# fields and currents

Discussion in 'General Science' started by fila, May 12, 2011.

1. ### fila Thread Starter Member

Feb 14, 2011
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Hi guys, this thread made me join this forum so I want to ask a few questions about prof. Lewin's lecture and his lecture supplement ( View attachment sup.PDF ).

Open the PDF (read it a few times and watch the video) document on page 1 and look at the picture. Imagine you close the switch and wait a while until the induction effects die out. There will be a current in the circuit I = E/R where R is the resistance of a resistor assuming all the wires are ideal conductors (superconductors cooled down to 0 K, haha). Now think what are the fields inside the circuit elements. The induction in that large loop (circle) had died out and we can apply the concept of potential. So there is a voltage drop across the resistor and because E = -grad(potential) there must be an electric field inside it in the same direction as the conventional current. But in the ideal wires there can't be an electric field because there isn't any voltage drop across them.
(I found an interesting question on the internet about this latter conclusion. How can a current flow when the electric field is zero? I explained this to myself with the concept of force and dynamics. If the electron is already moving then you don't have to apply any electric field (force) to keep him moving. Its Newtons first (second ???) law. An object will have constant speed if there are no forces.)
So to conclude, the electric field is only inside the resistor, confined only in that location. But this is also the case if we now include the induction effects (induced EMF/electric field) because of the changing current.

On page 4 prof. Lewin tries to explain this. He makes an ''experiment'' by inducing EMF inside an uniform loop (same resistance everywhere) and concludes that the induced electric field will also be uniform at every point. I can accept that because of the symmetry of the problem. Now comes the fun stuff. What if the loop isn't uniformly resistive, one side having lower resistance than the other (that's the second picture). If you read paragraph 2 of page 4 he says: (the current) must be the same on both sides by charge conservation... bla bla and than in paragraph 4: ''Nature accomplishes the reduction in E1 compared by E2 by charging up the junctions separating the wire segments.'' and he continues with his explanation to yield a conclusion that I already mentioned.

Is this legal???

Last edited: May 12, 2011

Apr 20, 2004
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3. ### magnet18 Senior Member

Dec 22, 2010
1,232
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Welcome to the site!
I'm not 100% sure what youre asking, but I would think that a coil with un-uniform resistance would likewise have un-uniform field and un-uniform voltage drop. I guess this voltage drop can be modeled by looking at the junctions.
Current must be the same, electrons in=electrons out.

4. ### fila Thread Starter Member

Feb 14, 2011
64
5
I didn't mean to hijack anything!

magnet18 I am confused when he says charge conservation and later that the junctions are charging up. Obviously there's some charge that is left in those junctions! Is that really what's happening?
But my intention was to talk about the electric fields in circuits, this was just an introductory part. I don't know if anyone will read the first pdf attachment.
If you read prof. Lewin's document you will see that he says that electric fields (induced and non-induced) only exist in the resistive parts of the circuit. On page 1 he says: ''There is no electric field in this loop if the resistance of the wire making up the loop is zero (this may bother you - if so, see the next section).''
Of course it bothered me!!! So I skipped to page 3 and 4 where he explained this by charging up the junctions where wires of different resistance touch each other. He says that's ''natural''. But this was my first question when I opened the thread.
OK, let's accept the ''fact'' that there is no electric field in the wire with zero resistance. Now look at page 1: first equation (Faraday's Law) and the picture. If you close the switch the current will rise and because it is changing there will be some flux change in the big loop and because of that an induced EMF in the loop countering the flux.
Now I ask you what are the electric fields in the circuit?
On the bottom of page 1 prof. Lewin names these fields:
1) electric field in the battery directed from the positive to the negative terminal;
2) electric field in the resistor with resistance R.
Electric field in the battery exists because of some chemical reactions (am I right?) but the second electric field in the resistor puzzles me. Far as I understand it consists of: induced electric field + electric field formed by charges at the junctions (see page 4) + ''normal'' electric field due to voltage drop across every resistor. Prof. Lewin didn't explicitly say this and I would like if someone could correct me if I am wrong.

This is all fine theory and the equation 1 on page 2 is OK but then I ask you this:
1) open the attachment View attachment circuit.pdf
2) what are the electric fields in the circuit,
3) how would you apply Faradays Law on the attached circuit?

Edit: One more thing. In Faraday's Law equation (page 1) does E stand for all electric fields (induced and non-induced)? I suppose so.

Last edited: May 13, 2011
5. ### steveb Senior Member

Jul 3, 2008
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You are referencing Faraday's Law in integral form. In this equation, you can consider E to be the induced field (Ei), or the sum of Ei AND the non-induced field (Ec). Generally, any non-induced field will be a conservative field, and the line integral of a conservative field around a closed loop is zero.

Hence, using either Ei or the sum Ei+Ec gives you the same result.

Last edited: May 13, 2011
6. ### magnet18 Senior Member

Dec 22, 2010
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I believe he meant charging up to be voltage, whereas charge conservation is current.

7. ### steveb Senior Member

Jul 3, 2008
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Is this directed at me? If so, can you explain? I don't understand your comment. Above, I'm just answering one of his many questions?

8. ### fila Thread Starter Member

Feb 14, 2011
64
5
Yeah, that's correct. Did you try solving the circuit problem I attached? I am a student of electrical engineering and we have been taught that there is a ''voltage drop'' across an inductive element (coil) . So by applying the Kirchhoff's rule you get E (source emf) = L*di/dt.

But how do you get this equation by applying Faraday's law. The problem is that there are no electric fields in ideal wires! When the coil induces emf there must be an induced electric field, but if the wires are ideal induced electric field is zero so there is no induced emf. But that is contradictory!

