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\markright{{\sf Lecture 12 (Ch 5 part 2) Dipoles and magnetization}}
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\begin{document}

\chapter{Ch. 5, part 2}

\bigskip
\bigskip

\textsc{Lecture 12 - Dipoles and Magnetization}

\newpage\ 

We have seen that the electrostatic field of a localized charge distribution
can be expanded in electric multipoles. Similarly, the magnetic field of a
stationary current distribution can be expanded in magnetic multipoles when
the currents are localized. The general expansion involves vector spherical
harmonics, which are left for later (Jackson Ch. 16). The multipole
expansion is valid in the region external to a sphere that contains the
currents.

\section{Magnetic dipole fields}

The leading term, the magnetic dipole, can be obtained directly with the use
of certain differential identities for tensors and of the condition $\mathbf{%
\nabla \cdot J}=0.$

\subsection{Tensor identities for stationary currents}

The identities we need all follow from the simple fact that the integral of
a perfect differential is zero and are valid for Cartesian components only,
or for vectors and tensors considered as abstract geometric quantities. We
use the convention that a sum over repeated indices is implied.

\subsubsection{The integral of $\mathbf{J}$ vanishes}

We start from 
\begin{equation}
\partial _i(J_ix_j)=\left( \partial _iJ_i\right) \,x_j+J_i(\partial
_ix_j)=J_j  \label{d0}
\end{equation}
which is obtained by using 
\[
\partial _iJ_i\equiv \mathbf{\nabla \cdot J}=0 
\]
\[
\partial _ix_j\equiv \frac{\partial x_j}{\partial x_i}=\delta _{ij} 
\]
Integrating (\ref{d0}) we obtain the useful identity 
\begin{equation}
\int J_j(\mathbf{x)\,}d^3x=0  \label{i0}
\end{equation}

\subsubsection{The first moment of $\mathbf{J}$ gives the magnetic moment}

We want to prove that 
\begin{equation}
\frac 1c\int J_ix_j\mathbf{\,}\,d^3x=-\varepsilon _{ijk\,}m_k  \label{jm}
\end{equation}
where $\varepsilon _{ijk}$ is the totally antisymmetric tensor with $%
\varepsilon _{123}=1$ (for a right-handed Cartesian system) and $\mathbf{m}$
is by definition the magnetic dipole moment 
\begin{equation}
\mathbf{m=}\frac 1{2c}\int \mathbf{x\times J}(\mathbf{x})\,\,d^3x  \label{m}
\end{equation}

Proceeding as in the derivation of (\ref{i0}), we integrate 
\[
\partial _i(J_ix_jx_k)=\left( \partial _iJ_i\right) x_j\,x_k+J_i(\partial
_ix_j)x_k+J_ix_j(\partial _ix_k)=J_jx_k+J_kx_j
\]
and obtain 
\begin{equation}
\int \left( J_jx_k+J_kx_j\mathbf{\,}\right) \,d^3x=0  \label{i1}
\end{equation}
We recognize that the integrand is twice the symmetric part of the first
moment tensor $J_jx_k.$ We can then decompose this tensor into symmetric and
antisymmetric parts 
\[
J_jx_k=\frac 12\left( J_jx_k+J_kx_j\mathbf{\,}\right) +\frac 12\left(
J_jx_k-J_kx_j\mathbf{\,}\right) 
\]
and obtain the equation 
\begin{equation}
\int J_jx_k\mathbf{\,}\,d^3x=\frac 12\int \left( J_jx_k-J_kx_j\mathbf{\,}%
\right) \,d^3x  \label{i2}
\end{equation}
This does not look like progress at first, but it reduces the 9 components
of $J_jx_k\mathbf{\,}$ to just 3, and these 3 can be identified with the
components of $\,\mathbf{J\times x}$ according to 
\[
\left( \mathbf{J\times x}\right) _i\equiv \varepsilon _{ijk}J_jx_k=\frac
12\varepsilon _{ijk}\left( J_jx_k-J_kx_j\mathbf{\,}\right) 
\]
\begin{equation}
J_jx_k-J_kx_j=\varepsilon _{jkl}\left( \mathbf{J\times x}\right) _l
\label{a1}
\end{equation}
These equations can easily be verified directly, because it is enough to
verify the $_{12}$ component and invoke antisymmetry. Collecting the results
(\ref{i2}), (\ref{a1}) and the definition (\ref{m}): 
\[
\frac 1c\int J_jx_k\mathbf{\,}\,d^3x=-\varepsilon _{jkl\,}m_l
\]
which is the same as (\ref{jm}), completing the proof. (The names of tensor
indices can be changed with impunity, and must often be changed when making
substitutions).

