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Hello, I’m Dirk Van Meirvenne andwe are going to talk about a special experiment of magnetism and electricity which was shown by the world famouse prof Walter Lewin. An experiment that looks like a mystery and meanwhile became a subject of heavy discussions. It was a colleague, who is following the prof’s courses who made me aware about the fact that I could watch a video on youtube from this man. And in particular the last part, that starts after 34 minutes. For those ones who want to watch this part first I give the Youtube coordinates,so you can watch it. What is this about? The prof… this is a picture of that Youtube video. The professor showed us this circuit Where he demonstrated the second law of Kirchhoff, in an easy way. This law says: “The directed sum of the electrical potential differences (voltage) around any closed network is zero”. So what did he showed? He made a drawing,on a blackboard from the circuit with two resistors. He gave this resistor a value of 100 ohm and this one a value of 900 ohm. He placed a battery in the circuit. and then he showed us how Kirchhoff’s law works in this circuit. He gave the battery a voltage of 1 volt. this is plus and this is minus The higher potential has a direction flow to the lower potential so this is the direction how the current will flow. Which voltages will be over the resistors? He demonstrated simply with ohms lawthat if you put a voltmeter over this resistance you will measure 0.1V and the second voltmeter gives us 0.9V There is clearly one thing missingto demonstrate Kirchhoff’s law; the polarities. The plus goes this way, so this means that this side is “plus” the current goes this way also on this point we have “plus”the current goes this way The current keeps on going this way and over the battery we have a potential in this wayplus and minus If we simply apply Kirchhoff’s law, we can write: +0.1V +0.9V minus the battery voltage: 1V is equal to zero volt. Very simple, this is Kirchhoff’s law. Then the prof made a new drawing exact the same as before, but without the battery He considered what would happen if in this closed circuit an increasing magnetic pulse would come out of the blackboard through the open surface. And that magnetic pulse creates “by changes” also one volt in the circuit. He calculated which voltages would be over the resistors. He said: “Here we have 0.1V and here we have 0.9V. He indicated the wire above as D and below as A How are we going from D to A? There are two paths he said. And he find something strange with the polarities this was plus and minus and on this voltmeter he had minus and plus. His students were very surprised. because there were 2 cupper wires which connecting the resistors. and on one side the voltmeter shows us 0.1Vand on the other side the voltmeter shows us 0.9V! And there is also something strange with the polarities! We have here minus and plus and on the other side it switches in plus and minus and this on the same wires! The prof didn’t approach this experiment only theoretically with Faradays law. He said that these result couldn’t be explained with Kirchhoff’s law. thus KIRCHHOFF IS FOR THE BIRDS” he said. Now I will show you the scope pictures,because he did the experiment in practice too And indeed, I can see a negative pulse of 0.1V over the 100 ohm resistor over the 100 ohm resistor and a positive pulse of 0.9V over the 900 ohm resistor. The prof said that for many scientists this was something incomprehensible. And also the many reactions shown on Youtube were clear even people with high college grades didn’t understand it.