9. ### magnet18 Senior Member

Dec 22, 2010
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No, sorry, I was directing it towards this comment-

Though I admit I'm slightly confused also.

10. ### fila Thread Starter Member

Feb 14, 2011
64
5
magnet18
Charge conservation in its mathematical form is dQ/dt = 0, but if we have some junctions charging up that means that at some time dQ/dt isn't zero. But maybe that happens in a few nanoseconds and after that dQ/dt = 0. So we can live with that. Nothing is simple.

But did you try solving the circuit I posted? (magnet18, steveb)
Maybe if we are dealing with ideal wires than we can't use the fact that there is no electric field in a wire with zero resistance. When dealing with wires with R = 0 we can't use the equation V = IR. Imagine if we had a torus-like superconducting wire, and we induce an EMF over that torus. We can't compute the current with equation I = E/R (R is zero!). We need a different approach. Am I wright or am I wright?

Last edited: May 14, 2011
11. ### nsaspook AAC Fanatic!

Aug 27, 2009
3,557
3,639
KVL is only an very restrained approximation of field theory.
As others have said:
You can't solve field problems with circuit theory.

12. ### fila Thread Starter Member

Feb 14, 2011
64
5
I understand that but how would you apply Faradays Law in the attached problem .View attachment circuit.pdf "> View attachment circuit.pdf
And how do electric fields look like inside those wires?
Maybe I should have asked this right away when I opened the thread.

Unrelated question: Why can't I edit my old posts?

P.S. Big thanks to the guys who are commenting on this thread. If you manage to solve my problem I will be able to continue normally with my life.

Last edited: May 14, 2011
13. ### steveb Senior Member

Jul 3, 2008
2,433
469
You can get the coil equation by applying Faraday's Law with the definition of inductance.

$emf =-{{d\Lambda}\over{dt}}$ is Faraday's Law, with $\Lambda$ representing the flux.

The definition of inductance is $L={{\Lambda}\over{I}}$

Assuming the geometry and material properties of the inductor are not time dependent, you end up with the following from Faraday's Law.

$emf =-{{d\Lambda}\over{dt}}=-{{dLI}\over{dt}}=-L{{dI}\over{dt}}$

Then the negative sign gets absorbed into the definition of coil voltage drop, so V=-emf.

So, what about the mystery of no fields in an ideal inductor? Since the ideal inductor has zero resistance, we conclude that the induced electric field is outside the inductor, and the terminal voltage is the line integral of electric field for any external path that closes the loop at the inductor terminals.

Remember that Faraday's Law only talks about the emf developed around a fully closed path. An inductor, by itself from terminal to terminal, is not a closed path. So, to apply Faraday's law, you must close the path into a loop. Faraday's Law does not tell you where the electric field is around that closed path. It only tells you the net accumulation (line integral) of the field, which it the total emf. So, if you know that the electric field is zero in the inductor, then the electric field must be nonzero outside the inductor, and the inductor terminal voltage is the line integral of that external field. Any, and every path that closes the inductor (even through the air or vacuum, or wires, or resistors etc) must have a net emf determined by the negative time rate of change of flux enclosed by the loop.

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14. ### fila Thread Starter Member

Feb 14, 2011
64
5
steveb
So if we take our circuit as a closed loop then only the electric field inside the battery will contribute to the net emf and we have emf = -L*di/dt.
But if we take some random loop which includes the inductor then some other electric field will contribute to the net emf which will have the same value emf = - L*di/dt.

EDIT: One more thing that I find interesting. The EMF of the source (battery) controls the induced EMF which must have the same value.

Last edited: May 14, 2011
15. ### amilton542 Active Member

Nov 13, 2010
496
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Have you viewed Walter louin's online MIT videos on electricity and magnetism that back up the MIT reading material? Everything your asking are in the following links

http://www.youtube.com/watch?v=w8D3ikuyaGo Moving charge in B fields

16. ### magnet18 Senior Member

Dec 22, 2010
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My guess would be that current would be infinite.
Or at least the maximum possible amount allowed by things like the speed of light and the number of electrons in the circuit or whatever.
As for everything else, most of it flew over my head, I've never had a circuitry class.

17. ### fila Thread Starter Member

Feb 14, 2011
64
5
amilton542 I watched almost every video but this is so complex that I would like to talk to prof. Lewin in person. That's impossible so I came to this forum.
steveb gave a good explanation.

18. ### Wendy Moderator

Mar 24, 2008
20,993
2,731
I haven't contributed because there wasn't anything to say until now.

Beware of throwing the number infinite around. Electrons can be considered particles, and as such there is a finite but large number of them. In any circuit this number will likely not be nearly as large as you think it is.

19. ### fila Thread Starter Member

Feb 14, 2011
64
5
I don't think so. I think the current would be just high enough to counter the flux change. Steveb what are your thoughts about this?

20. ### steveb Senior Member

Jul 3, 2008
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I think that is a valid point of view. But we have to be a little careful here because we are assuming perfect inductors with zero resistance. With a perfect inductor, there will be no limit to the current, and the emf will equal L dI/dt. But, in the real world, there is a finite resistance. Eventually, the resistance of wire drops all the emf.

Point taken Bill and magnet18 ! Sometimes considering ideal condition is profitable for learning and understanding, but it can also cause confusion and get us into conceptual problems, if we are not careful.

Yes, that's right. The standard inductor works by self inductance; so, there is no external flux source. The battery is supplying the current. That current generates a flux. Because of the geometry of the coil, plus the circuit external to the coil, the coil will enclose a good portion of the flux. Hence when the current changes, then the flux changes by just the right amount to generate the counter emf that matches the battery terminal emf. (it does this to obey Faraday's Law)