# Inconsistencies in finding magnetic vector potentials

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!

• One mistake seems to be that you integrated your expression for the exterior field over the interior (down to $r=0$). – G. Smith May 1 at 14:10
• 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. – G. Smith May 1 at 14:13
• Another is that the magnetic flux through your loop is zero because it is tangential and thus does not pass through the loop. – G. Smith May 1 at 14:16

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