\documentclass[10pt]{article}
\usepackage{graphicx, multicol,wrapfig,exscale,epsfig,fancybox,fullpage}
\pagestyle{empty}

\parindent=0pt
\parskip=.1in

\newcommand\hs{\hspace{6pt}}

\begin{document}

\newcommand{\bmit}[1]{\mathchoice
        {\mbox{\boldmath$\displaystyle#1$}}
        {\mbox{\boldmath$\textstyle#1$}}
        {\mbox{\boldmath$\scriptstyle#1$}}
        {\mbox{\boldmath$\scriptscriptstyle#1$}}
        }
\newcommand{\vf}[1]{\vec{\bmit#1}}
\newcommand{\VF}[1]{\kern-0.5pt\vec{\kern0.5pt\bmit#1}}
\newcommand{\Hat}[1]{\kern1pt\hat{\kern-1pt\bmit#1}}
\newcommand{\HAT}[1]{\kern-0.25pt\hat{\kern0.25pt\bmit#1}}
\newcommand{\Prime}{{}\kern0.5pt'}

\renewcommand{\AA}{\vf A}
\newcommand{\BB}{\vf B}
\newcommand{\EE}{\vf E}
\newcommand{\FF}{\vf F}
\newcommand{\GG}{\vf G}
\newcommand{\HH}{\vf H}
\newcommand{\II}{\vf I}
\newcommand{\JJ}{\kern1pt\vec{\kern-1pt\bmit J}}
\newcommand{\KK}{\vf K}
\newcommand{\gv}{\VF g}
\newcommand{\rr}{\VF r}
\newcommand{\rrp}{\rr\Prime}
\newcommand{\vv}{\VF v}
\newcommand{\grad}{\vec\nabla}
\newcommand{\zero}{\kern-1pt\vec{\kern1pt\bf 0}}

\newcommand{\ii}{\Hat\imath}
\newcommand{\jj}{\Hat\jmath}
\newcommand{\kk}{\Hat k}
\newcommand{\nn}{\Hat n}
\newcommand{\rhat}{\HAT r}
\newcommand{\xhat}{\Hat x}
\newcommand{\yhat}{\Hat y}
\newcommand{\zhat}{\Hat z}
\newcommand{\that}{\Hat\theta}
\newcommand{\phat}{\Hat\phi}

\newcommand{\LargeMath}[1]{\hbox{\large$#1$}}
\newcommand{\INT}{\LargeMath{\int}}
\newcommand{\OINT}{\LargeMath{\oint}}
\newcommand{\LINT}{\mathop{\INT}\limits_C}

\newcommand{\Int}{\int\limits}
\newcommand{\dint}{\mathchoice{\int\!\!\!\int}{\int\!\!\int}{}{}}
\newcommand{\tint}{\int\!\!\!\int\!\!\!\int}
\newcommand{\DInt}[1]{\int\!\!\!\!\int\limits_{#1~~}}
\newcommand{\TInt}[1]{\int\!\!\!\int\limits_{#1}\!\!\!\int}

\newcommand{\Bint}{\TInt{B}}
\newcommand{\Dint}{\DInt{D}}
\newcommand{\Eint}{\TInt{E}}
\newcommand{\Lint}{\int\limits_C}
\newcommand{\Oint}{\oint\limits_C}
\newcommand{\Rint}{\DInt{R}}
\newcommand{\Sint}{\DInt{S}}

\newcommand{\Partial}[2]{{\partial#1\over\partial#2}}

\newcommand{\Item}{\smallskip\item{$\bullet$}}

\renewcommand{\thesubsection}{\arabic{subsection}}

\section*{Activity 4: Solution for magnetic vector potential}
\subsection*{Find the 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}