# Singularity of $B$-field in a Dirac String

I was assigned this question related to Dirac strings:

Given a vector potential $\vec{A}= \frac{1-\cos(\theta)}{r\sin(\theta)}\hat{\phi}$, show that there is a singularity in the B field proportional to $\Theta(-z)\delta(x)\delta(y)$ on the z axis. ($\Theta(x-x_0)$ is the step function with its jump at $x_0$) Find the proportionality constant.

My attempt at a solution:

So showing there is a singularity simply results from the fact that at $\theta=0$, the vector potential explodes because the denominator goes to zero. The same holds true for r. My question becomes quantifying the magnitude of this singularity via this proportionality constant. I'm assuming it's essentially a measure of how quickly the field increases close the the z axis, but I'm not sure how to procede from here. I've looked at quite a bit of literature on the matter, including Dirac's original paper, but they all simply state there is a singularity, and make no statement about the size of it. Any insight that can be offered about the nature of this singularity would be deeply appreciated!

• Careful: at $\theta=0$ the numerator also goes to zero, so it's not immediately obvious that it should blow up there. – probably_someone May 21 '18 at 14:19

## 1 Answer

HINT: integrate $\vec{A}$ around a small loop of size $\epsilon$ centered along the $z$-axis. Use the fact that the magnetic flux through this loop is $\Phi_B = \oint \vec{A} \cdot d \vec{l}$.