\subsection{The expansion of $\mathbf{A}$ and $\mathbf{B}$}

Now we have the tools to compute the vector potential 
\[
A_i(\mathbf{x})=\frac 1c\int \frac{J_i(\mathbf{x}^{\prime })}{\left| \mathbf{%
x-x}^{\prime }\right| }\,d^3x^{\prime }
\]
when $\left| \mathbf{x}\right| >>\left| \mathbf{x}^{\prime }\right| \,$ for
all points $\mathbf{x}^{\prime }$ where $\mathbf{J}$ is non-negligible. We
expand as usual 
\[
\frac 1{\left| \mathbf{x-x}^{\prime }\right| }=\frac 1{\left| \mathbf{x}%
\right| }+\frac{x_jx_j^{\prime }}{\left| \mathbf{x}\right| ^3}+\dots 
\]
and we find 
\[
A_i(\mathbf{x})=\frac 1{c\left| \mathbf{x}\right| }\int J_i(\mathbf{x}%
^{\prime })\,d^3x^{\prime }+\frac{x_j}{c\left| \mathbf{x}\right| ^3}\int J_i(%
\mathbf{x}^{\prime })x_j^{\prime }\,d^3x^{\prime }+\dots 
\]
Using (\ref{i0}) and (\ref{jm}) this gives the desired result 
\[
A_i(\mathbf{x})=-\frac{x_j}{\left| \mathbf{x}\right| ^3}\varepsilon
_{ijk}m_k+\dots 
\]
or 
\[
\mathbf{A}(\mathbf{x})=\frac{\mathbf{m\times x}}{\left| \mathbf{x}\right| ^3}%
+\dots 
\]

The magnetic field is $\mathbf{B=\nabla \times A},$ or $B_i=\varepsilon
_{ijk}\partial _jA_k$ with 
\[
A_k=-\varepsilon _{kpq}\frac{x_p}{\left| \mathbf{x}\right| ^3}%
m_q=\varepsilon _{kpq}\left( \partial _p\frac 1{|\mathbf{x}|}\right) m_q 
\]
Thus, using $\varepsilon _{ijk}\varepsilon _{kpq}=\delta _{ip}\delta
_{jq}-\delta _{iq}\delta _{jp},$%
\[
B_i=\varepsilon _{ijk}\varepsilon _{kpq}\left( \partial _j\partial _p\frac
1{|\mathbf{x}|}\right) m_q=\left( \partial _j\partial _i\frac 1{|\mathbf{x}%
|}\right) m_j-\left( \nabla ^2\frac 1{|\mathbf{x}|}\right) m_i 
\]
The second term on the r.h.s. vanishes, except at $\mathbf{x}=0$ where the
expansion is not to be used anyhow. The first term gives 
\[
B_i=\partial _i\left( m_j\partial _j\frac 1{|\mathbf{x}|}\right)
=-\,\partial _i\left( \frac{x_jm_j}{\left| \mathbf{x}\right| ^3}\right) =%
\frac{3x_ix_jm_j-\left| \mathbf{x}\right| ^2m_i}{\left| \mathbf{x}\right| ^5}
\]
or 
\[
\mathbf{B=\nabla }\left( \mathbf{m\cdot \nabla }\frac 1{|\mathbf{x}|}\right)
=-\mathbf{\nabla }\left( \frac{\mathbf{x\cdot m}}{\left| \mathbf{x}\right| ^3%
}\right) =\frac{3\mathbf{x}\left( \mathbf{x\cdot m}\right) -\left| \mathbf{x}%
\right| ^2\mathbf{m}}{\left| \mathbf{x}\right| ^5} 
\]
in complete analogy to the $\mathbf{E}$ field for an electric dipole $%
\mathbf{p}.$

\subsection{Magnetic moments of a plane loop and an orbit}

For a current $I$ flowing in a loop eq. (\ref{m}) becomes 
\[
\mathbf{m}=\frac I{2c}\oint \mathbf{x\times dl} 
\]
For a plane loop, $\mathbf{m}$ is perpendicular to the plane of the loop and
has magnitude 
\[
m=\frac{IA}c 
\]
where $A$ is the area of the loop (see Jackson Fig. 5.7).

