then adds all 4 voltages (going the same direction) in the loop, the result will be close to zero.

But mistakes still could have been made,and i think they were.

Walter Lewin has been at this sort of stuff for years on end,and he knows his stuff.
You also know your way around electronics,that is for sure.
But if the great Walter Lewin can make a mistake,is it no possible you could have as well?.

I would like to draw your attention to the following test i carried out.
The circuit is hard to draw out,due to multiple loops going through a toroid,so i will explain it instead.

If it is true that all voltage sums around a loop equal 0,then the same should be true for multiple loops, as 1 x 0 or 10 x 0 =0

In the video,i took two identical lengths of wire of the same size.
I soldered the 100 ohm resistor to one end of one length of wire. I glued the resistor to the former,and then wound the wire around the former. This gave me 15 turns. I then soldered the 1K ohm resistor to the end of that wire. I then soldered the second length of wire to the other end of the 1K resistor,and wrapped the second length of wire around the former in the same direction as the first. This of course gave me another 15 turns,where i then soldered that end to the leftover end of the 100 ohm resistor.

Now we have a total of 30 turns going through the hole of the toroid,which means 15 turn on each side of each resistor.

As we have 2 lots of 15 turns,the voltages across each of those 15 turns will cancel each other out,and give us a value of 0v. This leaves only the two resistor voltages,which must also sum to 0v in order for Kirchhoff's loop rule to hold. A visual method was used instead of using the scope,in that an LED was placed across each resistor.
As you can clearly see in the video,1 LED lights brightly,and the other dose not.

Using your measurement method at each of the 4 measurement point's,we can clearly see that the sum voltage is not 0,and Kirchhoff's loop rule dose not apply.

https://www.youtube.com/watch?v=Lm0gGQQtcws


Brad

Quote from: TinMan
As we have 2 lots of 15 turns,the voltages across each of those 15 turns
will cancel each other out,and give us a value of 0v.

Providing the two windings do in fact develop equal voltages.
Was a measurement of those voltages taken at some point to assure
complete cancellation?

What other explanation could there possibly be?

Quote from: muDped on 2019.06.23, 09:56:24
Providing the two windings do in fact develop equal voltages.
Was a measurement of those voltages taken at some point to assure
complete cancellation?

What other explanation could there possibly be?

Yes the voltages have been measured,and there is a 4mVrms variation between the two.
That is very close when the voltages measured are 1.32VRMS.
But why would we expect any different,when both windings have the 1100 ohm series resistance,both windings have the same amount of turns ,using the same length of wire of the same size.

So 1.326 minus 1.322 leaves us with 4mV.
I think we can safely say there is more than a 4mV variation across the two LED/resistor combo's--don't you think ?,as the 10mm LEDs need 2.6 volts across them before they start to conduct.
We can clearly see one pumping out the light,while the other is completely dead.
Of course we can also place the scope across each LED to see what the voltage actually is.
But the point here is that all the sumed voltages in the loop do not equal 0v,nor dose it come even close.
ATM,we have around +2.71VRMS-somehow  ???


Brad

There is nothing wrong with the conclusion that KVL holds in Lewin's experiment.

There is nothing wrong with the test methodology I used and put forward. The results clearly indicate that KVL holds. I have proven it on my bench.

You are welcome to disagree, but I am not going down the rabbit hole being offered.

Peace out.

Quote from: Matt Watts on 2014.03.18, 17:12:57

When dealing with B fields, your position of measurement matters.



Just noticed, you guys are on top of this:
http://www.overunityresearch.com/index.php?topic=739.0

Precisely.

Quote from: poynt99 on 2019.06.23, 19:23:57
Precisely.

Only with my setup,there are no B fields that can interact with any of the scope probes to cause measurement error.


Brad

You should listen intently to ION, as he is 100% correct, and you are leading yourself astray once again.

Here are some corrections for you to consider:

a) In Lewin's original setup, and the one I tested, the E field is what is directly inducing the loop current, not the B field.
b) In your toroid setup, the B field is what is inducing the loop current.
c) When going around the loop to reconcile all the measured voltages, we ADD them, not subtract them as you did in your first video. In other words, the SUM of the series voltages in a loop is zero. The SUM of your two resistor measurements is 80.3mV. (see e) below)
d) Proper measurement of the wire part of the loop is done around the outside of the toroid, and if you account for the polarity and voltage, you will find that when added to the other two resistor voltages, you will be left with something close to zero.
e) Your "proper" measurement of the toroid section of the wire was about 81mV. The two resistor voltages summed is 73mV + 7.3mV = 80.3mV. Summed with the series-subtracting wire voltage leaves you with 0.7mV.

KVL holds.

Quote from: poynt99 on 2019.06.24, 00:20:31
You should listen intently to ION, as he is 100% correct, and you are leading yourself astray once again.

Here are some corrections for you to consider:

a) In Lewin's original setup, and the one I tested, the E field is what is directly inducing the loop current, not the B field.
b) In your toroid setup, the B field is what is inducing the loop current.
c) When going around the loop to reconcile all the measured voltages, we ADD them, not subtract them as you did in your first video. In other words, the SUM of the series voltages in a loop is zero. The SUM of your two resistor measurements is 80.3mV. (see e) below)
d) Proper measurement of the wire part of the loop is done around the outside of the toroid, and if you account for the polarity and voltage, you will find that when added to the other two resistor voltages, you will be left with something close to zero.
e) Your "proper" measurement of the toroid section of the wire was about 81mV. The two resistor voltages summed is 73mV + 7.3mV = 80.3mV. Summed with the series-subtracting wire voltage leaves you with 0.7mV.

KVL holds.

Im going to disagree with you Poynt,and i think you have it backwards.
In Lewins setup,it is the B field cutting through the loop.
In my setup,the B field is contained within the toroid.

I have provided video's showing the loop shielded from any magnetic field by way of steel tubes placed through the center hole of the toroid,and the loop wire then threaded through that steel tube.

The drawing i peovided clearly shows the electric field of a toroid,and also shows the B field is contained within the toroid core.


Brad

Brad

Take a good look at the Loopy looped.jpg you have supplied.

It is assumed there is a loop current of 1 ampere, OK we can agree on that.

We can also agree that the loop current produces the designated voltages across the resistors per Ohms Law, even without measuring them.

Now meditate on this:

How does the 1 ampere come to exist  in the loop and what is the actual voltage produced across the two wire segments?

Hint: you show a "primary coil", now where is(are) the secondary coil(s)?

Regards

p.s. this is my last post for this evening, hitting the sack, maybe see you in the morning after a few cups of coffee..

Quote from: ion on 2019.06.24, 02:16:24
Brad

Take a good look at the Loopy looped.jpg you have supplied.

It is assumed there is a loop current of 1 ampere, OK we can agree on that.

We can also agree that the loop current produces the designated voltages across the resistors per Ohms Law, even without measuring them.

Now meditate on this:

How does the 1 ampere come to exist  in the loop and what is the actual voltage produced across the two wire segments?

Hint: you show a "primary coil", now where is(are) the secondary coil(s)?

Regards

p.s. this is my last post for this evening, hitting the sack, maybe see you in the morning after a few cups of coffee..

The 1 ampere is produce via the magnetic field cutting through the loop,which consists of A ,B, and the two resistors.

As Lewin clearly states,the two connecting wires (A and B) are super conductive,and so with 1 ampere flowing through them,there will be 0 volts across them.

As stated above,the secondary coil consist of A,B, and the two resistors.

The 9 ohm resistor will have 9V across it,and the 1 ohm resistor will have 1V across it.
So how dose this sum to 0 volts around the loop ?.

If the measurement leads are on the same plane as the loop,they become the 3rd and 4th loop,and will develope there own voltage due to the magnetic field also cutting through those loops,which will give incorrect voltage values across the resistors.

This is my very point when i say you cannot place the scope probe and ground around the toroid when trying to measure the voltage across the connecting wire that passes through the toroid,as you are doing the very same thing,and forming a second loop that the electric field is passing through. So in doing this,like the lewin test,you are reading an incorrect voltage across that small piece of wire.

I know i am right on this,and have proven that.
I also have other ways of showing the scope dose not read the correct voltage in doing it that way.
All you are doing is forming a second loop,which is the short piece of wire through the toroid,and the scope it self.

The voltage around lewins circuit dose not sum to 0,nor dose the voltage in any of my toroid circuits--ohms law clearly shows that.

Kirchhoffs loop rule fails,but Faradays law holds,as dose ohms law.


Brad

 author=TinMan link=topic=2369.msg75510#msg75510 date=1561350574]

QuoteThe 1 ampere is produce via the magnetic field cutting through the loop,which consists of A ,B, and the two resistors.

We agree on this

QuoteAs Lewin clearly states,the two connecting wires (A and B) are super conductive,and so with 1 ampere flowing through them,there will be 0 volts across them.

First assumption a fail, A and B are not superconductive, they are transformer secondaries and will have a voltage on them, they are active potential sources. Even if they were superconductive transformer secondaries they could still have voltage induced in them.

