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      <text>13.6.1.6 Engineering example</text>
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      <text><![CDATA[<h3><span>Magnetic flux</span><br></h3>

<p>The magnitude of the magnetic flux density on the axis of a solenoid, as in the Figure below, can be found by the integral: \[B= \int_{\beta_1}^{\beta_2} \frac{\mu_0 n I}{2} \sin\beta \, \mathrm{d}\beta\] where \(\mu_0\) is the permeability of free space
    (\(\approx 4\pi\times 10^{-7} \mbox{Hm}^{-1}\)), \(n\) is the number of turns and \(I\) is the current.</p>
<figure class="HELM_figure">
    <img src="@@PLUGINFILE@@/13.6%20Solenoid.svg" alt="Solenoid and the angles defining its extent" width="387" height="96" class="img-responsive atto_image_button_text-bottom">
  <figcaption>A solenoid and angles defining its extent.</figcaption>
</figure>
<strong>Problem in words</strong>
<p>Predict the magnetic flux in the middle of a long solenoid.</p>
<strong>Mathematical statement of the problem</strong>
<p>We assume that the solenoid is so long that \(\beta_1\approx 0\) and \(\beta_2\approx \pi\) so that \[B = {\displaystyle\int_{\beta_1}^{\beta_2}\dfrac{\mu_0 n I}{2}\sin \beta \, \mathrm{d}\beta}\approx {\displaystyle\int_0^{\pi}\dfrac{\mu_0 n I}{2}\sin\beta
    \, \mathrm{d}\beta}.\]</p>

<strong>Mathematical analysis</strong>
<p>The factor \(\dfrac{\mu_0 n I}{2}\) can be taken outside the integral, i.e. \[\begin{aligned} B &amp;= \dfrac{\mu_0 n I}{2}{\displaystyle\int_0^{\pi}\sin\beta \, \mathrm{d}\beta} \\ &amp;= \dfrac{\mu_0 n I}{2}\Bigg[ -\cos\beta\Bigg]_0^{\pi} \\ &amp;=\dfrac{\mu_0
    nI}{2}(-\cos\pi +\cos 0) \\ &amp;=\dfrac{\mu_0 n I}{2}(-(-1)+1) = \mu_0 nI.\end{aligned}\]</p>

<strong>Interpretation</strong>
<p>The magnitude of the magnetic flux density at the midpoint of the axis of a long solenoid is predicted to be approximately \(\mu_0 nI\), i.e. proportional to the number of turns and proportional to the current flowing in the solenoid.</p>]]></text>
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