For a stream of particles of charge $q$ and mass $m_q,$ the current density
is 
\[
\mathbf{J}(\mathbf{x})=q\sum_n\mathbf{v}_n\delta (\mathbf{x-x}_n)
\]
where $\mathbf{v}_n=d\mathbf{x}_n/dt$. If the particles are in a closed
orbit, or somehow confined, the magnetic moment is 
\[
\mathbf{m}=\frac q{2c}\sum_n\mathbf{x}_n\mathbf{\times v}_n
\]
On the other hand, the angular momentum is 
\[
\mathbf{L}=m_q\,\sum_n\mathbf{x}_n\mathbf{\times v}_n
\]
so that 
\[
\mathbf{m}=\frac q{2m_qc}\,\mathbf{L}
\]
The ratio $q/2m_qc$ is called the \emph{gyromagnetic ratio. }In a quantum
system, any component of $\mathbf{L}$ is quantized in units of $\hbar .$ For
electrons in atoms then, the unit of magnetic moment is the \emph{Bohr
magneton }$\mu _B=$\emph{\ }$|e|\hbar /2m_ec=0.927\cdot 10^{-20}$ $\mathrm{%
gauss\cdot cm}^{\mathrm{3}}.$

As already discussed at the beginning of Lecture 11, electrons also have a
magnetic moment $\mu _B$ (actually, 1.00116 $\mu _B$) associated with their
spin angular momentum $\frac 12\hbar .$ Thus the gyromagnetic ratio for
intrinsic spin is (essentially) twice the orbital value.

\section{Forces and torques}

\subsection{Force on a dipole}

Consider a current distribution $\mathbf{J}$ localized near the point $%
\mathbf{x.}$ There is the question: how to choose $\mathbf{x}$? Sometimes
there is a natural choice, for instance the center of the circle for a
circular current loop. But for an elliptical current loop either the center
of the ellipse, or one of the foci could be the ``natural'' choice for $%
\mathbf{x.}$ Let us leave the question aside for the moment and write the
current at the point $\mathbf{x}^{\prime }$ in the form $\mathbf{J}(\mathbf{x%
}^{\prime }-\mathbf{x})\,$ to stress that $\mathbf{J}$ vanishes for large
values of $\left| \mathbf{x}^{\prime }-\mathbf{x}\right| .$ According to
Amp\`ere's force law, this current in a field $\mathbf{B}$ is subject to the
force $\mathbf{F}$ of components 
\[
F_i(\mathbf{x})=\frac 1c\,\varepsilon _{ijk}\int J_j(\mathbf{x}^{\prime }-%
\mathbf{x})B_k(\mathbf{x}^{\prime })\,d^3x^{\prime } 
\]
Expanding 
\[
B_k(\mathbf{x}^{\prime })=B_k(\mathbf{x})+\left( x_p^{\prime }-x_p\right)
\partial _pB_k(\mathbf{x})+\dots 
\]
and using from (\ref{i0}) 
\[
\int J_j(\mathbf{x}^{\prime }-\mathbf{x})\,d^3x^{\prime }=\int J_j(\mathbf{x}%
^{\prime \prime })\,d^3x^{\prime \prime }=0 
\]
we obtain 
\[
F_i(\mathbf{x})=\frac 1c\,\varepsilon _{ijk}\left( \partial _pB_k(\mathbf{x}%
)\right) \int J_j(\mathbf{x}^{\prime }-\mathbf{x})\,\left( x_p^{\prime
}-x_p\right) d^3x^{\prime }+\dots 
\]
Further, using (\ref{jm}) in the form 
\[
\frac 1c\int J_j(\mathbf{x}^{\prime \prime })x_p^{\prime \prime
}\,d^3x^{\prime \prime }=-\varepsilon _{jpq}m_q 
\]
and $-\varepsilon _{ijk}\varepsilon _{jpq}=\varepsilon _{jik}\varepsilon
_{jpq}=\delta _{ip}\delta _{kq}-\delta _{iq}\delta _{kp},$ we arrive at 
\[
F_i(\mathbf{x})=\left( \delta _{ip}\delta _{kq}-\delta _{iq}\delta
_{kp}\right) m_q\partial _pB_k(\mathbf{x})=m_k\partial _iB_k(\mathbf{x}%
)-m_i\partial _kB_k(\mathbf{x}) 
\]
or, noting that $\partial _kB_k\equiv \mathbf{\nabla \cdot B}=0,$%
\[
F_i(\mathbf{x})=\partial _i\left( m_kB_k(\mathbf{x})\right) \,\quad \quad
\quad \mathbf{F}(\mathbf{x})=\mathbf{\nabla }\left( \mathbf{m\cdot B}(%
\mathbf{x})\right) 
\]
in complete analogy with the electrostatic case. This formula is exact for
any current distribution if $\mathbf{B}$ varies linearly. In particular,
there is no net force on a current loop in a uniform field, even if the loop
is not small. Returning to the question of the choice of $\mathbf{x,}$ we
note that $\mathbf{m}$ is actually independent of this choice, but $\partial
_iB_k$ is not.

