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Discrepancy bet. maxwell + circuit notation w/ inductors?

  1. Jun 1, 2009 #1
    Well...

    Consider a basic RL circuit. Maxwell tells us that the induced EMF=-L dI/dt. The voltage drop across the resistor is of course RI, but why does this mean that RI =L dI/dt as every physics textbook says. That statement relies on the fact that path integral around the circuit is = 0, but its not? According to maxwell that path integral is approximately (ignoring fringe fields, etc.) -L dI/dt. The path ingral of E around the curve is only equal to 0 in electrostatics...

    I understand that voltage notation is a convenient way of describing circuits, but is not always rigorous (i.e. there doesn't exist a scalar potential in a non-static EB field). But, I do not understand why that formula is correct.

    that formula being RI=L dI/dt.
     
  2. jcsd
  3. Jun 1, 2009 #2

    atyy

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    Yes, circuit theory is not correct according to Maxwell's equations. When you change the current in an inductor, the changing B field should cause energy loss through radiation. Circuit theory is an approximation when such things can be ignored, usually when the wavelength of the radiation is long compared to the circuit.
     
  4. Jun 2, 2009 #3
    Basically, in this case you are dealing with lumped circuit elements. The effect of the inductor is determined with Faraday's law, and the resulting EMF is considered a voltage drop in the circuit equation (Kirchoff's Voltage Law). This works as an approximation because the flux change is mostly cutting the circuit in the coil itself and very little leaks out into the rest of the circuit.

    You are correctly noting that the path integral of electric field around a closed path is not zero, but is equal to the negative rate of change of flux. This is basically equivalent to the statement RI =L dI/dt. The discrepancy with the minus sign is just a sign convention (flux=-LI) used with the coil equation V=L dI/dt. Normally EMFs are like voltage sources, but the EMF from self inductance of the coil acts like a voltage drop. This is a good thing, since the minus sign keeps the coil from exploding in a positive feedback loop.
     
  5. Jun 2, 2009 #4
    Given an inductor (maybe in an RL circuit, whatever) that starts at point A and ends at point B. If the voltage drop across the inductor is indeed -L dI/dt then ohms law should imply that -L dI/dt = RI, where R := the resistance of the inductor. This implies that -L dI/dt=0 which is definitely not correct.

    SO, either ohms law doesn't work for inductors (?) or the voltage drop across the inductor from A to B is not -L dI/dt.

    I think if I knew the answer to that question a lot of this would be cleared up.

    atyy: Although it is definitely not accounted for in classic circuit theory, I don't see how energy loss due to radiation clears this particular problem up.

    elect_eng: 'You are correctly noting that the path integral of electric field around a closed path is not zero, but is equal to the negative rate of change of flux. This is basically equivalent to the statement RI =L dI/dt'

    True, True...But if RI=L dI/dt then in this RL circuit the total path integral of E would be zero (if indeed the inductor has a L dI/dt voltage drop across it)

    **thanks for the quick responses. this was my first time using physics forum, could turn out to be an invaluable resource.
     
  6. Jun 2, 2009 #5
    What you wrote, -L dI/dt =0 is correct for a RL circuit with no voltage source. However, I=0 so dI/dt also zero, so you have -L * zero = zero which is a valid math expression given that there is no current flowing in the circuit or through the inductor.

    Now if you assume that some current initially exists (a switch was thrown or something), then you have a differential equation that you can integrate which will contain a constant of integration that will be defined by the current at time zero. You will find that the current decays exponentially and ends up at zero. The same with voltage across the inductor.
     
  7. Jun 2, 2009 #6
    The inductor EMF is due to the rate of change of flux, not due to the integral of E over the length of the inductor. So you can't say that the integral of E is zero around this closed path. There is often confusion on this issue. Some people mis-state Kirchoff's voltage law as the integral of electric field around loop is zero. Kirchoff's voltage law says that the sum of the resistances times the currents (i.e. resistive drops) around a loop is equal to the EMFs around that loop. Faraday's law says something similar, but it is more precise about what those EMFs are. A capacitor has EMF as integral of electiric field and an inductor (or generator) has EMF as time rate of change of flux. Faraday's law requires you to distinguish the difference while Kirchoff's law doesn't care what the source of the EMF is.
     
  8. Jun 2, 2009 #7

    Dale

    Staff: Mentor

    An ideal inductor has no resistance, or equivalently has purely imaginary impedance. If you want to model a real inductor a little more accurately the usual way is with an ideal inductor and an ideal resistor in series.
     
  9. Jun 2, 2009 #8

    atyy

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    Something like this. Electrostatics: path integral of E around a closed loop is zero. Electrodynamics: electromagnetic radiation. If we can ignore electromagnetic radiation, then electrostatics is approximately true, and we can take path integral of E around a closed loop to be zero.

    That's just the rough idea. In the real way there are usually several regimes/approximations of Maxwell's equations: electrostatics, magnetostatics, quasistatic etc, where people state exactly what approximations are being made.