QuoteAs stated above,the secondary coil consist of A,B, and the two resistors.

Go back and meditate on what voltage you would actually expect the segments A and B to have developed in them.(Hint: not 0)

QuoteThe 9 ohm resistor will have 9V across it,and the 1 ohm resistor will have 1V across it.
So how dose this sum to 0 volts around the loop ?.

Before you get to that question, you must answer how the resistors come to have a voltage impressed on them.

When you identify the voltage sources that develop the current in the loop, it will be clear.

QuoteIf the measurement leads are on the same plane as the loop,they become the 3rd and 4th loop,and will develope there own voltage due to the magnetic field also cutting through those loops,which will give incorrect voltage values across the resistors.

Forget about measurement leads it is not important for this exercise as Ohms Law is all that is needed.

QuoteThis is my very point when i say you cannot place the scope probe and ground around the toroid when trying to measure the voltage across the connecting wire that passes through the toroid,as you are doing the very same thing,and forming a second loop that the electric field is passing through. So in doing this,like the lewin test,you are reading an incorrect voltage across that small piece of wire.

Then how do you normally measure the voltage on the secondary of a transformer. I have asked this before,but you did not reply?

QuoteI know i am right on this,and have proven that.

Don't be so sure

You have rigged the test to comply with the first erroneous premise of Lewin, that the segments are superconductive and will have zero volts on them. You actually made the correct measurement of 82mV but discarded this truthful measurement to comply with Lewin's error.

QuoteI also have other ways of showing the scope dose not read the correct voltage in doing it that way.
All you are doing is forming a second loop,which is the short piece of wire through the toroid,and the scope it self.

Forget all this and just demonstrate how you would normally read the secondary voltage of a toroidal transformer with a single turn secondary.
Surely you don't thread the scope ground lead also through the toroid or all toroidal transformers would read zero volts using your method.

QuoteThe voltage around Lewins circuit dose not sum to 0,nor dose the voltage in any of my toroid circuits--ohms law clearly shows that.

Kirchhoffs loop rule fails,but Faradays law holds,as dose ohms law.


Brad

It does not sum to zero in Lewin's case because of Lewin's erroneous assumption.  In following the erroneous assumption  Kirchhoff fails. When you get rid of that assumption it works.

Over and Out

Quote from: TinMan on 2019.06.24, 04:29:34
...
The 9 ohm resistor will have 9V across it,and the 1 ohm resistor will have 1V across it.
So how dose this sum to 0 volts around the loop ?.
...

A battery imposes a voltage, this potential difference generates the electric field, the electric field generates the force on the electrons and therefore the current.

An induced voltage is the opposite, a force, contrary to what its unit suggest, the EMF.  We have the force on the electrons first, so the current. Then, what the electrons encounter in their path, such as a resistance, will oppose the force and thus create potential differences along their path.
If the circuit is open, the resistance is infinite, then a potential difference equal to the EMF is measured. When you have 10v between the two open terminals, you have also -10v along the conductor loop, not zero.

The surprise of the measurements comes from there, it is the mistake to reason as if the EMF was the voltage of a battery. It's the force on the electrons and indirectly, a voltage.

 author=ion link=topic=2369.msg75534#msg75534 date=1561385624]
author=TinMan link=topic=2369.msg75510#msg75510 date=1561350574]

QuoteGo back and meditate on what voltage you would actually expect the segments A and B to have developed in them.(Hint: not 0)

As there is 10 ohms of resistors in the loop,i would expect the voltage across A and B to be very low,as there resistance value would be very low.

QuoteBefore you get to that question, you must answer how the resistors come to have a voltage impressed on them.

Because they get a buildup of charge on one side,and a lesser amount on the other.
Same as a water pipe that has water flowing through it,with a restrictor in the center of it.

QuoteWhen you identify the voltage sources that develop the current in the loop, it will be clear.

It is the current in the loop that caused the voltage across the resistors and A and B.
The voltage source is my SG. The electric field is what causes a current to flow in the loop.
This flow of current in turn creates a magnetic field,as in all transformers.

QuoteForget about measurement leads it is not important for this exercise as Ohms Law is all that is needed.

Indeed.

QuoteThen how do you normally measure the voltage on the secondary of a transformer. I have asked this before,but you did not reply?

Not by wrapping the scope leads around the transformer core to form another loop,where the leads are placed across only a small portion of the transformers turns.

QuoteYou have rigged the test to comply with the first erroneous premise of Lewin, that the segments are superconductive and will have zero volts on them. You actually made the correct measurement of 82mV but discarded this truthful measurement to comply with Lewin's error.

That 82mV shows exactly the voltage that would be had with a single loop.
In order for that 82mV to be subtracted from the 82mV of the circuit,so as we end up with our 0 volts,it would have to be of the opposite polarity to that of the single loop,so as when added to the sum of the two resistor's,it will be close to 0,but it is of the same polarity,not opposite..

On that note,with the scope probe through the toroid,and clipped to the ground lead of the probe,we get 82mV  C.C. Now,what is the scope measuring if it is not measuring it's own circuit voltage due to the now newly formed looped circuit--which is the scope and leads them self.

Quote Poynt: d) Proper measurement of the wire part of the loop is done around the outside of the toroid, and if you account for the polarity and voltage, you will find that when added to the other two resistor voltages, you will be left with something close to zero.

So this means that the polarity should be opposite to that of the loop.
When the resistor voltages are in the + side of the wave,then the probe looped around the toroid should be in the - side of the wave,so as when added they sum to 0.

Now,using the differential measurement method,where the ground leads of the scope are floating,i can accurately measure the voltage across all points of the loop using two channels of the scope.
The voltage measured across A and B are 2mV and 3mV
Across the two resistors,we have for the 1k 78mV,and for the 100R 7.8mV
So i ask again,how dose that all add to 0 volts around the loop?.

QuoteForget all this and just demonstrate how you would normally read the secondary voltage of a toroidal transformer with a single turn secondary.
Surely you don't thread the scope ground lead also through the toroid or all toroidal transformers would read zero volts using your method.

Is the secondary looped on it self as the one we are testing,or is it open ?
I answered this question above.

QuoteIt does not sum to zero in Lewin's case because of Lewin's erroneous assumption.  In following the erroneous assumption  Kirchhoff fails. When you get rid of that assumption it works.

Forget Lewin for a moment,and let's look at my circuit.
Below is a diagram.
Please fill in the voltages for all test points,so as i can see how you get them to all sum to 0.

Thanks


Brad

Here is the proper solution to the Loop loop circuit and the questions I posed in reply #32 and  34. Note that the segments A and B are transformer secondaries and will have induced in them the EMF which creates the loop current. They do not have zero volts, rather they have the missing EMF's that solve KVL=0 around the loop.

Whoever came up with the idea that a superconducting secondary of a transformer must have zero volts is completely wrong and does not understand induction. If it was Lewin, then too bad for him. Does not matter how low the resistance of a transformer secondary is, it can still have a large voltage induced into it and EMF appears. For a simple understanding just consider it to be charge separation within the low resistance conductor.

Note that for this case we assume the resistors to be to be very thin disc type so we do not have to consider the small amount of induction directly into the resistors. (lumped element mode). If we don't do this then we would have to apportion lengths of conductors into the solution, which could get messy but still yields KVL=0.

You might also consider this video by Mehdi:
https://www.youtube.com/watch?v=Q9LuVBfwvzA

Excellent, and of course 100% correct ION!

O0

Quote from: ion on 2019.06.25, 15:28:23
Here is the proper solution to the Loop loop circuit and the questions I posed in reply #32 and  34. Note that the segments A and B are transformer secondaries and will have induced in them the EMF which creates the loop current. They do not have zero volts, rather they have the missing EMF's that solve KVL=0 around the loop.

Whoever came up with the idea that a superconducting secondary of a transformer must have zero volts is completely wrong and does not understand induction. If it was Lewin, then too bad for him. Does not matter how low the resistance of a transformer secondary is, it can still have a large voltage induced into it and EMF appears. For a simple understanding just consider it to be charge separation within the low resistance conductor.

Note that for this case we assume the resistors to be to be very thin disc type so we do not have to consider the small amount of induction directly into the resistors. (lumped element mode). If we don't do this then we would have to apportion lengths of conductors into the solution, which could get messy but still yields KVL=0.

You might also consider this video by Mehdi:
https://www.youtube.com/watch?v=Q9LuVBfwvzA

Ion

I knew the A and B measurement you would write in before i even posted the question. I assumed you would make the loop voltages sum to 0.

On paper it may look good,but it is an incorrect assumption.
There is no 5v across A and B,nor dose there have to be in reality.


Brad

Quote from: ion on 2019.06.25, 15:28:23
Here is the proper solution to the Loop loop circuit and the questions I posed in reply #32 and  34. Note that the segments A and B are transformer secondaries and will have induced in them the EMF which creates the loop current. They do not have zero volts, rather they have the missing EMF's that solve KVL=0 around the loop.

