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The Law of Biot and Savart again

  1. Feb 10, 2010 #1
    The magnetic field of a steady current in a loop is given by the Biot and savart integral which is

    1/4pi Integral[((x-y)/|x-y|^3) x dy] = B(x)

    What is the corresponding formula for the vector potential?
  2. jcsd
  3. Feb 10, 2010 #2
    Although i can't decrypt the formula you have stated, i give it a try

    \vec A(\vec r) = \frac{\mu_0 I}{4\pi} \oint \limits_{\mathcal{C}} \mathrm d\vec r^{\, \prime} \, \frac{1}{|\vec r - \vec r^{\, \prime}|}

    The curve [tex]\mathcal{C}[/tex] is parameterized through [tex]\vec r^{ \, \, \prime}(t)[/tex] !

    It is necessary to mention, that the curve [tex]\mathcal{C}[/tex] must be closed, otherwise the integral diverges!

    Best regards...
  4. Feb 10, 2010 #3
    thanks I will try to prove it works.

    BTW: how do you do the math notation?
  5. Feb 10, 2010 #4
    Thanks that works.
    What about if you have an arbitrary divergence free field defined in space minus possibly a finite number of loops?

    For instance if I have two magnetic fields generated by two non-linking current loops their cross product is divergence free. If there an integral formula for the vector potential of the cross product?

    Or - suppose the magnetic field is confined to the interior of a closed tube as in a magnetic filament.
    Last edited: Feb 10, 2010
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