Find vector field given curl

As John Hughes already mentioned, we require $\nabla \cdot \vec J=0$. Under that restriction, we proceed.

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Since the curl of the gradient is zero ($\nabla \times \nabla \Phi=0$), then if

$$\nabla \times \vec B =\mu_0 \vec J$$

for the magnetic field $\vec B$, then we also have

$$\nabla \times (\vec B+\nabla \Phi) =\mu_0 \vec J$$

for any (smooth) scalar field $\Phi$. This means that there is not a unique solution to the problem since $\vec B +\nabla \Phi$ is also a solution for any (smooth) $\Phi$.

However, if we also specify the divergence of the magnetic field (we know that it is zero), then we can pursue a unique solution. For example,

$$\nabla \times \nabla \times \vec B =-\mu_0 \nabla \times \vec J$$

whereupon using the vector identity $\nabla \times \nabla \times \vec B= \nabla ( \nabla \cdot \vec B)-\nabla^2 \vec B$ and exploiting $\nabla \cdot \vec B=0$ gives

$$\nabla^2 \vec B=-\mu_0 \nabla \times \vec J$$

which has solution

$$\vec B(\vec r)=\mu_0 \int_V \frac{\nabla' \times \vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV'$$

where the volume integral extends over all space where $\vec J=\ne 0$. We can integrate by parts in three dimensions by using the vector product rule identity $\nabla \times (\Phi \vec A) = \Phi \nabla \times \vec A+\nabla \Phi \times \vec A$ to write

$$\begin{align} \vec B(\vec r)&=\mu_0 \int_V \frac{\nabla' \times \vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV'\\\\ &=\mu_0 \int_V \left(\nabla' \times \left(\frac{\vec J(\vec r')}{4\pi |\vec r-\vec r'|}\right) -\nabla' \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') \right)dV'\\\\ &=\mu_0 \oint_S \frac{\hat n' \times \vec J(\vec r')}{4\pi |\vec r-\vec r'|}dS'- \mu_0 \int_V \nabla' \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') dV' \end{align}$$

Now, we may extend the integration region to all of space. Then, if $\vec J=0$ outside a finite region, then the surface integral vanishes and we have

$$\begin{align} \vec B(\vec r)&= -\mu_0 \int_V \nabla' \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') dV'\\\\ &=\mu_0 \int_V \nabla \left(\frac{1}{4\pi |\vec r-\vec r'|}\right) \times \vec J(\vec r') dV'\\\\ &=\nabla \times \left( \mu_0 \int_V \frac{\vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV' \right) \end{align}$$

where the first equality is effectively the Biot-Savart law and the last equality reveals that $\vec B =\nabla \times \vec A$ for the vector potential

$$\vec A(\vec r) = \mu_0 \int_V \frac{\vec J(\vec r')}{4\pi |\vec r-\vec r'|}dV'$$