\subsection{Torque on a dipole}

To compute the leading term of the torque $\mathbf{N,}$ it is enough to keep
the leading term in the expansion $\mathbf{B}(\mathbf{x}^{\prime })=$ $%
\mathbf{B}(\mathbf{x})+\cdots $. The torque about $\mathbf{x}$ is 
\[
\mathbf{N}=\frac 1c\int \left( \mathbf{x}^{\prime }-\mathbf{x}\right) \times
\left[ \mathbf{J}(\mathbf{x}^{\prime }-\mathbf{x})\times \mathbf{B}(\mathbf{x%
})\right] \,d^3x^{\prime } 
\]
\[
N_i=\frac 1c\int x_j^{\prime \prime }\left[ B_j(\mathbf{x})J_i(\mathbf{x}%
^{\prime \prime })-J_j(\mathbf{x}^{\prime \prime })B_i(\mathbf{x})\right]
\,d^3x^{\prime \prime } 
\]
Using again (\ref{jm}) and the trace of (\ref{i1}) we obtain 
\[
N_i=-\varepsilon _{ijk}B_j(\mathbf{x)}m_k\quad \quad \quad \mathbf{N=-B}%
\left( \mathbf{x}\right) \times \mathbf{m=m\times B}\left( \mathbf{x}\right) 
\]
This formula is exact for a uniform field, and also for a field that is
uniform over a current loop, but varies perpendicularly to it.

\subsection{Energy of a permanent dipole}

Both the force and the torque can be derived from the potential energy 
\[
U(\mathbf{x},\theta \mathbf{)}=-\mathbf{m\cdot B}\left( \mathbf{x}\right)
=-mB(\mathbf{x})\cos \theta 
\]
as follows: 
\[
\mathbf{F}=-\mathbf{\nabla }U(\mathbf{x})
\]
and, with the angle $\theta $ oriented from $\mathbf{B}$ to $\mathbf{m,}$ 
\[
N=-\partial U/\partial \theta 
\]
The dipole tends to orient itself parallel to the field and to be attracted
to regions of high flux density. In cyclotron motion, the orbit is a spiral
with an effective dipole moment that is antiparallel to the field. Thus the
particles are repelled from regions of high flux density.

\section{Hyperfine interaction in atoms}

A nuclear magnetic moment $\mathbf{\mu }^{(N)}$ interacts with the magnetic
field $e\mathbf{L/}mcr^3$ that is caused by the orbiting motion of the
electrons in a state with angular momentum $\mathbf{L.}$ The interaction
energy is described by 
\[
\mathcal{H}_{HF}^{(orbit)}=-\frac e{mc}\,\mathbf{L\cdot \mu }^{(N)}\int 
\frac{\rho (\mathbf{r})\,}{r^3}d^3r 
\]
This ``orbital'' contribution to the hyperfine interaction vanishes for 
\textsl{s} states ($\mathbf{L=0}$).