brainteasereven people with high college grades didn’t understand it. Thanks to my colleague,I have also been able to view that video and it became clear to me that something was wrong. I‘ve sent a few messages to the profwith my opinion about this experiment. But his answer was thatI was mistaken and that I didn’t understand it. I then build a test setup and with the results it became clear to me that I wasn’t mistaken at all! I then have sent an answer to the professorthat I would prove that I was right and put it on a new Youtube film, what I am doing now. I will first explain now how I worked and why I did it in that way. The core was that we had to study a single magnetic pulsethrough a closed circuit with open surface. And to make this possible we needed a memory scope that could hold this one picture shot It was also necessary to push manuallya button for every pulse I wanted to study I started this way, but soon it becameclear to me that this wasn’t practical at all. And with all the experiments I had to do,this would consume too much of my time. For that reason, I made an electronic system. which created the pulses automatically just as if I pushed every timea button to create that pulse. So we let the electronics do that job. But there was a little problem! Because this pulse whichwe can see on the video of prof Lewin. is created at the moment the current is switched on. What we didn’t see on that video is what happened when the button is switched off. Then you’ve got also this negative pulse!But we can’t do anything with that pulse So the electronica was made that waythat only this positive pulse remains. because this shows us also the polarity Here I have a solenoid and I have put a pulse train on it. You can view the pulse here. For all the experiments you will see,this electronic system will be used. Now a coil or solenoid like this, there is a problem with it! There are magnetic fields all around it. And when I was doing my experiments it became clear to me that a coillike this wasn’t so practically. I will show you what I mean. These are oscilloscope probes. This is the minus and this is the plus. I make a short circuit now between minus and plus This way, the probe isn’t able anymore to do a measurement. Now we are going to watch what will happen if we put these short circuited scope wiresthrough the magnetic field of this coil, where at this moment a pulse train is generated on. What can we see. while I have a short circuited probe that normally can’t measure anything? On every side of the coil,my wires pick up the magnetic fields all around. That is something we can miss like the plaque We only want to see whatis generated by the experiment. We can’t have it that the magnetic fieldinfluences our results by inducing our measuring wires. For that reason,I had to find another solution. After some experimenting and research my conclusion was that a transformer wasan excellent solenoid to create magnetic pulses. Why is a transformer better than this coil? Here you can see how the magneticfield lines goes around the coil. They come out of the coil’s centregoes around like this. and flow into the centre on the other side of the coil.This way we have a magnetic field all around the coil. That‘s the reason whymy oscilloscope wires picked up this field. With a transformer, it’s completely different.

brainteaserWith a transformer, it’s completely different. Here we have the coil, with an iron core in the middle. And here on both sides we have iron magnetic field conductors On this drawing I removedthe upper part and projected the plan view. Which path takes the magnetic field in this case? It goes through the middle of the coil and follows its path throughthe iron magnetic field conductors. The magnetic field that –without the iron core-would chose normally also this path. doesn’t manifest (almost) And that is very important because when our volt- or scope wires passes through this area we don’t pick up the magnetic field.It would influenceour measurements and we certainly have to avoid that. This is a transformer. Normally it is used for transforming the main voltage in the primary side to 36V on the secondary side. We ain’t gonna use it that way at all. In this part we sent a huge magnetic… sorry, we will put on this side a voltage which creates a large magnetic pulse. Let’s take a view how it looks like. Here we have our pulse which I offer to the transformer. The sensitivity of the oscilloscope is very low because we are viewing a large voltage which createa large current through the transformer. Lets take a look if there is indeed no magnetic field around the transformer and if it is really concentrated on theinner side as I have drawn on the white board I restore my scope setting on a high sensitivity. We have set it on 20 millivoltper division, that’s very sensitive. Let’s go around the transformer and wesee only a very small magnetic pulse. but that is negligible comparedto what we have seen around the coil. Neither on the frontnor at the back, a pulse appears. So now we are sure that there is no magneticfield that could induce our measuring wires. Let us now take a look if thereis a magnetic field on the inner side. I put a simple wire in the transformer,between the space of the transformers coiland the iron side. And now we have a scope viewfrom the pulse that is induced in the wire. Our result is not influenced by a magnetic field into the measuring wires. Now you are going to watch the main experimentas professor Lewin did at the end of his video. with the two scope views. How did I do it? I used the same schematic drawing. But the professor’s drawing had 2voltmeters and mine has four voltmeters. And the results are just what they are!We are going to measure voltages and polarities I’ve indicated this corners as A, B, C and D. The blue and red wire aregoing through the transformer’s magnetic field. They form together with the 100 ohmand 900 ohm resistor a closed loop. Also these corners are indicated with A, B, C and D. The measuring wires are also indicated with these letters. Also these wires are indicated with A, B, C and D. First we are going to watch the voltages over the resistors. and we will compare the measurementswith the results of professor Lewin. There is still one important thing I have to say. In a moment you will see four scope views,two on every scope. But the earth or minus from of each scope have an inside connection. That means that two measuring probes have a common minus. We see the same with the other scope and that is the reason why the 2 minus connections from eachscope need to be connected at the same measuring point. In order of the polarities,we have to pay attention to that. The black clips are the minus. Now we will view the voltagesover the 100 ohm and 900 ohm resistor. This is the voltage over the 900 ohm resistor.

