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It is a Electromagnetics problem ~

  1. Nov 29, 2005 #1
    It is a Electromagnetics problem.. plz help~

    In certain experiments it is desirable to have a region of constant magnetic
    flux density. This can be created in an off-center cylindrical cavity that is
    cut in a very long cylindrical conductor carrying a uniform current density.
    The uniform axial current density is J=az j
    Find the magnitude and direction of B in the cylindrical cavity whose axis
    is dispalced that of the conducting part by a distance d


    I am depressed because of this problem. I have no know idea...... ㅜㅜ

    so... I submit this forum. plz help~



    reference : David K. Cheng, Field and Wave Electromagnetics, Second Edition, Addison-Wesley(1989)
    Problem 6-15
     

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  2. jcsd
  3. Dec 5, 2005 #2

    berkeman

    User Avatar

    Staff: Mentor

    Hi fufu, Sorry it took me a while to get back to your question -- probably you already have the answer from your prof by now. But in case you're still working on this problem, I'll say a couple of things:

    -- My initial reaction was to use Ampere's Law to figure out the B field distribution in the enclosed hollow cylinder that is caused by the uniform J current density in the conducting part of the cylinder. But Ampere's Law works better for symmetric geometries (like coaxial cable), so that's probably not the way to go on this problem.

    -- Instead, I'd suggest solving the Biot-Savart Law integral for each point in the hollow cylinder slice, and using the following trick. Think of the geometry of this question as two conducting cylinders, superimposed on one another. The outer cylinder is a uniform conductor with radius b, and the current density is uniformly J across it. Then superimpose a conductor of radius a, displaced from the 1st conductor's center by the distance d as shown in your figure, but its current density is uniformly -J. So inside the outer cylinder's wall the current density is J, but inside the inner cylinder wall, the total current density adds up J-J=0. This is a way to set up your Biot-Savart integrals so that they are easier to do. Just figure out the field at each point inside the inner cylinder as the sum of all the current source elements throughout a uniform outer cylinder with +J, plus the current source elements throughout a uniform inner cylinder with -J. Using this trick to set up the integrations makes them a lot easier to do, with reasonable limits in cylindrical coordinates.

    -- Even easier is probably to do the initial simplification from Biot-Savart for calculating the B field next to an infinitely long wire (uI/2PI*R), and use that with the opposing -J trick above to calculate the B field in the hollow cylinder area as the integral of all the little current elements in the overall cylinder(s). Hope that makes sense.

    Good luck! Let us know if you get the right answer. -Mike-
     
    Last edited: Dec 5, 2005
  4. Dec 6, 2005 #3
    Thank you for your reply.
    I want to write more better English well
    But I can't English well...
    Anyway thanks
     
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