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It is known that if a vector field $\vec{B}$ is divergence-free, and defined on $\mathbb R^3$ then it can be shown as $\vec{B} = \nabla\times\vec{A}$ for some vector field $A$.

Is there a way to find $A$ that would satisfy this equation? (I know there are many possibilities for $A$)


Note: I want to find it without using the explicit formula for $B_x(x,y,z), B_y(x,y,z), B_z(x,y,z)$, but maybe with a formula involving surface/curve integrals. For example, I've found that in the 2D case (if $B_z=0$ and $\vec{B}=\vec{B}(x,y)$) then $A$ can be shown as:

$$\vec{A}(x,y)=\hat{z}\int_{\vec{R_0}}^{\vec{r}} (\hat{z}\times\vec{B})\cdot\vec{dl}$$

I am looking for something similar in the general case.

3 Answers 3