brainteaser This is the voltage over the 900 ohm resistor.

brainteaserThis is the voltage over the 900 ohm resistor. And that is the pulse over the 100 ohm resistor. What are the voltages? The small pulse is over the red…excuse meit is this red one. This is A-D. When I disconnect the probe it is gone. Each division is 100 milivolt. This pulse is 100 millivolt, 0.1V. What is the plus and the minus? The minus is D. and the plus is A. OK. We will do the same over the 900 ohm resistor B and C. When I pull out this probe, the view disappears,so this is indeed connecting point B and C. We see a much larger pulse,too large for this view, because the setting is the same as the scope above. We are going to lower sensitivity. One, two, three, four and a half, multiplied with 0.2 volt is equal to 0.9 volt. OK, let us now take a look to the polarities. This grey probe is over B and C… disconnecting… OK, the plus is C and the minus is B. At this moment we are sure that the results of professor Lewin are exact the same with our measurements. We have over the two cupper wires a voltage of 0.9 voltand a voltage of 0.1 volt and also with changed polarities. The mystery isn’t solved yet! Till now we only have the explanation as given by mister Walter Lewin. But we still have to do the measurementover connections A and B and over D and C. We expect that we wouldn’t measure anything becausethese are cupper wires with a resistance of maybe 0.1 milliohm. Lets have a look. This one is the red probe over D and C. Yes… and what is the voltage over D and C? Five scale divisions and each division is 0.2V… sorry 0.1 volt.Five divisions multiplied with 0.1V is equal to 0.5V. 0,5 volt over C and D. We also need to know the polarities. The pulses rises to the upper side. this means that D is the minus and C the plus. We do the same over A and B. We disconnect the lower probe so we get a better view. We see five scale divisions. multiplied with 0.1 volt is equal to 0.5 volt. This measuring is done by the red probe. B is the minus. And A is the plus. In other words, it is clear now. The magnetic pulse, coming out of the white board… sorry in this case it goes into the white board, but that changes only all polarities. we see that the two wires in the magnetic field. are acting like power supplies or voltages sources. And if we look how the currents flow from the two “power supplies” I think that some things are very clear now. Taking Kirchhoff’s low in account. 0.5 volt and 0.5V from the wiresand 0.1V and 0.9V from the resistors… If then you take the polarities in account, this is corresponding 100% with Kirchhoff’s law. In this case, Kirchhoff is still alive and kicking! The prof said: “KIRCHHOFF IS FOR THE BIRDS!” I think we can say:“WE GET KIRCHHOFF BACK OUT OF THE BIRDSHIT!”. I haven”t put the results in the Kirchhoff formula yet. We are going to do that now. We see the direction of battery potentials and we also see the opposite direction over the resistors. Now we can write: +0.1V – 0.5V +0.9V -0.5V = zero volt.

