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Magnetic field of a finite-length wire

  1. Feb 20, 2009 #1
    Let's say I want to calculate the magnetic field at a distance d from the center of a wire of finite length L, carrying a current I. Why would it be wrong to apply Ampere's law to a circular path of radius d centered on the wire, and say that the integral of B.dl is simply B times 2pi*d? (obviously it gives the wrong answer...)

    Is the magnetic field not constant along this circular path? I would say so - the problem obviously has cylindrical symmetry.
    Is it not parallel to the path at all points? I would think so - from Biot-Savart's law applied to every small element of the wire.
    So where is the mistake in this logic?

    Thanks,
     
  2. jcsd
  3. Feb 20, 2009 #2

    Delphi51

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    That makes a lot of sense, Chen. Do you happen to know what B is around a "short" current carrying wire? It would be interesting to see how it is different from the B for an infinitely long one. I would expect a slightly smaller answer due to the "missing" length on either side.

    Looks to me like the finite length L in the Biot-Savard Law gives you a factor of
    L/sqrt(L^2 + r^2). Anyhow it goes to 1 as L goes to infinity.

    Interesting, but it does not answer your question of why Ampere's Law cannot be applied to the finite length! If you find the answer, I hope you will let me know.
     
  4. Feb 20, 2009 #3
    There is a formula here:
    http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/MagneticField/MFStraitWire.html [Broken]

    If we limit our discussion only to the center of the wire, theta=phi and you obtain the result of an infinite wire, times cos(theta). And I am not at all sure where in Ampere's law this factor comes from. There is probably some fine print that I'm missing here.
     
    Last edited by a moderator: May 4, 2017
  5. Feb 20, 2009 #4

    Delphi51

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    dB = u*i/(4*pi)*sin(A)*dx/r^2 where x runs from -L/2 to L/2, sin(A) = R/r
    and r = sqrt(x^2 + R^2)
    Here R is the distance from the wire center to the point we are computing B for.
    B = u*i/(4*pi)*integral R*dx/(x^2 + R^2)^1.5
    B = u*i/(4*pi*R)*[x/sqrt(x^2 + R^2)] evaluated between -L/2 and L/2.
    B = u*i/(2*pi*R)*L/sqrt(L^2 + R^2)

    This is from the standard derivation for the infinite wire, with only the integral limit being changed to -L/2 to L/2.
     
  6. Feb 21, 2009 #5
    Thanks. But I know how to calculate this with Biot-Savart's law. I want to know where the application of Ampere's law to this problem fails.
     
  7. May 18, 2009 #6
    Ampere's Law is only useful in calculating B for situations with very high symmetry. When calculating B of an infinite wire using Ampere's law we choose our path to be a circle around the wire so that B will always be tangent to our path. If the length of the wire isn't large compared to the distance r where we calculate B, this assumption is no longer valid (the B field is no longer symmetric in this way).
     
  8. May 18, 2009 #7
    In short your assumption that a finite wire's B field has cylindrical symmetry is not correct--the effects of the ends make the field lines shaped oddly and not in a way Ampere's Law can work for us.
     
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