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Kirchhoff's 2nd law wrongly used or not? Lewin vs all college books

  1. May 4, 2004 #1
    MIT Professor vs. all college books

    Hi there!

    I'm a first year student and found quite interesting the MIT on-line video lectures on electromagnetism.
    http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoLectures/index.htm [Broken]

    In one these lectures given by Dr. Walter Lewin it is explained how all college books and teachers wrogly use Kirchhoff's 2md law in order to calculate the current through a RL circuit.

    That is emf-IR-L (dI/dt)=0. Dr.Lewin says that this is the right answer but physically the procedure is not. In his lecture he states that Faraday's law is misinterpreted and not taken into account. He thinks that everybody knows that that's the differential equation, and just because they see a zero automatically we are using Kirchoff's law.

    Lewin follows saying that Kirchoff tells us that the sum of V through a closed loop is 0, but how can it be zero if we have a change in magnetic flux. and therefore the closed integral of E.dl =-dflux/dt.

    I couldn't convince my teacher. I am totally confused and don't know what to think.

    What do you guys think? Is Lewin right? How can I convince my teacher?



    P.S.: here's the lecture in PDF
    http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/LectureNotes/index.htm [Broken]
    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Jun 17, 2004 #2
    go Lewin

    Lewin is correct.
    E.dl =-dflux/dt is the correct law, but when using it in a circuit -dflux/dt is usually -L(di/dt) which is the potential over that component. However the flux part isnt really part of the closed circuit. Think of a closed circuit with no components, when the current changes theres a change in the flux and faradays law must be used, that is theres a contribution from L(di/dt) even thought its not seen in the circuit.

    I dont think Lewin agrees that kirchoff should get so much credit for using faradays law.
  4. Jun 18, 2004 #3


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    Whatever Lewin says I would go with. I managed to get an A in my E&m class even while missing half my professors lectures by watching him online and paying close attention. I wouldn't recommend anyone do the same, but if you're going to be a slacker anyway, just do it! :wink:
  5. Jun 18, 2004 #4


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    You will notice as you proceed in your education that there are a lot more of these things going on - where the procedure gives you a correct (or at least an accurate enough) answer, but if you look "under the covers", there conceptual problems with it. A case in point is the force on a dielectric slab in an electric field.

    Let's say you have a parallel plate capacitor maintained at a constant potential. You have a dielectric slab that is being pushed into the capacitor, where the front end of the slab is already in the capacitor while part of the rear of it still sticks out. The typical standard textbook procedure of finding the force on the slab starts off by using the principle of virtual work via

    [tex]\vec{F}(\vec{r}) =- \nabla U(\vec{r})[/tex]

    If you write down properly the electrostatic energy of the system using the assumption that the separation of the plates are very small when compared to the rest of its dimension, then you'll end up at the end with a force on the slab having only a component perpendicular to the parallel plates of the capacitor.

    Already, there is a conceptual problem here. The E-field of the parallel capacitor is uniform going from one plate to the other. Yet, the NET FORCE on the dielectric is PERPENDICULAR to this field! The standard procedure in finding the force in this case produces the correct answer. But a student who is thinking about this long enough will notice a problem.

    If one wishes to see what is the "under the cover" explanation of this, which is ignored in practically every E&M text, I highly suggest reading Ref. [1] below.


    [1] S. Margulies, Am. J. Phys. v.52, p.515 (1984).
    Last edited: Jun 18, 2004
  6. Jun 22, 2004 #5
    I dont quite follow. The U in gradU to get the force is not the electrostatic potential. It is the electrostatic energry. Isn't it. I dons see why it's the wrong procedure.
  7. Jun 22, 2004 #6


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    Taking the gradient of a scalar field gives you a vector field. Electrostatic energy is not a vector field.
  8. Jun 23, 2004 #7
    Taking the negative of the gradient of the electrostatic energy in a capacitor gets you the force on a dielectric put in it.THe energy is not the same inside the dielectric the outside. Right? In my contribution I didn't say electrostatic energy was a vector field. Obviosly, energy is a scalar
    Last edited: Jun 23, 2004
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