An electron also carries a magnetic moment $\mathbf{\mu }_e$ and associated
with the electron density $\rho $ there is a (spin) magnetization density $%
\mathbf{M}=\rho \mathbf{\mu }_e$ and a vector potential 
\[
\mathbf{A}(\mathbf{x})=\int \frac{\mathbf{M}(\mathbf{x}^{\prime })\mathbf{%
\times }\left( \mathbf{x-x}^{\prime }\right) }{\left| \mathbf{x-x}^{\prime
}\right| ^3}\,d^3x^{\prime }=\int \mathbf{M}(\mathbf{x}^{\prime })\times 
\mathbf{\nabla }^{\prime }\frac 1{\left| \mathbf{x-x}^{\prime }\right|
}\,d^3x^{\prime } 
\]
Integrating by parts, this becomes 
\[
\mathbf{A}(\mathbf{x})=\int \frac{\mathbf{\nabla }^{\prime }\times \mathbf{M}%
(\mathbf{x}^{\prime })}{\left| \mathbf{x-x}^{\prime }\right| }\,d^3x^{\prime
} 
\]
showing that there is an effective magnetization current 
\[
\mathbf{J}_M=c\mathbf{\nabla \times M}=c\mathbf{\nabla }\rho \times \mathbf{%
\mu }_e 
\]
which causes a magnetic field 
\[
\mathbf{B}(\mathbf{x})=\int \left( \mathbf{\nabla }^{\prime }\rho (\mathbf{x}%
^{\prime })\times \mathbf{\mu }^{(e)}\right) \times \frac{\mathbf{x-x}%
^{\prime }}{\left| \mathbf{x-x}^{\prime }\right| ^3}\,d^3x^{\prime } 
\]
Using tensor notation and $r=\left| \mathbf{x-x}^{\prime }\right| ^{-1}$%
\[
B_i=\int \varepsilon _{ijk}\,\varepsilon _{jpq}\left( \partial _p^{\,\prime
}\rho \right) \mu _q^{(e)}\,\partial _k^{\,\prime }\frac 1r\,d^3x^{\prime
}=\int \varepsilon _{ikj}\varepsilon _{jpq}\rho \,\mu _q^{(e)}\partial
_k^{\,\prime }\partial _p^{\,\prime }\frac 1r\,d^3x^{\prime } 
\]
where we have integrated by parts and used $\varepsilon _{ikj}=-\varepsilon
_{ijk}.$ Now we use 
\[
\varepsilon _{ikj}\varepsilon _{jpq}=\delta _{ip}\delta _{kq}-\delta
_{iq}\delta _{kp} 
\]
and obtain 
\[
B_i(\mathbf{x})=\int \rho (\mathbf{x}^{\prime })\,\left( \mu
_k^{(e)}\partial _k^{\,\prime }\partial _i^{\,\prime }\frac 1r-\,\mu
_i^{(e)}\nabla ^{\prime \,2}\,\frac 1r\right) d^3x^{\prime } 
\]
The tensor $\partial _k^{\,\prime }\partial _i^{\,\prime }(1/r)$ has a
singularity at $r=0$ in its isotropic part $\frac 13\,\delta _{ki}\nabla
^2\left( 1/r\right) $ (The traceless part averages to zero over a small
sphere and its singular behavior does not affect the integral). Thus the
parenthesis in the integrand is conveniently written 
\[
\mu _k^{(e)}\left( \partial _k^{\,\prime }\partial _i^{\,\prime }-\frac
13\,\delta _{ki}\nabla ^{\prime \,2}\right) \frac 1r-\frac 23\mu
_i^{(e)}\nabla ^{\prime \,2}\,\frac 1r 
\]
and, with no ambiguity at $r=0,$ it becomes 
\[
\mu _k^{(e)}\left( \frac{3r_ir_k}{r^5}-\frac{\delta _{ik}}{r^3}\right) +%
\frac{8\pi }3\mu _i^{(e)}\delta (\mathbf{r}) 
\]
where $\mathbf{r}=\mathbf{x-x}^{\prime }.$ Finally 
\[
B_i(\mathbf{x})=\int \rho (\mathbf{x}^{\prime })\,\left[ \mu _k^{(e)}\left( 
\frac{3r_ir_k}{r^5}-\frac{\delta _{ik}}{r^3}\right) +\frac{8\pi }3\mu
_i^{(e)}\delta (\mathbf{r})\right] \,d^3x^{\prime } 
\]
The hyperfine interaction $\mathcal{H}_{HF}$ in an atom is $-\mu
_i^{(N)}B_i(0),$ where $\mu _i^{(N)}$ is the magnetic moment of the nucleus.
From the above we get 
\begin{eqnarray}
\mathcal{H}_{HF}^{(spin)} &=&\int \rho (\mathbf{r})\,\left( \frac{\mathbf{%
\mu }^{(e)}\!\cdot \mathbf{\mu }^{(N)}}{r^3}-\frac{3\left( \mathbf{r\cdot
\mu }^{(e)}\right) \left( \mathbf{r\cdot \mu }^{(e)}\right) }{r^5}\right)
d^3r  \nonumber \\
&&\ \ -\frac{8\pi }3\,\mathbf{\mu }^{(e)}\!\!\cdot \mathbf{\mu }^{(N)}\rho (%
\mathbf{0})\,\,
\end{eqnarray}
The first line of this equation is the standard external interaction between
two dipoles; the second line must be added when the two dipoles are on top
of each other. It is this ``contact term'' that is responsible for hyperfine
splitting in $l=0$ atomic states, including the splitting of the \textsl{1s}
state of atomic hydrogen that gives rise to radio absorption at 21$\,\mathrm{%
cm}$ in interstellar space.

