If B is uniform,show that $\mathit{A}\left(r\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\left(r\times B\right)$works. That is, check that $\mathbf{\nabla }\mathbf{.}\mathit{A}\mathbf{=}\mathbf{0}$and$\mathbf{\nabla }\mathbf{\times }\mathbf{A}\mathbf{=}\mathbf{B}$. Is this result unique, or are there other functions with the same divergence and curl?

The curl is mainly used for quantifying the circulation of a particular electric field. It mainly measures the tendency of a particular fluid that swirls around a point.

The equation of the dot product of the curl and the function A is expressed as:

$\nabla .A=-\frac{1}{2}\nabla .\left(\overrightarrow{r}\times \overrightarrow{B}\right)$

Here,$\nabla$is the curl, is the function,$\overrightarrow{r}$is the radius of the magnetic field and $\overrightarrow{B}$is the magnetic field.

Calculating the above equation

$$\begin{gathered} \nabla .A=-\frac{1}{2}\left[\overrightarrow{B}.\left(\nabla \times \overrightarrow{r}\right)-\overrightarrow{r}.\left(\nabla \times \overrightarrow{B}\right)\right] \\ =0 \end{gathered}$$

The equation of the dot product of the curl and the function A is expressed as:

$\nabla \times A=-\frac{1}{2}\nabla \times \left(\overrightarrow{r}\times \overrightarrow{B}\right)$

Here,$\nabla$is the curl,A is the function,$\overrightarrow{r}$is the radius of the magnetic field and $\overrightarrow{B}$is the magnetic field.

Calculating the above equation

$\nabla \times A=-\frac{1}{2}\left[\left(\overrightarrow{B}.\nabla \right)\overrightarrow{r}-\left(\overrightarrow{r}.\nabla \right)\overrightarrow{B}+\overrightarrow{r}\left(\nabla .\overrightarrow{B}\right)-\overrightarrow{B}\left(\nabla .\overrightarrow{r}\right)\right]$

Substitute 3 for$\nabla .\overrightarrow{r}$and $\overrightarrow{B}$for$\left(\overrightarrow{B}.\nabla \right)\overrightarrow{r}$ in the above equation.

$$\begin{gathered} \nabla \times A=-\frac{1}{2}\left[\overrightarrow{B}-0+0-3\overrightarrow{B}\right] \\ =\overrightarrow{B} \end{gathered}$$

The above solution shows that the vector potential does not produce a uniform magnetic field.

The solution is not a unique solution as the vector $\overrightarrow{r}$can be replaced with$\overrightarrow{r}+\overrightarrow{d}$where$\overrightarrow{d}$is a constant vector and the same result will come.

Thus, $A\left(r\right)=-\frac{1}{2}\left(r\times B\right)$works. The solution is not a unique solution as the vector $\overrightarrow{r}$can be replaced with $\overrightarrow{r}+\overrightarrow{d}$where$\overrightarrow{d}$ is a constant vector and the same result will come.