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      <text>29.2.3.8 Current Continuity Equation Engineering Example</text>
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      <text><![CDATA[<h4 class="HELM_example">Engineering Example</h4>

<h5>The Current Continuity Equation</h5>
<p><strong>Introduction</strong></p> 
<p>When an electric current flows at a constant rate through a conductor, then the <strong>current continuity equation</strong> states that \(\displaystyle \oint_S \underline{J} \cdot \underline{dS} = 0\) where \(\underline{J}\) is the current density (or current flow per unit area) and \(S\) is a closed surface. The equation is an expression of the fact that, under these conditions, the current flow into a closed volume equals the flow out.</p>

<p><strong>Problem in words</strong></p>
<p>A person is standing nearby when lightning strikes the ground. Find the potential difference between the feet of that person.
</p>
<p></p><img src="@@PLUGINFILE@@/lightening.svg" alt="Lightning: a current dissipating into the ground " width="189" height="169" class="img-fluid atto_image_button_text-bottom"><p></p>

<p><strong>Mathematical statement of problem</strong></p>
<p>The current from the lightning dissipates radially (see above figure).</p>
<ol>
  <li> Find a relationship between the current \(I\) and current density \(J\) at a distance \(r\) from the strike by integrating the current density over the hemisphere \(\displaystyle I=\int_S{\underline J}\cdot {\underline {dS}}\)</li>
  <li> Find the field \({\underline E}\) from the equation \(\displaystyle E=\dfrac{\rho I}{2\pi^2 r}\) where \(E=|\underline E|\) and \(I\) is the current.</li>
  <li> Find \(V\) from the integral \(\displaystyle \int_{R_1}^{R_2}{\underline E}\cdot {\underline {dr}}\)</li>
</ol>

<p><strong>Mathematical analysis</strong></p>
<p>Imagine a hemisphere of radius \(r\) level with the surface of the ground so that the point of lightning strike is at its centre. By symmetry, the pattern of current flow from the point of strike will be uniform radial lines, and the magnitude of \(\underline{J}\) will be a constant, i.e. over the curved surface of the hemisphere \(\underline{J} = J \underline{\hat{r}}\). Since the amount of current entering the hemisphere is \(I\), then it follows that the current leaving must be the same i.e.
\[\begin{aligned}
I &amp;= \int_{S_c} \underline{J} \cdot \underline{dS} \mbox{ (where \(S_c\) is the curved surface of the hemisphere)}\qquad\qquad\\
&amp;= \int_{S_c} (J \underline{\hat{r}}) \cdot (dS \; \underline{\hat{r}}) \\
&amp;= J \int_{S_c} dS \\
&amp;= 2 \pi r^2 J [={\text{surface area}}\ (2\pi r^2)\times {\text{flux}}\ (J)]
\end{aligned}\]
since the surface area of a sphere is \(4\pi r^2\). Therefore \(\displaystyle J =\frac{I}{2\pi r^2}\) Note that if the current density \(\underline{J}\) is uniformly radial over the curved surface, then so must be the electric field \(\underline{E}\), i.e. \(\underline{E} = E \underline{\hat{r}}\). Using Ohm’s law \(\displaystyle \underline{J} = \sigma \underline{E}  \mbox{ or }  E = \rho J\) where \(\sigma\) = conductivity = \(1/\rho\). Hence \(\qquad\displaystyle E = \frac{\rho I}{2\pi r^2}\) The potential difference between two points at radii \(R_1\) and \(R_2\) from the lightning strike is found by integrating \(\underline E\) between them, so that
\[\begin{aligned}
V &amp;= \int_{R_1}^{R_2} \underline{E} \cdot \underline{dr} \\
&amp;= \int_{R_1}^{R_2} E \; dr \\
&amp;=  \frac{\rho I}{2\pi} \int_{R_1}^{R_2} \frac{dr}{r^2} \\
&amp;=  \frac{\rho I}{2\pi} \left[ \frac{-1}{r} \right]_{R_1}^{R_2} \\
&amp;= \frac{\rho I}{2\pi} \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \left( = \frac{\rho I}{2\pi} \left( \frac{R_2-R_1}{R_1 R_2}\right) \right)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
\end{aligned}\]
</p>

<p><strong>Interpretation</strong></p>
<p>Suppose the lightning strength is a current \(I=10,000\) A, that the person is 12 m away with feet 0.35 m apart, and that the resistivity of the ground is 80 \(\Omega\) m. Clearly, the worst case (i.e. maximum voltage) would occur when the difference between \(R_1\) and \(R_2\) is greatest, i.e. \(R_1\)=12 m and \(R_2\)=12.35 m which would be the case if both feet were on the same radial line. The voltage produced between the person’s feet under these circumstances is
\[\begin{aligned}
V &amp;= \frac{\rho I}{2 \pi} \left[ \frac{1}{R_1} - \frac{1}{R_2} \right]\\
&amp;= \frac{80 \times 10000}{2 \pi} \left[ \frac{1}{12} - \frac{1}{12.35} \right] \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\\
&amp;\approx 300\ \text{V}
\end{aligned}\]
</p>]]></text>
<file name="lightening.svg" path="/" encoding="base64">[encoded data, 8620 characters]</file>
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