Whoever came up with the idea that a superconducting secondary of a transformer must have zero volts is completely wrong and does not understand induction. If it was Lewin, then too bad for him. Does not matter how low the resistance of a transformer secondary is, it can still have a large voltage induced into it and EMF appears. For a simple understanding just consider it to be charge separation within the low resistance conductor.

Note that for this case we assume the resistors to be to be very thin disc type so we do not have to consider the small amount of induction directly into the resistors. (lumped element mode). If we don't do this then we would have to apportion lengths of conductors into the solution, which could get messy but still yields KVL=0.

You might also consider this video by Mehdi:
https://www.youtube.com/watch?v=Q9LuVBfwvzA

I agree also. I wrote this in the other thread but then this thread popped up....

"If the wires between the resistors are the 'source' to the circuit by being induced by flux from the driver coil, then there will also be a voltage measured across each wire due to the pressures of charge induced in the wires met with the impedance of the resistors, whether the conductors are normal or superconducting. A shift of electrons from one end of the wire to the other, between the resistors, will show voltage across that wire. Just because the wires are a current source, does not mean that they cannot have voltage read across them when in the circuit.

And, if it measures 0v between the center of the length of one wire to the center of the length of the other wire, then we have another thing going on in the wires, if the resistors have different voltages at that point in time.  One half length of the wire should have more voltage across it than the other, in order for there to be 0v measured between the 2 centers of the wires. With the smaller value resistor, the 2 halves of the induced wires connected to it will have higher voltage values than the 2 halves connected to the higher value resistor, all in order to measure 0v any where around the circuit as per Kirchhoff's experiment."

Its prob not all correct, but close. ^-^

Mags

Quote from: poynt99 on 2019.06.25, 16:35:30
Excellent, and of course 100% correct ION!

O0

In the case of my setup,unfortunately not.


Brad

Here are the actual voltages across my circuit.
Area's A,B and C where measured via normal scope probing,as area D traverses the toroid center.

Confirmation--
Area's A,B,C and D where measured using the differential probing method.

Each method agree's with the other.


Brad

Quote from: TinMan on 2019.06.26, 11:31:10
Here are the actual voltages across my circuit.
Area's A,B and C where measured via normal scope probing,as area D traverses the toroid center.

Confirmation--
Area's A,B,C and D where measured using the differential probing method.

Each method agree's with the other.


Brad

Brad, now you are mixing apples and oranges. You are now stating that you are using a toroid induction which does not agree with a solenoid induction as shown in your present drawing (circle in the center of drawing). The voltages I referred to in the corrected drawing were for solenoid induction per Lewin. Naturally the induction distribution will be different for the toroid method vs. solenoid method.

Provide a picture of your actual test setup and actual probing points as the latest drawing is different than reality.

Quote from: ion on 2019.06.26, 14:00:57
Brad, now you are mixing apples and oranges. You are now stating that you are using a toroid induction which does not agree with a solenoid induction as shown in your present drawing (circle in the center of drawing). The voltages I referred to in the corrected drawing were for solenoid induction per Lewin. Naturally the induction distribution will be different for the toroid method vs. solenoid method.

Provide a picture of your actual test setup and actual probing points as the latest drawing is different than reality.

Yes
Sorry about the drawing.
Forgot to change it.

I will change it tonight after the work day is done.
But for now,just remove the solenoid,and place the toroid around link D. You now have my circuit and voltage values for each part of the circuit.

Whether by solenoid of toroid,both induce current flow through the secondary loop.

Brad

Why is it thought that it is necessary to thread the meter leads through the centre of the toroid in order to avoid influence from the fields and to get an accurate reading?

After some reflection on the toroid configuration, I am in agreement that the E field is the action mechanism on the wire loop. However, in the solenoid configuration (the original one), again it is the E field that acts upon the loop.

With the toroid configuration, and with the E fields in mind, spanning the toroid from the outside avoids influence on the measurement leads. Threading the lead through the toroid is actually what does influence the measurement.

Why? A second loop is formed by the leads and part of the loop, and since the E field is now acting simultaneously on both halves of this second loop, of course the result is that 0V is induced in the loop, and 0V will be measured.

Imagine taking your loop with the resistors (no voltmeter leads) and elongating it so it could be threaded through the centre of the toroid. Do you think any emf will be induced in the loop? I hope not. Well, same scenario occurs with the measurement leads when they are thread through; you are nulling out the measurement by feeding both "sides" of the loop into the toroid.

Measuring across (not through) the toroid is the correct form to avoid influence from the fields.

Quote from: poynt99 on 2019.06.27, 02:09:49
Why is it thought that it is necessary to thread the meter leads through the centre of the toroid in order to avoid influence from the fields and to get an accurate reading?

After some reflection on the toroid configuration, I am in agreement that the E field is the action mechanism on the wire loop. However, in the solenoid configuration (the original one), again it is the E field that acts upon the loop.

With the toroid configuration, and with the E fields in mind, spanning the toroid from the outside avoids influence on the measurement leads. Threading the lead through the toroid is actually what does influence the measurement.

Why? A second loop is formed by the leads and part of the loop, and since the E field is now acting simultaneously on both halves of this second loop, of course the result is that 0V is induced in the loop, and 0V will be measured.

Imagine taking your loop with the resistors (no voltmeter leads) and elongating it so it could be threaded through the centre of the toroid. Do you think any emf will be induced in the loop? I hope not. Well, same scenario occurs with the measurement leads when they are thread through; you are nulling out the measurement by feeding both "sides" of the loop into the toroid.

Measuring across (not through) the toroid is the correct form to avoid influence from the fields.

I am glad you agree that it is the E field that is the source of induction in the secondary loop. Thats one issue sorted.

Now,the measured voltages around the loop.

I maintain that trying to measure link D by having the scope probes around the toroid is wrong. Doing that you are not measuring the voltage across link D,you are measuring the voltage of a complete new loop around the toroid,and this second loops voltage is governed by the load placed on the toroid by the first loop,and where that second loops voltage will be the same as the first loops voltage.

Using the differential method,we can obtain accurate voltages around the loop. These voltages conform to the voltages read using normal scope probing. At no point do the sum of the voltages sum to 0,and that is a verifiable fact.

I do not understand as to why you think this is impossible,as this can be achieved in many ways,like a looped water circuit and pumps,where the pressures around the loop do not sum to 0.

Brad

There is no field interaction when the voltmeter or scope leads are outside the toroid. When inside the toroid, there is. Surely you see that?

When you are doing the differential measurement, do you also somehow put one scope probe through the toroid?

Quote from: poynt99 on 2019.06.27, 03:24:44
There is no field interaction when the voltmeter or scope leads are outside the toroid. When inside the toroid, there is. Surely you see that?

When you are doing the differential measurement, do you also somehow put one scope probe through the toroid?

Yes,putting the scope ground through the toroid is incorrect,as is having it around the toroid.

And no,neither of the scope probes are !somehow! through the toroid,as both ground leads are floating.
The fact that both links have very nearly the same voltage when using the differential probing method,and there resistance value is almost exact to one another,and the fact that the current flowing through the loop is the same at any two points,is proof that the correct voltage is being displayed via the differential method across the bottom link that passes through the toroid.
No matter how you try,the summed voltages around the loop will never sum to 0.

Im sure you have a small toroid inductor and a couple of resistors lying around Poynt. The circuit takes only a couple of minutes to set up. If not,then i will take the time tonight and shoot a video of my test methods for you to see--which i will do regardless.

At this point,i would be interested in seeing what a sim says.
Is it possible to set the sim up to induce a current through the loop via a toroid as in my setup ?.


Brad

Quote from: poynt99 on 2019.06.24, 00:20:31
You should listen intently to ION, as he is 100% correct, and you are leading yourself astray once again.

Here are some corrections for you to consider:

a) In Lewin's original setup, and the one I tested, the E field is what is directly inducing the loop current, not the B field.
b) In your toroid setup, the B field is what is inducing the loop current.
c) When going around the loop to reconcile all the measured voltages, we ADD them, not subtract them as you did in your first video. In other words, the SUM of the series voltages in a loop is zero. The SUM of your two resistor measurements is 80.3mV. (see e) below)
d) Proper measurement of the wire part of the loop is done around the outside of the toroid, and if you account for the polarity and voltage, you will find that when added to the other two resistor voltages, you will be left with something close to zero.
e) Your "proper" measurement of the toroid section of the wire was about 81mV. The two resistor voltages summed is 73mV + 7.3mV = 80.3mV. Summed with the series-subtracting wire voltage leaves you with 0.7mV.

KVL holds.

Poynt

I need to address these issue's with you,as i think you really need to take another look at this.

So far,here are some of the events that have taken place in this thread.

In reply 28,you highlighted a quote from Matt-Watts -->When dealing with B fields, your position of measurement matters. To which you replied-->Precisely.

