example of vector potential

If the solenoidal vector  $\overrightarrow{U}=\overrightarrow{U}(x,y,z)$  is a homogeneous function of degree $\lambda$ ($\ne -2$),  then it has the vector potential

$\overrightarrow{A}={\displaystyle \frac{1}{\lambda +2}}\overrightarrow{U}\times \overrightarrow{r},$(1)

where  $\overrightarrow{r}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$  is the position vector.

Proof.  Using the entry nabla acting on products, we first may write

$$\nabla \times (\frac{1}{\lambda +2}\overrightarrow{U}\times \overrightarrow{r})=\frac{1}{\lambda +2}[(\overrightarrow{r}\cdot \nabla )\overrightarrow{U}-(\overrightarrow{U}\cdot \nabla )\overrightarrow{r}-(\nabla \cdot \overrightarrow{U})\overrightarrow{r}+(\nabla \cdot \overrightarrow{r})\overrightarrow{U}].$$

In the brackets the first product is, according to Euler’s theorem on homogeneous functions, equal to $\lambda \overrightarrow{U}$.  The second product can be written as  $U_x\frac{\partial \overrightarrow{r}}{\partial x}+U_y\frac{\partial \overrightarrow{r}}{\partial y}+U_z\frac{\partial \overrightarrow{r}}{\partial z}$, which is $U_x\overrightarrow{i}+U_y\overrightarrow{j}+U_z\overrightarrow{k}$, i.e. $\overrightarrow{U}$.  The third product is, due to the sodenoidalness, equal to  $0\overrightarrow{r}=\overrightarrow{0}$.  The last product equals to $3\overrightarrow{U}$ (see the first formula (http://planetmath.org/PositionVector) for position vector).  Thus we get the result

$$\nabla \times (\frac{1}{\lambda +2}\overrightarrow{U}\times \overrightarrow{r})=\frac{1}{\lambda +2}[\lambda \overrightarrow{U}-\overrightarrow{U}-\overrightarrow{0}+3\overrightarrow{U}]=\overrightarrow{U}.$$

This means that $\overrightarrow{U}$ has the vector potential (1).