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Recently I've been studying for my electromagnetism finals and I reached a question about magnetic vector potentials. If I have a wire with constant current distribution, what is the magnetic vector potential inside and outside the wire.

For simplicity sake, I will only include the case of outside the wire. Using Ampere's Law, I have found that the magnetic field at a distance $r$ away from the wire is

$$B=\frac{\mu_{0}I_{total}}{2\pi r}$$

The first method I attempted was to use

$$\oint{\vec{A}}{d\vec{l}}=\int{\vec{B}}{d\vec{S}}$$ In this case I am taking a closed loop with radius equal to $r$. $\oint{d\vec{l}}$ should just give the circumference and since $\vec{S}$ is $\pi r^{2}$, $d\vec{S}$ should be $2\pi rd\vec{r}$.

Or at least I think it is so, because if I plug in these values, I get that the magnetic vector potential outside is a constant.

$$\vec{A}(2\pi r)=\int_{0}^{r}{\frac{\mu_{0}I_{tot}}{2\pi r'}2\pi r'dr'}$$ $$\vec{A}=\frac{\mu_{0}I_{total}}{2\pi}$$

The second method I attempted was to say that

$$\vec{B}=\nabla\times\vec{A}$$ And since $\vec{B}$ is in the $\hat{\phi}$ direction, and assuming $\vec{A}$ only varies in the $\hat{r}$ direction, doing the cross product just results in

$$\frac{\mu_{0}I_{total}}{2\pi r}=-\frac{\partial A_{z}}{\partial r}$$

Which after integrating gives me a logarithm.

Doing research online, I found that the magnetic vector potential outside of a wire is supposed to give a logarithm, which means my first method is flawed. The problem is I don't know where. I feel like I'm misunderstanding something about the vectors involved. Any help would be appreciated. Thank you!

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  • $\begingroup$ One mistake seems to be that you integrated your expression for the exterior field over the interior (down to $r=0$). $\endgroup$ – G. Smith May 1 at 14:10
  • $\begingroup$ Another is that in the first method you assume the vector potential is tangential, while the second method tells you that it is parallel to the wire. $\endgroup$ – G. Smith May 1 at 14:13
  • $\begingroup$ Another is that the magnetic flux through your loop is zero because it is tangential and thus does not pass through the loop. $\endgroup$ – G. Smith May 1 at 14:16
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When I mark electromagnetism exams and homework at undergraduate level, writing any formula of the form

$$\vec{A}=\frac{\mu_{0}I_{total}}{2\pi},$$

specifically, with a vector on the left-hand-side equaled to a scalar on the right-hand side, or vice versa, is immediate cause to lose all of the marks for that question. The reason is that, unless and until you know exactly what you're doing, playing fast and loose with scalar-vs-vector notation is the fastest route to making a catastrophic mistake that will sink your work. In this case, this is precisely what's happened to you.

For the configuration you've chosen, the magnetic field is azimuthal (along $\hat{\varphi}$), and you've chosen a gaussian surface with an axial surface normal (so $d\vec S = dS\,\hat z$), which means that the two vectors are orthogonal, and you get $$ \int{\vec{B}}{d\vec{S}} = \int{\vec{B}}\cdot {d\vec{S}} = 0. $$ Similarly, for this configuration the line integral has an azimuthal line element (so $d\vec \ell = d\ell \, \hat\varphi$) and the vector potential is axial (i.e., for the most convenient gauge, $\vec A$ is along $\hat z$), which means that $$ \oint{\vec{A}}{d\vec{\ell}} = \oint{\vec{A}}\cdot{d\vec{\ell}} = 0. $$ In other words, your equation is correct, and it reads, $$ 0 = 0, $$ and it tells you absolutely nothing about the field configuration. (OK, not quite. If you start with an arbitrary azimuthal component $A_\varphi \hat{\varphi}$ of the vector potential, then this tells you that it needs to vanish. But you could have deduced that from the symmetry.)

The gaussian surface that you need to use for this case is different. To get it right, make sure that you correctly handle all the vector notation in the problem.

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