RL Circuit: Kirchoff's Law is Wrong - Faraday's Law is Correct

[Wannabeagenius]

There is no difference between considering the ideal case and using the approximations. That is just two ways of saying the same thing.

The ideal case that you mention is the same as the third assumption
3) no magnetic coupling between components

But note carefully that he also uses the other two assumptions in his work:

He does not consider the finite speed of light so he is assuming
1) the circuit is small relative to the wavelengths involved (lumped-parameter system)

He does not consider the self-capacitance of the wires so he is assuming
2) no net charge on any component at any time

Something tells me that we are getting lost in the details and missing the big picture!

 

[Upisoft]

Professor Lewin is an awesome professor and the lecture that you viewed was from MIT

I've read part of the attached document. At the beginning he counts the electric fields in the circuit claiming there is none in the loop inductor.

there is no electric field in this loop if the resistance of the wire making up the loop is zero

Then he is doing some calculations without mentioning "electric field" anywhere, until suddenly it magically appears:

This is the effect of having a non-zero inductance in a circuit, i.e., of taking into account the induced electric fields due to time changing $\vec{B}$ fields.

 

[Wannabeagenius]

I've read part of the attached document. At the beginning he counts the electric fields in the circuit claiming there is none in the loop inductor.


Then he is doing some calculations without mentioning "electric field" anywhere, until suddenly it magically appears:

Let me reread it.

Please remember that Professor Lewin took the writeup of another professor at MIT, Professor Belcher. This writeup apparently justified Professor Lewin's position and Professor Lewin modifed it slightly. On this, I'm sure Professor Lewin is correct.

That means this is the position of two MIT professors.

Bob

 

[stevenb]

Perhaps he would regard it as a challenge by amateurs. I just don't know.

Any thoughts on that idea?

Some of us here are professionals too (MIT professors haven't cornered that market), and amateurs have the right to challenge ideas. I think someone should make him aware that there are many professionals who don't agree with his quoted definition of KVL. I would challenge you or him to find any book from the early to middle 20'th century which conflicts with Maxwell's definition. Why should we suddenly change our definition because of a handful of recent books that "get it wrong" and leave us with a definition that violates fundamental physical laws that we all accept?

Still, let's not let a disagreement about a definition tarnish the great respect Prof. Lewin deserves for his vast knowledge, great teaching style and his commitment to education. As thinking people, we can isolate semantics from substance. The substance (i.e. the physics) of what Prof. Lewin says is sound in my view. I just disagree with a definition he chooses to use for KVL.

 

[Dale]

The substance (i.e. the physics) of what Prof. Lewin says is sound in my view. I just disagree with a definition he choses to use for KVL.

I agree with that. I particularly like how he explained the physics of a loop with non-uniform resistance. My only disagreement with him is his assertion that everyone else is wrong to use KVL in this context.

 

[Upisoft]

Yes he is correct, in sense. The induced electric fields, he is speaking about can be measured. If he has created a very bad inductor, the induced electric fields can be everywhere, even in the TV set of your neighbor. Let's assume this is not the case, then the professor is correct again. There will be induced emf in the resistor, the battery, the loop and the wires connecting them, totaling the expected emf. So, he is correct, great. What he fails to explain, is that usually people don't make inductors that leak magnetic field all around. And when this is taken into account the expected emf will be found on the inductor itself.

 

[Dale]

That means this is the position of two MIT professors

This repeated appeal to authority is pointless. Lewin and another MIT professor (unnamed) are less authoritative than Maxwell, Kirchoff, Serway, Irwin, Nilsson, Riedel, Rizzoni, and many others not found on my bookshelf.

 

[Andrew Mason]

Have a look at the part of Prof. Lewin's lecture where he disconnects the battery. Current keeps going for a brief time. The reason? The induced emf in the inductor = -LdI/dt drives current through the resistor. The positive potential difference from one end of the coil to the other due to the negative rate of change of current means there is an electric field within the coil (the only place in the circuit where an electric field exists) such that for the circuit:

$$\int_0^{length}\vec{E}\cdot d\vec{l} = -L\frac{dI}{dt} = \oint \vec{E}\cdot d\vec{l}$$

So a volt meter will register a potential difference cross the coil and an equal and opposite potential drop across the resistor. The sum of these readings equals 0: IR + (-LdI/dt) = 0, which is Kirchoff's voltage law. Or you could say that

$$\oint E\cdot dl = -L\frac{dI}{dt}= - IR$$

So if one applies Kirchoff's voltage law correctly and Faraday's law one gets the same result.