brainteaserNow we can write: +0.1V – 0.5V +0.9V -0.5V = zero volt. But final, the drawing isn’t entirely correct. It would have been more correct if Professor Walter Lewin should have made a drawing were he replaces the wires by coil symbols. Because the wires act like coils and they become the power supplies in this circuit. You can also replace the coils by batteries which are almost empty. If you have batteries that are almost empty and a you connect a low resistor on them.they will discharge immediately It will create a similar pulse curvebecause the little amount of energy will be discharged in a short while, and ends at zero volt. And now it probably will be very clear to you. If we would replace the two coil symbolsby that kind of “almost empty batteries” you see then two times 0.5V and in the other direction 0.1V and 0.9V. This fully complies with Kirchhoff’s law. In other words, Kirchhoff’s law remainsfor the fully 100% in Mr. Lewin’s experiment and in the drawing of the circuit as he showed on the blackboard. What has really happened? Let’s look back to our transformer to see what has happened. I have put around the transformer’s coil, one loop. We are going to measure the voltage over the wire. One scale division is 0.1 volt.One, two, three, four, five, six… we have six times 0.1V. 0.6 volt is generated in this wire. The interesting thing aboutsuch a pulse is that we can see a polarity. I will show it by switching my measuring probes. Here we have the same six scale division pulse,but with a switched polarity. Now we can ask us the next question. What would happen if we cut the wire in two pieces. Let’s do it. This is about the middle. We disconnect the scope probes. Now we have two wires. Let us do a measurement over one wire, if there is still a voltage. We indeed see a voltage over the wire. Three scale divisions, while in our last measurement we had six scale divisions. You will probably know already what willbe measured over the other piece of wire. Also three scale divisions of course but with a reversed polarity! How are the polarities over these two wires against each other? This wire has a minus and a plus like this. The other wire has a plus and a minus like this. Do you start to see the equations with professor Lewin’s circuit? Finally it is very normal that you see the polarities like this. Because if I connect the two halve wires each with a voltage of 0.3V, to one loop again. then we have a voltage of 0.6, what we have measured the first time. This is only possible if the polaritiesof the halve wires are reversed. What if the polarities were equal? You can compare it with a circuit of two batteries. you will recognize them, such 1.5 volt batteries. If these 1.5 volt batteries have the plus on the same side. And we make a connection between the two minus poles. The potential between the two plus poles will be zero volt.when you connect a voltmeter The batteries have to be connected this way. When you make this connection, you will measure 3 volt. These are just some basics.. You can compare this with theprocess in the coils and transformers. Imagine what would happen if a comparableprocess did not took place in a transformer. If we have a transformer loop… we’re gone draw just one transformer loop. We have here one loop and imagine that we cut it in two pieces. And at a certain moment in time you should have herethe minus and the plus in this piece of wire and suppose that we had in the other pieceof wire the plus and the minus at these points in that case the two voltages would bring themselves to zero. In the next transformers loops would happens the same. In other words, a transformer can’t work that way. The only possible way that a transformer will work is if the polarities of the two halve loops are reversed as I just demonstrated In a main voltage circuit,the polarity switched 50 times each second. This is very basic. This is what also happens in our circuit when a magnetic pulse is going throughthe two wires of the closed circuit. But some people think that when a single magnetic pulse is created with DCthis pulse is also DC You have to know that every voltageincrease or voltage decrease is an AC voltage. If you have a flashlight with a 3 volt DC battery. the moment you switch it on that moment correspondents to laws of accelerating current. Every moment when you switch it on and off, and on and off,… this moment is according the laws of accelerating current. Switching power supplies are also working that way. They sometimes have a DC voltage over the 300V. And in an electronic way, the 300V DC is cuteach second in about 20.000 to 50.000 pieces. It goes this way: on – off – on – off… Computers, televisions etc. have this kind of switching power supplies where a DC voltage is cut in pieces. The reason why they want it this way, isbecause the voltage is given to a transformer to create the voltages they want to use. And that is what also has happenedwith the magnetic pulse in this circuit. The single magnetic pulse that goes through this closed circuit, has