\section{Magnetic potential of a loop}

We have seen that an infinitesimal loop of area $d\mathcal{A}$ has the
magnetic scalar potential of a dipole $dm=(I/c)d\mathcal{A}.$ A large loop
can be built up from many small loops and has the magnetic potential of a
dipole layer bounded by the loop. The shape of the dipole layer is
arbitrary. For instance, for a circular loop, we can choose the fictitious
dipole layer in the form of any spherical cap bounded by the loop, or the
flat circle inside the loop. We know from electrostatics that $\Phi
_M=I\Omega /c,$ where $\Omega $ is the solid angle subtended by the loop. In
general, to compute $\Omega $ is easier said than done. We turn to an
example. \emph{\ }

\section{Circular current loop}

The magnetic field of a circular loop of radius $a$ carrying a current $I$\/
can be computed in several ways. One can compute the vector potential, which
has only the component $A_\phi $, or one can compute the magnetic potential $%
\Phi _M$, which is just the solid angle subtended by the loop. One can work
directly with elliptic integrals, or use cylindrical coordinates and work
with Bessel functions, or use spherical polar coordinates and spherical
harmonics. Because of cylindrical symmetry, computations can be restricted
to the $xz$ plane ($\phi =0$). The geometry is defined by 
\[
\mathbf{x}=r\,(\sin \theta ,0,\cos \theta )=(\rho ,0,z)\quad \text{field
point}
\]
\[
\mathbf{x}^{\prime }=a\,(\cos \phi ^{\prime },\sin \phi ^{\prime },0)\quad 
\text{source point}
\]
\[
\left| \mathbf{x-x}^{\prime }\right| ^2=r^2+a^2-2ra\sin \theta \cos \phi
^{\prime }
\]
\[
d\mathbf{l}^{\prime }=a\,(-\sin \phi ^{\prime },\cos \phi ^{\prime },0)d\phi
^{\prime }\quad \text{wire element}
\]
Starting formulas are 
\[
\mathbf{A}(\mathbf{x})=\frac Ic\oint \frac{d\mathbf{l}^{\prime }}{\left| 
\mathbf{x-x}^{\prime }\right| }
\]
\begin{equation}
A_\phi (\rho ,z)=\frac Ic\int_0^{2\pi }\frac{a\cos \phi ^{\prime }\,d\phi
^{\prime }}{\sqrt{r^2+a^2-2a\rho \cos \phi ^{\prime }}}  \label{A}
\end{equation}
\begin{equation}
\mathbf{B}(\mathbf{x})=-\,\frac Ic\oint \frac{\left( \mathbf{x-x}^{\prime
}\right) \times d\mathbf{l}^{\prime }}{\left| \mathbf{x-x}^{\prime }\right|
^3}  \label{B}
\end{equation}

