fields and currents

Thread Starter

fila

Joined Feb 14, 2011
64
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).

OK, for you who want to read this OLD thread:
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???

P.S: I have so many questions about this particular problem and generally about confusion that arises from wrongly applied physics in engineering.
 
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magnet18

Joined Dec 22, 2010
1,227
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.
 

Thread Starter

fila

Joined Feb 14, 2011
64
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.
 
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steveb

Joined Jul 3, 2008
2,436
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.
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.
 
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Thread Starter

fila

Joined Feb 14, 2011
64
Hence, using either Ei or the sum Ei+Ec gives you the same result.
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!
 

Thread Starter

fila

Joined Feb 14, 2011
64
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? :)
 
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nsaspook

Joined Aug 27, 2009
13,087
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!
KVL is only an very restrained approximation of field theory.
As others have said:
You can't solve field problems with circuit theory.
http://forum.allaboutcircuits.com/showpost.php?p=128143&postcount=19
 

Thread Starter

fila

Joined Feb 14, 2011
64
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. :)
 
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steveb

Joined Jul 3, 2008
2,436
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!
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.
 

Thread Starter

fila

Joined Feb 14, 2011
64
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.
 
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amilton542

Joined Nov 13, 2010
497
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=3omwHv3Cmog Electric charges
http://www.youtube.com/watch?v=OsWDUqJQcpk Electric field
http://www.youtube.com/watch?v=XaaP1bWFjDA Electric flux
http://www.youtube.com/watch?v=ldJhMDuOGxY Electrostatic potential
http://www.youtube.com/watch?v=qaZQzIXv2RQ Electrostatic shielding
http://www.youtube.com/watch?v=9501V-D-SM4 HV breakdown
http://www.youtube.com/watch?v=7NUbsQt-G9U Capacitance
http://www.youtube.com/watch?v=E185G_JBd7U Polarization
http://www.youtube.com/watch?v=cJSp8v0YJrA Currents
http://www.youtube.com/watch?v=RQX8I9ZWtPQ Batteries
http://www.youtube.com/watch?v=qqkUeQ0nsF8 Magnetic field
http://www.youtube.com/watch?v=kXH8ZW6apVc Review exam
http://www.youtube.com/watch?v=w8D3ikuyaGo Moving charge in B fields
http://www.youtube.com/watch?v=MZOaVXmK5zk Biot-savart law
http://www.youtube.com/watch?v=q_v2NpgEpws Amperes law
http://www.youtube.com/watch?v=G3eI4SVDyME Faradays law
http://www.youtube.com/watch?v=qxuGDEz8wDg Motional EMF
http://www.youtube.com/watch?v=FGaSeSP0tCM Displacement current
http://www.youtube.com/watch?v=njIRAKW5WHc Vacation special
http://www.youtube.com/watch?v=UpO6t00bPb8 Inductance RL
http://www.youtube.com/watch?v=SNDqAuxYOQ8 Magnetic materials
http://www.youtube.com/watch?v=ddU6HBFlvEk Hysteresis
http://www.youtube.com/watch?v=HSN2hSULTi4 Review exam
http://www.youtube.com/watch?v=LDtDNHMveBQ Transformers
http://www.youtube.com/watch?v=QwUgYBzdcrM Driven LRC circuits
http://www.youtube.com/watch?v=AuYFcjBmMJ0 Traveling waves
http://www.youtube.com/watch?v=7T0DD9KilmM Resonance
http://www.youtube.com/watch?v=XtHsVSW2W3E Refraction
http://www.youtube.com/watch?v=ab0czJ6_OF8 Snell's law
http://www.youtube.com/watch?v=rnkBgp7ZtEA Polarizers
http://www.youtube.com/watch?v=p9iB2PALVeY Rainbows
http://www.youtube.com/watch?v=js9SLJ2TU2c Review exam
http://www.youtube.com/watch?v=r9IyXxICJOA Double slit interference
http://www.youtube.com/watch?v=EjPIL9_YpKU Gratings
http://www.youtube.com/watch?v=rQ2kUpeDly0 Doppler effect
http://www.youtube.com/watch?v=faJ8RXQkk3o Farewell Walter Louin
 

magnet18

Joined Dec 22, 2010
1,227
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? :)
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.
 

Thread Starter

fila

Joined Feb 14, 2011
64
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.
 

Wendy

Joined Mar 24, 2008
23,415
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.
 

Thread Starter

fila

Joined Feb 14, 2011
64
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.
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?
 

steveb

Joined Jul 3, 2008
2,436
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.
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.

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