created in our wires a very short time an AC voltage. And it is that voltage of two times 0.5 volt as we saw, that acts like a small power supply over the two resistors. You have seen how we created a magnetic pulse with a transformer and sent a magnetic pulse through one loop. which created 0.6V. Then we cut the wire into two halve pieces andagain we let a magnetic pulse pass through it. The result was that we had over each piece of wire 0.3V and the polarities as showed on this white board. Let us now take a look to the circuit that the prof has used. At a certain moment the prof said: “It depends on which pathwill be taken to go from D to A.” And he drew two arrows on the blackboard in order if this is answeringto Kirchhoff’s law and Faradays law. Normally the left path can’texist when you take a look to the polarities. Only the clockwise path exist in this circuit,when taking the current flow in account. Now you can ask yourself if it is possible that two paths exist in this circuit at the same moment? Yes, it’s certain possible! In that case the magnetic pulse should be manifested on another place. Now he’s in the middle of the circuit but if the pulse had been here, then you would have two paths from D to A. But which result can weexpect in this case from our measurements? Lets’ take a look. We’re gone use our transformer again. where the red and blue wire is stillon the left and right side of the coil. And as I have shown, on each wire we had 0.3 volt. Though with reversed polaritiesas I have drew on the white board. If we connect the wires back again together to one loop we had 0.6V. But now we are going to bring themagnetic pulse on one side of the wires. I take the blue wire on the left and place it together with the red wire I am going to connect my probes on the blue wire.this moment you see voltage of 0.3V on the red wire. and now I am going to connect my probes on the blue wire. And as expected , we see also a voltageof 0.3V but with the same polarity now. So we have two wires with a magnetic pulse on this side. And as we have measured, are both plusseson this side and both minus on this side. With a voltage of 0.3V each. Now we are going to connect both wires Let us take a look what we will measure on the other side of the wires? and we connect our voltmeteron this side of the wires. Zero! There is no voltage at all! But this is very normal. As I gave an example with the two paralleled batteries the voltages neutralise each other because they have the same phase or polarity. In other words if you have a circuit where both paths arecreated with a magnetic pulse on one side. then you would see a result as I just demonstrated. There are also other possibilities, e.g. if in this closed circuit with open surface there would be two magnetic pulses at the same time Suppose one pulse goes into the whiteboard and another comes out at the same moment! If the magnetic pulses have exactly the same strength the result would be zero volt.There would be no current flow. If either of magnetic pulses were stronger, then the resultant determined the result. The strongest pulse will dominate. E.g. if this pulse creates an EMF of“0.8V” and the other one “minus 0.5V” you would have a current…, sorry a voltage of 0.3V. Take in account that in these examples the bothpulses were at exact the same moment of time. What would happen if one of both pulses appears some milli- or microseconds later than the other? In that case you have to calculate it. This is such a calculation example. We have a large positive pulse.. I first put the paper on the white board,if the camera can zoom the picture. So we have a positive large pulse and some milliseconds later, we see a smaller negative pulse. In that case it will be the resultant which determinate the voltage at each momentin time, created in the loop. This is what happen when you have two reversedpulses given at a different moment in time. Anyway, at each moment in time Kirchhoff’s law applies. E.g. if at a certain moment in time, the resultant would be 0.3V at that moment the whole circuitwill act as we have seen with the 1V experiment, but now with 0.3V. In other words, if you have one or more magnetic pulsesin the inner side of the closed circuit. or pulses next to the circuit,Kirchhoff’s law remains applicable. In experimenting with closedloops there are some special cases. I am going to show such a special case. Suppose I have a copper wire with a length of one meter. I create a closed loop with this wire. And again we sent an increasing magnetic pulse through the open surface. What would happen? Is it possible to do a measurement onsuch a circuit? What do I have to imagine in this case? Every kind of metal has a specific resistance. In this circuit we have a copper wire and copper has an extreme low specific resistance. You can calculate it’s resistance when you have one meter and a certain diameter. But that doesn’t matter now. We suppose thatthe diameter is equal over the whole length. Suppose that we sent a magnetic pulsethrough this loop and it creates