Magnetic Potential of Long Straight Wire: 65 Chars

[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?

 

[gabbagabbahey]

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

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

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?

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)

 

[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?

 

[gabbagabbahey]

\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?

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)

now my trouble is explaining why \mathbf{J}=\frac{I}{\pi R^2} \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?

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.

 

[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.