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\newpage
\section{Magnetostatics}

\emph{Magnetostatics} is the study of the magnetic
field generated by a stationary current
distribution.
\begin{align*}
    \nabla \times \bf{B}
    &= \mu_0 \bf{J} \tag{M4'} \\
    \nabla \cdot \bf{B}
    &= 0 \tag{M2}
\end{align*}
(M4') $\implies \nabla \cdot \bf{J} = 0$, the
time-independent equation of charge conservation.

\subsection{Amp\`ere's Law}

Consider a closed curve $C$ that is the boundary
of an open surface $S$. Integrate (M4') over $S$
and apply Stoke's theorem to obtain Amp\`ere's
law:
\[ \boxed{\int_C \bf{B} \cdot \dd \bf{x} = \mu_0 I} \]
where $I = \int_S \bf{J} \cdot \dd \bf{S}$ is
the total current through $S$.
\begin{center}
    \includegraphics[width=0.2\linewidth]
    {images/36fb71a4a2da11ed.png}
\end{center}
Since $\nabla \cdot \bf{J} = 0$, the same current
$I$ flows through \emph{any} open surface $S$ such
that $\partial S = C$. Amp\`ere's law is the
integral version of (M4') and is valid provided
that $\pfrac{\bf{E}}{t} = \bf{0}$.
\[ \text{circulation of magnetic field around
loop} \propto \text{total current through loop} \]
In special situations we can use Amp\`ere's law
together with symmetry to deduce $\bf{B}$ from
$\bf{J}$. A cylindrically symmetric situation
could involve
\begin{itemize}
    \item An axial current distribution
        \[ J_z(r) \bf{e}_z \qquad (r, \phi, z) \]
    \item an azimuthal current distribution
        \[ J_\phi(r) \bf{e}_\phi \]
\end{itemize}
or a combination. ($\nabla \cdot \bf{J} = 0$
excludes a radial current).

\myskip
The same applies to $\bf{B}$. The curl in (M4')
implies:
\begin{itemize}
    \item $B_\phi$ is linearly related to $J_z$
    \item $B_z$ is linearly related to $J_\phi$
\end{itemize}

\subsubsection*{Long straight wire}

A cylindrical wire of radius $R$ carries a total
current $I$ parallel to its axis. To find
$B_\phi(r)$ generated by $J_z(r)$, apply
Amp\`ere's law to a circle $C$ of radius $r$ ($S$
is a disc).
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/17ad986ca2db11ed.png}
\end{center}
If $r > R$ then
\begin{align*}
    \int_C \bf{B} \cdot \dd \bf{x}
    &= B_\phi(r) \int_C \bf{e}_\phi \cdot \dd
    \bf{x} \\
    &= B_\phi(r) \int_C \dd l \\
    &= B_\phi(r) 2\pi r \\
    &= \mu_0 I
\end{align*}
Thus, outside the wire,
\[ \bf{B} = \frac{\mu_0 I}{2\pi r} \bf{e}_\phi \]

\subsection{Solenoid}

A thin wire is coiled around a cylindrical tube of
radius $R$. An \emph{ideal solenoid} is infinitely
long and tightly wound, having cylindrical
symmetry and purely azimuthal current.

\myskip
The wire carries current $I$ and has $N$ turns per
unit length of the tube.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/386a88a6a2dd11ed.png}
\end{center}
To find $B_z(r)$ generated by $J_\phi(r)$, apply
Amp\`ere's law to a rectangular loop $C$. Taking
$a < b < R$ or $R < a < b$ gives
\[ L(B_z(a) - B_z(b)) = 0 \]
Taking $a < R < b$ gives
\[ L(B_z(a) - B_z(b)) = \mu_0 NLI \]
Assuming that $B_z(r) \to 0$ as $r \to \infty$, we
deduce that
\[ B_z(r) \begin{cases}
    \mu_0 NLI & r < R \\
    0 & r > R
\end{cases} \]

\subsubsection*{Surface current}

The ideal solenoid is an example of a
\emph{surface current}, here is of the form
\[ J_\phi(r) = K_\phi \delta(r - R) \]
with $K_\phi = NI$. Generally, a \emph{surface
current density} $\bf{K}$ produces a discontinuity
in the tangential magnetic field:
\[ [\bf{n} \times \bf{B}] = \mu_0 \bf{K} \]
\newpage