Then in reply 30 you state--
a) In Lewin's original setup, and the one I tested, the E field is what is directly inducing the loop current, not the B field.
It is my belief that both the E and B field played a part in the Lewin circuit,and i will get to that later,after i have assembled Lewins original test setup along side mine.

In post 11 i state-->the secondary loop is induced by the electric field,and not the magnetic field.

In post 30 you also state this--
b) In your toroid setup, the B field is what is inducing the loop current.

In post 31 i state--Im going to disagree with you Poynt,and i think you have it backwards.
In Lewins setup,it is the B field cutting through the loop.
In my setup,the B field is contained within the toroid.

Then in reply 45 you state-After some reflection on the toroid configuration, I am in agreement that the E field is the action mechanism on the wire loop. However, in the solenoid configuration (the original one), again it is the E field that acts upon the loop.

I am happy that you agree with me about it being the E field that is inducing the current in my loop,as that means all my hard work and time in gaining an understanding into all forms of induction, by way of being self taught and my time spent on the bench,and my own thought's is paying off.

Now,while i wish to remain focused mainly on my toroid setup,i would like to reflect a little on Lewins original setup.

You state it is the E field inducing the single loop,but you also seem to agree with Matt as well when he says-- When dealing with B fields, your position of measurement matters

Now,the way i see it(these are my thought's),the E field is inducing the current through the secondary loop of Lewins setup,and the B field is inducing the voltage in the loops of the measuring equipment leads.
I confirmed this on my bench-->the E field will not induce a current in the secondary loop when that secondary loop is not around the circumference of the primary coil. If the secondary loop is sitting beside the primary,and not wrapped around the primary coil,no current will be induced via the E field.
Now,this is not the case with the B field,where the secondary loop will be induced if that secondary loop is sitting beside the primary coil,and not wrapped around it.

In saying all that,here is the problem i see in the testing of Lewin's circuit.
The secondary loop is induced by the E field(as you state),but the measuring equipment leads were induced by the B field !apparently!,in accordance with your reply to Matts comment regarding the B field-->When dealing with B fields, your position of measurement matters,to which you reply !Precisely!
Now the biggest problem here is the phase relationship between the induced voltage of the secondary loop by the E field,and the induced voltage of the measurement equipment leads by the B field.
These two induced voltages should have been 90* out of phase,but in your tests,they were all in phase  ???

How do we know this?
Well if it is the E field inducing the EMF in the secondary,then the primary and secondary voltage phase relationship should be in phase. If it is the B field inducing the EMF in the secondary,then voltage phase relationship between the primary and secondary should be 90*,as the magnetic field in the primary is a result of current flow,and we know the current lags the voltage by 90* in an inductor.

So i have to wonder,which voltage the measurement equipment was actually measuring?.
Was it the voltage across the resistors in the secondary loop induced by the E field,or the induced voltage in the third loop formed by the measurement equipments lead's which was induced by the B field?.
And what is the outcome of mixing two different circuits,where 1 has an in phase relationship with the primary inductor's voltage,and the other has a 90* phase relationship with the primary inductor's voltage.

If i am wrong about all this,could you please explain as to why you agree with Matts comment about the B field,but state it is the E field that is inducing the EMF in Lewin's experiment ?.

Thanks


Brad

Quote from: TinMan on 2019.06.27, 04:50:42


At this point,i would be interested in seeing what a sim says.
Is it possible to set the sim up to induce a current through the loop via a toroid as in my setup ?.


Brad

Brad,

Good question?  The induction process is normally internal in LtSpice but the electric and magnetic properties of transformer induction can be modeled with gyrator/capacitor networks.  I'm just not sure if I'm up to doing that! 

Poynt?

I think a bench replication would be much easier!!

Pm

FWIW

I replicated Brad's toroid induction method this morning on the bench using a Fluke 77 TRMS meter, probing directly across the item of interest

I set the input level to the toroid such that I had 75 mV measured across the 1K

I then measured 7.5 mV across the 100 Ohm

I then measured 82.5  mV across segment A

I measured +/-0 mV across segment B

82.5mV total across the resistors minus (82.5mV segment A plus 0 mV segment B) = 0mV

KVL holds

With segment A looked at as the secondary of a transformer delivering 82.5mV and using the voltage divider rule:

Voltage across 100 Ohm = 100/1100 * 82.5mV= 7.5mV

Voltage across 1000 Ohm = 1000/1100 * 82.5mV = 75mv

Voltage across segment B =+/- 0mV

Again using current

82.5mV/1100 Ohms = 75 uA in the loop

75uA * 1000 Ohms = 75mV

75uA * 100 Ohms = 7.5mV

Before I address Brad's post in any detail, I'd like to speak to the question of simulating the Lewin experiment.

This immediately leads me to the following question: Can we all agree that both the solenoid and toroid configurations of the experiment can be modeled, and are essentially equivalent to, a loosely-coupled air-core transformer?

I did simulate the solenoid configuration years ago, and as with the toroid configuration, would entail using a loosely-coupled air core transformer. There is no easy way that I am aware of to model the B and E fields from a physics perspective. SPICE uses the well-known equations to obtain its results. In order to simulate this experiment in a more realistic manner, I think an electromagnetic simulation tool (https://www.ansys.com/products/electronics/ansys-maxwell) would be required. However, if all can agree regarding the above question, there really is no need to simulate anything.

So let's see what answers arise.

Quote from: ion on 2019.06.27, 18:40:50
FWIW

I replicated Brad's toroid induction method this morning on the bench using a Fluke 77 TRMS meter, probing directly across the item of interest

I set the input level to the toroid such that I had 75 mV measured across the 1K

I then measured 7.5 mV across the 100 Ohm

I then measured 82.5  mV across segment A

I measured +/-0 mV across segment B

82.5mV total across the resistors minus (82.5mV segment A plus 0 mV segment B) = 0mV

KVL holds

With segment A looked at as the secondary of a transformer delivering 82.5mV and using the voltage divider rule:

Voltage across 100 Ohm = 100/1100 * 82.5mV= 7.5mV

Voltage across 1000 Ohm = 1000/1100 * 82.5mV = 75mv

Voltage across segment B =+/- 0mV

Again using current

82.5mV/1100 Ohms = 75 uA in the loop

75uA * 1000 Ohms = 75mV

75uA * 100 Ohms = 7.5mV

So no current flow through segment B  ???

Brad

Quote from: poynt99 on 2019.06.27, 21:41:25
Before I address Brad's post in any detail, I'd like to speak to the question of simulating the Lewin experiment.

This immediately leads me to the following question: Can we all agree that both the solenoid and toroid configurations of the experiment can be modeled, and are essentially equivalent to, a loosely-coupled air-core transformer?

I did simulate the solenoid configuration years ago, and as with the toroid configuration, would entail using a loosely-coupled air core transformer. There is no easy way that I am aware of to model the B and E fields from a physics perspective. SPICE uses the well-known equations to obtain its results. In order to simulate this experiment in a more realistic manner, I think an electromagnetic simulation tool (https://www.ansys.com/products/electronics/ansys-maxwell) would be required. However, if all can agree regarding the above question, there really is no need to simulate anything.

So let's see what answers arise.

Well the Lewin experiment is an aircore setup,but the toroid setup contains the B field within the core,and so it would seem not to be an air core setup as such.

Brad

Quote from: poynt99 on 2019.06.27, 21:41:25
Before I address Brad's post in any detail, I'd like to speak to the question of simulating the Lewin experiment.

This immediately leads me to the following question: Can we all agree that both the solenoid and toroid configurations of the experiment can be modeled, and are essentially equivalent to, a loosely-coupled air-core transformer?

I did simulate the solenoid configuration years ago, and as with the toroid configuration, would entail using a loosely-coupled air core transformer. There is no easy way that I am aware of to model the B and E fields from a physics perspective. SPICE uses the well-known equations to obtain its results. In order to simulate this experiment in a more realistic manner, I think an electromagnetic simulation tool (https://www.ansys.com/products/electronics/ansys-maxwell) would be required. However, if all can agree regarding the above question, there really is no need to simulate anything.

So let's see what answers arise.

I agree with the first part:

"Can we all agree that both the solenoid and toroid configurations of the experiment can be modeled"

so Yes.

Then: "and are essentially equivalent to, a loosely-coupled air-core transformer?"

As for the second part it really depends on how you define loose. An air core transformer can be tightly coupled if the construction between primary and secondary is e.g. made coaxially. As you depart from truly coaxial, and spacing between primary and secondary increase, then so does the leakage inductance (uncoupled inductance) and then it can be loose, but that is a matter of degree that we put a number on (the K factor). Of course you already know all this, as you aided my understanding of it regarding Spice simulations.

So what number would we be using for the definition of "loose"? And how tightly or loosely coupled can we make the Lewin experiment, provided we want to experiment over a range of coupling factors. Or did I misunderstand the question?

