- #1

latentcorpse

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

[itex]\mathbf{B}=\frac{\mu_0 I r}{2 \pi R^2} \mathbf{\hat{\phi}}[/itex]

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:

[itex]\mathbf{B}=\nabla \wedge \mathbf{A}[/itex]

But [itex]\mathbf{A(r)}=\frac{\mu_0}{4 \pi} \int dV' \frac{\mathbf{J(r-r')}}{|\mathbf{r-r'}|}[/itex]

so clearly it's parallel to [itex]\mathbf{J}[/itex] which is in the z direction so we conclude that

[itex]A_r=A_{\phi}=0[/itex] and that [itex]A_z[/itex] is non zero.

Taking the curl in cylindrical polars,

[itex]\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}}[/itex]

which whittles down to:

[itex]\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}}[/itex]

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

[itex]-\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[/itex]

and the otehr term gives:

[itex]\frac{\partial{A_z}}{\partial{\phi}}=0 \Rightarrow A_z[/itex] 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?