Magnetic Vector Potential

Vector Potential

The vector potential $\textbf{A}$ is defined as $$\bbox[5px,border:2px solid #666] { \textbf{B} = \nabla \times \textbf{A} }$$

Checkpoint (Griffiths Third Edition Q5.25)

Find the vector potential a distance $s$ from an infinite straight wire carrying a current $I$.

Use cylndrical coordinate. The magnetic field is $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I \implies \vec{B}=\frac{\mu_0I}{2\pi s}\hat{\phi}$$ So, $$\begin{align} \vec{B}=\nabla\times\vec{A}&=\frac{\mu_0I}{2\pi s}\hat{\phi}\\ -\frac{\partial A}{\partial s}\hat{\phi} &= \frac{\mu_0I}{2\pi s}\hat{\phi}\\ \frac{\partial A}{\partial s} &= -\frac{\mu_0I}{2\pi s}\\ A&=-\frac{\mu_0I}{2\pi}\ln\left(\frac{s}{a}\right) \end{align}$$ As $\vec{B}=\nabla\times\vec{A}$, $\vec{A}$ should be along z-axis. $$\vec{A}(\vec{r})=-\frac{\mu_0I}{2\pi s}\ln\left(\frac{s}{a}\right)\hat{z}$$

Divergence of Vector Potential

Let$\textbf{A}$ be the vector potential of the magnetic field $\textbf{B} = \nabla \times \textbf{A}$ Suppose $\nabla\cdot\textbf{A} = k$ where $k$ is non-zero.
Let $\textbf{F}$ be a vector field such that $\nabla\cdot\textbf{F} = k$ and curl-free $\nabla\times\textbf{F} = 0$. Poisson's equation guarantees the existence of such a vector field. Then the vector field $\textbf{A}' = \textbf{A} - \textbf{F}$ has zero divergence $$\nabla\cdot\textbf{A}' = \nabla\cdot\textbf{A} - \nabla\cdot\textbf{F} = k - k = 0$$ and has curl $$\nabla\times\textbf{A}' = \nabla\times\textbf{A} - \nabla\times\textbf{F} = \textbf{B} - 0 = \textbf{B}$$ So, $$\textbf{B} = \nabla \times \textbf{A}'$$ Therefore, we can always choose the vector potential such that $$\bbox[5px,border:2px solid #666] { \nabla \cdot \textbf{A} = 0 }$$

Curl of Vector Potential

By definition, the curl of magnetic vector potential of a magnetic field is the magnetic field itself.

Solving Vector Potential

By Ampere's law, $$\begin{align} \mu_0\textbf{J} = \nabla \times \textbf{B} &= \nabla \times (\nabla \times \textbf{A}) \\ &= \nabla(\nabla\cdot\textbf{A}) - \nabla^2\textbf{A} \\ &= -\nabla^2\textbf{A} \end{align}$$ So, $$\bbox[5px,border:2px solid #666] { \nabla^2\textbf{A} = - \mu_0\textbf{J} }$$ Previously, we know that $$\nabla^2V = \frac{\rho}{\epsilon_0}$$ and $$V = \frac{1}{4\pi\epsilon_0} \int\frac{\rho}{d}d\tau$$ So, $$\bbox[5px,border:2px solid #666] { \textbf{A}(\textbf{r}) = \frac{\mu_0}{4\pi} \int \frac{\textbf{J}(\textbf{r}')}{d}d\tau }$$ Similarly, $$\textbf{A} = \frac{\mu_0}{4\pi} \int \frac{\textbf{I}}{d}dl = \frac{\mu_0}{4\pi} \int \frac{\textbf{K}}{d} da$$

Example (Griffiths Third Edition Q5.22)

Find the magnetic vector potential of an finite segment of straight wire, carrying a current $I$.

Solution:

$$\begin{align} \vec{A} &= \frac{\mu_0}{4\pi}\int \frac{I\hat{z}}{d}dz\\ &= \frac{\mu_0I}{4\pi}\int^{z_2}_{z_1}\frac{dz}{\sqrt{z^2+s^2}}\\ &= \frac{\mu_0I}{4\pi} \hat{z}\left[\ln(z+\sqrt{z^2+s^2})\right]^{z_2}_{z_1}\\ &= \frac{\mu_0I}{4\pi}\ln\left[\frac{z_2+\sqrt{z_2^2+s^2}}{z_1+\sqrt{z_1^2+s^2}}\right]\hat{z} \end{align}$$