\noindent
Follows from Amp\`ere's law applied to
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/940cb65ca2dd11ed.png}
\end{center}
(M2) implies normal component is continuous
\[ [\bf{n} \cdot \bf{B}] = 0 .\]

\subsection{The magnetic vector potential}

(M2) implies that $\bf{B}$ can be written in terms
of a \emph{magnetic vector potential}
$\bf{A}(\bf{X})$:
\[ \boxed{\bf{B} = \nabla \times \bf{A}} \]
$\bf{A}$ is not unique. If we make a \emph{gauge
transformation}, replacing $\bf{A}$ with
\[ \bf{A}' = \bf{A} + \nabla \chi \]
where $\chi(\bf{x})$ is an arbitrary scalar field,
then $\bf{B}$ is unchanged:
\[ \bf{B} = \nabla \times \bf{A} = \nabla \times
\bf{A}' .\]
A convenient gauge for many calculations is the
\emph{Coulomb gauge} in which $\nabla \cdot \bf{A}
= 0$. We can assume this condition without loss of
generality. If $\nabla \cdot \bf{A} \neq 0$ then
we can make a gauge transformation such that
$\nabla \cdot \bf{A}' = 0$ by choosing $\chi$ to
be the solution of Poisson's equation
\[ -\nabla^2 \chi = \nabla \cdot \bf{A} \]
In terms of $\bf{A}$, (M4') becomes
\[ \nabla \times (\nabla \times \bf{A}) = \mu_0
\bf{J} \]
Using the identity
\[ \nabla \times (\nabla \times \bf{A}) =
\nabla(\nabla \cdot \bf{A}) - \nabla^2 \bf{A} \]
and assuming Coulomb gauge ($\nabla \cdot \bf{A} =
0$) we obtain Poisson's equation in vector form:
\[ \boxed{-\nabla^2 \bf{A} = \mu_0 \bf{J}} \]
(compare with $-\nabla^2 \Phi = \frac{\rho}{\eps_0}$)

\subsection{The Biot-Savat Law}

The solution of Poisson's equation is (integrating
over al space, assuming decay at infinity)
\[ \boxed{\bf{A}(\bf{x}) = \frac{\mu_0}{4\pi} \int
\frac{\bf{J}(\bf{x}')}{|\bf{x} - \bf{x}'|} \dd^3
\bf{x}'} \]
We should check that the solution satisfies the
assumed Coulomb gauge condition:
\begin{align*}
    \nabla \cdot \bf{A}(\bf{x})
    &= \frac{\mu_0}{4\pi} \int_V \nabla \cdot
    \left( \frac{\bf{J}(\bf{x}')}{|\bf{x} -
    \bf{x}'|} \right) \dd^3 \bf{x}' \\
    &= \frac{\mu_0}{4\pi} \int_V \bf{J}(\bf{x}')
    \cdot \nabla \left( \frac{1}{|\bf{x} -
    \bf{x}'|} \right) \dd^3 \bf{x}' \\
    &= -\frac{\mu_0}{4\pi} \int_V \bf{J}(\bf{x}')
    \cdot \nabla' \left( \frac{1}{|\bf{x} -
    \bf{x}'|} \right) \dd^3 \bf{x}' \\
    &= -\frac{\mu_0}{4\pi} \int_V \nabla' \cdot
    \left( \frac{\bf{J}(\bf{x}')}{|\bf{x} -
    \bf{x}'|} \right) \dd^3 \bf{x}'
    &&\text{using $\nabla \cdot \bf{J} = 0$} \\
    &= -\frac{\mu_0}{4\pi} \int_{\partial V}
    \frac{\bf{J}(\bf{x}') \cdot \dd
    \bf{S}'}{|\bf{x} - \bf{x}'|}
\end{align*}
Thus $\nabla \cdot \bf{A} = 0$, as assumed, if the
current is contained in some finite volume and we
take $V$ to be at least as large, or if $\bf{J}$
decays sufficiently as $|\bf{x}| \to \infty$.
