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  <question type="description">
    <name>
      <text>28.3.3.3 Cylindrical polar coordinates</text>
    </name>
    <questiontext format="html">
      <text><![CDATA[<h3>Cylindrical polar coordinates</h3>
<p>In cylindrical polar coordinates \((\rho,\phi,z)\), the three unit vectors are \(\underline{\hat{\rho}}\), \(\underline{\hat{\phi}}\) and \(\underline{\hat{z}}\) (see Figure 20(b) on page 38) with scale factors <br>
\(\qquad\displaystyle h_{\rho}=1\), \(h_{\phi} = \rho\), \(h_z=1\).<br>
The quantities \(\rho\) and \(\phi\) are related to \(x\) and \(y\) by \(x=\rho \cos \phi\) and \(y=\rho \sin \phi\). The unit vectors are \(\underline{\hat{\rho}}=\cos \phi \underline{i}+\sin \phi \underline{j}\) and \(\underline{\hat{\phi}}=-\sin \phi\underline{i}+\cos \phi\underline{j}\). In cylindrical polar coordinates,<br>
grad \(\displaystyle f = {\underline \nabla} f = \frac{\partial f}{\partial \rho} \underline{\hat{\rho}} + \frac{1}{\rho} \frac{\partial f}{\partial \phi} \underline{\hat{\phi}} + \frac{\partial f}{\partial z} \underline{\hat{z}}\)<br>
The scale factor \(\rho\) is necessary in the \(\phi\)-component because the derivatives with respect to \(\phi\) are distorted by the distance from the axis \(\rho = 0\).

If \(\quad\underline{F} = F_{\rho} \underline{\hat{\rho}} + F_{\phi} \underline{\hat{\phi}} + F_z \underline{\hat{z}}\quad\) then<br>
div \(\displaystyle \underline{F} = {\underline \nabla} \cdot \underline{F} = \frac{1}{\rho} \left[ \frac{\partial}{\partial \rho}(\rho F_{\rho}) + \frac{\partial}{\partial \phi}(F_{\phi}) + \frac{\partial}{\partial z}(\rho F_z) \right]\)<br>
curl \(\displaystyle \underline{F} = {\underline \nabla} \times \displaystyle \underline{F} = \frac{1}{\rho} \left|\begin{array}{ccc} \underline{\hat{\rho}} &amp; \rho \underline{\hat{\phi}} &amp; \underline{\hat{z}} \\ \\ \displaystyle \frac{\partial}{\partial \rho} &amp; \displaystyle \frac{\partial}{\partial \phi} &amp; \displaystyle \frac{\partial}{\partial z} \\ \\ F_{\rho} &amp; \rho F_{\phi} &amp; F_z \end{array} \right|\).</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Working in cylindrical polar coordinates, find \({\underline \nabla} f\) for \(f = \rho^2 + z^2\)</p>
<h4 class="HELM_solution">Solution</h4>
<p>If \(f= \rho^2 + z^2\) then
\(\displaystyle \frac{\partial f}{\partial \rho} = 2 \rho\),\(\displaystyle \frac{\partial f}{\partial \phi} = 0\) and\(\displaystyle \frac{\partial f}{\partial z} = 2 z\) so \({\underline \nabla} f =
2 \rho \underline{\hat{\rho}} + 2 z \underline{\hat{z}}\).</p>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Working in cylindrical polar coordinates find</p>
    <ol class="HELM_parts">
      <li>\({\underline \nabla}f\) for \(f=\rho^3\sin\phi\)</li>
      <li> Show that the result for (a) is consistent with that found working in Cartesian coordinates. </li>
    </ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>If \(f = \rho^3 \sin \phi\) then \(\displaystyle \frac{\partial f}{\partial \rho} = 3 \rho^2 \sin \phi\),
\(\displaystyle \frac{\partial f}{\partial \phi} = \rho^3 \cos
\phi\) and \(\displaystyle \frac{\partial f}{\partial z} = 0\) and
hence, \({\underline \nabla} f = 3 \rho^2 \sin \phi \underline{\hat{\rho}} +
\rho^2 \cos \phi \underline{\hat{\phi}}\).</li>

