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\section*{Magnetic Vector Potential for Ring}
\section*{Instructor's Guide}

Keywords: Upper-division, E and M, Magnetic Vector Potential,
Symmetry, Ring

\subsection*{Brief overview of the activity}

In this activity, students work in small groups to write the
magnetic potential everywhere in space due to a ring of charge $Q$
and radius $R$ spinning with period $T$.

\subsection*{This activity brings together student understanding of:}
\begin{enumerate}
\item Electrostatic potential

\item Spherical and cylindrical coordinates

\item Superposition
\item Integration as "chopping and adding"
\item Linear charge density
\item 3-dimensional geometric reasoning
\item Power series expansion
\end{enumerate}
\subsection*{Student prerequisite skills}

This activity is may be used as the fourth in a sequence,
following the electric field activity, or may be used on its own.
Students will need understandings of:
\begin{enumerate}
\item The prerequisites addressed in the \texttt{electrostatic
field activity}. \item Spherical and cylindrical coordinates.
\texttt{Link to spherical and cylindrical coordinates activity}.
\item Integration as chopping and adding. \texttt{Link to
Integration activity. \item Linear charge density}
\end{enumerate}
\subsection*{Props}
\begin{itemize}
\item Hula hoop or other thin ring
\item Balls to represent point charges
\item Voltmeter
\item Coordinate system (e.g. with straws or Tinkertoys)
\item Poster-sized whiteboards
\item markers
\item whiteboards around room. \texttt{Link to room set-up}.
\end{itemize}
\section*{The activity - Allow 50 minutes.}

\subsection*{Overview}



\subsection*{What the students will be challenged by and how to facilitate their
learning}


\subsection*{Highlights of the activity}

Working in small groups students are asked to consider a ring with
charge Q, and radius R rotating about its axis with period T and
create an integral expression for the vector potential caused by
this ring everywhere in space. Students also develop the power
series expansion for the potential near the center or far from the
ring.

\subsection*{Reasons to spend class time on this activity:}
\begin{enumerate}

\item Surprisingly, it is non-trivial for students to start with
Q, R, and T and determine the current density. Although the
process does not take long, having students clarify these
relationships is important for building understanding. For
physicists the picture generated in one's mind can be make this
appear so simple that is easy to overlook the important
connections that students need to make regarding current density
before dealing with other aspects of the problem.

\item This activity allows students to build upon prior
understandings and apply them specifically to vector potentials.
Even though students had prior experience with this geometry, the
instructor found herself discussing with several students the
difference between r and r' and which variables could be held
constant during integration.

\end{enumerate}

In this problem students must deal with the concept of current
density and with the concept of a vector potential. Students will
often find it surprising that their intuition about electrostatic
potential is not directly applicable to vector potentials.

Instructors may see it as so simple that they wouldn't consider that
it would require considerable mental effort from students to find
the linear current density from a ring of charge $Q$ and radius $R$
rotating with period $T$. The reality is that the concept of current
density is sufficiently new and unfamiliar that students must spend
time grappling with the concept in this context. The understanding
gained during this problem ``pays off'' when students face future
problems involving linear, surface and volume current densities.

In addition to determining the magnitude of the current, students
will need to consider direction.  This may be the first time
students have had to consider the vector nature of current beyond
simply using the ``right hand rule.'' The understandings gained
here will also be needed for the fifth activity where students are
required to find the magnetic field.

The concept of magnetic vector potential is new to most students.
Since most people try to understand something new by comparing it
with something familiar, students will often try to use intuitions
about electric potential to understand magnetic vector potential.
Because expressions for both potentials contain $1\over {|\rr -
\rrp|}$, students may make the assumption that both potentials
basically the same thing with a different constant in front. This
activity will force students to confront some of the important
differences between a scalar potential and a vector potential.

One example of a situation where students must confront this
difference is with the magnetic vector potential along the axis of
the ring. Whereas the electric potential is positive for all finite
values along this axis, the magnetic vector potential is always
zero. For many students this result will be counterintuitive and
will cause them to think more deeply about the differences between
scalar potentials and vector potentials.



\section*{Solution for magnetic vector potential in all space due to a ring with total charge $Q$ and radius $R$ rotating with a period $T$}
\begin{eqnarray}
\vf A(\rr) = {\mu_0\over 4\pi} \int\limits_{\hbox{ring}}
       {\vf I(\rrp)\,dl' \over|\rr-\rrp|}
\end{eqnarray}
where $\rr$ denotes the position in space at which the magnetic
vector potential is measured and $\rrp$ denotes the position of the
current segment.

For the current
\begin{eqnarray}
\vf I(\rrp)&=& \lambda(\rr)\vf v = {Q\over{2\pi}}{{2\pi R}\over
T}{\hat\phi} = {QR\over T}{\hat\phi}\\\noalign{\smallskip} &=&
{QR\over T} {(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat j})}
\end{eqnarray}
In cylindrical coordinates, $dl' = R\,d\phi'$, and, as discussed in
previous solutions,
\begin{equation}
|\rr-\rrp| = \sqrt{r^2 - 2rR\cos(\phi - \phi') + R^2 + z^2}
\end{equation}
Thus
\begin{eqnarray}
\vf A(\rr) = {\mu_0\over 4\pi}  \int\limits_0^{2\pi}
       {QR\over T}{{{(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat
       j})}{R d\phi'}}\over{\sqrt{r^2 - 2rR\cos(\phi - \phi') + R^2 + z^2}}}
\end{eqnarray}
\begin{eqnarray}
\vf A(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{{(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat
       j})}{d\phi'}}\over{\sqrt{r^2 - 2rR\cos(\phi - \phi') + R^2 + z^2}}}
\end{eqnarray}

\subsection{The $z$ axis}
For points on the $z$ axis, $r = 0$ and the integral simplifies to
\begin{eqnarray}
\vf A(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{{(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat
       j})}{d\phi'}}\over{\sqrt{R^2 + z^2}}}
\end{eqnarray}
Doing the integral results in
\begin{eqnarray}
\vf A(\rr) = 0
\end{eqnarray}
\subsection{The $x$ axis}
For points on the $x$ axis, $z = 0$ and $\phi = 0$, so the integral
simplifies to
\begin{eqnarray}
\vf A(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{{(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat
       j})}{d\phi'}}\over{\sqrt{r^2 - 2rR\cos\phi' + R^2}}}
\end{eqnarray}
This results in a very similar situation as the case for electric
field on the $x$ axis, except that now we will address the $\ii$
component instead of the $\jj$ component. Using the same process we
let $u =x^2 - 2xR\cos\phi' + R^2$, then $du = 2xR\sin\phi'd\phi'$,
and for the $\ii$ component the integral becomes
\begin{eqnarray}
\vf A_x(\rr) = {1\over 4\pi\epsilon_0} {Q\over 2\pi}{{-1}\over {2x}}
\int\limits_0^{2\pi} {{du\jj}\over{u^{1/2}}}
\end{eqnarray}
Doing the integral, we find
\begin{eqnarray}
\vf A_x(\rr) = 0
\end{eqnarray}
Thus the $\ii$ component disappears and we are left with an elliptic
integral with only a $\jj$ component
\begin{eqnarray}
\vf A(\rr) = {1\over 4\pi\epsilon_0} {Q\over 2\pi}
\int\limits_0^{2\pi} {{\cos\phi'\,\jj\,d\phi'}\over{\sqrt{r^2 -
2rR\cos\phi' + R^2}}}
\end{eqnarray}
 \vfill\eject
\end{document}