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.
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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
\(\int\limits_{C_i}\boldsymbol{\vec{B}}\cdot d\boldsymbol{\vec{r}}\).
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\(C_1\), 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.
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\(C_4\), 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 \((5,0)\).
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\(C_6\), 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?