a current of 10A. And the 1 meter copper wire has a resistance of 1 milliohm. One thing is sure,this kind of closed loop is a short circuit. It’s not recommended to put such a closed loop aroundthe core of a transformer. the high current, because of theshort circuit, would heat it up very fast. Imagine that the wire is divided in ten parts. And we put a voltmeter over such a segment of 0.1 meter. The voltage over this littlepiece of wire will be 0.001V. Over each part of 10 cm we will measure the same voltage of 0.001V. You can ask yourself if this is still according Kirchhoff’s law? I can assure you that Kirchhoff’s law still applies. If we remember the comparable example I have given with that almost empty battery and we apply the symbol of that kind of battery then we see that over each piece of wire with a resistance of 0.1 milliohm. will be a voltage of 0.001V. In other words every piece ofwire is a battery- and resistor circuit, it looks like the inner resistance of the battery. Anyway, if the battery voltage is 0.001V. and over the resistor we have 0.001V. taking the polarities in account,plus 0.001V. minus 0.001V. is zero. You can do now the same over a fifth or half the wire, the same system applies. Even this special caseis according Kirchhoff’s law! Let’s see the answers which prof Lewin has given to me. “You are mistaken, this demo is done now by many teachers in the world”. “Since the wire at A and D have nearly ZERO resistance.” “You can connect the voltmeter anywhere you want to”.” It is clear that the prof isn’t aware about the way how the voltage in his circuit is created. It is the induced magnetism in the wires and also a very small part in the resistorsthat creates the voltage in the circuit. “however that is not at all a necessary condition” “As the resistance of the common copper wire ismuch much lower than the resistances of R1 and R2.” Thus my diagram on the blackboard is fine.”. So the way he has drawn it onthe blackboard and as I’ve build it– this is exact the same as his blackboard drawing- and which I put over the core of the magnetic pulse creating transformer. The reaction of the prof is that you can measure anywhere on these copper wires. because the resistance is nearly zero. What he said was correct, the wires are just a few milliohm. But he ignores that these wires are alsothe medium that pick up the induction and create the voltage. “Do not feel bad if you don’t get it,very few people have a deep understanding of Faraday’s Law, even though it runs our entire economy. I don’t want to waste my time by arguing with anyone about this as it is to basic. This is therefore my only response to you. I suggest you read the Lecture Notes onthis lecture and that you also use google.” I didn’t read his Lecture notes and I didn’t use google.. I just have done the experiments as you have seen them and watched the produced results. There still is one more thingthat needs our attention. The prof says that he didn’t used resistors… sorry… that he didn’t used wires at all. As he wrote: “The length of wire A and wire D was zero.” I will show you a few resistors. This are resistors. I have no Idea how the prof couldconnect these resistors without using wires. or using the resistor wires themselves. There do exist smaller and larger resistors,even very small resistors. The very small ones are SMD’sand have the size of about one millimetre. They don’t have wires, but if you would connect these resistors in a closed circuit together you should have no open surface anymore. In that case your experiment has to be done on a few square millimetre. So the prof will not have done it that way. So I have been searching if I could find resistors without wires and I found them. I will show you. This is a plastic carbon tape. The resistance of this carbon tape is evenly distributed. If I make a loop with this, then I have indeeda closed loop with open surface, without wires I made such one and we will watch the resultswhen an increasing magnetic pulse is sent through this open surface. You can see connections, to connect my voltmeters.And to close the loop properly, I made these copper connections. This to be sure that it is a good loop whenwe sent our magnetic pulse through it. How are we going to do this? We have a closed loop with open surfaceand this loop has no wires at all. This loop is only made from carbon tape And I glued it to cardboard to have some firmness. I’ve indicated the four corners as A, B, C and D. We will do about six measurements. The transformer is located in the middle. Over these lengths we will do three measurements and the same on the other side. The different with the previous experiment is that in this case is that all resistors are equal. But this doesn’t matter at all. Later on, you will see that this is no issue to understand what really happens. I only do this kind of experiment because the prof said that he didn’t used wires. Ok, no problem, now we have a circuit without wires. We connect the voltmeter next to the transformer and connection B. We see two