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Solve \nabla x A = B for A

  1. Feb 1, 2009 #1

    Is there a general, analytic way to obtain a vector potential for a prescribed magnetic field? I.e. to solve

    [tex]\vec \nabla \times \vec A = \vec B[/tex]

    for [tex]\vec A[/tex]?

    I'm looking for something like what e.g. the Green's function method is for Poisson's equation, so the answer might be a complicated integral but it is a definite, closed solution.

  2. jcsd
  3. Feb 1, 2009 #2
    Yup, we can write:

    \vec A(x) = \int d^3x \frac{\vec \nabla \times \vec B(x')}{|x-x'|}


    where the curl in the integrand is with respect to the primed coordinates, and I have neglected the displacment current proportional to the time derivative of the electric field.
    Last edited: Feb 1, 2009
  4. Feb 1, 2009 #3
    Some time ago I was wondering the some problem. I had one difficulty with it, so I posted this integral formula for vector potential in hope of getting some help. However, I solved it myself quickly, posted a clarifying reply myself, and seemingly nobody considered it useful to respond anymore.

    It seems that confinement's formula was lacking a [itex]4\pi[/itex] constant somewhere, but now I started to wonder my own minus sign there too. Is my formula in contradiction with the formula in Wikipedia because of the minus sign?
  5. Feb 1, 2009 #4
    Jostpuur is right that there should be a 1/4Pi in front of the integral in the formula I gave, although I am reasonably confident that there is no missing minus sign.
  6. Feb 1, 2009 #5
    Griffiths suggests this: compare [tex]\vec\nabla \cdot \vec A = 0, \vec\nabla \times \vec A = \vec B[/tex] with Maxwell's equations for B (in magnetostatics): [tex]\vec\nabla \cdot \vec B = 0, \vec\nabla \times \vec B = \mu_0 \vec J[/tex]. Then, in the same way the Biot-Savart law gives

    [tex]\vec B(\vec r) = \frac{\mu_0}{4\pi} \int \frac{\vec J(\vec r') \times (\vec r - \vec r')}{\lvert \vec r - \vec r' \rvert^3} \,d^3\vec r',[/tex]

    by analogy we should have

    [tex]\vec A(\vec r) = \frac{1}{4\pi} \int \frac{\vec B(\vec r') \times (\vec r - \vec r')}{\lvert \vec r - \vec r' \rvert^3} \,d^3\vec r'.[/tex]
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