The phenomenon is exactly the same.
The only spot i disagree with Dr. Lewin is the use of KVL on a circuit that has not been modeled to include the magnetic properties of wires.
So what is always true is that the sum of the electric field E and the cross product of the velocity of the conductor and the magnetic field B—which is the total force on a unit charge—must have the value zero inside the conductor:
F/unit charge = E+v×B = 0 (in a perfect conductor), (22.12)
where v represents the velocity of the conductor. Our earlier statement that there is no electric field inside a perfect conductor is all right if the velocity v of the conductor is zero;
Once they are modeled everything works fine and gives identical results without any sort of paradox.
The phenomenon is exactly the same.
You say that circuit of Dr.Lewin's experiment at RF frequencies like 3GHz acts same way as 300Hz?
You say that circuit of Dr.Lewin's experiment at RF frequencies like 3GHz acts same way as 300Hz?
Let me see. To explain Lewin's circuit, we need Maxwell's equations. To explain how a loop antenna works, we need Maxwell's equations. I'm starting to see a coincidence there.
Your BS answer actually proves my point
The only spot i disagree with Dr. Lewin is the use of KVL on a circuit that has not been modeled to include the magnetic properties of wires.The wires do not have "magnetic properties". The only way for a wire to "react" to magnetic field is if it is moving in relation to a frame of reference.
This is stated clearly in Feynman's chapter 22:
From the point of view of circuit analysis the wire is a magnetic component because the current inside it interacts with the magnetic field, much like a capacitor is an electrostatic component (Even if its just parasitic capacitance between two wires rather than an actual component with parallel plates inside)
To prove my point that you can measure 0V or 250mV depending on what you want to see i have put together some examples:
So there you go, that's how you measure 250mV in there, if you don't believe me go on and try building one of these circuits to see for yourself.
Circuit analysis stays perfectly consistent when you model things correctly.
If its 0V that you want to see just look at circuits in figures 2a 2b 2c and see it is indeed 0V. Happy now?
There is no paradox with this circuit as long as its correctly modeled.
So everything is in one plane, 2 dimensional.
Now what do you measure? Does it matter if the scope is on the left side or the right side?
What if you measure a quarter turn of the wire? Do you get 0V, or something else?
I don't want to see 0V. Nature shows it is 0V.
Mehdi and Mabilde reconstructed Lewin's circuit. Mabilde got even to the point of recreating Lewin's solenoid.
They thought like you: there must be a voltage in the wires. Lewin goofed it up. He bad probed the whole thing. He doesn't know how to lump model his circuit.
When they measured exactly zero volts they got puzzled, and invented each their own completely different convoluted pseudo-scientific theory to explain what the Maxwell's equations predict with simplicity and elegance: there are no voltages in the wires, nor in the probes, Kirchhoff doesn't hold for varying magnetic fields in a circuit, Faraday's law is what accounts for the voltages across the resistors, nothing else.
QuoteThere is no paradox with this circuit as long as its correctly modeled.
The only thing your circuit is modeling is the lack of understanding of electromagnetism.
Your version contradicts Mehdi's version which contradicts Mabilde's version. Which shows that the assumption that Kirchhoff always holds for varying magnetic fields is self-contradictory, i.e. is a paradox. Which is what Lewin brilliantly showed.
Since it is impossible to lump model Lewin's circuit nor any other circuit with varying magnetic fields in them, the paradox ceases to exist when you realize that the only way to solve them is to apply Faraday's law and give Kirchhoff and his doggone "law" to what it deserves: the birds.
If you know enough about circuit meshes you also know that you can't have any voltage jumps within the same net, hence why a component is introduced into the mesh to represent the voltage jump, this component is what represents the potential across the ends of the wire. There is Faradays law itself inside that component.
All i did was show that Dr. Lewins circuit can be lump modeled just fine and that the resulting circuit mesh behaves identically while causing no paradoxes.
So everything is in one plane, 2 dimensional.
Now what do you measure? Does it matter if the scope is on the left side or the right side?
What if you measure a quarter turn of the wire? Do you get 0V, or something else?
Typical coax cable doesn't really give any shielding to magnetic fields so the scope would show different values depending on what side it is on. If the shielding material on the coax cable is made of a infinite permeability material then you should get the same value no matter where you put the scope. Same difference as Figures 1 and 2 in my post, its a different magnetic path.
So everything is in one plane, 2 dimensional.
Now what do you measure? Does it matter if the scope is on the left side or the right side?
What if you measure a quarter turn of the wire? Do you get 0V, or something else?
Typical coax cable doesn't really give any shielding to magnetic fields so the scope would show different values depending on what side it is on. If the shielding material on the coax cable is made of a infinite permeability material then you should get the same value no matter where you put the scope. Same difference as Figures 1 and 2 in my post, its a different magnetic path.
There are no magnetic fields where the coax is. You yourself simulated the field pattern for a solenoid. There is only an electric field. The coax should have some effect with an electric field present.
Yes the conductive shield around a coax does make it immune to electrostatic fields as the charges in the shield will redistribute to perfectly oppose it. However the non conservative field caused by the vicinity of a magnetic field is not the same thing.
The reason Romer uses coax cable...
Yes so you take the dotted Γ line in a path that encloses no net field, everything else is solved using Maxwell and everything is fine.
If this is the wrong way to apply circuit analysis can you then explain why all the voltmeters still show correct values regardless if the type of analysis applied is Faradays loop equation or just circuit analysis as a transformer? If the voltmeters are supposed to show something else feel free to point it out.
(Regarding to this post: https://www.eevblog.com/forum/chat/does-kirchhoffs-law-hold-disagreeing-with-a-master/msg2182160/#msg2182160 )
If its wrong you should be able to make it spit out wrong results too (at least using some special case circuit if not in general).
It is wrong because you added "compensation" to your probes to account for a non existing voltage in the wires.
You know, this is the problem with using Spice as a learning tool. Spice is a stupid software with no critical thinking. If your premises are wrong it will accept them as truth and will sheepishly confirm whatever wrong conclusions you have drawn.
Ditch this devilish program at once and learn electromagnetism as it should: studying the classics, like Feynman's lectures and such.
Where is the compensation? I'm just treating the probe wires the same way as the circuit wires since they follow the same path. Should wires that connect voltmeters be treated in some special way?
And what is wrong with SPICE?
The difference between a purely electrostatic field and one induced by a magnetic field is the non conservative part. This gives the field ability to push electrons around closed loops of conductor while a electrostatic field can only redistribute them but not sustain a current apart from the very brief transient as they redistribute.
Since the inner conductor forms a continuous loop trough the voltmeter, it means that the the field can push the electrons around it and create a current that the voltmeter detects as voltage across its internal resistance. The shield however does not form a continuous loop and as such can't experience any current trough it.
...My point is that the coax shield in this configuration has no effect.