scale divisions… first making a good connection… We see two scale divisions, and lets suppose that2 devisions represent 2V it could be also 20 millivolt or 200 millivolt…But it doesn’t matter So let us say this is 2 volt. The polarity is minus on thisside and plus on the other side. We are going to do the same measurements over the other tape resistors. We’re going to work a little faster. Over B and C, about two volt…minus on this side, plus on the other side. Now the connection next to the transformer and C. Also 2 volt. We could expect this, because I just said that everycentimetre of this carbon tape has about the same resistance. The minus in the corner at C and the plus next to the transformer. The structure of the circuit becomes a little clear now watch the polarities; “minus – plus / minus – plus / minus – plus”!It is in line with the expectations. that we will see the same here. Indeed 2 volt, but with a reversed polarity. I switch my measuring probes so we see the positive probe at the plus side that makes the notes more easy. We see 2 volt now with the plus on this side. The last results can’t surprise us anymore. Between D and A also 2 volt, plus at A and minus at D. And the last measurement between A and the point next to the transformer about two volt, a little more, with the polaritiesminus at A and plus on the other side. And now I am going to express me as professor Walter Lewin did: “ What you will see now, this will blow your mind.”. Let us take a look t the polarities of this side: “minus – plus – minus –plus”. Which result do you expect to see on this common voltmeter above the two others between connection A and B? If you ask me, I would answer that we will see 4 volt. And if you ask me what do you expect about the polarity? I would answer: “It can’t be more clear is we go from A to B we see minus – plus –minus –plus. Thus the minus must be at A and the plus at B. We will check if it is right. Hey, this is strange! A connectionhas plus and at B connection minus! And we measure about 2V instead of 4 volt! I can’t believe my eyes, so to be sure about the polarities, I am going to switch my measuring probes. Connection B has now the minus clips. And indeed we see at A the plus and again about 2V. I can believe that most of you don’t understand this phenomenon,but finally these are the measuring results! Let us do now the same over D and C. Also this polarity is reversed,compared to the two voltmeters above it. I ask now to everyone who want to approach this mystery withWalter Lewins theory, to stop this video. Also the other people; stop the video now! Try to give first an explanation about what you have seen with your own eyes!We resume after this break. I hope you have solved it. I can tell you this, those ones who don’t understand Kirchhoff’s law, will not have solved it. Only with Kirchhoff’s law we can solve this mystery. I have told you that the magnetic pulseis located here at this place. And not outside my transformer,to avoid affecting our measuring wires. Now we will do measurements on A B but next to the transformer. About six division on the scope, a little more. In our first measurements we had seen that the 2 volts were also a little more. I switch my probes this way to see a positive pulse on the screen. I put the positive probe on the A side, and the negative probe on the B side,but both next to the transformer. We have 6 volt, with the plus on the leftand the minus on the right. We can expect the same on the other side. Indeed, a pulse of the same size. At this place we see also 6 volt and the polarity on theleft is minus and on the right is plus. I think it becomes clear now. I am going to use the battery symbols again, thosekind of batteries that are almost empty as we have seen before. On the upper side we have this polarity and 6 volt. At the lower side we have 6V and a reversed polarity. Both batteries together creates 12V, equals 6 times 2V. And let’s find out what creates the mistery with the polarities. We’re just going to calculate the upper circuit, than it will be clear for the rest. If we take a measure point next to the transformer on the left side, over to connection B, taking the polarities and potentials in account then we have to write:” -6V +2V = -4V”. This is 100% Kirchhoff’s law. This was a theoretical approach.Let us see if it’s correct in practice. We connect next to the transformer and connection B. We see minus 4 volt. We can do the same calculation over each side of the circuit. I have made a drawing and those one who want to fill in all the values and calculate them, can do that. If we put the 6V battery symbols here, … with the plus and minus this way you can calculate over each sidewith Kirchhoff’s formulae the results. And this is the reason why the two voltmetersover the resistors had reversed polarities as measured over A and B. We had in the middle a kind of power source which Icompared with the almost empty battery, that produces 6V. of course there is no battery it is pure induction that has created it. Anyway, as you have