\subsection{ Elementary results}

Directly from eq. (\ref{B}) on the $z$ axis ($\rho =0$) $B_\rho =0$ and 
\begin{equation}
B_z=\frac{2\pi I}{ca}\,\frac{a^3}{\left( a^2+z^2\right) ^{3/2}}  \label{Bz}
\end{equation}
From eq. (\ref{A}) in the limit where $r^2+a^2>>2a\rho $%
\begin{equation}
A_\phi (\rho ,z)\simeq \frac Ic\int_0^{2\pi }\frac{a\cos \phi ^{\prime }}{%
\sqrt{r^2+a^2}}\left( 1+\frac{a\rho \cos \phi ^{\prime }}{r^2+a^2}\right)
\,d\phi ^{\prime }=\frac Ic\,\frac{\pi a^2\rho }{\left( r^2+a^2\right) ^{3/2}%
}  \label{Ad}
\end{equation}
\[
B_\rho =-\frac{\partial A_\phi }{\partial z}\simeq \frac{\pi Ia^2}c\frac{%
3\rho z}{\left( r^2+a^2\right) ^{5/2}} 
\]
\[
B_z=\frac 1\rho \frac{\partial \left( \rho A_\phi \right) }{\partial \rho }%
\simeq \frac{\pi Ia^2}c\,\frac{2\left( r^2+a^2\right) -3\rho ^2}{\left(
r^2+a^2\right) ^{5/2}} 
\]
In particular along the $z$ axis ($\rho =0$) we recover (\ref{Bz}).

Far from the loop ($r>>a$) we recover the dipole field: 
\[
A_\phi (r,\theta )\simeq \frac{\pi Ia^2}c\,\frac{\sin \theta }{r^2} 
\]
\[
B_\theta =-\frac 1r\frac{\partial \left( rA_\phi \right) }{\partial r}=\frac{%
\pi Ia^2}c\,\frac{\sin \theta }{r^3} 
\]
\[
B_r=\frac 1{r\sin \theta }\frac{\partial \left( \sin \theta \,A_\phi \right) 
}{\partial \theta }=\frac{\pi Ia^2}c\,\frac{2\cos \theta }{r^3} 
\]

\subsection{By elliptic integrals\ }

We work out this method in detail and leave the method of spherical
harmonics for reading in Jackson.

\subsubsection{The charged ring}

First we review the similar problem of computing the electrostatic potential
of a uniformly charged ring, which is 
\[
\Phi (\rho ,z)=\frac Q\pi \int_0^\pi \frac{d\phi ^{\prime }}{\sqrt{%
r^2+a^2-2a\rho \cos \phi ^{\prime }}} 
\]

The integral is transformed to standard form by the change of variable $\phi
^{\prime }=\pi -2\varphi $, $\cos \phi ^{\prime }=2\sin ^2\varphi -1,$ so
that 
\[
\sqrt{r^2+a^2-2a\rho \cos \phi }=\sqrt{a^2+r^2+2a\rho }\sqrt{1-\frac{4a\rho 
}{a^2+r^2+2a\rho }\sin ^2\varphi } 
\]
The result is 
\[
\Phi =\frac{2Q}{\pi \sqrt{a^2+r^2+2a\rho }}K\left( \frac{4a\rho }{%
a^2+r^2+2a\rho }\right) 
\]
where $K(m)$ is the complete elliptic integral of the first kind 
\begin{equation}
K(m)=\int_0^{\pi /2}\frac{d\varphi }{\sqrt{1-x\sin ^2\varphi }}  \label{K1}
\end{equation}

