How do we deduce the vector potential for a constant magnetic field? How do we show that for a constant magnetic field $\vec B = const$, the vector potential is $\vec A = \frac12 \vec r \times \vec B$?
 A: We know that the magnetic field $\vec B$ is derived via a curl of the magnetic vector potential $\vec A$ by the formula:
\begin{equation}
\vec B (\vec r) = \vec \nabla \times \vec A (\vec r). \label{potfld} \tag{1}
\end{equation}
Naturally, this would mean that upon using (\ref{potfld}) the magnetic flux passing through a surface $\mathcal S$ bounded by the loop $\mathcal C = \partial {\mathcal S}$, according to Stoke's Theorem can be written as:
\begin{equation}
\mathcal I = \int_{\mathcal S} d \vec S . \vec B = \int_{\mathcal S} d \vec S . \left( \vec \nabla \times \vec A \right) = \int_{\mathcal C} d \vec l . \vec A (\vec r). \label{stokes} \tag{2}
\end{equation}
Now, at the same time, since the magnetic field $\vec B$ is constant, we can write:
\begin{equation}
\mathcal I = \int_{\mathcal S} d \vec S . \vec B = \vec B . \int_{\mathcal S} d \vec S. \label{areaint} \tag{3}
\end{equation}
Here, we shall consider an arbitrary flat surface containing the origin, and we shall divide the surface $\mathcal S$ into infinitesimally narrow triangular segments, with all their vertices at the origin. This will allow us to write the surface integral in (\ref{areaint}) as a boundary line integral:
\begin{equation}
\mathcal I = \vec B . \int_{\mathcal S} d \vec S = \vec B . \int_{\mathcal C} \left( \frac12 d \vec l \times \vec r \right) = \int_{\mathcal C} \vec B . \left( \frac12 d \vec l \times \vec r \right) = \int_{\mathcal C} d \vec l . \left( \frac12 \vec r \times \vec B \right). \label{boundint} \tag{4}
\end{equation}
Thus, comparing (\ref{stokes}) and (\ref{boundint}), we shall have:
$$\int_{\mathcal C} d \vec l . \vec A (\vec r) = \int_{\mathcal C} d \vec l . \left( \frac12 \vec r \times \vec B \right)$$
\begin{equation}
\Rightarrow \qquad \boxed{\vec A (\vec r) = \frac12 \vec r \times \vec B.} \label{crosspot} \tag{5}
\end{equation}
A: $\def\vA{\vec A} \def\vB{\vec B} \def\vnab{\vec\nabla}$ 
You may not deduce $\vA$ from $\vB$, since eq. (1) is not a one-to-one
mapping. There are an infinity of $\vA$'s giving rise to the same
$\vB$. If $\vA_0$ is a solution and $f$ is any scalar function, then
$$\vA = \vA_0 + \vnab f \tag1$$
is another solution, as a consequence of 
$$\vnab \times \vnab f = 0.$$
This is a special case of gauge invariance: in electromagnetism
fields determine potentials up to a gauge transformation. You can
fix $\vA$ by imposing an additional condition, e.g.
$$\vnab \cdot \vA = 0$$
(this is known as Coulomb gauge). Taking the div of my eq. (1) you
have
$$\nabla^2\!f = 0$$
and $f$ is fixed by suitable boundary conditions, e.g. $f=0$ at
infinity.
A: This is sort of a blended answer that also touches upon possible ways to invert a curl under special conditions.
$\vec{B}=\nabla \times \vec{A}$
Take curl of both sides.
$\mu_0\vec{J}=\nabla(\nabla\cdot \vec{A})- \nabla^2\vec{A}$.
The left side by Amperes' law, the right by vector identity.
Assume $\nabla \cdot \vec{A}=0$, by the Coulomb Gauge.
We have $\nabla^2\vec{A}=-\mu_0\vec{J}$.
With $\vec{J}=0$ in empty space.
Now we can proceed by solving Laplace's equation component by component respecting applicable boundary conditions.  
A: Can be checked by taking curl of the proposed answer :
$$ \nabla \times (\vec{r} \times \vec{B})= (\nabla\cdot\vec{B})\vec{r}-(\nabla\cdot\vec{r})\vec{B}
$$
Remembering that in both the terms the nabla operator acts on both the subsequent arguments and that $\vec{B}$ is a constant vector we get $-2\vec{B}$.