seen,this is 100% according Kirchhoff’s law. I’ve also done other experiments for myself. This is also a loop made from carbon tape. I made some holes in both sides to have higher resistances values. Thus these resistors became different from these ones. I can assure you, we could again applyKirchhoff’s law for the fully 100%. We’re going to end this video,but first I am going to explain just one more thing! There were so many questions about one particular aspect in the youtube video of mr. Lewin. The question was what if both voltmeters were attached exactly in the middle of wire D and A? The prof his answer was every time the same:”. It doesn’t matter where you connect them, thetwo values (-0.1V and +0.9V) will be seen.even if the length of the wires are ZERO!.” But some people asked, which result would be seen if a third or fourth voltmeter were connected over A and D? The prof didn’t answer that. We’re going to check if the answers of the prof were right. For the experiment, I need to use this coil. Around the coil is the closed loop with the 100 and 900 ohm resistors. My probes of the two voltmeters are attachedexactly in the middle of wire D and A. I disconnect my oscilloscope connections and switch them. So you can see that both probes have exact the same readings. This is what I answered every time(scope shows 2 x 0.4V) when both voltmeters are attached at exact the same connection, you will see the same result twice. I think the proof has been delivered…the connection has to be proper of course this are very small voltages. In other words; if you connect both voltmeters over thisresistor or both over the other resistor or both at the exact same point over these wires, you will see the same result over both meters.I can assure you, I’ve tried it all! What are my conclusions? I think for those ones who have seen the experiment in the prof his video and compare it with the experiment in this video, have a new view on this case. About the prof himself.I can tell you he’s one of the best teachers in the world! Meanwhile I have seen a few other of his lessons on Youtube and I can tell you that, if all teachers gave theirlessons in such an enthusiastic way. it would be a better world! Now we are going to end with a new riddle. And maybe I will make a new videowith the answer on the riddle. When you watch the prof his video, at about 29 minutes you will see that he wraps a loop around a solenoid. He says that the shape of that loop doesn’t matter. That’s right, shape doesn’t matter. “It is the size of the surface that counts” he said.“ And the place doesn’t matter, where the magnetic pulse of the same strength will go through the open surfaceof that closed loop it will create always the same EMF in that loop. I have made such a loop. And I connect myvoltmeter over a piece of the copper wire. Sorry, this measuring wire seems to gives a bad connection. We will use the other one. We will do our measurement over this short piece of wire. My voltmeter has to be set on an extremely sensitive setting. It is so sensitive now thatother small interferences come in. But that doesn’t matter. Now I again make this coilproducing serious magnetic pulses and let us watch the effecton the close loop with open surface. OK, I’ll turn the coil this way,so we see the pulses upwards. We indeed see the coil’s induction coming into the loop. But we will see something very strange. When the coil is moving to the middle, the pulses become smaller. Sorry for the moving image on the screen, because triggeringis very difficult with such small signals. But that’s no problem,because it is the size that matter now. I am going back to the edge of the loop and the size of the pulses become larger again while I am still in the surface of the closed loop. The prof also said “shape doesn’t matter”. Let us watch what happensif we change the shape this way. The pulse becomes much larger now, while the size off the open surface doesn’t change. If you watch the prof’s video at 29 minutes you will see that he runsthe experiment with the single loop twice. He uses such a voltmeter with a needle in the middle. And in his first experiment you see his needle points about plus and minus 14 to 16 µA. The second time you can see that hisloop is a little closer to his solenoid. And if you watch carefully, you will see that the needlepoints a value of about plus and minus 20µA. So…. this is not according the laws. Maybe I have to make a new law? I don’t know. Maybe you’ll see me again with a new video? It is easy to do this experiment at home. Very simple, if you have an 100 meter extension wire,which you use e.g. for your lawn mower. This can be used to make many large loops. Close this lope by putting avery sensitive current meter between it. Then you need a strong magnet,e.g. a magnet used in speakers and move it through the open surface of the many loops. You will see that the current meter will show a larger current whenyou are close to the edge and smaller in the middle. It’s a new riddle that I give to you and maybe we meet again!