<li>\(f = \rho^3\sin\phi =\rho^2 \rho \sin \phi = (x^2 + y^2)y = x^2 y + y^3\) so
\({\underline \nabla} f = 2 x y \underline{i} + (x^2 + 3 y^2) \underline{j}\).<br>
Using cylindrical polar coordinates, from (a) we have
\begin{eqnarray*} {\underline \nabla} f
&amp;=&amp; 3 \rho^2 \sin \phi \underline{\hat{\rho}} + \rho^3 \cos \phi
\underline{\hat{\phi}} \\ &amp;=&amp; 3 \rho^2 \sin \phi (\cos \phi
\underline{i} + \sin \phi \underline{j}) + \rho^2 \cos \phi (-
\sin \phi \underline{i} + \cos \phi \underline{j}) \\
&amp;=&amp; \left[3 \rho^2 \sin \phi \cos \phi - \rho^2 \sin \phi \cos
\phi \right]\underline{i} + \left[3 \rho^2 \sin^2 \phi  + \rho^2
\cos^2 \phi \right]\underline{j} \\
&amp;=&amp; \left[2 \rho^2 \sin \phi \cos \phi  \right]\underline{i} +
\left[3 \rho^2 \sin^2 \phi  + \rho^2 \cos^2 \phi
\right]\underline{j}= 2 x y \underline{i} + (3y^2 + x^2) \underline{j}\qquad\qquad
\end{eqnarray*} So the results using Cartesian and cylindrical polar coordinates are consistent. </li>

</ol>

<hr>
<h4 class="HELM_example">Example</h4>
<p>Find \({\underline \nabla} \cdot \underline{F}\) for \(\underline{F} = F_{\rho} \underline{\hat{\rho}} + F_{\phi} \underline{\hat{\phi}} + F_z \underline{\hat{z}} = \rho^3 \underline{\hat{\rho}} + \rho z \underline{\hat{\phi}} + \rho z \sin \phi \underline{\hat{z}}\). Show that the results are consistent with those found using Cartesian coordinates.</p>
<h4 class="HELM_solution">Solution</h4>
<p>Here, \(F_{\rho} = \rho^3\), \(F_{\phi} = \rho z\) and \(F_z= \rho z \sin \phi\) so
\begin{eqnarray*}{\underline \nabla} \cdot \underline{F} &amp;=&amp; \frac{1}{\rho} \left[
\frac{\partial}{\partial \rho}(\rho F_{\rho}) +\frac{\partial}{\partial \phi}(F_{\phi}) +
\frac{\partial}{\partial z}(\rho F_z) \right] \\&amp;=&amp; \frac{1}{\rho} \left[ \frac{\partial}{\partial \rho}(\rho^4) +
\frac{\partial}{\partial \phi}(\rho z) +\frac{\partial}{\partial z}(\rho^2 z \sin \phi) \right] \\
&amp;=&amp; \frac{1}{\rho} \left[ 4 \rho^3 + 0 + \rho^2 \sin \phi \right]\\ &amp;=&amp; 4 \rho^2 + \rho \sin \phi\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\end{eqnarray*}
Converting to Cartesian coordinates,
\begin{eqnarray*} \underline{F} &amp;=&amp; F_{\rho} \underline{\hat{\rho}} +
F_{\phi} \underline{\hat{\phi}} + F_z \underline{\hat{z}} = \rho^3
\underline{\hat{\rho}} + \rho z \underline{\hat{\phi}} + \rho z
\sin \phi \underline{\hat{z}} \\
&amp;=&amp; \rho^3 (\cos \phi \underline{i} + \sin \phi \underline{j}) +
\rho z (-\sin \phi \underline{i} + \cos \phi \underline{j}) + \rho
z \sin \phi \underline{k} \\
&amp;=&amp; (\rho^3 \cos \phi - \rho z \sin \phi) \underline{i} + (\rho^3
\sin \phi + \rho z \cos \phi) + \rho z \underline{k} \\
&amp;=&amp; \left[\rho^2  (\rho \cos \phi) - \rho \sin \phi z \right]
\underline{i} + \left[ \rho^2 (\rho \sin \phi) + \rho \cos \phi 
z\right]\underline{j} + \rho \sin \phi z \underline{k} \\
&amp;=&amp; \left[(x^2 + y^2) x  - y z \right] \underline{i} + \left[ (x^2
+ y^2) y + x z \right]\underline{j} + y z \underline{k} \\
&amp;=&amp; (x^3 + x y^2  - y z ) \underline{i} + ( x^2 y + y^3 + x z )\underline{j} +
y z \underline{k}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\end{eqnarray*}
So \begin{eqnarray*} {\underline \nabla} \cdot \underline{F} &amp;=&amp;
\frac{\partial}{\partial x}(x^3 + x y^2  - y z ) +
\frac{\partial}{\partial y}(x^2 y + y^3 + x z ) +
\frac{\partial}{\partial z}(y z ) \\
&amp;=&amp; (3x^2 + y^2) + (x^2 + 3 y^2) + y = 4 x^2 + 4 y^2 + y\\
&amp;=&amp;4(x^2+y^2)+y \\
&amp;=&amp; 4 \rho^2 + \rho \sin \phi\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\end{eqnarray*}
So \({\underline \nabla}
\cdot \underline{F}\) is the same in both
coordinate systems.</p><p>

