# Solve \nabla x A = B for A

1. Feb 1, 2009

### Nikratio

Hello,

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

$$\vec \nabla \times \vec A = \vec B$$

for $$\vec A$$?

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. Feb 1, 2009

### confinement

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
3. Feb 1, 2009

### jostpuur

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 $4\pi$ 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?

4. Feb 1, 2009

### confinement

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.

5. Feb 1, 2009

### adriank

Griffiths suggests this: compare $$\vec\nabla \cdot \vec A = 0, \vec\nabla \times \vec A = \vec B$$ with Maxwell's equations for B (in magnetostatics): $$\vec\nabla \cdot \vec B = 0, \vec\nabla \times \vec B = \mu_0 \vec J$$. Then, in the same way the Biot-Savart law gives

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

by analogy we should have

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

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