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Table of Contents

1. Introduction

Magnetostatics is a little bit of an oxymoron; the magnetic field is created by a moving current of charges and a magnetic field for a point charge is therefore hard to model or often wrong; because magnetostatics assumes a steady current, a point charge moving cannot be replaced with another charge. However, for some continuous current distribution, the magnetic field is:

\begin{align*} \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J} \times \hat{r}}{r^{2}}dV \end{align*}

Which is the Bio-Savart law. Later, we will derive this from the axioms of and .

2. Divergence of B

The divergence of B is given by:

\begin{align*} \vec{\nabla} \cdot \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{\nabla} \cdot (\vec{J} \times \hat{r})}{r^{2}}dV = \frac{\mu_{0}}{4\pi}\int_{V}\vec{\nabla} \cdot (\vec{J} \times \frac{\hat{r}}{r^{2}})dV \\ \end{align*}

Now we want to evaluate \(\vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}}\) using the del operator rules:

\begin{align*} \vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}} = (\vec{\nabla} \times \vec{J}) \cdot \frac{\hat{r}}{r^{2}} - \vec{J} \cdot (\vec{\nabla} \times \frac{\hat{r}}{r^{2}}) \end{align*}

And while \(B\), and by extension \(\vec{\nabla}\) is dependent on \(\vec{r}\), the radial distance between two charges, \(\vec{J}\) is a function of \(r'\), the position vector to a given charge. This, of course, means that \(J\) does not depend on any of the variables that we are taking the derivative over. Thus:

\begin{align*} \vec{\nabla} \times \vec{J} = 0 \end{align*}

Also, due to the inverse square law, the curl of \(\frac{\hat{r}}{r^{2}}\) is zero, and therefore:

\begin{align*} \vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}} = 0 \end{align*}

And therefore:

\begin{align*} \vec{\nabla} \cdot \vec{B} = 0 \end{align*}

This is one of Maxwell's Equations.

3. Curl of B

The curl of \(\vec{B}\) is given by:

\begin{align*} \vec{\nabla} \times \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{\nabla} \times (\vec{J} \times \hat{r})}{r^{2}}dV \\ = \frac{\mu_{0}}{4\pi}\int_{V}\vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}})dV \end{align*}

where \(\vec{\nabla} \times \vec{J} \times \hat{r}\) is given by the del operator identity:

\begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = \vec{J}(\vec{\nabla} \cdot \frac{\hat{r}}{r^{2}}) - \frac{\hat{r}}{r^{2}}(\vec{\nabla} \cdot \vec{J}) + (\frac{\hat{r}}{r^{2}} \cdot \vec{\nabla})\vec{J} - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} \end{align*}

Due to the inverse square law, we know that \(\vec{\nabla} \cdot \frac{\hat{r}}{r^{2}} = 4\pi\delta(\vec{r})\); From the section above we know that \(\vec{\nabla} \cdot \vec{J} = 0\); hence:

\begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) + (\frac{\hat{r}}{r^{2}} \cdot \vec{\nabla})\vec{J} - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} \end{align*}

The first directional derivative is zero because \(\vec{J}\) does not depend on the same coordinates as \(\vec{\nabla}\) with the same reasoning as for the divergence, so we have:

\begin{align*} \vec{\nabla} \times (\vec{J} \times \frac{\hat{r}}{r^{2}}) = 4\pi\vec{J}(\vec{r'})\delta(\vec{r}) - (\vec{J} \cdot \vec{\nabla})\frac{\hat{r}}{r^{2}} \end{align*}

The second term reduces to zero for some reason (reason coming later; I currently do not know why). Now plugging this back into the original equation:

\begin{align*} \vec{\nabla} \times \vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}4\pi\vec{J}(r')\delta(\vec{r})dV \\ = \mu_{0}\int_{V}\vec{J}(r')\delta(\vec{r'} - \vec{r''})dV \end{align*}

Where \(r''\) is the location of the test particle. Now the distribution propreties under an integral:

\begin{align*} \vec{\nabla} \times \vec{B} = \mu_{0}\vec{J}(\vec{r''}) \end{align*}

or simply:

\begin{align*} \vec{\nabla} \times \vec{B} = \mu_{0}\vec{J} \end{align*}

This is for magnetostatics only; Maxwell's Equations offer a correction to this equation.

4. The Vector Potential

We can define a vector potential \(\vec{A}(\vec{r})\) such that:

\begin{align*} \vec{B} = \vec{\nabla} \times \vec{A} \end{align*}

which is consistent with the fact that:

\begin{align*} \vec{\nabla} \cdot \vec{B} = 0 \end{align*}

By taking the divergence of both sides in the first equation in this section. When \(\vec{J}\) is zero at infinity:

\begin{align*} \vec{A} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J}}{r}d\tau \end{align*}

The reasoning for this is not obvious, even to me. One could analogize this to the scalar potential for .

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