%% LyX 2.1.2.2 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[12pt,american]{article}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\usepackage[letterpaper]{geometry}
\geometry{verbose,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
\setlength{\parskip}{\medskipamount}
\setlength{\parindent}{0pt}
\usepackage{amssymb}
\usepackage{graphicx}

\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
% make units in Roman type with space before
\newcommand{\sunit}[1]{\mbox{ #1}}

\makeatother

\usepackage{babel}
\begin{document}

\section{Angular momentum and magnetic moment}

The Rutherford model of the atom has electrons moving in orbit around
a small, positively charged nucleus. Thus, in general electrons have
both angular momentum and magnetic moments. These magnetic moments
are important for understanding the magnetic properties of matter.
Even though the Rutherford model has been superceded by quantum mechanics,
the classically derived relationship between angular momentum and
magnetic moment remains useful.

\begin{figure}
\noindent \begin{centering}
\includegraphics[scale=0.6]{gyromag}
\par\end{centering}

\protect\caption{\label{fig:gyromag} Charge of mass $M$ and charge $q$ moving in
a circle of radius $R$ at speed $v$.}


\end{figure}
Imagine a particle with mass $M$ and charge $q$ moving in a circle
of radius $R$ with speed $v$. Treating the motion of this particle
classically, with speed much less than the speed of light, the orbital
angular momentum of the particle is $L=MvR$. The time for the particle
to make one revolution is $T=2\pi R/v$, so the circular current represented
by this moving charge is $i=q/T=qv/(2\pi R)$. The magnetic moment
generated by this circular current is the current times the area of
the circle, or $m=iA=i\pi R^{2}$. The ratio of the magnetic moment
to the angular momentum of the particle
\begin{equation}
\gamma=\frac{m}{L}=\frac{q}{2M}\label{eq:gyromagnetic_ratio}
\end{equation}
is called the \emph{gyromagnetic ratio}.

This equation turns out to be generally true, even if particles are
not moving in circular orbits. However, elementary particles have
an intrinsic angular momentum that has nothing to do with orbital
motion. For the electron in particular, the gyromagnetic ratio is
(to within a tenth of a percent)
\begin{equation}
\gamma_{e}=\frac{e}{M_{e}},\label{eq:electron_gr}
\end{equation}
where $e$ is the charge on the electron and $M_{e}$ is its mass,
or twice the gyromagnetic ratio of the electron in orbital motion.
(Technically, the gyromagnetic ratio of particles with negative charge
is negative, though this sign is often ignored.) Since the spin angular
momentum of the electron is $\hbar/2$, the intrinsic magnetic moment
associated with the electron spin is
\begin{equation}
m_{B}=\frac{\hbar\gamma_{e}}{2}=\frac{\hbar e}{2M_{e}}.\label{eq:electron_mm}
\end{equation}
This is sometimes called the \emph{Bohr magneton}, hence the subscripted
$B$. The factor of $2$ missing from equation (\ref{eq:electron_gr})
in comparison to equation (\ref{eq:gyromagnetic_ratio}) results from
relativistic factors, but the derivation is rather complex, so we
will not present it here.

Magnetic dipoles from orbital and spin angular momenta have important
practical consequences; they are central to the phenomena of \emph{paramagnetism}
and \emph{ferromagnetism}. Paramagnetism occurs when atoms have an
odd number of electrons, which means that the angular momentum of
these electrons does not add up to zero (see next chapter). In this
case, there is a net magnetic dipole moment for each atom. If a magnetic
field is imposed, this dipole tends to align with the magnetic field,
which in turn reinforces the imposed field, resulting in a stronger
total field. If the imposed magnetic field is taken away, the atoms
reorient themselves randomly, resulting in zero field.

A few materials, such as iron, exhibit a more complex phenomenon called
ferromagnetism. In this case the imposition of an external magnetic
field causes very strong dipole alignment effects to the point where
interactions between the newly aligned dipoles prevent the reversion
to random orientation when the imposed field is removed. ``Permanent''
magnets may be created this way. As the temperature increases, the
thermal jiggling of atoms increases in proportion. If this jiggling
becomes strong enough, the tendency of neighboring dipoles to maintain
a common orientation is overcome and the material becomes paramagnetic.
The temperature at which this occurs for a particular material is
called the \emph{Curie temperature}.

If there are no unpaired electrons in a material, then there are no
pre-existing dipoles to align. However, the magnetic field still distorts
the motions of electrons via the Lorentz force. The result is the
production of induced magnetic dipoles that anti-align with the field,
as happens with the electrons in a plasma. The resulting induced magnetic
field opposes the original field, resulting in a total field that
is less than originally imposed. Such a material is called \emph{diamagnetic}.

\begin{figure}
\noindent \begin{centering}
\includegraphics[scale=0.5]{Stern-Gerlach_experiment}
\par\end{centering}

\protect\caption{\label{fig:stern_gerlach} Stern-Gerlach experiment. Licensed from
Wikimedia Commons under Creative Commons BY-SA 3.0.}


\end{figure}



\section{Problems}
\begin{enumerate}
\item Consider the hydrogen atom, which has one electron circling the nucleus.

\begin{enumerate}
\item Suppose that the electron moves in a circle of radius $R$ with velocity
$v$, as shown in figure \ref{fig:gyromag}. (This can only occur
in an excited state of the atom.) Use Amp\'{e}re's law to estimate
the magnetic field produced at the nucleus. Hint: Derive an equation
for the average magnetic field over a magnetic circulation loop of
radius $R$ that circles the trajectory of the electron. Assume that
the field at the nucleus is somewhat greater than this average value.
(Pick a number.)
\item The radius of the hydrogen atom in an excited state is of order $10^{-10}\,\mathrm{m}$.
Compute the orbital velocity of the electron for a circular orbit,
given the electron mass and the charges on the electron and the proton
nucleus.
\item From the above information compute an approximate value for the magnetic
field at the nucleus caused by the electron's orbital motion.
\end{enumerate}
\item The Stern-Gerlach experiment demonstrated the quantization of angular
momentum by passing a beam of neutral silver atoms (approximately
108 times mass of proton) through a magnetic field with a strong gradient.
(See figure \ref{fig:stern_gerlach}.) Since silver has one unpaired
electron, the magnetic dipole moment is non-zero. Assuming a vertical
gradient in the vertical component of the magnetic field, the silver
atoms will be deflected either up or down, depending on the orientation
of the dipole moment.

\begin{enumerate}
\item For a given value of $\partial B_{z}/\partial z$, derive an expression
for the $z$ component of the force on a silver atom, assuming that
the dipole moment equals one Bohr magneton aligned (1) in the $+z$
and (2) in the $-z$ directions.
\item If the silver atom has mass $M$ and initial speed $v$, compute the
angular deflection up or down of the trajectory of the silver atom
as it passes through the field of the magnet, assumed to have length
$L$.
\item Assuming $\partial B_{z}/\partial z=10^{2}\,\mathrm{T/m}$, $v=300\,\mathrm{m/s}$,
and $L=10\,\mathrm{cm}$, compute the above deflection angle.
\item Explain why the classical result is different from the quantum result
as illustrated in figure \ref{fig:stern_gerlach}.\end{enumerate}
\end{enumerate}

\end{document}
