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Ampere's Law and Magnetic Fields

  1. Jan 26, 2013 #1
    1. The problem statement, all variables and given/known data
    I have a coaxial cable with current density Jo in the center, with radius a, going in -z_hat direction. This generates a magnetic field. The outside of the cable, radius c, also carries a current density Jout going in the +z_hat direction. This generates its own magnetic field.

    Find the value of the outside current density to make the magnetic field 0 for r > c

    2. Relevant equations
    I used Ampere's equation.

    3. The attempt at a solution

    I do not use Ampere's equation explicitly (starting from dot product and such), but conclude that H = 0 if the net current = 0. and then find out Jout in terms of Jo.

    Is this correct thinking? I mean, if I drew a loop that was outside of the coaxial cable, the Inet would be 0, but there is still an H-field being contributed by both the current densities.

    How else would I go solving this problem then?

  2. jcsd
  3. Jan 27, 2013 #2

    rude man

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    We need to know the cross-sectional area of the outide conductor.

    You are right in assuming the net current = 0 for H to be 0. If the inside conductor cross-sectional area = a then you know the inside current is Joπa2 but we also need to know the outside area & it can't be πc2 obviously.
  4. Jan 27, 2013 #3
    Ok thank you! This was more of a conceptual question than anything. So if the two current densities are in opposite direction but equal to each other in magnitude, this will cancel out the magnetic field entirely?

    I'm lost in the understanding because if we take an amperian loop outside of the coaxial cable and assume that the currents were different (meaning they each generate their own magnetic field, which does not cancel each other out), the enclosed current will be 0, but that does not necessarily guarantee the magnetic field is 0 right?
  5. Jan 27, 2013 #4

    rude man

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    It's not the current densities that cancel each other to attain H = 0, it's the currents. That's why you have to know the cross-sectional area of the outer conductor. Then (area of inner conductor) * Jo = (area of outer conductor) * Jout to give ∫Hds = I = net current = 0 so by symmetry H = 0 everywhere along any closed loop outside the outer conductor.
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