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Question about Faraday's Law

  1. Apr 8, 2007 #1
    Consider a copper disk rotating with constant angular velocity, as shown in the figure. A bar magnet is directed normal to the surface of the disk, as shown in the following figure: feynman disk.JPG

    If a galvanometer is used to measure the induced current in the outer rim of the disk, a nonzero induced current is detected. Using the differential form of Faraday's law and the Lorentz force law, this is easily explainable. Consider a small charge dq on the outer rim of the disk. At any given time, it has a nonzero velocity because of the rotation of the disk. For this reason, the bar magnet exerts a magnetic force on the charge. This makes the charge move differently than the disk itself. Some of these charge will go through the galvanometer, and thus the galvanometer will indicate the existence of a current.

    But wait a minute. Let's try applying the integral form of Faraday's Law: the induced emf along a closed loop is equal to the rate of change of the magnetic flux through the loop. But the magnetic flux through surface of the copper disk is constant, and thus the rate of change of magnetic flux is zero. This is because both the magnetic field and the area are both constant. So we have a strange situation in which the induced emf is nonzero even though the rate of change of magnetic flux is zero.

    Feynman gives this example in the Feynman Lectures on Physics Volume 2 Chapter 17-2. He states that you shouldn't worry about this, and that when the integral form or "flux rule" doesn't seem to work, you should just "go back to the fundamental equations," the Lorentz force law and the differential form of Faraday's Law as I did above. But what's wrong with using the integral form in this case?
  2. jcsd
  3. Apr 10, 2007 #2


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    Simple answer: Electric potential in undefined for Electric fields induced by magnetic fields.
  4. Apr 10, 2007 #3
    I just wanted to clarify my question by describing how the induced current will be measured. An ammeter is embedded in the outer rim of the copper disk and rotates along with it. The reason I'm doing this is that rotation of the disk already causes a current even without the magnet since current is nothing but the motion of charge. But if the ammeter is also rotating, then a charge will only go through it if it has a nonzero relative velocity with respect to the disk. Also, let's assume this ammeter is allowed to detect current going through it in any direction.
  5. Apr 10, 2007 #4
    Well, if your statement is taken literally, it is obviously false. Just define the emf around a closed loop to be the line integral of E.dl.

    What I think you mean is that this induced emf shouldn't be called a potential because the corresponding electric force is nonconservative. But that doesn't matter. It's not relevant to the problem that the emf is not a potential.
  6. Apr 10, 2007 #5


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    Notice that I said electric potential.
    In that case your integral, would always be zero for E field induced by B fields.
    Last edited: Apr 10, 2007
  7. Apr 10, 2007 #6
    Well, you have as many negative charges as positive ones in the metal. The net current is zero in absence of magnet. When you put a magnet, negative charges can move, but not the positive ones and you obtain a current.
  8. Apr 10, 2007 #7
    Sorry, Hootenanny. You are wrong and lugita is right. The line integral E.dl is lighting your lamps and feeding your computer.
  9. Apr 10, 2007 #8
    This seems a rather stupid phrase "Feynman was right". Go back to how Faraday's Law is derived from the differential form coming from Maxwell laws.
    You integrate the time derivative of B.ds. In the other side of the equation you transform the surface integral of the curl of E in the line integral E.dl using the Stokes theorem. In the example that you give, this integral is zero (no change in magnetic flux). But you must add the other part of the problem; that is, the forces acting on the charges. This is what is missing and why Feynman was right.
  10. Apr 10, 2007 #9
    Both of Feynman's examples have loops that are moving, but he only uses partial time derivatives. You should use total time derivatives (convective derivatives) if the loops you are considering are moving. I'm reasonably certain that's why this isn't working out.
  11. Apr 10, 2007 #10
    If the copper disk was simply translating with constant velocity, then one could use convective derivatives. However, the copper disk's velocity is itself changing direction because of the centripetal force which is making the disk spin. Therefore, since convective derivatives can't be used on spinning systems like this, that can't be the source of the difficulty.
  12. Apr 10, 2007 #11
    I can only think of one explanation. Unlike the differential form of maxwell's equation, 'flux rule' deals with magnetic flux change through the area defined by closed path. I think such flux should be due to all sources of magnetic field so that,

    Total Flux = Flux due to magnet + Flux due to induced current.

    So, if change in flux due to induced current is not zero, change in total flux can't be zero.

    Also, the induced current I talk about is only due to v X B law.
  13. Apr 10, 2007 #12
    I disagree. The magnetic flux due to the induced current is constant. This is because the magnetic flux due to the induced current is dependent only on the induced current and the area of the loop. Therefore, there is no change of magnetic flux due to an induced current
  14. Apr 10, 2007 #13
    But the thing is that Maxwell's equations have the property of completeness. What I mean is that if you consider any quantity that occurs in Maxwell's equations which is uniquely determined by other quantities which also occur in Maxwell's equations, then the Maxwell equations are sufficient to find the first quantity in terms of the other quantities. This property applies to the set of Maxwell's equations, in both their differential form and in their integral form.

    What all this means is that it should be possible to find the induced current without appealing to the Lorentz Force Law.
  15. Apr 11, 2007 #14
    I do not know the origin of your assertion about completeness. If you take only the four classical Maxwell equations with curls and divergences of E and B, you are far from completeness. You need also the force acting on charges:
    [tex]F=qE+q\times v\times B[/tex]
    Without this equation, you can't even predict that an electric charge in an electric field will move.
  16. Apr 11, 2007 #15
    First thing: what is the definition of electric and magnetic field? Or, in a thought experiment, how do you measure de electrical and magnetically fields in a point of space?
    For the electrical field there is no doubt. The definition is:
    [tex]E={F\over q}[/tex]
    (I don't write the limit)
    For the magnetic field the definition can be more indirect, using sometimes forces exerted on currents. But it comes finally to [tex]F=q\times v\times B[/tex].

    On the other hand, you can find the completeness of Maxwell equations if you include the Lorentz force (no need to name it). The curl of B terms comes to be:
    [tex]\nabla\times B =\mu_\circ\left[j_{true}+\rho_{true}v+{\partial D\over \partial t}+\nabla\times\left(P\times v \right) \right][/tex]
    [tex] j_{true}= \sigma(E+v\times B) [/tex]
    [tex]P=\varepsilon_\circ(\kappa -1)(E+v\times B) [/tex]

    I think that you have noticed the term [tex]E+v\times B[/tex] and that it is the Lorentz force, even if it is not named.

    These formulas comes from the chapter 9 Motion of a Conductor in a Magnetic Field from the Classical Electricity and Magnetism, W. Panofsky and M. Phillips. Addison-Wesley.

    If you can read this chapter, I think it can help answer your question.
  17. Apr 12, 2007 #16
    Unfortunately, I don't have access to the book you mentioned.
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