AM

 

[Wannabeagenius]

Have a look at the part of Prof. Lewin's lecture where he disconnects the battery. Current keeps going for a brief time. The reason? The induced emf in the inductor = -LdI/dt (the positive potential difference from one end of the coil to the other due to the negative rate of change of current means there is an electric field within the coil) drives current through the resistor. So a volt meter will register a potential difference cross the coil and an equal and opposite potential difference across the resistor. The sum of these readings equals 0. IR + (-LdI/dt) = 0. Or you could say that IR = LdI/dt.

So if one applies Kirchoff's voltage law correctly and Faraday's law one gets the same result.

AM

I was thinking about this myself.

If you put the probes of an oscilloscope across the inductor, I'm sure that you will get the trace of a voltage. I've never actually done it but how can it not be?

Perhaps there is something much deeper to this. I say this because how could Professor Lewin be so wrong on what superficially seems like such a simple issue?

Really guys! I don't get it!

Bob

 

[Upisoft]

I was thinking about this myself.

If you put the probes of an oscilloscope across the inductor, I'm sure that you will get the trace of a voltage. I've never actually done it but how can it not be?

Perhaps there is something much deeper to this. I say this because how could Professor Lewin be so wrong on what superficially seems like such a simple issue?

Really guys! I don't get it!

The professor is right in the most general case. However usually the inductor will have a magnetic core that will keep the magnetic field from inducing any significant electric field in the surrounding circuit. Thus the emf will be found on the inductor ends.

In his example there is an loop inductor with air core, that does not help much, thus you end up with electric fields induced in all parts of the circuit.

 

[Wannabeagenius]

The professor is right in the most general case. However usually the inductor will have a magnetic core that will keep the magnetic field from inducing any significant electric field in the surrounding circuit. Thus the emf will be found on the inductor ends.

In his example there is an loop inductor with air core, that does not help much, thus you end up with electric fields induced in all parts of the circuit.

Are you saying that in the most general case, if you took a voltage measurement across the leads of an inductor, you would get zero?

At this point I am getting very confused!

 

[Upisoft]

Are you saying that in the most general case, if you took a voltage measurement across the leads of an inductor, you would get zero?

At this point I am getting very confused!

No, you will usually get the best part of the EMF on the inductor, because the L of the inductor is a lot more that L of the rest of the circuit. However there will be slight emf induced in the other parts of the circuit, due to the leakage of the magnetic field outside of the inductor.

 

[Wannabeagenius]

No, you will usually get the best part of the EMF on the inductor, because the L of the inductor is a lot more that L of the rest of the circuit. However there will be slight emf induced in the other parts of the circuit, due to the leakage of the magnetic field outside of the inductor.

I think you guys are overanalyzing the situation. This was not an Engineering course but a freshman level Physics course.

Just an ideal circuit. No coupling of coils magnetic field to other components, etc.

Bob

 

[Upisoft]

I think you guys are overanalyzing the situation. This was not an Engineering course but a freshman level Physics course.

Just an ideal circuit. No coupling of coils magnetic field to other components, etc.

Bob

Then take it as he tells it. He just doesn't care where and how the emf is induced.

 

[sophiecentaur]

I have a feeling that the OP has never actually looked across an inductor in a circuit like this and never seen a voltage or he would try to reconcile what the guy in the movie is actually saying and not think that he is saying there are no volts developed across the L. It's great to 'spot the silly mistake' and he seems to think he has actually spotted one; he hasn't; it's just a way of analysing the circuit and not 'reality'. You can see / measure an induced emf every time. No one is actually wrong or disagreeing about the observed facts.

 

[Wannabeagenius]

Hi Guys,

I just want to thank everyone for helping me wrestle with this issue.

Bob