Regards

Quote from: poynt99 on 2019.06.27, 21:41:25
Before I address Brad's post in any detail, I'd like to speak to the question of simulating the Lewin experiment.

This immediately leads me to the following question: Can we all agree that both the solenoid and toroid configurations of the experiment can be modeled, and are essentially equivalent to, a loosely-coupled air-core transformer?

I did simulate the solenoid configuration years ago, and as with the toroid configuration, would entail using a loosely-coupled air core transformer. There is no easy way that I am aware of to model the B and E fields from a physics perspective. SPICE uses the well-known equations to obtain its results. In order to simulate this experiment in a more realistic manner, I think an electromagnetic simulation tool (https://www.ansys.com/products/electronics/ansys-maxwell) would be required. However, if all can agree regarding the above question, there really is no need to simulate anything.

So let's see what answers arise.

I would like to add that i !think! you will find that there is no such thing as a loosely-coupled secondary in a toroid transformer.

The voltages across the risistors will be the same if the secondary is tightly wound around the toroid as if the secondary loop had a much larger diameter-as in my setup.

Although the E field gets weaker the further you go out in the E field radius,you have a higher volume of conductor in the secondary loop.

Hope that made sense ?.


Brad

Quote from: TinMan on 2019.06.27, 22:14:37
So no current flow through segment B  ???

Brad

Loop current of 75uA through a piece of wire a fraction of an Ohm,  this voltage is below the AC resolution of my meter. Lets say the wire was 0.1 Ohm then it would have 7.5uV across it. This is at the level of thermoelectric effects where a slight temperature difference and dissimilar materials can sway the readings. Thus the experiment would have to be carefully constructed if you want to resolve meaningful uV levels.

Quote from: ion on 2019.06.27, 23:15:24
Loop current of 75uA through a piece of wire a fraction of an Ohm,  this voltage is below the AC resolution of my meter. Lets say the wire was 0.1 Ohm then it would have 7.5uV across it. This is at the level of thermoelectric effects where a slight temperature difference and dissimilar materials can sway the readings. Thus the experiment would have to be carefully constructed if you want to record accurate uV levels.

So an unmeasurable voltage across segment B,but 80+ mV across segment A that has roughly the same resistive value as segment B?


Brad

If the resistors are 1000 ohm and 100 ohm, what is the impedance of the scope or meter? 

Has it been tested that there really are issues with induction of the test leads?

Just askin

Mags

Gents,

I think the coupling between two closely-spaced coaxial air coils would still be increased by using a core, and I'm not 100% certain but I think a K factor of 1 is almost impossible to achieve with an air-core transformer. On the other hand, it is quite common place to achieve K factors of 0.999 with a soft iron core for example.

And I don't disagree about the emf generation; I know the line integral of E along the circumference, whatever it may be, is always the same.

Perhaps we could call the toroid configuration a semi-air-core transformer; the primary is tightly coupled to the core, and the secondary is loosely coupled to the core. But I think we are drifting from the point and getting bogged down in unnecessary details. The K factor is irrelevant, and even the core is somewhat irrelevant. The can experiment work with many parameter values.

Let me then re-phrase the question such that hopefully we can either simply agree or disagree:

Are we in agreement that the Lewin experiment, configured either with a solenoid exciter or toroid exciter, can be simplified and modeled by the attached illustration?

Quote from: Magluvin on 2019.06.28, 03:05:44
If the resistors are 1000 ohm and 100 ohm, what is the impedance of the scope or meter? 

Has it been tested that there really are issues with induction of the test leads?

Just askin

Mags

Yes

Just put the scope probe through the toroid,clip the earth lead to it,and you have the very same voltage being read from the scope.

As i said,the scope is reading the voltage across its own looped circuit.

This means it is not reading the voltage across the link that passes through the toroid.

Common sense and ohms law both tell us that both conductor links,having the same resistance value,with the same current flowing through them,will have the same voltage across them.

One cannot have 80mV+ across it while the other has 0v across it.
But it would seem common sense and ohms law is being tossed aside,just so as KVL holds.

Despite our differences in other subjects Mags,i know you would search for the actual answer yourself--just as you did with the cap to cap transfer subject,and then again against MH in regards to the ignition system resonance. You stood your ground,and you were correct both times. Perhaps this is another one of those times.


Brad

Quote from: poynt99 on 2019.06.28, 03:07:32
Gents,

Are we in agreement that the Lewin experiment, configured either with a solenoid exciter or toroid exciter, can be simplified and modeled by the attached illustration?

No,as that circuit has only 1 position of measurement,where at least 2 values are needed in order to do any sort of math.

Place a 1 ohm resistor inside the hole of the toroid.
Then using your measuring method,measure the voltage across that 1 ohm resistor. Now,using ohms law,you will have the value of current flowing through the circuit !apparently!.

Once again,using ohms law,you can now calculate the voltage that should be across you other resistor !apparently!

So give that a try,and see how things pan out.

Brad

Trying to convey information is difficult this way to say the least.

If I had 10 minutes with you at a bench, I think it would be a lot easier, and more fruitful.

You seem very fixated and aren't willing to or aren't aware of a bigger picture.

For instance, why are you so fixated on the notion that the wire segment that is excited can not have a voltage induced (and measured) across it, just because the other wire segment (that is not excited btw) does not? If your logic is limited simply to "because Ohm's law tells me so", then you are choosing to be oblivious to a whole other dimension of knowledge and awareness, which is unfortunate.

I can't spend any more time on this when those asking the questions refuse to open their minds.

Quote from: poynt99 on 2019.06.28, 04:18:33
Trying to convey information is difficult this way to say the least.

If I had 10 minutes with you at a bench, I think it would be a lot easier, and more fruitful.

You seem very fixated and aren't willing to or aren't aware of a bigger picture.

For instance, why are you so fixated on the notion that the wire segment that is excited can not have a voltage induced (and measured) across it, just because the other wire segment (that is not excited btw) does not? If your logic is limited simply to "because Ohm's law tells me so", then you are choosing to be oblivious to a whole other dimension of knowledge and awareness, which is unfortunate.

I can't spend any more time on this when those asking the questions refuse to open their minds.

Poynt

It would seem the same applies both ways.

I have now posted 3 video's showing the wire segment pasing through the hole of the toroid being well shielded,up to 70% of the complete loop,and the voltage across the load resistor never dropped 1mV,even in my large 3 watt setup.

There is no reason to pack up and leave because you think i am not listening,as it would seem to be the same my end,where no one is paying attention to what i am showing--but im still here.

And all my tests have shown that the loop is induced via the whole loop,not just the small portion through the loop.

If you can explain as to how that very small section of wire passing through the toroid can have a larger voltage across it to that of the opposite link inductor,then i will listen to what you have to say.

I am also waiting for your answers on my post a few posts back,regarding Matts comment on the B field-and such.

So yes,i would like to know why ohms law dose not work in this circuit.


Brad

Let me leave you with one last bit of information that you may wish to contemplate.

- Don't confuse (as Lewin does) the resulting E field (from the uniform B field) with the induced emf. They are different animals, and is one of the reasons Lewin's lecture is erroneous.
- The resulting E field is non-conservative, meaning it does not equal 0 when traveling the entire loop. No one disagrees with this.
- The circular E field is evenly distributed around the circumference if the path has uniform resistance.
- A circular wire path with two resistors causes the E field to bunch or concentrate at the nodes of least charge carriers, i.e. the resistor terminals. The E field induces emf's in the adjacent wire segments (solenoid configuration).
- There is no E field across an inductor (hence why when the measuring probes are across the wire segments and in the same plane, the measurement is 0V (solenoid config)).
- The E field causes an induced emf in the material with the most charge carriers, i.e. the excited wire segment(s).
- The total induced emf is equal to, but opposite polarity to the total E field.
- Current in a closed loop can not exist without a source of emf.
- Current does not just appear out of nowhere.
- Current in the loop, either the toroid or solenoid configs, is a result of the induced emf.
- The induced emf(s) in a conductive loop (or segments of) is measurable.
- The E field distribution and resistor voltage drops due to the induced emf current are equal.
- The solenoid config is eloquent, in that it is uniform in all ways and symmetrical. All conductive material produces an induced emf (the vast majority being in the wire segments).
- The toroid config is non-uniform, and asymmetrical. The emf is induced in the one wire segment that threads the toroid.
- The E field in the toroid config is limited to the area near the excited wire segment, and mostly on the inner side of the toroid.
- The E field around the toroid itself is circular and uniform, but from the perspective of one side of the hole to the other side of the hole (the path the loop wire travels), the E field is concentrated and presents an influence to the charge carriers in the wire segment running through its centre. This is what induces the source of emf in that wire segment for the loop current.
- With the solenoid configuration, care must be taken when measuring the 4 voltages. In the horizontal plane, all the measurements are of the E field only, and are therefore non-conservative (KVL does not apply, nor do we try to make it so). With the leads in the vertical plane, all the measurements are of the induced emf(s), and resistor voltage drops due to the induced current. These measurements are conservative, i.e. KVL applies and holds.
- In the toroid configuration, all the measurements are of the emf, and resistor voltage drops due to the induced emf. It is only if the measurement lead is fed through the toroid centre that the measurement of this wire segment is of the E field only, and since an inductance can not have an E field across it, 0V will be measured.
- The voltage must be measured without threading the lead through the toroid, otherwise only the E field will be measured for that wire segment, and KVL does not apply to E fields alone (and we don't try to make it so). It must include the induced emf(s) as well.