\subsubsection{Vector potential of the loop}

Now, for the circular current loop we have, recalling eq. (\ref{A}), 
\[
A_\phi (\rho ,z)=\frac{2I}c\int_0^\pi \frac{a\cos \phi ^{\prime }\,d\phi
^{\prime }}{\sqrt{r^2+a^2-2a\rho \cos \phi ^{\prime }}}
\]
Using the same change of variables as above 
\[
A_\phi (\rho ,z)=\frac{2I}c\frac{2a}{\sqrt{r^2+a^2+2a\rho }}\int_0^{\pi /2}%
\frac{2\sin ^2\varphi -1}{\sqrt{1-k^2\sin ^2\varphi }}\,d\varphi 
\]
where we have defined for convenience 
\[
k^2=\frac{4a\rho }{a^2+r^2+2a\rho }=\frac{4a\rho }{\left( a+\rho \right)
^2+z^2}
\]
Writing$\ $%
\[
\sin ^2\varphi =\frac{1-\left( 1-k^2\sin ^2\varphi \right) }{k^2}
\]
the integral becomes 
\[
\int_0^{\pi /2}\frac{2\sin ^2\varphi -1}{\sqrt{1-k^2\sin ^2\varphi }}%
\,d\varphi =\left( \frac 2{k^2}-1\right) K(k^2)-\frac 2{k^2}E(k^2)
\]
where $E(m)$ is the complete elliptic integral of the second kind 
\begin{equation}
E(m)=\int_0^{\pi /2}\sqrt{1-m\sin ^2\varphi }\,d\varphi   \label{K2}
\end{equation}
The final result for the vector potential is 
\begin{equation}
A_\phi (\rho ,z)=\frac{4Ia}{c\sqrt{\left( a+\rho \right) ^2+z^2}}\left[ 
\frac{(2-k^2)K(k^2)-2E(k^2)}{k^2}\right]   \label{Ae}
\end{equation}
The elementary results can be recovered from this.

\subsubsection{Magnetic field}

We now compute 
\[
B_\rho =-\frac{\partial A_\phi }{\partial z}\quad \quad B_z=\frac 1\rho
\,\frac \partial {\partial \rho }\left( \rho A_\phi \right) 
\]
with the help of the formulas (Arfken, page 326) 
\[
\frac{dE}{dm}=\frac{E-K}m\quad \quad \frac{dK}{dm}=\frac E{m(1-m)}-\frac Km 
\]
The result is 
\[
B_\rho =\frac{2Iaz}{c\rho \sqrt{\left( a+\rho \right) ^2+z^2}}\left[ \frac{%
a^2+\rho ^2+z^2}{\left( a-\rho \right) ^2+z^2}E(k^2)-K(k^2)\right] 
\]
\[
B_z=\frac{2Ia}{c\sqrt{\left( a+\rho \right) ^2+z^2}}\left[ \frac{a^2-\rho
^2-z^2}{\left( a-\rho \right) ^2+z^2}E(k^2)+K(k^2)\right] 
\]
The simple results on the $z$ axis, for instance, are recovered using $%
K(0)=E(0)=\pi /2.$

\section{Magnetization}

Because the electrons carry a magnetic moment, there are actually two
contributions to the magnetization, the orbital magnetization 
\[
\mathbf{M}_{orbit}\mathbf{(x})=\frac 1{2c}\,\mathbf{x\times J}(\mathbf{x})
\]
that we have already discussed, and the spin magnetization 
\[
\mathbf{M}_{spin}(\mathbf{x)=}\sum_i\rho _i\mathbf{(x)\mu }_i^{(e)}
\]
where $\mathbf{\mu }_i^{(e)}$ is the spin magnetic moment of the i-th
electron. The integral of $\mathbf{M}_{orbit}$ over the volume of an atom
(or whatever) gives its orbital magnetic moment 
\[
\mathbf{m}_{orbit}=\frac 1{2c}\int \mathbf{x\times J}(\mathbf{x})\,d^3x
\]
in agreement with our earlier definition (\ref{m}) (but we use $\mathbf{m}%
_{orbit}$ to be more precise). It is also true that $\mathbf{M}_{orbit}%
\mathbf{\ }$is the orbital magnetic moment per unit volume, and $\mathbf{M}%
_{spin}$ is the spin magnetic moment per unit volume.

With an atom density $N$ of almost $10^{23}/\mathrm{cm}^{\mathrm{3}},$ which
is typical of a solid, one Bohr magneton per atom corresponds to a
magnetization $M$ of about $1\,\mathrm{kilo{}gauss}$ or to $B\simeq 10$ $%
\mathrm{kgauss}=1\emph{\,}\mathrm{Tesla}$ (we are anticipating $B=4\pi M$).

The spontaneous magnetization of Fe is due to the alignment of the spins,
and the same is generally true for ferromagnetic materials. The actual value
of $M$ for Fe at room temperature is $1.7\,\mathrm{kgauss,}$ in agreement
with the order-of-magnitude arguments of the preceding paragraph. Orbital
magnetism is usually quite weak. We discuss magnetization more in detail in
the next lecture.

\end{document}
