%% -*- latex -*- %/* \input ../macros/Header \if\Book0 \usepackage{exscale} \bibcite{Harvard}{3} \begin{document} \setcounter{section}{6} \setcounter{page}{100} \fi %*/ %*{{page>wiki:headers:hheader}} %* Navigate [[..:..:activities:link|back to the activity]]. %/* \Lab{The Wire} \SecMark \label{theta} %*/ %*==== THE WIRE ==== %/* \subsection{Essentials} %*/ %*=== Essentials === %/* \subsubsection{Main ideas} %*/ %*== Main ideas == %/* \Goal{ %*/ %/* \begin{itemize} \item %*/ %*

Calculating (vector) line integrals. %*
Use what you know! %*

%/* \end{itemize} %*/ %/* } %*/ %/* \subsubsection{Prerequisites} %*/ %*== Prerequisites == %/* \Req{ %*/ %/* \begin{itemize} \item %*/ %*

Familiarity with $d\rr$. %*
Familiarity with %/* ``Use what you know'' %*/ %*"Use what you know" strategy. %*

%/* \end{itemize} %*/ %/* } %*/ %/* \subsubsection{Warmup} %*/ %*== Warmup == This activity should be preceded by a short lecture on (vector) line integrals, which emphasizes that $\INT\FF\cdot d\rr$ represents chopping up the curve into small pieces. Integrals are sums; in this case, one is adding up the component of $\FF$ parallel to the curve times the length of each piece. A good warmup problem is %/* \S18.2:6 in MHG~\cite{Harvard}. %*/ %*§18.2:6 in MHG [3]. %/* \subsubsection{Props} %*/ %*== Props == %/* \begin{itemize} \item %*/ %*

whiteboards and pens %*

%/* \end{itemize} %*/ %/* \subsubsection{Wrapup} %*/ %*== Wrapup == %/* \begin{itemize} \item %*/ %*

Emphasize that students must express everything in terms of a single variable prior to integration. %*
Point out that in polar coordinates (and basis vectors) \begin{eqnarray*} \BB = {\mu_0 I\over2\pi} {\phat\over r} \end{eqnarray*} so that using $d\rr = dr\,\rhat + r\,d\phi\,\phat$ quickly yields $\BB\cdot d\rr$ along a circular arc (${\mu_0 I\over2\pi}\,d\phi$) or a radial line ($0$), respectively. %*

%/* \end{itemize} %*/ %/* \newpage %*/ %/* \subsection{Details} %*/ %*=== Details === %/* \subsubsection{In the Classroom} %*/ %*== In the Classroom == %/* \begin{itemize} \item %*/ %*

Sketching the vector field takes some students a long time. If time is short, have them do this before class. %*
Students who have not had physics don't know which way the current goes; they may need to be told about the right-hand rule. %*
Some students may confuse the wire with the paths of integration. %*
Students working in rectangular coordinates often get lost in the algebra of Question 2b. Make sure that nobody gets stuck here. %*
Students who calculate $\BB\cdot d\rr={dy\over x}$ on a circle need to be reminded that at the end of the day a line integral must be expressed in terms of a single variable. %*
Some students will be surprised when they calculate $\BB\cdot d\rr=0$ for radial lines. They should be encouraged to think about the directions of $\BB$ and $d\rr$. %*
Most students will either write everything in terms of $x$ or $y$ or switch to polar coordinates. We discuss each of these in turn. %/* \begin{itemize} \item %*/ %*
- This problem cries out for polar coordinates. Along a circular arc, $r=a$ yields $x=a\cos\phi$, $y=a\sin\phi$, so that $d\rr=-a\sin\phi\,d\phi\,\ii+a\cos\phi\,d\phi\,\jj$, from which one gets $\BB\cdot d\rr = {\mu_0 I\over2\pi}\,d\phi$. %*
- Students who fail to switch to polar coordinates can take the differential of both sides of the equation $x^2+y^2=a^2$, yielding $x\,dx+y\,dy=0$, which can be solved for $dx$ (or $dy$) and inserted into the fundamental formula $d\rr=dx\,\ii+dy\,\jj$. Taking the dot product then yields, %/* %\begin{eqnarray*} %*/ $ \BB\cdot d\rr = {\mu_0 I\over2\pi} {dy\over x} $. %/* %\end{eqnarray*} %*/ Students may get stuck here, not realizing that they need to write $x$ in terms of $y$. The resulting integral cries out for a trig substitution --- which is really just switching to polar coordinates. %*
%/* \end{itemize} %*/ %*
In either case, sketching $\BB$ should convince students that $\BB$ is tangent to the circular arcs, hence orthogonal to radial lines. Thus, along such lines, $\BB\cdot d\rr=0$; no calculation is necessary. (This calculation is straightforward even in rectangular coordinates.) %*
Watch out for folks who go from $r^2=x^2+y^2$ to $d\rr = 2x\,dx\,\ii + 2y\,dy\,\jj$. %*
Working in rectangular coordinates leads to an integral of the form $\DS\int-{dx\over y}$, with $y=\sqrt{r^2-x^2}$. Maple integrates this to $\DS-\tan^{-1}\left({x\over y}\right)$, which many students will not recognize as the polar angle $\phi$. If $r=1$, Maple instead integrates this to $-\sin^{-1}x$; same problem. One calculator %/* (the TI-89?)\ appears %*/ %*(the TI-89?) appears to use arcsin in both cases. %*

%/* \end{itemize} %*/ %/* \subsubsection{Subsidiary ideas} %*/ %*== Subsidiary ideas == %/* \Sub{ %*/ %/* \begin{itemize} \item %*/ %*

Independence of path. %*

%/* \end{itemize} %*/ %/* } %*/ %/* \subsubsection{Homework} %*/ %*== Homework == %/* \HW{ %*/ %/* \begin{itemize} \item %*/ %*

Any vector line integral for which the path is given geometrically, that is, without an explicit parameterization. %*

%/* \end{itemize} %*/ %/* } %*/ %/* \subsubsection{Essay questions} %*/ %*== Essay questions == %/* \Essay{ %*/ %/* \begin{itemize} \item %*/ %*

Discuss when $\LINT\BB\cdot d\rr$ around a closed curve will or will not be zero. %*

%/* \end{itemize} %*/ %/* } %*/ %/* \subsubsection{Enrichment} %*/ %*== Enrichment == %/* \Rich{ %*/ %/* \begin{itemize} \item %*/ %*

This activity leads naturally into a discussion of path independence. %*
Point out that $\BB\sim\grad\phi$ everywhere (except the origin), but that $\BB$ is only conservative on domains where $\phi$ is single-valued. %*
Discuss %/* \textit{winding number}, %*/ %*//winding number//, perhaps pointing out that $\BB\cdot d\rr$ is proportional to $d\phi$ along %/* \textit{any} curve. %*/ %*//any// curve. %*
Discuss %/* \textit{Amp\`ere's Law}, %*/ %*Ampère's Law, which says that $\LINT\!\BB\cdot d\rr$ is ($\mu_0$ times) the current flowing %/* \textit{through} %*/ %*//through// $C$ (in the $z$ direction). %*

%/* \end{itemize} %*/ %/* } %*/ %/* \input ../macros/footer %*/