Quote from: poynt99 on 2019.06.28, 05:53:29
Let me leave you with one last bit of information that you may wish to contemplate.

Quote- The resulting E field is non-conservative, meaning it does not equal 0 when traveling the entire loop. No one disagrees with this.

O0

Quote- The circular E field is evenly distributed around the circumference if the path has uniform resistance.
- A circular wire path with two resistors causes the E field to bunch or concentrate at the nodes of least charge carriers, i.e. the resistor terminals. The E field induces emf's in the adjacent wire segments (solenoid configuration).

I agree,as i stated this in a previous post.

Quote- There is no E field across an inductor (hence why when the measuring probes are across the wire segments and in the same plane, the measurement is 0V (solenoid config)).

I agree with this as well,as the E field would be around the inductor,on the same plane as the windings.

Quote- The E field causes an induced emf in the material with the most charge carriers, i.e. the excited wire segment(s).
- The total induced emf is equal to, but opposite polarity to the total E field.

I would say no.
If we look at a transformer where the B field of the primary induces current flow in a loaded secondary,then that secondary's magnetic field will appose that of the primary. Oddly enough,the two fields have to be of the same polarity (E.G,N and N)to appose each other.
I would say the same applies when the secondary is induced via the E field,and there polarities will be the same and in phase.

Quote- Current in a closed loop can not exist without a source of emf.

In every day stuff like we do,then i agree.
But if a magnet is placed on top of a toroid ring,and then that toroid ring is cryocooled to a super conductive state,and then the magnet is removed,then a current will continue to flow around that toroid for as long as it remains in a super conductive state.

Quote- Current in the loop, either the toroid or solenoid configs, is a result of the induced emf.

I some what agree.
But F6FLT had a more accurate description.

Quote- The induced emf(s) in a conductive loop (or segments of) is measurable.

Yes,except for the example i gave above,regarding the superconductive ring.

Quote- The E field distribution and resistor voltage drops due to the induced emf current are equal.

I agree that the E field distribution is equal ,but the voltage drops across different value resistors will not be equal.

Quote- The solenoid config is eloquent, in that it is uniform in all ways and symmetrical. All conductive material produces an induced emf (the vast majority being in the wire segments).

Agreed.

Quote- The toroid config is non-uniform, and asymmetrical. The emf is induced in the one wire segment that threads the toroid.

Certainly do not agree with that.
See video below.

Quote- The E field in the toroid config is limited to the area near the excited wire segment, and mostly on the inner side of the toroid.

I don't agree with that either.

Quote- The E field around the toroid itself is circular and uniform, but from the perspective of one side of the hole to the other side of the hole (the path the loop wire travels), the E field is concentrated and presents an influence to the charge carriers in the wire segment running through its centre. This is what induces the source of emf in that wire segment for the loop current.

Nor do i agree with that.
See video below.

Quote- With the solenoid configuration, care must be taken when measuring the 4 voltages. In the horizontal plane, all the measurements are of the E field only, and are therefore non-conservative (KVL does not apply, nor do we try to make it so). With the leads in the vertical plane, all the measurements are of the induced emf(s), and resistor voltage drops due to the induced current. These measurements are conservative, i.e. KVL applies and holds.

Not to sure on this one yet.'
I will wait until i have tested my new setup this weekend before i agree.

Quote- In the toroid configuration, all the measurements are of the emf, and resistor voltage drops due to the induced emf. It is only if the measurement lead is fed through the toroid centre that the measurement of this wire segment is of the E field only, and since an inductance can not have an E field across it, 0V will be measured.
- The voltage must be measured without threading the lead through the toroid, otherwise only the E field will be measured for that wire segment, and KVL does not apply to E fields alone (and we don't try to make it so). It must include the induced emf(s) as well.

Ok,KVL only holds if the voltage around the loop equal 0.
I have also measured the voltage across the conductor that threads the toroid via the differential method. This means that the measuring lead was not also threaded through the toroid. The voltage value was exactly the same as that of the other resistors link.

So as i understand it,you agree that the voltages around the actual loop do not sum to zero. In order for it to sum to 0,we have to subtract the total induced EMF from the sumed voltage around the loop.
This makes no sense at all,nor is it right.

https://www.youtube.com/watch?v=cqXC7Qjrh1Y


Brad

Here is my latest test,where i shield the portion of the conductor passing through the center of the toroid by means of using coax cable,where the outer braid is also grounded. This and my other 3 tests confirm that the induction of the secondary loop is not limited to the small portion of the conductor passing through the toroid.

https://www.youtube.com/watch?v=ozngW0DMp30

Plus my previous 3 tests

https://www.youtube.com/watch?v=UkUNZ5ik4go

https://www.youtube.com/watch?v=n84bPUWLLLY

https://www.youtube.com/watch?v=cqXC7Qjrh1Y


Brad

Brad,

FWIW, here is a sim of a toroid driven current loop including an approximate equivalent circuit of the transformer.  The calculations show an agreement with ION and Poynt.

IMO, the key to realize is that the secondary L2 that passes thru the core will always have a finite amount of inductance and in this case is 10uH.  This inductance from A to D therefore has an induced EMF from the primary and is the source voltage for the remaining parts of the loop.  The simulation supports this and clearly shows that KVL holds true here.

Regards,
Pm

Quote from: partzman on 2019.06.28, 13:35:05
Brad,

FWIW, here is a sim of a toroid driven current loop including an approximate equivalent circuit of the transformer.  The calculations show an agreement with ION and Poynt.

IMO, the key to realize is that the secondary L2 that passes thru the core will always have a finite amount of inductance and in this case is 10uH.  This inductance from A to D therefore has an induced EMF from the primary and is the source voltage for the remaining parts of the loop.  The simulation supports this and clearly shows that KVL holds true here.

Regards,
Pm

In my previous post,i show 4 video's of actual tests that say that is not correct.

Watch the first in the list at least.
It was the one i made just a couple of hours ago.

As a sim is based around !what they think! is known,then would you expect it to give any other result?.


Brad

Quote from: poynt99 on 2019.06.28, 03:07:32
Gents,

I think the coupling between two closely-spaced coaxial air coils would still be increased by using a core, and I'm not 100% certain but I think a K factor of 1 is almost impossible to achieve with an air-core transformer. On the other hand, it is quite common place to achieve K factors of 0.999 with a soft iron core for example.

And I don't disagree about the emf generation; I know the line integral of E along the circumference, whatever it may be, is always the same.

Perhaps we could call the toroid configuration a semi-air-core transformer; the primary is tightly coupled to the core, and the secondary is loosely coupled to the core. But I think we are drifting from the point and getting bogged down in unnecessary details. The K factor is irrelevant, and even the core is somewhat irrelevant. The can experiment work with many parameter values.

Let me then re-phrase the question such that hopefully we can either simply agree or disagree:

Are we in agreement that the Lewin experiment, configured either with a solenoid exciter or toroid exciter, can be simplified and modeled by the attached illustration?

Agreed  O0, sorry if I confused the issue by bringing up the coupling coefficient.

No worries Ernie, and no mal intent.

Just wanted to keep things focused.
:)

Quote from: TinMan on 2019.06.28, 02:18:27
So an unmeasurable voltage across segment B,but 80+ mV across segment A that has roughly the same resistive value as segment B?
Brad

Correct, segment B has no induced EMF in the toroid model. Segment B's fractional resistance value alone determines the voltage measured across it. Quite different than the induced segment A. You need to wrap your head around the fact that transformer induction continues to work on segment A even if it's resistance is close to zero or even a superconductor.

drawing attached, can you identify segment A and segment B?

Quote from: TinMan on 2019.06.28, 13:56:08
In my previous post,i show 4 video's of actual tests that say that is not correct.

Watch the first in the list at least.
It was the one i made just a couple of hours ago.

As a sim is based around !what they think! is known,then would you expect it to give any other result?.


Brad

Brad,

IMO, your 1st video above does not prove that no induction is occurring in the wire in the center of the coax.  There is no shielding action from the outer shield on the coax and in fact, what you now have are two secondaries.  The inner wire or 1st secondary, is loaded with the resistor loop and the 2nd secondary or the coax shield, is open circuit and in the same phase.  If you place a scope probe on the un-grounded end of the coax, you will see a voltage that is very close (slightly higher) to the sum of the two resistor voltages.  If this is not the case, then I am totally incorrect and everything I stated above is false.   :-[ 

I will watch the other videos.

