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Equivalent current source

  1. Oct 4, 2009 #1
    Hi,

    Lets say the electric and magnetic fields in an closed surface (2-D) are known. Is it possible to derive electric/magnetic currents that can create these fields? We can assume that the closed surface is homogenous with constant permittivity and permeability.

    Is this a well known solvable problem with known techniques?

    Really appreciate some guidance.

    Thanks
     
  2. jcsd
  3. Oct 5, 2009 #2

    Born2bwire

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    Yeah, this is known as the equivalence principle. http://books.google.com/books?id=PJ...epage&q=equivalence principle current&f=false

    Basically, if we know the total field in a given volume, then the fields can be reproduced by exciting currents on the surface enclosing the volume and then suppressing the original fields. The currents are

    [tex] \mathbf{J} = \hat{n}\times\mathbf{H}[/tex]
    [tex] \mathbf{M} = -\hat{n}\times\mathbf{E}[/tex]

    where n-hat is the outside normal of the surface. If we wish to solve for the scattered fields that arise using this method, then we use the PMCHWT moment method. This is where we use the appropriate boundary conditions to setup a set of linear equations relating the currents and the excitation fields. You could do this just using electric or magnetic currents but the problem because ill-conditioned because of possible resonant cavity modes in the interior of the enclosed volume. However, a moment method is unnecessary if you already know the total field of your system.
     
  4. Oct 5, 2009 #3
    Thanks a lot.

    Before reading the book i'll just make some points to clarify what you meant.

    - If E and H fields are known in a given area (or volume), the equivalent electric and magnetic currents can be calculated on the boundary (or surface).
    - These can be excited to get the E, H fields. Thus they are the sources for E, H
    - If we want to observe scattering of these fields, the scatter can be enclosed within the area (or volume) and excite the same currents.
    - Then to get the scattered field we can subtract the initial fields from the total field in the area (or volume)
     
  5. Oct 5, 2009 #4

    Born2bwire

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    Yeah, except we cannot calculate the scattered field directly in closed form for most problems. So instead, we estimate it using a finite basis and set up a matrix problem to solve for the excited currents. The fields from the excited currents are the scattered fields, add those to the incident fields to find the total field.
     
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