Suppose, we are given a magnetic field $\vec{B}$ as: $$\vec{B} = \phi \delta(x)\delta(y)\hat{e}_z$$ where $\phi$ is some constant and $\delta$ is dirac delta function.
How do we find the corresponding Magnetic vector potential?


2 Answers 2


You have the definition of the vector potential. $$\mathbf{B}=\nabla \times \mathbf{A}$$ According to Stokes' theorem this is equivalent to $$\iint_S \mathbf{B}\ d\mathbf{S} = \oint_{\partial S} \mathbf{A}\ d\mathbf{r}$$ where $S$ is any surface area and $\partial S$ is its boundary line.

Now choose for $S$ a circle around the $z$-axis. Then the integral on the left is trivial, it is just the constant $\phi$. And the integral on the right, done in cylindrical coordinates ($\rho,\varphi,z$), is a round-trip over $\varphi$: $$\phi=\int_0^{2\pi} \mathbf{A}\ \hat{\varphi}\ \rho\ d\varphi$$

It is easy to see that a solution is $$\mathbf{A}(\rho,\varphi,z) = \frac{\phi}{2\pi\rho}\,\hat{\mathbf{\varphi}}$$


One way to do this (though I wouldn't blame you for accusing me of cheating) is to "remember" the identity (in Cylindrical Coordinates):

$$\nabla \times \left(\frac{\hat{\mathbf{\varphi}}}{\rho}\right) = \hat{\mathbf{z}}\,\, 2 \pi \,\delta^2(\rho).\tag{1}\label{1}$$

This is reminiscent of the divergence identity (in Spherical Polar Coordinates): $$\nabla \cdot \left(\frac{\hat{\mathbf{r}}}{r^2}\right) = 4\pi \delta^3(r).\tag{2}\label{2}$$

Using the fact that $\delta^2(\rho) = \delta(x)\delta(y)$, you should be able to see that you can write the first equation in terms of your field $\mathbf{B}$ as:

$$\nabla \times \left(\frac{\phi}{2\pi\rho}\,\hat{\mathbf{\varphi}}\right) = \mathbf{B}.$$

By using the definition of the vector potential $\nabla \times \mathbf{A} = \mathbf{B},$ you should be able to see that one solution for $\mathbf{A}$ is clearly $$\mathbf{A}(\rho,\varphi,z) = \frac{\phi}{2\pi\rho}\,\hat{\mathbf{\varphi}}.$$

All other solutions for $\mathbf{A}$ can be obtained by adding the gradient of any scalar field (call it $f$), so the general solution is $$\mathbf{A}'= \mathbf{A} + \nabla f.$$ Since the curl of the gradient is zero, $$\nabla \times \mathbf{A}' = \nabla \times \mathbf{A} = \mathbf{B}.$$ This is just a reflection of gauge-invariance.

  • 1
    $\begingroup$ A good reason why this identity is worth remembering in this context is that $\mathbf{j} = I\cdot \delta(x)\cdot \delta(y)\cdot \mathbf{e}_z$ is the current field of a thin wire carrying the total current $I$, and it's well known that the magnetic field (per Ampère's law) around such a wire is $\frac{\mu_0\cdot I}{2\cdot\pi\cdot\rho}\mathbf{e}_\varphi$. Talking about $\mathbf{A}$ and $\mathbf{B}$ is analogous to talking about $\mathbf{B}$ and $\mathbf{j}$. $\endgroup$ Sep 21, 2021 at 18:45
  • $\begingroup$ Thanks for accepting this answer, @LifelongLearner, but I honestly think the answer by ThomasFritsch is better than mine! It's much simpler, clearer, and doesn't require you to remember any strange identities that are too difficult to prove. ;) $\endgroup$
    – Philip
    Nov 9, 2021 at 19:47

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