</p><hr>
<h4 class="HELM_example">Example</h4>
<p>Find \({\underline \nabla} \times \underline{F}\) for \(\underline{F} = \rho^2 \underline{\hat{\rho}} + z \sin \phi \underline{\hat{\phi}} + 2 z \cos \phi \underline{\hat{z}}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>\begin{eqnarray*}
{\underline \nabla} \times \displaystyle \underline{F} &amp;=&amp; \frac{1}{\rho}
\left|\begin{array}{ccc}
   \underline{\hat{\rho}} &amp; \rho \underline{\hat{\phi}} &amp;  \underline{\hat{z}}
   \\ \\
   \displaystyle \frac{\partial}{\partial \rho} &amp; \displaystyle \frac{\partial}{\partial \phi} &amp;
   \displaystyle \frac{\partial}{\partial z} \\ \\
  F_{\rho} &amp; \rho F_{\phi} &amp; F_z
\end{array} \right|
= \frac{1}{\rho}\left|\begin{array}{ccc}
   \underline{\hat{\rho}} &amp; \rho \underline{\hat{\phi}} &amp;  \underline{\hat{z}}
   \\ \\
   \displaystyle \frac{\partial}{\partial \rho} &amp; \displaystyle \frac{\partial}{\partial \phi} &amp;
   \displaystyle \frac{\partial}{\partial z} \\ \\
  \rho^2 &amp; \rho z \sin \phi &amp; 2 z \cos \phi
\end{array} \right| \\
&amp;=&amp; \!\frac{1}{\rho} \left[ \underline{\hat{\rho}} \left[
\frac{\partial}{\partial \phi}2 z \cos \phi -
\frac{\partial}{\partial z}\rho z \sin \phi \right]\! +\! \rho
\underline{\hat{\phi}} \left[ \frac{\partial}{\partial z}\rho^2 -
\frac{\partial}{\partial \rho }2 z \cos \phi \right]\! +\!
\underline{\hat{z}} \left[ \frac{\partial}{\partial \rho}\rho z
\sin \phi
- \frac{\partial}{\partial \phi}\rho^2 \right] \right] \\
&amp;=&amp; \frac{1}{\rho} \left[ \underline{\hat{\rho}} (-2 z \sin \phi -
\rho \sin \phi ) + \rho \underline{\hat{\phi}} (0) +
\underline{\hat{z}} (z \sin \phi) \right] \\
&amp;=&amp; -\frac{(2 z \sin \phi)}{\rho} \underline{\hat{\rho}}  +
\frac{z \sin \phi}{\rho} \underline{\hat{z}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\end{eqnarray*}</p>