    Try section 8.4 of http://ocw.mit.edu/OcwWeb/Electrical-Engineering-and-Computer-Science/6-641Spring-2005/TextbookwithVideos/index.htm [Broken]. "Voltage at the Terminals of a Perfectly Conducting Coil .......... In fact, as we now take care to define the circumstances required to make the terminal voltage of a coil a well defined variable"
     
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  10. Jun 2, 2009 #9
    To all: I guess what i'm looking for is a translation from circuit theory to reality. So, quite simply, if i give you a resistor, an inductor and i guess an ac source (to generate some current) what will the E field be...

    1. along the inductor (which has negligable resistance =Ro and inductance=L)
    2. along the resistor

    ********************************************************************
    hopefully, when i actually calculate the path integral of E over this I will indeed get -L dI/dt, but at the same time if I look at the voltage drop across the resistor I will get the same thing that circuit theory predicts (i.e. E-L dI/dt=RI)
    ********************************************************************
    Elect_Eng: Thank you for your restatement of kirchoff's law.

    atyy: i looked at the clip for that lecture and it was pretty uninformative, except for the cool surface they used to bound the coil curve.
     
  11. Jun 2, 2009 #10
    Wait. atyy: are you saying that the path integral of E around a closed curve C is approximately 0, if C is a circuit containing an inductor??
     
  12. Jun 2, 2009 #11

    atyy

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    Yes. It is approximately true if the rate of change of current is small, ie. the wavelength of radiation emitted is long compared to the size of the circuit. When the wavelength of the radiation emitted is infinite, we should recover electrostatics.

    In fact, one cannot even define an inductance that is a property of the geometry of the conductor alone, except in strict magnetostatics. The concept of inductance being purely geometrical is itself an approximation once changing currents are present.

    BTW, I meant the pdf, not the clip. http://ocw.mit.edu/ans7870/resources/zahn/08.pdf
     
  13. Jun 2, 2009 #12
    OK, it's important to know if the AC source is present or not, and if it is there, it's nice to know how it operates. For example, the AC source could be induced by a flux change cutting through your circuit loop; or, you may have an EMF generator actually wired in series. The first case can be analyzed with Faraday's Law. The second case is better handled by Kirchoff's Voltage Law if you don't know the generator design and only know the generator EMF.

    Let's look at both cases:

    1. Kirchoff's Law: The voltage drops around the loop is simply IR. The EMF in the coil is -L dI/dt, and the EMF from the generator we can call E. Applying Kirchoff's voltage law as I defined it above; (by the way this definition is directly from Maxwell himself in his famous Treatise on Electricity and Magnetism)

    IR=-L dI/dt+E .

    2. Faraday's Law: There is a gradient of voltage potential through the resistor, but the rest of the loop is a perfect conductor so there is no voltage gradient and no electric field. The potential drop around the loop is simply IR and is equal to the line integral of E through the resistor. The coil and generator are both generating EMF due to negative flux change.

    So we get IR=-dFl/dt - dFg/dt .

    where Fl is the flux in the coil Fl=LI
    and Fg is the flux cutting the circuit loop E=-dFg/dt.

    this boils down to the same thing: IR=-L dI/dt +E

    So both methods agree and answer your question. The electric field is in the resistor only. You will recall that the electric field is zero in a perfect conductor, so this makes sense.

    By the way, this is an EXTREMELY non-intuitive thing to think about. It confuses everyone from students through professors. We are saying that the potential drop is across the resistor, but the leads of the resistor are shorted through a perfect conductor by the rest of the loop, so how can there be a potential drop across a wire, right? It is even more nonintuitive if we take out the inductor, because then the perfect conductor is no longer a coil, but now just a wire! So again, how do we have a potential across the resistor if the leads of the resistor are shorted? Well, that's just Faraday's Law in action, you must accept it because that's what the experimental evidence shows.
     
    Last edited: Jun 2, 2009
  14. Jun 2, 2009 #13
    So, elect_eng if the voltage gradient is fully concentrated in the resistor that means that the E-field along the inductor is practically 0 (since its a perfect conductor). I just want to make sure thats what your saying.
     
  15. Jun 2, 2009 #14
    sections 8.3 and 8.4 are awesome. thanks atyy. i have never seen the scalar magnetic potential and the rest of it in any e+m text
     
  16. Jun 2, 2009 #15
    Yes, electric field in a perfect conductor is zero. However, a real inductor does have some resistance, so there is an electric field in the inductor completely unrelated to the (self inductance) counter EMF generated.


    I thought I would provide a quote directly from Maxwell's Treatise on Electricity and Magnetism which defines Kirchoff's voltage law (KVL) and current law (KCL). I think Maxwell's interpretation and translation of Kirchoff's Laws are the best available. The current law is usually interpreted correctly, but the voltage law is usually not. Look in all of your electrical books and on-line for a definition of KVL and you will see many variations. Some are basically correct but worded in a confusing way. Other definitions are just not right, such as the "the integral of electric field around a loop is zero". Maxwell, was very precise in his statement and avoids mention of potential which is a confusing term to use in this context. The definition deals with ohms law, (current times resistance) and how it relates to EMFs (batteries, capacitors, inductors, generators etc). This definition forces you to interpret things in a manner more consistent with Maxwell's equations. (or at least Faraday's Law which is one of those equations. Obviously, KVL can not capture all EM effects)
     

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    Last edited: Jun 2, 2009
  17. Jun 2, 2009 #16
    thanks guys. i got my answer and ive been thinking about this question for a while.
     
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