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Magnetic field/simple current loop

  1. Oct 7, 2008 #1
    given a current loop, centered at the origin, and tilted 90 degrees so that it enters at x=1 and exits at x=-1, carrying a unit amount of current, and completely disregarding the z axis.

    is the strength of the magnetic field at any point in the xy plane proportional to 1/(distance from 1,0)^2 - 1/(distance from -1,0)^2

    in other words does it have an inverse square relation to the current passing through those two points.

    I know there are better ways to calculate the net field but I'm looking to understand what is happening here at an intuitive level.

    I need the whole field. not just the far field or some sort of approximation.
    Last edited: Oct 7, 2008
  2. jcsd
  3. Oct 7, 2008 #2
    what I mean is:

    can I find the contribution of each of the 2 currents separately by simply finding the vector which is at a right angle to the vector from that point to the current and making its magnitude equal to 1/(distance to the current)^2

    and is adding those 2 vectors all that i need to do to calculate the total field
  4. Oct 7, 2008 #3


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    If it is a closed loop, B does not go like 1/r^2.
  5. Oct 7, 2008 #4
  6. Oct 7, 2008 #5
    ok. so evidently thet wont work.

    thinking 3 dimensionally, if we look at only one of the 2 currents (due to current along line element dL) then we can see from the symmetry that its field is confined to a wedge extending from the origin through the endpoints of that line element.

    I have no idea how to go about calculating the resulting field. I've never seen or done anything like it.
  7. Oct 7, 2008 #6

    Ben Niehoff

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    To find the magnetic field of a finite loop, you will need to use elliptic functions. It's complicated.

    However, it can be proven in general that

    [tex]\vec B(\vec r) \propto \nabla \Omega(\vec r)[/tex]

    where [itex]\Omega(\vec r)[/itex] is the (oriented) solid angle subtended by the current loop at the observation point.
  8. Oct 7, 2008 #7
    any connection to the scalar potential of a magnetic field?
  9. Oct 8, 2008 #8


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    Yes, Omega is the magnetic scalar potential, which happens to equal the solid angle subtended by a current loop of any shape.
    For a circular current loop, the MSP can be found be either an elliptic integral, or by a Legendre polynomial, partial wave expansion.
  10. Oct 8, 2008 #9
    so the msp isnt so much discontinuous as multi-valued (like an angle)?
    Last edited: Oct 8, 2008
  11. Oct 8, 2008 #10

    Ben Niehoff

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    It's multivalued, or discontinuous if you make a branch cut.

    However, in free space (in the absence of any currents), it is single-valued, as in this case (there is no J at the observation point).
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