Regards,
Pm

Quote from: partzman on 2019.06.28, 14:45:42
Brad,

IMO, your 1st video above does not prove that no induction is occurring in the wire in the center of the coax.  There is no shielding action from the outer shield on the coax and in fact, what you now have are two secondaries.  The inner wire or 1st secondary, is loaded with the resistor loop and the 2nd secondary or the coax shield, is open circuit and in the same phase.  If you place a scope probe on the un-grounded end of the coax, you will see a voltage that is very close (slightly higher) to the sum of the two resistor voltages.  If this is not the case, then I am totally incorrect and everything I stated above is false.   :-[ 

I will watch the other videos.

Regards,
Pm

Quote: A Faraday cage is an enclosure made of conductive materials which is capable of blocking external electric fields. In other words, it is a hollow conductor capable of keeping the charge or radiation on the external surface of the cage. Faraday cages are used in a wide range of applications, including protection of electronic devices from electrostatic discharge and external radio frequency interference.
A Faraday cage is also known as a Faraday shield.

Brad

I would not be convinced that any shielding is taking place until I completely "shielded" the internal loop.

If one could devise a way of getting some indication that the loop has current, or does not, then one would know with 100% certainty shielding is taking place.

One idea might be to fully encase a wire loop where the wire loop terminates on a very small piezo buzzer. Anything that can make an audible noise would work. I'm sure there are other ideas that could work.

Is this a conclusive test? If there is no current, is this because shielding is taking place, or is there no current because of the big shorted ring?
OR, if there IS still current, is that because there is no shielding, or is it because of coupling between the shorted ring and the inner loop?

I suspect that even with 100% coverage of the "shield", an emf would still be induced in the inner loop. In other words, the so-called "shielding" in this case is not effective. This means that either the E field is inducing the current at the wire segment threading the toroid (as already proposed), or there is another elusive penetrating field or potential responsible for the induction (the Magnetic Vector Potential, A Field?)


Also, there is much talk about the toroid's E field making its way around to other parts of the loop to cause the induced emf. May I ask you to draw out your vision of what the E field looks like around the toroid coil and associated wire loop?

Quote from: TinMan on 2019.06.28, 14:52:49
Quote: A Faraday cage is an enclosure made of conductive materials which is capable of blocking external electric fields. In other words, it is a hollow conductor capable of keeping the charge or radiation on the external surface of the cage. Faraday cages are used in a wide range of applications, including protection of electronic devices from electrostatic discharge and external radio frequency interference.
A Faraday cage is also known as a Faraday shield.

Brad

I understand Faraday shielding.

I ran a quick bench test using a powdered iron toroid with a single shielded coax loop connected to a precision 1 ohm and 10 resistor which are connected in series to complete the loop.

The first scope pix has the scope grounds of both CH2 and CH3 connected at the junction of the two resistors and the resulting voltages are shown.  The Math channel shows the sum (actually difference to compensate for phasing) of the two channels to be 294.1mv rms.  Grounding the shield or leaving it unconnected makes no difference in the measurements. 

The second scope pix shows CH2 measuring the voltage across both resistors to be 293.1mv rms which is in close agreement with the Math channel above.  CH3 shows the voltage measured across the shield to be 295.5mv rms which is in phase and slightly higher than the loaded loop.

I find no evidence of shielding in this test.

Regards,
Pm

   

Brad,

Here is another test using the same toroid but with a 4" diameter outer shield formed out of 1/8" diameter steel brake line with an insulated 18ga wire inserted for the main loop.  As can be seen from the pix, the ends of the loop are as close as possible and connect to a 10 ohm 1% resistor.

The scope pix shows the measurements with CH1(yel) being the input voltage to the toroid, CH2(blu) is the voltage across the 10 ohm resistor, and CH3(pnk) is the voltage across the outer shield.

If we consider the shield is actually working to shield the E field even though it appears as another secondary loop, then what is the method of induction for the inner wire?

Anyone?

Regards,
Pm

Nice test PM.

I guess you've confirmed my assertion.

I would also assert that if you shorted across the tubing and re-ran the test, you would still have an induced emf in the inner loop, but the outer tubing would now be "hot" as well, emanating its own magnetic field.

Yes, with the outer tube or shield shorted with a jumper that has some resistance, the voltage across the shield is 59.55mv rms and the voltage across the 10 ohm is 60.52mv rms.  This would indicate a rather tight coupling between the shield and inner wire.

Pm

Back in the late 90s I was a sounstream car audio dealer, when it was the good stuff, and we carried monster cable. It was all coaxial rca cables back then.

I chose to buy the rca cable on rolls with, gold plated connectors that were soldered on, and made the cable to length for each system to avoid the wires from being tight or rolling up excess if the premade were too long or short.

The cable on the roll had the center signal wire, then the braided shield over that and an alum foil shield that was to be only connected on the source end connector.

But at times we still might get alternator noise(engine whine) in the audio. It became a must to ground the head unit and all other equipment to a single ground point instead of the radio grounded somewhere near the radio and the amps grounded to chassis in the area they were installed to avoid ground loops within the chassis. But still the cables were an issue at times in vehicles. These days I wire the power and gnd of the amps all the way to the battery. Helps in many ways.

But some years after that some companies came out with twisted pair rca cables. No shielding. And it has been a beautiful thing ever since and most all companies use that now. You can run your rcas with the large power and gnd amp wires and come out unscathed with noise issues.

So Im wondering if that would be a good approach here with test equipment to avoid these em interference issues.

Mags

Quote from: partzman on 2019.06.28, 19:31:48
Brad,

Here is another test using the same toroid but with a 4" diameter outer shield formed out of 1/8" diameter steel brake line with an insulated 18ga wire inserted for the main loop.  As can be seen from the pix, the ends of the loop are as close as possible and connect to a 10 ohm 1% resistor.

The scope pix shows the measurements with CH1(yel) being the input voltage to the toroid, CH2(blu) is the voltage across the 10 ohm resistor, and CH3(pnk) is the voltage across the outer shield.

If we consider the shield is actually working to shield the E field even though it appears as another secondary loop, then what is the method of induction for the inner wire?

Anyone?

Regards,
Pm

Thank you for doing these tests Pm.

So i ask,is it not possible to shield the inner conductor from the E field?
-->Or,is it possible that i am right,and that the loop can be induced as long as 1 small part of it is exposed to the E field?.

Would you have not expected even a small reduction in voltage when you covered almost all of the secondary loop in a steel tube?.


Brad

Quote from: poynt99 on 2019.06.28, 16:16:08



Also, there is much talk about the toroid's E field making its way around to other parts of the loop to cause the induced emf. May I ask you to draw out your vision of what the E field looks like around the toroid coil and associated wire loop?

See diagram below.


Brad

Could have used this picture as well.

Quote from: TinMan on 2019.06.29, 00:21:54
is it not possible to shield the inner conductor from the E field?

Apparently not when the E field is created via a changing B field, and the wire is coincident with the E field.

Quote
-->Or,is it possible that i am right,and that the loop can be induced as long as 1 small part of it is exposed to the E field?.

In your opinion, if the loop was made 100 meters long (200 meters of wire) and elongated and pulled out taught, and all 100 meters from the toroid to the end was shielded, would there be an induced emf? (assuming essentially no wire is exposed to any significant E field)

Quote
Would you have not expected even a small reduction in voltage when you covered almost all of the secondary loop in a steel tube?.

I wouldn't.

Quote from: poynt99 on 2019.06.29, 00:45:32
Could have used this picture as well.

I did  use that picture as well.
Perhaps you missed it?


Brad

Quote from: partzman on 2019.06.28, 20:45:01
Yes, with the outer tube or shield shorted with a jumper that has some resistance, the voltage across the shield is 59.55mv rms and the voltage across the 10 ohm is 60.52mv rms.  This would indicate a rather tight coupling between the shield and inner wire.

Pm

OR, might this indicate that each loop has its own independently induced emf? (the likely answer in my opinion)

Quote from: partzman on 2019.06.28, 20:45:01
Yes, with the outer tube or shield shorted with a jumper that has some resistance, the voltage across the shield is 59.55mv rms and the voltage across the 10 ohm is 60.52mv rms.  This would indicate a rather tight coupling between the shield and inner wire.

Pm

Pm

With your outer shield open (not shorted),do you see any load placed on the primary when you place a load on the secondary loop ?.


Brad

Quote from: TinMan on 2019.06.29, 00:21:54
Thank you for doing these tests Pm.

So i ask,is it not possible to shield the inner conductor from the E field?
-->Or,is it possible that i am right,and that the loop can be induced as long as 1 small part of it is exposed to the E field?.

I'm not sure about being able to shield the inner conductor.  I've been trying to think of a way to build a shield that would cancel any induced field in the shield itself and yet allow an induced field in the inner conductor.

In my mind, any part of a loop that passes thru the center hole of a toroid coil will experience induction where the E field is concentrated. 

Quote
Would you have not expected even a small reduction in voltage when you covered almost all of the secondary loop in a steel tube?.

