Consider the vector field given by (\(\mu_0\) and \(I\) are constants):
\(\boldsymbol{\vec{B}}
= {\mu_0 I\over2\pi} \left({-y\,\boldsymbol{\hat{x}}+x\,\boldsymbol{\hat{y}}\over x^2+y^2}\right)
= {\mu_0 I\over2\pi} \, {\boldsymbol{\hat{\phi}}\over r}
\)
\(\boldsymbol{\vec{B}}\) is the magnetic field around a wire along the \(z\)-axis carrying a constant current \(I\) in the \(z\)-direction.
Ready:
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Determine \(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\) on any radial line of the form \(y=mx\), where \(m\) is a constant.
-
Determine \(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\) on any circle of the form \(x^2+y^2=a^2\), where \(a\)
is a constant.
You may wish to express the equations for these curves in polar
coordinates.
Go:
For each of the following curves \(C_i\), evaluate the line integral
\(\int\limits_{C_i}\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\).
-
\(C_1\), the top half of the circle \(r=5\), traversed in a
counterclockwise direction.
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\(C_2\), the top half of the circle \(r=2\), traversed in a
counterclockwise direction.
-
\(C_3\), the top half of the circle \(r=2\), traversed in a
clockwise direction.
-
\(C_4\), the bottom half of the circle \(r=2\), traversed in a
clockwise direction.
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\(C_5\), the radial line from \((2,0)\) to \((5,0)\).
-
\(C_6\), the radial line from \((-5,0)\) to \((-2,0)\).
FOOD FOR THOUGHT
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Construct closed curves \(C_7\) and \(C_8\) such that this integral \(\int\limits_{C_i}\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\)
is nonzero over \(C_7\) and zero over \(C_8\).
It is enough to draw your curves; you do not need to
parameterize them.
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Ampère's Law says that, for any closed curve \(C\), this integral is (\(\mu_0\) times) the current flowing through \(C\) (in the \(z\) direction). Can you use this fact to explain your results to part (a)?
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Is \(\boldsymbol{\vec{B}}\) conservative?
Main ideas
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Calculating (vector) line integrals.
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Use what you know!
Prerequisites
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Familiarity with \(d\boldsymbol{\vec{r}}\).
-
Familiarity with “Use what you know” strategy.
Warmup
This activity should be preceded by a short lecture on (vector) line
integrals, which emphasizes that \(\int_C\boldsymbol{\vec{F}}\cdot d\boldsymbol{\vec{r}}\) represents chopping up
the curve into small pieces. Integrals are sums; in this case, one is adding
up the component of \(\boldsymbol{\vec{B}}\) parallel to the curve times the length of each
piece.
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*}
\boldsymbol{\vec{B}}= {\mu_0 I\over2\pi} {\boldsymbol{\hat{\phi}}\over r}
\end{eqnarray*}
so that using \(d\boldsymbol{\vec{r}} = dr\,\boldsymbol{\hat{r}} + r\,d\phi\,\boldsymbol{\hat{\phi}}\) quickly yields \(\boldsymbol{\vec{B}}\cdot
dd\boldsymbol{\vec{r}}\) 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, or consider using MATLAB or similar technology to plot the field. Still, it's important to plot a few vectors by hand.
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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.
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Some students may confuse the wire with the paths of integration.
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Students working in rectangular coordinates often get lost in the algebra of
Question 2b. Make sure that nobody gets stuck here.
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Students who calculate \(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}={dy\over x}\) 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 \(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}=0\) for
radial lines. They should be encouraged to think about the directions of
\(\boldsymbol{\vec{B}}\) and \(d\boldsymbol{\vec{r}}\).
-
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
\(d\boldsymbol{\vec{r}}=-a\sin\phi\,d\phi\,\boldsymbol{\hat{x}}+a\cos\phi\,d\phi\,\boldsymbol{\hat{y}}\), from which one gets
\(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}} = {\mu_0 I\over2\pi}\,d\phi\).
-
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
\(d\boldsymbol{\vec{r}}=dx\,\boldsymbol{\hat{x}}+dy\,\boldsymbol{\hat{y}}\). Taking the dot product then yields,
\(
\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}= {\mu_0 I\over2\pi} {dy\over x}
\).
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 \(\boldsymbol{\vec{B}}\) should convince students that \(\boldsymbol{\vec{B}}\) is tangent
to the circular arcs, hence orthogonal to radial lines. Thus, along such
lines, \(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}=0\); 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
\(d\boldsymbol{\vec{r}} = 2x\,dx\,\boldsymbol{\hat{x}} + 2y\,dy\,\boldsymbol{\hat{y}}\).
-
Working in rectangular coordinates leads to an integral of the form
\(\int-{dx\over y}\), with \(y=\sqrt{r^2-x^2}\). Maple integrates this to
\(-\tan^{-1}\left({x\over y}\right)\), 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.
Homework
-
Any vector line integral for which the path is given geometrically, that is,
without an explicit parameterization.
Essay questions
-
Discuss when \(\oint\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\) around a closed curve will or will not be
zero.
Enrichment
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This activity leads naturally into a discussion of path independence.
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Point out that \(\boldsymbol{\vec{B}}\sim\nabla\phi\) everywhere (except the origin), but that
\(\boldsymbol{\vec{B}}\) is only conservative on domains where \(\phi\) is single-valued.
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Discuss winding number, perhaps pointing out that \(\boldsymbol{\vec{B}}\cdot d\boldsymbol{\hat{r}}\) is
proportional to \(d\phi\) along any curve.
-
Discuss Ampère's Law, which says that \(\oint\!\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\) is
(\(\mu_0\) times) the current flowing through \(C\) (in the \(z\)
direction).