Suppose we have two vectorsfield A(x,y,z) and B(x,y,z).If we know B = (0,2,1) can we compute A if:
?
EDIT:
After some searching online I found that there are infinitely many vectors fields A(x,y,z) in order for :
to be true.However do all these A vectors have something in common with each other?
The curl operator $\nabla \times$ is not injective so you have indeed infinitely many solutions. However, if one particular solution $A_0$ is given such that :
$$ \nabla \times A_0 = B $$
then any other solution $A$ has to satisfy :
$$ \nabla \times (A - A_0) = 0 $$
by linearity of the curl operator. Considering $A - A_0$ instead of $A$, you just look for solutions of :
$$ \nabla \times A = 0 $$
In $\mathbb{R}^3$, it is equivalent to the existence of a scalar function $\varphi$ such that $A = \nabla \varphi$ (it may be different if you change the base space). This means that your set of solutions is of the form :
$$ \{ A_0 + \nabla \varphi, \varphi : \mathbb{R}^3 \to \mathbb{R}, \mathcal{C}^1 \} $$
Note that the existence of $A_0$ depends on $B$ : namely, $B$ should satisfy $\nabla \cdot B = 0$. See for instance the Biot-Savart law (here).