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Disk in uniform magnetic field.

  1. Jul 26, 2012 #1
    1. The problem statement, all variables and given/known data
    A cilindrical copper disk (radius R, thickness a) at time t=0 is spinning around its axis with angular speed w in a uniform magnetic field B parallel to its axis. The edge of the disk is connected to the center with a "wire" which we assume to have negligible resistance and cilindrical simmetry.
    Determine the current density in the disk.


    2. Relevant equations
    Maxwell equations, Lorentz force, maybe [itex]\vec J = \sigma \vec E[/itex].


    3. The attempt at a solution
    Several failed attempts which I won't extensively report here.

    One of them required defining an auxiliary difference of potential due to the Lorentz force, which can be easily found to be (in CGS, but feel free to use MKS in your answer) [itex]V = \frac{1}{2c} w R^2 B[/itex]. Then, I wanted to find the overall current [itex]I = V / Resistance[/itex], but the resistance of the disk turns out to be infinite (not too hard to show, and also easy to find by Google).

    Another one of them required solving the problem in the accelerated rotating frame of reference, but then I remembered that the field [itex]B[/itex] should also be transformed and the appearance of an electric field in this frame is likely, and since the new frame of reference isn't inertial the field transformations may be non trivial (read as: I don't know how to transform the fields in the general case of non-inertial reference frames, and I think it may be really complicated).

    Charge conservation requires the result to be in the form [itex]J = \alpha/r[/itex], so that when considering to cilindrical surfaces (parallel to the disk) the current [itex]2 \pi r a J[/itex] doesn't depend on r, but I couldn't determine [itex]\alpha[/itex] .


    4. Various information.
    This problem is from a weird exam my institution uses to impose on its students. The exam should be similar to the PHD General exam common in many universities, though it should be a little easier since aimed at students undertaking a master course instead of a PHD. Weird problems have often come out, but they never had no possible solution at all (the original problem had 4 more questions I didn't report here, so I don't think the answer to the first one can be "the problem can't be solved because the resistance is infinite", which would make most of the following questions pointless).

    Forgive me for the (possibly lots of) grammar errors.
     
    Last edited: Jul 26, 2012
  2. jcsd
  3. Jul 26, 2012 #2

    rude man

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    Think of the disk as being formed of very many radial strips, insulated from each other. The strips are all joined together at the disk's periphery and also at its center by the zero-resistance wire. What is the emf generated across one of those strips? The current? How do they add?

    Why is it legitimate to imagine that the strips are insulated from each other, when in fact we have a solid copper disk?

    PS I think you need to know the conductivity of the disk also to get a numerical answer.
     
  4. Jul 27, 2012 #3
    [where is the "itex" button? Do I have to write itex and /itex by myself every time?]

    The radial Lorentz force is (signs depend on the orientation of the vectors and I'll never care about them in the equations, and I'm doing this in MKS this time) [itex]F(r) = eBwr[/itex]. Hence [itex]EMF = \int F(r) dr / e = \beta R^2[/itex] where [itex]\beta[/itex] is some constant.

    To determine the current, we need the resistance.

    The resistance is obtained by:
    [itex]dR = \rho \frac{dr}{\Delta S} = \rho \frac{dr}{a r \Delta \theta}[/itex]
    To determine the resistance of the whole slice of angle [itex]\Delta \theta[/itex], we should integrate over [itex]r[/itex]. Since no minimum radius is given, the resistance of the slice results infinite (as [itex]ln(R_{max}/R_{min})[/itex]). Therefore, no current at all.

    I had already addressed this issue in the main post:
     
  5. Jul 27, 2012 #4

    rude man

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    Excellent point you bring up, but it's a pathological case. Any contact between two separate conductors must have a finite contact volume. For example, you could not connect wires to capacitor plates if the points of contact had no volume. So yes, a finite inner disk area must be assumed, then R = (1/2πσh)ln(b/a) where a = outer radius, b = assumed inner radius, h = thickness of disk, σ = conductivity. Note that R changes little as b is shrunk relative to a, however.

    In this case the finite contact volume is 2πbh of course.

    Interesting situation, had me going for a while and maybe someone else can add to our insight here.
     
  6. Jul 29, 2012 #5

    gabbagabbahey

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    Given that this is for a master's level exam, I don't think you can ignore the thickness of the disk and the role that eddy currents and mutual inductance play here. I think you may be expected to solve the magnetic diffusion equation [itex]\nabla^2 \mathbf{H} = \mu \sigma \frac{\partial \mathbf{H} }{\partial t}[/itex] inside the disk (with appropriate boundary conditions) and then determine the current density from Ampere's law.
     
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