Actually no, I didn't expect any reduction because I viewed the shield as another secondary with a rather high coupling factor.  However, I'm not sure if we are truly shielding the inner conductor because we do have induction in the outer shield and with the apparent tight coupling to the inner conductor, we would then see induction in that inner wire.  So in my mind at this point, the jury is still out!

Pm

Quote
Brad

Quote from: TinMan on 2019.06.29, 08:42:13
Pm

With your outer shield open (not shorted),do you see any load placed on the primary when you place a load on the secondary loop ?.


Brad

Brad,

There was no measurable change in the primary current when adding a 10 ohm load to the shield.  So, I removed both loads to see what the primary current measured and again no change which means the core has a high magnetization current compared to the reflected current from the secondary to the primary.

Pm

OK, here is a test that was an attempt to cancel the induced field in the shield by applying an equal and opposite current to the shield from another source.  The source in this case was another toroid with a single turn that was coupled 180 degrees out of phase.  The second toroid is an identical core but with a different wire size and turn count on the primary so there was a slight phase adjustment in the driving signals from the power amplifier.

We do see a significant reduction in the inner wire voltage on CH3(pnk) but it is higher than the shield voltage on CH2(blu) so this reduction may be a result of the tight coupling rather than a shielding effect.  So, these results are inconclusive IMO. 

Also note the current on CH4(grn) in the cancelling turn on toroid 2 and the unbalanced voltage across the shield is the 3rd harmonic.

Regards,
Pm

Been thinking about the Prof. Lewin video's since they were put on you tube, done the research, read the papers and done the tests. My conclusion is that a point source like a battery in a closed loop is in effect an electron pump creating a high/low electron density which must distribute itself along any length of resistance... equal and opposite. We could use a water pump and restrictions in a loop of pipe and produce the same results using the same line of reasoning.

Now we take the same conductor loop which is induced from an external source and the only real difference is that the Electromotive Force propelling the electrons along the conductor is now distributed along the entire length of the conductor versus a point source. In effect this would be equivalent to using an impeller within a circular casing and imparting a uniform force on every part of the water loop along the outer circular path of the casing. If we then added restrictions along this path then it would seem reasonable that the sum of pressure drops must be equal to the sum of pressure gains... equal and opposite.

In my mind all the equations and math, all the measurements and debate over what they mean is less important than the line of reasoning used to explain what we think we are seeing. Thus to say this system is non-conservative is to say something came about which was not present in the first place. An electromotive force or pressure on the free electrons came about which caused a change in the difference in potential which was measured across the resistances not supplied by a point source or uniformly distributed source. In conclusion, I say it cannot be non-conservative within a simple closed loop however we should acknowledge that Kirchhoff said the rule only applies to a closed loop with good reason.

Personally... I think Lewin, possibly about to retire, may have decided to throw a monkey wrench into the works just to see what would happen. I mean look at his clothes, that hair and how intelligent and witty he is for a man of his age. You have to respect the fact he's a troublemaker and isn't afraid to offend a few people. I think he just wanted to stir the pot and make a few people think and that is not necessarily a bad thing in this day and age. That or he's been smoking crystal meth, it's uncertain at this point.

Regards
AC

AC,

I am quite sure that Prof. Lewin is quite serious in his assertion.

He would be correct if he would separate the discussion into E field measurements, and emf/voltage drop measurements. They are two distinct animals, but he intermixes them in his lectures, and it has caused a great deal of confusion and misunderstanding with his audiences, both students and academics alike.

If one is talking about the induced E field and the measurement thereof (as in Lewin's experiment), then yes, the "voltages" are non-conservative. In this case KVL does not hold, and it is not expected to.

If one is concerned with the induced emf and voltage drops, then one must measure them accordingly by effectively isolating the measurement device from the experiment. In this case KVL holds, and it is expected to.

When discussing or lecturing on this experiment, one must be keenly aware of the differences between these two perspectives and not to get them confused with each other or equate one with the other. Unfortunately, Lewin does a poor job of this, and is not even aware of his error.

https://www.insidehighered.com/news/2015/03/10/despite-mits-decision-delete-walter-lewins-lecture-videos-live

I like this response...

QuoteThose who oppose taking the videos down, meanwhile, often bring up the transgressions of other famous scientists. "I don't care that Newton was a petty asshole, an alchemist and a religious nut; I don't care that Feynman was misogynist; I don't care that Heisenberg worked for the Nazis; and I don't care about Lewin either," another commenter wrote. "I care about their knowledge that they gave to the world. I am not responsible for their lives, that's their goddamn headache, but don't punish me for their sins by restricting my access to their material."

What a sorry state of affairs.

Indeed,

Leftist knee-jerk reactions and protests (and the not-well-thought-out knee-jerk responses) are creating unnecessary damage to many a reputation and society in general.

I also totally agree with what Allcanadian quoted.
Should we also delete the films of Roman Polanski and Woody Allen accused in a similar way?
Nobody's perfect. Men's faults must not make us destroy the works of their talented part, especially when it is not proven.

Quote from: poynt99 on 2019.07.05, 02:17:37
AC,

I am quite sure that Prof. Lewin is quite serious in his assertion.

He would be correct if he would separate the discussion into E field measurements, and emf/voltage drop measurements. They are two distinct animals...
...

I wouldn't say that. We have to go back to definitions.
An E field is a property of a point in space, where a charge q is subjected to a force F such that E=F/q (I can't put arrows but F and E are vectors).
There is therefore only one electric field, defined from a mechanical force.

Only the way to obtain it changes depending on whether it is by a potential difference, or by induction.

In the first case, the electric field is created through a conductor or dielectric by a potential difference, i. e. by forcing an imbalance of the charges between the two ends. The electric field E at a point therefore depends on the differences in charge accumulation at distant locations.

In the case of induction, the field is directly created in any point of space by moving charges, those of the current of the source of the magnetic field, because their motion creates an anisotropy of their coulombian field that do no more cancel the positive charges at rest.
The induced charges in a conductor move through it because they are subject to this resultant field. If there is a resistance in the induced circuit, then the displacement of the induced charges is no longer uniform and will create differences in the accumulation of charges that are seen as potential differences.

This is the opposite effect of the previous one: the induction field E generates the potential difference in the resistance, while in the other case the potential difference generates the field E through the resistance.

Quote from: poynt99 on 2019.07.06, 04:37:55
Indeed,

Leftist knee-jerk reactions and protests (and the not-well-thought-out knee-jerk responses) are creating unnecessary damage to many a reputation and society in general.

Agreed

From F6FLT

QuoteI also totally agree with what Allcanadian quoted.
Should we also delete the films of Roman Polanski and Woody Allen accused in a similar way?
Nobody's perfect. Men's faults must not make us destroy the works of their talented part, especially when it is not proven.

Agreed again.

I don't know the full truth of the matter, but from what I have read it seems IMO that MIT is caving in to a few politically correct feminist staff members yielding more power than they deserve to have.  Oh that's right we are supposed to believe at all costs a mentally ill person's word without proof even if it destroys the reputation and works of Lewin. Now where have we seen this pattern before?

We are in very sick times.

Quote from: F6FLT on 2019.07.07, 19:26:40
An E field is a property of a point in space, where a charge q is subjected to a force F such that E=F/q (I can't put arrows but F and E are vectors).

Several of my books on EM theory use bold characters to represent vectors, and you will have seen that I follow that practice.  Perhaps I should have made that clear before posting any formula.  Anyhow its easy to do here.
Smudge

What i find interesting is that we seem to be unable to shield the secondary from the electric field.
Dose no one else find this interesting ?.


Brad

Quote from: TinMan on 2019.07.08, 11:38:31
What i find interesting is that we seem to be unable to shield the secondary from the electric field.
Dose no one else find this interesting ?.


Brad

Well, I'm certainly no expert but it seems to me that the secondary induction is not caused by the primary's E field but rather by change in the magnetic vector potential or A field in the center of the toroid over time or dA/dt .  The question then would be- is the A field able to be shielded?

Those much more knowledgeable that I should refute or confirm this!  C.C

Pm

Quote from: partzman on 2019.07.08, 13:37:21
Well, I'm certainly no expert but it seems to me that the secondary induction is not caused by the primary's E field but rather by change in the magnetic vector potential or A field in the center of the toroid over time or dA/dt .

It isn't just the A field at the center of the toroid (although it has maximum value there) but the whole circular A field that goes around the core and passes through the center (as does the secondary conductor).  IOW there is an A field alternating in value hence an E field all along the secondary wire.

QuoteThe question then would be- is the A field able to be shielded?

No it can't be shielded in the normal sense of a shield.  But it can be almost negated and the easiest way to do this is to make the shield a shorted turn.  Then the current induced into that shorted turn produces an additional A field that opposes the original one, and that opposing A field has curl so there is also a B field opposing the applied B field in the core.  The poor old primary now has a hard time trying to drive any flux into the core.

Smudge

Thanks Smudge!

Pm