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Homework Help: Steady Current and Magnetic Field

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Question 3 of http://www.damtp.cam.ac.uk/user/examples/B10b.pdf .

    2. Relevant equations

    [itex]\nabla \times B = \mu_0 J[/itex]

    3. The attempt at a solution

    I know for a cylinder, J (vector) = J (scalar) * k (vector), where unit k is the vector in the z-direction. So [itex]\nabla \times B = (0, 0, \mu_0 J)[/itex], somewhere (not quite sure where this is true in terms of a, b, d). I think I need to find B for the big cylinder, then B for the small cylinder, and combine them somehow? I don't see why the field in between the cylinders is important; we're only interested in B for x² + y² < a².

  2. jcsd
  3. Feb 9, 2010 #2
    I think you have everything you need to do this problem. Write down explicitly the sum of the two fields, one from +J and one from -J, for any point in the current free region of interest. The x dependence should drop out.
  4. Feb 9, 2010 #3
    If I was not clear enough, the problem suggested what to do.
  5. Feb 10, 2010 #4
    What is the 'free region of interest'? If I take a +J thing from one cylinder and then subtract the J bit from the other cylinder, surely I'm going to end up with an expression for the stuff in between the two cylinders? The question asks for the field within the smaller cylinder, not between the two.

    My notes on a vaguely similar problem claim B = B(r)e_theta 'by symmetry'. Not quite sure why it's e_theta by symmetry.
  6. Feb 10, 2010 #5
    You break the complex problem down into two simple ones. If you had a single cylinder with current density J you could tell me the magnetic field for any radius r inside the conductor via Ampere's law. To this field you add the imaginary field from the current -J which flows in the smaller cylinder. The sum of J and -J = 0, flows in the small cylinder. Properly added the x component of the magnetic fields should cancel.

    If you are still stuck I'll try and add more.
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