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magnetostatics

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:

B→=μ04π∫VJ→×r̂r2dV\begin{aligned}\vec{B} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J} \times \hat{r}}{r^{2}}dV\end{aligned}

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

2. Divergence of B

The divergence of B is given by:

∇→⋅B→=μ04π∫V∇→⋅(J→×r̂)r2dV=μ04π∫V∇→⋅(J→×r̂r2)dV\begin{aligned}\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{aligned}

Now we want to evaluate ∇→⋅J→×r̂r2\vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}} using the del operator rules:

∇→⋅J→×r̂r2=(∇→×J→)⋅r̂r2−J→⋅(∇→×r̂r2)\begin{aligned}\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{aligned}

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

∇→×J→=0\begin{aligned}\vec{\nabla} \times \vec{J} = 0\end{aligned}

Also, due to the inverse square law, the curl of r̂r2\frac{\hat{r}}{r^{2}} is zero, and therefore:

∇→⋅J→×r̂r2=0\begin{aligned}\vec{\nabla} \cdot \vec{J} \times \frac{\hat{r}}{r^{2}} = 0\end{aligned}

And therefore:

∇→⋅B→=0\begin{aligned}\vec{\nabla} \cdot \vec{B} = 0\end{aligned}

This is one of Maxwell's Equations.

3. Curl of B

The curl of B→\vec{B} is given by:

∇→×B→=μ04π∫V∇→×(J→×r̂)r2dV=μ04π∫V∇→×(J→×r̂r2)dV\begin{aligned}\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{aligned}

where ∇→×J→×r̂\vec{\nabla} \times \vec{J} \times \hat{r} is given by the del operator identity:

∇→×(J→×r̂r2)=J→(∇→⋅r̂r2)−r̂r2(∇→⋅J→)+(r̂r2⋅∇→)J→−(J→⋅∇→)r̂r2\begin{aligned}\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{aligned}

Due to the inverse square law, we know that ∇→⋅r̂r2=4πδ(r→)\vec{\nabla} \cdot \frac{\hat{r}}{r^{2}} = 4\pi\delta(\vec{r}); From the section above we know that ∇→⋅J→=0\vec{\nabla} \cdot \vec{J} = 0; hence:

∇→×(J→×r̂r2)=4πJ→(r′→)δ(r→)+(r̂r2⋅∇→)J→−(J→⋅∇→)r̂r2\begin{aligned}\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{aligned}

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

∇→×(J→×r̂r2)=4πJ→(r′→)δ(r→)−(J→⋅∇→)r̂r2\begin{aligned}\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{aligned}

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:

∇→×B→=μ04π∫V4πJ→(r′)δ(r→)dV=μ0∫VJ→(r′)δ(r′→−r″→)dV\begin{aligned}\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{aligned}

Where r″r'' is the location of the test particle. Now the dirac delta distribution propreties under an integral:

∇→×B→=μ0J→(r″→)\begin{aligned}\vec{\nabla} \times \vec{B} = \mu_{0}\vec{J}(\vec{r''})\end{aligned}

or simply:

∇→×B→=μ0J→\begin{aligned}\vec{\nabla} \times \vec{B} = \mu_{0}\vec{J}\end{aligned}

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

4. The Vector Potential

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

B→=∇→×A→\begin{aligned}\vec{B} = \vec{\nabla} \times \vec{A}\end{aligned}

which is consistent with the fact that:

∇→⋅B→=0\begin{aligned}\vec{\nabla} \cdot \vec{B} = 0\end{aligned}

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

A→=μ04π∫VJ→rdτ\begin{aligned}\vec{A} = \frac{\mu_{0}}{4\pi}\int_{V}\frac{\vec{J}}{r}d\tau\end{aligned}

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