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### THE WIRE

#### Essentials

##### Main ideas

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

##### Prerequisites

- Familiarity with $d\rr$.
- Familiarity with “Use what you know” strategy.

##### 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 §18.2:6 in MHG [3].

##### Props

- whiteboards and pens

##### Wrapup

- 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.

#### Details

##### In the Classroom

- 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.
- 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, $ \BB\cdot d\rr = {\mu_0 I\over2\pi} {dy\over x} $. 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.

- 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 to use arcsin in both cases.

##### Subsidiary ideas

- Independence of path.

##### Homework

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

##### Essay questions

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

##### Enrichment

- 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
*winding number*, perhaps pointing out that $\BB\cdot d\rr$ is proportional to $d\phi$ along*any*curve. - Discuss
*Ampère's Law*, which says that $\LINT\!\BB\cdot d\rr$ is ($\mu_0$ times) the current flowing*through*$C$ (in the $z$ direction).