# Magnetic vector potential of an infinite wire

Using the integral $$A=\frac{\mu_0}{4 \pi} \int \frac{I \vec{dl}}{r}$$ for calculating magnetic vector potential of an infinite wire we get $$A = \left(\frac{\mu_0 I}{4 \pi}\right) \ln(\sec \theta + \tan \theta)$$ which diverges when the limits are from $-\pi$ to $\pi$. We can get around this by solving $B=\nabla \times A$ which gives us a finite answer.

My question is why does the first formula fail for this problem and is it fixable?

• The integral formula cannot be used when the current extends all the way to infinity. This is also the case in electrostatics: if you try to compute the electric potential of an infinite wire using the integral representation you get a divergent integral as well. Sep 8 at 14:46

Ok, so lets start with the basics, the answer we are expecting is given by: $$\vec B= \frac{\mu_0I}{2\pi r} \hat e_\theta$$ Which is from Ampere's law.

From this we can kind of backwards engineer, to show that: $$\vec A=-\frac{\mu_0I}{2\pi} \ln(r) \hat e_z$$ would work as the potential.

The reason I don't think your method works is because you are forcing the coulomb gauge (i.e. $\nabla \cdot \vec A=0$) onto the system and in this situation the integral diverges. In other words we have the freedom to chose $\vec A$ since: $$\vec A'=\vec A+\nabla(\phi)$$ For some function $\phi$ both satisfy: $$\vec B=\nabla \times \vec A=\nabla \times \vec A'$$ It happens that your choice of $\vec A$ is not well defined in this case, i.e. the coulomb gauge doesn't work.

• The vector potential written here is correct, but the statement that the Coulomb gauge doesn't work is incorrect. Just compute $\vec\nabla \cdot\vec A$ to see that it vanishes. The expression for $\vec A$ as an integral involving the current cannot be used because the wire extends to infinity. Sep 8 at 14:44

Below picture shows an infinite wire and an imaginary loop for calculating the vector potential super much easier.

\begin{align} B &= \nabla \times A \\ \iint B \cdot \mathrm{d}a &= \oint A \cdot \mathrm{d}l \\ \int_a^b \frac{\mu l}{2 \pi r} L \, \mathrm{d}r &= AL \\ A &= \frac{\mu l}{2 \pi} \ln{\frac{b}{a}} Z \end{align}

• Please type the math with MathJaX. It makes it easier for people on all devices to read it. Also, it would help if you could include a more detailed annotation.
– user191954
Jun 5 '18 at 11:46
• Welcome to SE.Physics! I converted the equations in your picture to $\mathrm{\TeX}$ (which is implemented through MathJax here), as @Chair recommended. If you try to edit your post, you can see the $\mathrm{\TeX}$ code to pick up on how it's done and make any changes that might improve the answer. In general, you can write $x$ to render $x$.
– Nat
Jun 5 '18 at 14:22