# Magnetic Potential

1. Feb 18, 2009

### latentcorpse

A long straight wire of radius R carries a unifrom current density $\mathbf{J}$ inside it.

In the first part of the question I worked out that the magnetic field inside the wire was

$\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}$

I am now asked to get the vector potential insde the wire and give the hint of taking the curl in cylindrical polars. So far I have:

$\mathbf{B}=\nabla \wedge \mathbf{A}$
But $\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r-r')}}{|\mathbf{r-r'}|}$
so clearly it's parallel to $\mathbf{J}$ which is in the z direction so we conclude that
$A_r=A_{\phi}=0$ and that $A_z$ is non zero.

Taking the curl in cylindrical polars,

$\nabla \wedge \mathbf{A}=\left(\frac{1}{r} \frac{\partial{A_z}}{\partial{\phi}}-\frac{\partial{A_{\phi}}}{\partial{z}} \right) \mathbf{\hat{r}} + \left(\frac{\partial{A_r}}{\partial{z}}-\frac{\partial{A_z}}{\partial{r}} \right) \mathbf{\hat{\phi}} + \frac{1}{r} \left(\frac{\partial}{\partial{r}}(rA_{\phi})-\frac{\partial{A_r}}{\partial{\phi}} \right) \mathbf{\hat{z}}$
which whittles down to:
$\nabla \wedge \mathbf{A}=\frac{1}{r} \frac{\partial{A_z}}{\partial{\phi}} \mathbf{\hat{r}} - \frac{\partial{A_z}}{\partial{r}} \mathbf{\hat{\phi}}$

We have that this must be equal to $\mathbf{B}$ which is given above, so by comparing terms we get

$-\frac{\partial{A_z}}{\partial{r}}=\frac{\mu_0 I r}{2 \pi R^2} \Rightarrow -\int dA_z=\frac{\mu_0 I}{2 \pi R^2} \int r dr \Rightarrow A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 + const$
and the otehr term gives:
$\frac{\partial{A_z}}{\partial{\phi}}=0 \Rightarrow A_z$ is a constant, but in the line above, we found it to have radial dependence - something isn't quite right here, can anybody help me?

2. Feb 19, 2009

### gabbagabbahey

The 'const' in this equation doesn't have to be a constant in all variables, it just can't have any $r$ dependence (otherwise the partial derivative w.r.t. $r$ of that "constant" term would be non-zero) It can however depend on $\phi$ and $z$ (For example, $\frac{\partial}{\partial r} 5z^2\cos\phi=0$ )

$$\implies A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2+ g(\phi,z)$$

Where $g$ is some unknown function of $\phi$ and $z$

Again, your constant need not be a constant, it just can't depend on $\phi$

$$\implies A_z=h(r,z)$$

Put the two conditions together and you find that $$A_z=-\frac{\mu_0 I}{4 \pi R^2}r^2 +f(z)$$ satisfies both equations simultaneously for any function $f(z)$....You are free to choose any $f(z)$ you like; as is typical since the vector potential is defined by a first order differential (the curl in this case) and so is only unique up to a 'constant' (in this case, your constant can depend on $z$ without affecting B)

3. Feb 19, 2009

### latentcorpse

ok. thanks.

next i'm asked to show explicitly that this satisfies Poisson's equation, i.e.

$\nabla^2 \mathbf{A}=-\mu_0 \mathbf{J}$

and also that $\nabla \cdot \mathbf{A}=0$.

For the first part I have:
$\nabla^2 \mathbf{A}=\nabla(\nabla \cdot \mathbf{A})-\nabla \wedge \nabla \wedge \mathbf{A}$ (taking the Laplacian of a vector.

then it makes more sense to show the divergence is 0 now to eliminate that term from the expansion of the laplacian.

$\nabla \cdot \mathbf{A}=\frac{\partial{A_z}}{\partial{z}}=\frac{\partial{f(z)}}{\partial{z}} \neq 0 \forall f(z) \neq 0$
bit confused here???

however assuming that works, we get $\nabla^2 \mathbf{A}=-\nabla \wedge (\nabla \wedge \mathbf{A})=-\nabla \wedge \mathbf{B}$

$\nabla \wedge \mathbf{B}=-\frac{\partial{B_{\phi}}}{\partial{z}}\mathbf{\hat{r}} + \frac{1}{r} \frac{\partial}{\partial{r}} (r B_{\phi}) \mathbf{\hat{z}}$ as $B_r=B_z=0$

now $\frac{\partial{B_{\phi}}}{\partial{z}}=0$ as $\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}$

so $\nabla \wedge \mathbf{B}=\frac{1}{r} \frac{\partial}{\partial{r}}(\frac{\mu_0 I r^2}{2 \pi R^2}) \mathbf{\hat{z}}=\frac{\mu_0 I}{\pi R^2}\mathbf{\hat{z}}$

$\Rightarrow\nabla^2 \mathbf{A}=-\mu_0 \frac{I}{\pi R^2}\mathbf{\hat{z}}$

now my trouble is explaining why $\mathbf{J}=\frac{I}{\pi R} \mathbf{\hat{z}}$. Clearly the direction is fine and the untis seem ok as we have amperes/metre^2 - but there must be some definition i'm meant to use that i'm just missing in my notes???

Last edited: Feb 19, 2009
4. Feb 19, 2009

### gabbagabbahey

Requiring $$\vec{\nabla}\cdot\vec{A}=0$$

corresponds to a specific gauge choice (it removes some of the freedom you normally have in chossing your 'constants'); so if you want to choose $$\vec{A}$$ in a way that makes this true, just choose an $f(z)$ with zero divergence (the simplest choice is just $f(z)=0$)

Well, what is the total current passing through a cross-section of the wire if the volume current $$\vec{J}$$ is uniform and runs in the z-direction?....since the total current is required to be $I$ equate the two expressions and solve for $J$.

5. Feb 20, 2009

### latentcorpse

$I=\int_S \mathbf{J} \cdot \mathbf{dA} \Rightarrow I=|\mathbf{J}| \pi R^2 \Rightarrow |\mathbf{J}|=\frac{I}{\pi R^2}$

cheers m8.