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\section*{Magnetic Field for Ring - Instructor's Guide}

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


\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 magnetic field caused by this
ring everywhere in space.

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

This is designed as the culminating activity for this unit which
allows students to connect much prior learning in a single
problem. Prior to upper division physics courses, students have
little experience in dealing with anything involving the
synthesizing or ``pulling together'' of so many things
simultaneously. Students need to use symmetry and geometric
understanding to be able to construct the integral using the
Biot-Savart Law. For many students, this will be the messiest
integral they have ever had to face (see Eq. 6 in the solutions
for this activity). Successfully unpacking (dealing with) this
integral requires that students first believe that they are
capable of tackling something like this.

The primary new piece to the problem is the vector cross product
in the numerator of the integrand. Although students have done
vector cross products in math classes, students will need to
realize that they can apply vector cross products to this context
and that doing so will help make the problem more manageable.
There is ample opportunity for algebraic errors while taking the
cross product, including sign errors, losing track of $\phi$ vs
$\phi'$, and failure to recognize the trigonometric identity
${\cos\phi}\,{\cos\phi'} + {\sin\phi}\,{\sin\phi'} = \cos(\phi -
\phi')$. However, students should be encouraged to recognize that
they have all the fundamental pieces to understanding this problem
along with the ability to put them together, and that they simply
need to work carefully in order to obtain a correct solution.

\section*{Solution for magnetic field in all space due to a ring with total charge $Q$ and radius $R$ rotating with a period $T$}
\begin{eqnarray}
\vf B(\rr) = {\mu_0\over 4\pi} \int\limits_{\hbox{ring}}
       {\vf I(\rrp) \times (\rr - \rrp) \,dl' \over|\rr-\rrp|^3}
\end{eqnarray}
where $\rr$ denotes the position in space at which the magnetic
field is measured and $\rrp$ denotes the position of the current
segment. As described in previous solutions,
\begin{eqnarray}
dl' &=& R\,d\phi'\\\noalign{\medskip}\vf I(\rrp)&=& {QR\over T}
{(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat
j})}\\\noalign{\medskip} \rr-\rrp &=& (r\cos\phi -
R\cos\phi\Prime)\ii + (r\sin\phi - R\sin\phi\Prime)\jj + (z -
z\Prime)\kk\\\noalign{\bigskip} |\rr-\rrp| &=& \sqrt{r^2 -
2rR\cos(\phi - \phi') + R^2 + z^2}
\end{eqnarray}
Thus $\vf B(\rr) =$
\begin{eqnarray}
{\mu_0\over 4\pi}{QR^2\over T}
\int\limits_0^{2\pi}{{{(-\sin\phi'{\bf \hat i} + \cos\phi'{\bf \hat
       j})}\times{[(r\cos\phi -
R\cos\phi\Prime)\ii + (r\sin\phi - R\sin\phi\Prime)\jj + z\kk]}{
d\phi'}}\over{\sqrt{r^2 - 2rR\cos(\phi - \phi') + R^2 + z^2}}}
\end{eqnarray}
\begin{eqnarray}
\vf B(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{({z\sin\phi'{\bf \ii} + z\cos\phi'{\bf \jj} + {[R + \cos(\phi - \phi')]}{\bf \hat
       k}}){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 $\phi$ can be arbitrarily
taken as zero. Thus, the integral simplifies to
\begin{eqnarray}
\vf B(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{[{z\sin\phi'{\bf \ii} + z\cos\phi'{\bf \jj} + {(R + \cos\phi')}{\bf \hat
       k}}]{d\phi'}}\over{\sqrt{R^2 + z^2}}}
\end{eqnarray}

Doing the integral results in
\begin{eqnarray}
\vf B(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  {{2\pi R}
       \over{\sqrt{R^2 + z^2}}}
\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 B(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{[{z\sin\phi'{\bf \ii} + z\cos\phi'{\bf \jj} + {(R + \cos\phi')}{\bf \hat
       k}}]{d\phi'}}\over{\sqrt{r^2 - 2rR\cos\phi' + R^2}}}
\end{eqnarray}
using the same process as the previous two solutions, the $\ii$ and
the component disappears and the remaining elliptic integral is
\begin{eqnarray}
\vf B(\rr) = {\mu_0\over 4\pi}\,{QR^2\over T}  \int\limits_0^{2\pi}
       {{[{z\cos\phi'{\bf \jj} + {(R + \cos\phi')}{\bf \hat
       k}}]{d\phi'}}\over{\sqrt{r^2 - 2rR\cos\phi' + R^2}}}
\end{eqnarray}
 \vfill\eject
\end{document}