<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><b>Divergence of a magnetic field</b></p>
<p><strong>Introduction</strong> A magnetic field \({\underline B}\) must satisfy \({{\underline \nabla}}\cdot {\underline B}=0\). An associated current is given by: \(\qquad\displaystyle {\underline I}=\dfrac{1}{\mu_0}({ {\underline \nabla}}\times{\underline B})\)</p>
<p><strong>Problem in words</strong> For the magnetic field (in cylindrical polar coordinates \(\rho,\phi, z)\) \(\displaystyle {\underline B}=B_0\dfrac{\rho}{1+\rho^2}{\underline {\hat\phi}}+\alpha {\underline {\hat z}}\) show that the divergence of \(\underline B\) is zero and find the associated current. </p>
<p><strong>Mathematical statement of problem</strong> We must (a) show that \({ {\underline \nabla}}\cdot{\underline B}=0.\) (b) find the current \({\underline I}=\dfrac{1}{\mu_0}({ {\underline \nabla}}\times{\underline B})\)
</p>
<p>
  <strong>Mathematical analysis</strong>
</p>
    <ol>
      <li> Express \({\underline B}\) as \((B_{\rho},B_{\phi}, B_z)\); then 
\[\begin{aligned}
{{\underline \nabla}}\cdot {\underline B}&amp;=\dfrac{1}{{\rho}}\left[\dfrac{\partial}{\partial {\rho}}({\rho}B_{\rho}) +\dfrac{\partial }{\partial \phi}(B_{\phi}) +\dfrac{\partial }{\partial z}({\rho}B_z)\right]\\ 
&amp;=\dfrac{1}{{\rho}}\left[\dfrac{\partial }{\partial {\rho}}(0)+\dfrac{\partial} {\partial\phi}\left(B_0\dfrac{{\rho}}{1+{\rho}^2}\right)+{\rho}\dfrac{\partial}{\partial z}(\alpha)\right]\\
&amp;=\dfrac{1}{{\rho}}[0+0+0]=0\qquad\text{ as required.}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\end{aligned}\]</li>
      <li> To find the current evaluate \[\begin{aligned} {\underline I}=\dfrac{1}{\mu_0}({ {\underline \nabla}}\times{\underline B})&amp;=\dfrac{1}{\mu_0}\dfrac{1}{{\rho}}\left| \begin{matrix}{\underline {\hat {\rho}}}&amp;{\rho}{\underline {\hat\phi}}&amp;{\underline {\hat z}}\\\\ \dfrac{\partial}{\partial {\rho}}&amp;\dfrac{\partial}{\partial\phi}&amp;\dfrac{\partial}{\partial z}\\\\ B_{\rho}&amp;{\rho}B_{\phi}&amp;B_z \end{matrix}\right|=\left| \begin{matrix}{\underline {\hat {\rho}}}&amp;{\rho}{\underline {\hat\phi}}&amp;{\underline {\hat z}}\\\\ \dfrac{\partial}{\partial {\rho}}&amp;\dfrac{\partial}{\partial\phi}&amp;\dfrac{\partial}{\partial z}\\\\ 0&amp;B_0\dfrac{{\rho}^2}{1+{\rho}^2}&amp;\alpha \end{matrix}\right|\\ &amp;=\dfrac{1}{\mu_0{\rho}}\left[0{\underline {\hat {\rho}}}+0{\rho}{\underline {\hat\phi}}+ B_0\dfrac{\partial}{\partial {\rho}}\left(\dfrac{{\rho}^2}{1+{\rho}^2}\right){\underline {\hat z}}\right]\\\\ &amp;=\dfrac{1}{\mu_0{\rho}}B_0\left[\dfrac{2{\rho}}{(1+{\rho}^2)^2}\right]{\underline {\hat z}}=\dfrac{2B_0}{\mu_0{(1+{\rho}^2)^2}}\;{\underline {\hat z}}\end{aligned}\]</li>
    </ol>
<p>
  <strong>Interpretation</strong> The magnetic field is in the form of a helix with the current pointing along its axis. Such an arrangement is often used for the magnetic containment of charged particles in a fusion reactor.</p><p>&nbsp;<img src="@@PLUGINFILE@@/helix.svg" alt="The magnetic field forms a helix" width="420" height="149" class="img-fluid atto_image_button_text-bottom">
</p>


<hr>
<h4 class="HELM_example">Example</h4>
<p>A magnetic field \(\underline{B}\) is given by \(\underline{B} = \rho^{-2} \underline{\hat{\phi}} + k \underline{{\hat z}}\). Find \({\underline \nabla} \cdot \underline{B}\) and \({\underline \nabla} \times \underline{B}\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>\(\displaystyle {\underline \nabla} \cdot \underline{B} =
\frac{1}{\rho}\left[\frac{\partial}{\partial \rho}0 +
\frac{\partial}{\partial \phi}\rho^{-2} + \frac{\partial}{\partial
z}k\rho \right] = \frac{1}{\rho} \left[ 0
+ 0 + 0 \right] = 0 \) 
\(\displaystyle\qquad 
{\underline \nabla} \times \displaystyle \underline{B} = \frac{1}{\rho}
\left|\begin{array}{ccc}
   \underline{\hat{\rho}} &amp; \rho \underline{\hat{\phi}} &amp;  \underline{\hat{z}}
   \\ \\
   \displaystyle \frac{\partial}{\partial \rho} &amp; \displaystyle \frac{\partial}{\partial \phi} &amp;
   \displaystyle \frac{\partial}{\partial z} \\ \\
  B_{\rho} &amp; B_{\phi} &amp; B_z
\end{array} \right|
= \frac{1}{\rho}\left|\begin{array}{ccc}
   \underline{\hat{\rho}} &amp; \rho \underline{\hat{\phi}} &amp;  \underline{\hat{z}}
   \\ \\
   \displaystyle \frac{\partial}{\partial \rho} &amp; \displaystyle \frac{\partial}{\partial \phi} &amp;
   \displaystyle \frac{\partial}{\partial z} \\ \\
  \rho^{-2} &amp; 0 &amp; k
\end{array} \right|\)
\(\displaystyle = \underline{0}\) <br>
All magnetic fields satisfy \({{\underline \nabla}}\cdot{\underline B}=0\) i.e. an absence of magnetic monopoles. 
There is a class of magnetic fields known as potential fields that satisfy 
\({{\underline \nabla}}\times{\underline B}={\underline 0}\)</p>]]></text>
<file name="helix.svg" path="/" encoding="base64">[encoded data, 13816 characters]</file>
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