Solution of simple problems using only Maxwell equations in differential form

Solve simple electrostatic or magnetostatic problems using only Maxwell equations. For example:

In every book there is an excercise to find a magnetic field outside a thin wire of radius $a$ with current $I$. The usual approach is Biot-Savart law or Ampere law. I know you can derive Biot-Savart law from Maxwell equations or use integral form of Ampere law to solve this easily, but I'm interested in solution involving vector potential $A$ and a Poisson equation. Then solving the equation by separation of variables. What would be the boundary conditions?

EDIT:

Consider it like this: You know noting but these two magnetostatic equations:

$\nabla\cdot \textbf{B} = 0$ and $\nabla \times \textbf{H} = \textbf{J}$

and you now about Coulomb gauge $\nabla \cdot \textbf{A} =0$ and $\textbf{B}=\nabla \times \textbf{A}$ and that $\textbf{H}$ and $\textbf{B}$ are simply related by $\textbf{B}= \mu \textbf{H}$

What differential equation does this produce and what boundary conditions would you use for this specific problem?

This isn't really a "boundary conditon" problem in the sense that we aren't trying to take knowledge of $\vec{A}$ on some surface and extend it to a solution of Laplaces equation in some region, which has the surface as a boundary. The reason you can't do this is that you have no apriori reason to know how $\vec{A}$ should look for a given $\vec{J}$.

Rather you're trying to turn information about the sources, $\vec{J}$, into information about the field, $\vec{A}$. This requires actually inverting the $\nabla^2$ operator, which involves the use of Green's functions. For instance you could use the formula,

$$A_x(\vec{x}) = \frac{\mu}{4\pi} \int \frac{J_x(\vec{x}') d^3\vec{x}'}{\left| \vec{x}-\vec{x}' \right|} .$$

• @RobJeffries, thanks for pointing that out. Its been a while since I've worked in SI units. – Spencer May 28 '15 at 9:25
• I see, I did actually solved the last equation from post below, and got the right B field, but only because I knew how A should behave i.e. I used the symmeties of z and phi coord., which is weird,because I shouldn't know what A looks as you pointed out.I still don't get the whole idea behind those Green functions, but I now realise I can use Poisson eq. to solve things ONLY if know some information obut A or B i.e. the boundary conditions.This is all frustrating because I have to decide what method to use,for example the 5.13 excerc. in Jackson, some use Poisson eq. and some use that integral – nini May 28 '15 at 13:49

$$\mathbf{\bigtriangledown \times B = \mu J}$$ and $$\mathbf{B = \bigtriangledown \times A}$$ so $$\mathbf{\bigtriangledown \times \bigtriangledown \times A = \mu J}$$

From the definition of the vector Laplacian we have $$\mathbf{\bigtriangledown \times \bigtriangledown \times A} = \mathbf{\bigtriangledown}^{2}\mathbf{A} - \mathbf{ \bigtriangledown\left (\bigtriangledown\cdot A \right )}$$

The Coulomb gauge makes the second term on the RHS vanish, so we're left with $$\mathbf{\bigtriangledown}^{2}\mathbf{A} = -\mathbf{\mu J}$$ which is just the Poisson equation in vector form. To relate the vector potential to the radius of the wire and the current passing through it, you can integrate the current density over a cross-section of the wire.