Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    [tex]
    \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}|}
    [/tex]

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook