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    <name>
      <text>29.2.2.7 Magnetic Flux Engineering Example</text>
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    <questiontext format="html">
      <text><![CDATA[<h4 class="HELM_example">Engineering Example</h4>
<h5>Magnetic Flux</h5>
<p><strong>Introduction</strong></p>
<p>The magnetic flux through a surface is given by \(\displaystyle\ \int \!\!\! \int_S \underline{B} \cdot \underline{dS}\) where \(S\) is the surface under consideration, \(\underline{B}\) is the magnetic field and \(\underline{dS}\) is the vector normal to the surface.</p>
<p><strong>Problem in words</strong></p>
<p>The magnetic field generated by an infinitely long vertical wire on the \(z\)-axis, carrying a current \(I\), is given by: \(\displaystyle \underline{B} = \frac{\mu_0 I}{2 \pi} \left(\frac{- y \underline{i} + x \underline{j}}{x^2 + y^2}\right)\) Find the flux through a rectangular region (with sides parallel to the axes) on the plane \(y=0\).</p>
<p><strong>Mathematical statement of problem</strong></p>
<p>Find the integral \(\displaystyle\int\!\!\int_S{\underline B}\cdot{\underline {dS}}\) over the surface, \(x_1 \leq x \leq x_2\), \(\;\) \(z_1 \leq z \leq z_2\). (see the below figure which shows part of the plane \(y=0\) for which the flux is to be found and a single magnetic field line. The strength of the field is inversely proportional to the distance from the axis.) </p>
<p><img class="img-fluid" src="@@PLUGINFILE@@/infline.svg" alt="The surface S defined by x_1&lt; x &lt; x_2, z_1&lt; z&lt; z_2" width="350" height="220"></p>
<p><strong>Mathematical analysis</strong></p>
<p>On \(y=0\), \(\displaystyle \underline{B} = \frac{\mu_0 I }{2 \pi x} \underline{j} \mbox{ and } \underline{dS}= dx\ dz\ \underline{j}\) so \(\displaystyle \underline{B} \cdot \underline{dS}= \frac{\mu_0 I}{2 \pi x} \; dx \; dz\) <br>The flux is given by the double integral: \[\begin{aligned} \int_{z=z_1}^{z_2} \int_{x=x_1}^{x_2} \frac{\mu_0 I}{2 \pi x} \; dx \; dz &amp;= \frac{\mu_0 I}{2 \pi} \; \int_{z=z_1}^{z_2} \Bigl[ \ln x \Bigr]_{x_1}^{x_2} \; dz \\ &amp;= \frac{\mu_0 I}{2 \pi} \; \int_{z=z_1}^{z_2} \Bigl( \ln x_2 - \ln x_1 \Bigr) \; dz \\ &amp;= \frac{\mu_0 I}{2 \pi} \; \Bigl[ z \bigl( \ln x_2 - \ln x_1 \bigr) \Bigr]_{z=z_1}^{z_2}=\frac{\mu_0 I}{2 \pi} \; (z_2 - z_1) \ln\left( \frac{x_2}{x_1}\right) \end{aligned}\]</p>
<p><strong>Interpretation</strong></p>
<p>The magnetic flux increases in direct proportion to the extent of the side parallel to the axis (i.e. along the \(z\)-direction) but logarithmically with respect to the extent of the side perpendicular to the axis (i.e. along the \(x\)-axis).</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>If \(\underline{F} = x^2 \underline{i} + y^2 \underline{j} + z^2 \underline{k}\), evaluate \(\displaystyle\int\!\!\int_S \underline{F} \times \underline{dS}\) where \(S\) is the part of the plane \(z=0\) bounded by \(x = \pm 1\), \(y = \pm 1\).</p>
<h4 class="HELM_solution">Solution</h4>
<p>Here \(\underline{dS} = dx\, dy\, \underline{k}\) and hence \(\qquad \underline{F} \times \underline{dS} = \left| \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ \\ x^2 &amp; y^2 &amp; z^2 \\ \\ 0 &amp; 0 &amp; dx\, dy \end{array}\right| = y^2\, dx\, dy\,\underline{i} - x^2\, dx\, dy\,\underline{j}\) \(\qquad \displaystyle\int\!\!\int_S \underline{F} \times \underline{dS} = \displaystyle\int_{y=-1}^1 \displaystyle\int_{x=-1}^1 y^2\, dx\, dy\, \underline{i} - \displaystyle\int_{y=-1}^1 \displaystyle\int_{x=-1}^1 x^2 \, dx\, dy\, \underline{j}\) The first integral is \(\qquad\displaystyle \displaystyle\int_{y=-1}^1 \displaystyle\int_{x=-1}^1 y^2\, dx\, dy = \displaystyle\int_{y=-1}^1 \left[\phantom{\dfrac{1}{1}}\!\!\! y^2 x \right]_{x=-1}^1 dy = \displaystyle\int_{y=-1}^1 2 y^2 dy = \left[ \frac{2}{3}y^3 \right]_{-1}^1 = \frac{4}{3}\) Similarly \(\displaystyle\int_{y=-1}^1 \displaystyle\int_{x=-1}^1 x^2\, dx\, dy = \frac{4}{3}\). Thus \( \displaystyle\int\!\!\int_S \underline{F} \times \underline{dS} = \frac{4}{3}\underline{i} - \frac{4}{3}\underline{j}\)</p>
<p></p>
<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>(a) An integral of the form \(\displaystyle\int_S \underline{F}(\underline{r}) \cdot \underline{dS}\) evaluates to a scalar.</p>
<p>(b) An integral of the form \(\displaystyle\int_S \underline{F}(\underline{r}) \times \underline{dS}\) evaluates to a vector.</p>
<p>The vector function involved may be the gradient of a scalar or the curl of a vector.</p>
</div>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Integrate \(\displaystyle\int\!\!\int_S ({\underline \nabla} \phi).\underline{dS}\) where \(\phi = x^2 + 2 y z\) and \(S\) is the area between \(y = 0\) and \(y = x^2\) for \(0 \le x \le 1\) and \(z = 0\). (See below figure)</p>
<p></p>
<p><img class="img-fluid" src="@@PLUGINFILE@@/37_5_4.svg" alt="The area S between y=0 and y=x^2, for 0 &lt; x &lt; 1 and z=0" width="350" height="152"></p>
<p></p>
<h4 class="HELM_solution">Solution</h4>
<p>Here \({\underline \nabla} \phi = 2x \underline{i} + 2z \underline{j} + 2y \underline{k}\) and \(\underline{dS} = \underline{k} \; dy dx\). Thus \(({\underline \nabla} \phi).\underline{dS} = 2y dy dx\) and \begin{eqnarray*} \displaystyle\int\!\!\int_S ({\underline \nabla} \phi).\underline{dS} &amp;=&amp; \displaystyle\int_{x=0}^1 \displaystyle\int_{y=0}^{x^2} 2y \; dy dx \\ &amp;=&amp; \displaystyle\int_{x=0}^1 \left[\phantom{\dfrac{1}{1}}\!\!\! y^2 \right]_{y=0}^{x^2} dx = \displaystyle\int_{x=0}^1 x^4 dx\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \\ &amp;=&amp; \left[\frac{1}{5}x^5\right]_0^1 = \frac{1}{5} \end{eqnarray*}</p>
<p>For integrals of the form \(\displaystyle\int\!\!\int_S {\underline F} .\underline{dS} \), non-Cartesian coordinates e.g. cylindrical polar or spherical polar coordinates may be used. Once again, it is necessary to include any scale factors along with the unit normal</p>
<hr>
<h4 class="HELM_example">Example</h4>
<p>Using cylindrical polar coordinates, find the integral \(\displaystyle \int_S \underline{F}(\underline{r}) \cdot \underline{dS}\) for \(\underline{F} = \rho z \underline{\hat{\rho}} + z \sin^2 \phi \underline{\hat{z}}\) and \(S\) being the complete surface (including ends) of the cylinder \(\rho \le a\), \(0 \le z \le 1 \). (See Figure below.)</p>
   
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/37_5_5-1.svg" alt="A cylinder around the z-axis with r&lt;=a and height between 0 and 1." width="350" height="153">
 <figcaption>The cylinder \( \rho \leq \alpha, \,\, 0 \leq z \leq 1 \).</figcaption>
</figure>

<h4 class="HELM_solution">Solution</h4>
<p>The integral \(\displaystyle\int_S \underline{F}(\underline{r}) \cdot \underline{dS}\) must be evaluated separately for the curved surface and the ends.</p>
<p>For the curved surface, \(\underline{dS} = \underline{\hat{\rho}} a d\phi dz\) (with the \(a\) coming from \(\rho\) the scale factor for \(\phi\) and the fact that \(\rho = a\) on the curved surface.) Thus, \(\underline{F} \cdot \underline{dS} = a^2 z\; d\phi dz\) and</p>
<p>\begin{eqnarray*} \displaystyle \int\!\!\int_S \underline{F}(\underline{r}) \cdot \underline{dS} &amp;=&amp; \displaystyle \int_{z=0}^1 \displaystyle\int_{\phi = 0}^{2\pi} a^2 z\; d\phi dz \\ &amp;=&amp; 2 \pi a^2 \displaystyle \int_{z=0}^1 z\; dz = 2 \pi a^2 \left[\frac{1}{2}z^2\right]_0^1 = \pi a^2\qquad\qquad\qquad\qquad\qquad\qquad \end{eqnarray*}</p>
<p>On the bottom surface, \(z=0\) so \(\underline{F} = \underline{0}\) and the contribution to the integral is zero.</p>
<p>On the top surface, \(z = 1\) and \(\underline{dS} = \underline{\hat{z}} \rho \;d\rho d\phi\) and \(\underline{F} \cdot \underline{dS} = \rho z \sin^2\phi\; d\phi d\rho = \rho \sin^2\phi\; d\phi d\rho\) and</p>
<p>\begin{eqnarray*} \displaystyle \int\!\!\int_S \underline{F}(\underline{r}) \cdot \underline{dS} &amp;=&amp; \displaystyle \int_{\rho = 0}^a \displaystyle\int_{\phi = 0}^{2\pi} \rho \sin^2\phi\; d\phi d\rho\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\ &amp;=&amp; \pi\displaystyle\int_{\rho = 0}^a \rho\; d\rho =\frac{1}{2} \pi a^2 \end{eqnarray*}</p>
<p>So \(\qquad \displaystyle \int\!\!\int_S \underline{F}(\underline{r}) \cdot \underline{dS} = \pi a^2 + \frac{1}{2} \pi a^2 = \frac{3}{2} \pi a^2 \)</p>]]></text>
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