# Misunderstanding of the magnetic field

You can reconstruct a vector field $\mathbf F$ from its divergence $\nabla \cdot \mathbf F$ and curl $\nabla \times\mathbf F$ *and boundary conditions*. More precisely, if $R$ is some bounded region on which $\mathbf F$ is twice continuously differentiable, then

$$\mathbf F = \nabla \Phi + \nabla \times \mathbf G$$

where $$\Phi(\mathbf r) = \frac{1}{4\pi} \int_R \frac{\nabla' \cdot \mathbf F(\mathbf r')}{|\mathbf r - \mathbf r'|}d^3 r' -\frac{1}{4\pi}\oint_{\partial R} \frac{\hat n' \cdot \mathbf F(\mathbf r')}{|\mathbf r - \mathbf r'|}dS$$ and $$\mathbf G(\mathbf r) = \frac{1}{4\pi}\int_R \frac{\nabla'\times\mathbf F(\mathbf r')}{|\mathbf r - \mathbf r'|}d^3 r' - \frac{1}{4\pi}\oint_{\partial R} \frac{\hat n' \times \mathbf F(\mathbf r')}{|\mathbf r - \mathbf r'|}dS$$

where $\partial R$ is the boundary of $R$, and $\hat n$ is the normal vector to the surface at the point $\mathbf r'$. If the fields fall off faster that $1/|\mathbf r-\mathbf r'|$, then this can be extended to an unbounded region.

In the static case, Ampere's Law says that $$\nabla \times \mathbf B = \mu_0 \mathbf J$$

but this is not continuous at the boundary of the wire. As a result, the Helmholtz theorem applies **separately** to the interior region within the wire and the region outside it. Reconstructing the magnetic field *outside* the wire requires you to know the value of the magnetic field at the *surface* of the wire.

To recover ${\bf B}$ from $\nabla\times {\bf B}$ and $\nabla\cdot {\bf B}=0$ you need to know $\nabla\times {\bf B}$ *everywhere*, not just at the point where you want to know ${\bf B}$

The statement that *"We know that any field can be constructed using its divergence and rotational."* is wrong (I assume that you mean "curl" when you say *"rotational"*). An easy counterexample are the following two fields

\begin{alignat}{3} \mathbf E_1 &= 2\mathbf {\hat i}\quad &&;\quad&&&\mathbf E_2=3\mathbf{\hat j}\\ \nabla \cdot \mathbf E_1&=0 \quad &&;\quad\nabla \cdot &&&\mathbf E_2 =0\\ \nabla \times \mathbf E_1&=0\quad &&;\quad \nabla \times &&&\mathbf E_2 =0 \end{alignat}

As you can see that the curl and divergence of both $\mathbf E_1$ and $\mathbf E_2$ are the same, yet $\mathbf E_1 \not\equiv\mathbf E_2$. So, there isn't any way of determining a vector field just from its curl and divergence.