1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Solve \nabla x A = B for A

  1. Feb 1, 2009 #1
    Hello,

    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.

    Best,
    Nikratio
     
  2. jcsd
  3. Feb 1, 2009 #2
    Yup, we can write:
    [tex]

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

    [/tex]

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Solve \nabla x A = B for A
  1. Solve y=x^6 for x (Replies: 7)

  2. Solving for x (Replies: 4)

Loading...