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      <text>28.2.3.12 Curl of a vector field</text>
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      <text><![CDATA[<h3>Curl of a vector field</h3>
<p>The curl of the vector field given by \(\underline{F} = F_1 \underline{i} + F_2 \underline{j} + F_3 \underline{k}\) is defined as the vector field</p>
<p>\(\qquad\displaystyle \text{curl} \; \underline{F} = {\underline \nabla} \times \underline{F} = \left | \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ \\ F_1 &amp; F_2 &amp; F_3 \end{array} \right|\) \(\displaystyle \qquad\qquad\quad = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} \right) \underline{i} + \left( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} \right) \underline{j} + \left( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right) \underline{k}\) <br></p>

<p><strong>Physical significance of curl</strong><br></p>
<p>The divergence of a vector field represents the outflow rate from a point; however the curl of a vector field represents the rotation at a point.</p>
<p>Consider the flow of water down a river (in the figure below). The surface velocity \(\underline{v}\) of the water is revealed by watching a light floating object such as a leaf. You will notice two types of motion. First the leaf floats down the river following the streamlines of \(\underline{v}\), but it may also rotate. This rotation may be quite fast near the bank, but slow or zero in midstream. Rotation occurs when the velocity, and hence the drag, is greater on one side of the leaf than the other. </p>

<img src="@@PLUGINFILE@@/3727.svg" alt="Rotation of a leaf in a stream" width="231" height="103" class="img-fluid atto_image_button_text-bottom"><br></p>
 
<p>Note that for a two-dimensional vector field, such as \(\underline{v}\) described here, curl \(\underline{v}\) is perpendicular to the motion, and this is the direction of the axis about which the leaf rotates. The magnitude of curl \(\underline{v}\) is related to the speed of rotation.</p>
<p>For motion in three dimensions a particle will tend to rotate about the axis that points in the direction of curl \(\underline{v}\), with its magnitude measuring the speed of rotation.</p>
<p>If, at any point P, curl \(\underline{v} = {\underline 0}\) then there is no rotation at P and \(\underline{v}\) is said to be <strong>irrotational at P</strong>. If curl \(\underline{v} = {\underline 0}\) at all points of the domain of \(\underline{v}\) then the vector field is an <strong>irrotational vector field</strong>. </p>

<div class="HELM_keypoint">
<h4>Key Point</h4>
<p>Note that \(\underline{F}\) is a vector field and that curl \(\underline{F}\) is also a vector field.</p>
</div>

<hr>
<h4 class="HELM_example">Example</h4>
 <p>Find curl \(\underline{v}\) for the following two-dimensional vector fields</p>
<ol class="HELM_parts_inline">
<li>\(\underline{v} = x \underline{i} + 2 \underline{j}\)</li>
<li>\(\underline{v} = -y \underline{i} + x \underline{j}\)</li> 
</ol>
<p>If \(\underline{v}\) represents the surface velocity of the flow of water, describe the motion of a floating leaf.</p>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
  <li>\[\underline{\nabla} \times \underline{v} = \left | \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ \\ x &amp; 2 &amp; 0 \end{array} \right|\] \[= \left(\dfrac{\partial}{\partial y}(0) -\dfrac{\partial}{\partial z}(2) \right)\underline{i} + \left(\dfrac{\partial}{\partial z}(x) -\dfrac{\partial}{\partial x}(0) \right)\underline{j} + \left(\dfrac{\partial}{\partial x}(2) -\dfrac{\partial}{\partial y}(x) \right)\underline{k} = \underline{0}\]
<p>A floating leaf will travel along the streamlines without rotating.</p></li>
  <li>\[\underline{\nabla} \times \underline{v} = \left | \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ \\ -y &amp; x &amp; 0 \end{array} \right|\] \[= \left(\dfrac{\partial}{\partial y}(0) -\dfrac{\partial}{\partial z}(x) \right)\underline{i} + \left(\dfrac{\partial}{\partial z}(-y) -\dfrac{\partial}{\partial x}(0) \right)\underline{j} + \left(\dfrac{\partial}{\partial x}(x) -\dfrac{\partial}{\partial y}(-y) \right)\underline{k}\] \[= 0\underline{i} + 0\underline{j} +2\underline{k} = 2\underline{k}\]
<p>A floating leaf will travel along the streamlines (anti-clockwise around the origin ) and will rotate
anticlockwise (as seen from above). <br>
An analogy of the right-hand screw rule is that a positive (anti-clockwise) rotation in the xy plane
represents a positive z-component of the curl. Similar results apply for the other components.</p></li>
</ol>

<hr>
<h4 class="HELM_example">Example</h4>
<ol class="HELM_parts">
     <li> Find the curl of \(\underline{u} = x^2 \underline{i} + y^2 \underline{j}\). When is \(\underline{u}\) irrotational?</li>
     <li> Given \(\displaystyle \underline{F} = (xy-xz)\underline{i} + 3x^2 \underline{j} + yz \underline{k}\), find curl \(\underline{F}\) at the origin \((0,0,0)\) and at the point \(P = (1,2,3)\).</li>
 </ol>
<h4 class="HELM_solution">Solution</h4>
<ol class="HELM_parts">
<li>\[ \text{curl }\underline{u} =\underline{\nabla} \times \underline{F} = \left | \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ \\ x^2 &amp; y^2 &amp; 0 \end{array} \right|\] \[= \left(\dfrac{\partial}{\partial y}(0) -\dfrac{\partial}{\partial z}(y^2) \right)\underline{i} + \left(\dfrac{\partial}{\partial z}(x^2) -\dfrac{\partial}{\partial x}(0) \right)\underline{j} + \left(\dfrac{\partial}{\partial x}(y^2) -\dfrac{\partial}{\partial y}(x^2) \right)\underline{k}\]
\[= 0\underline{i} + 0\underline{j} + 0\underline{k} = \underline{0}\]
<p>curl \(\underline{u} = \underline{0}\) so \(\underline{u}\) is irrotational everywhere.</p></li>
  <li>\[\underline{\nabla} \times \underline{v} = \left | \begin{array}{ccc} \underline{i} &amp; \underline{j} &amp; \underline{k} \\ \\ \dfrac{\partial}{\partial x} &amp; \dfrac{\partial}{\partial y} &amp; \dfrac{\partial}{\partial z} \\ \\ xy-xz &amp; 3x^2 &amp; yz \end{array} \right|\] \[= \left(\dfrac{\partial}{\partial y}(yz) -\dfrac{\partial}{\partial z}(3x^2) \right)\underline{i} + \left(\dfrac{\partial}{\partial z}(xy-xz) -\dfrac{\partial}{\partial x}(yz) \right)\underline{j} + \left(\dfrac{\partial}{\partial x}(3x^2) -\dfrac{\partial}{\partial y}(xy-xz) \right)\underline{k}\] \[= z\underline{i} - x\underline{j} +5x\underline{k} = 2\underline{k}\]
<p>At the point \((0, 0, 0)\), curl \(\underline{F} = 0\). At the point \((1, 2, 3)\), curl \(\underline{F} = 3\underline{i} − \underline{j} + 5\underline{k}\).</p></li>
</ol>
<hr>
<h4 class="HELM_example">Engineering Example</h4>
<p><b>Current associated with a magnetic field</b></p>
 <p><strong>Introduction</strong> <br> In a magnetic field \(\underline B\), 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> <br> Given the magnetic field \({\underline B}=B_0x{\underline k}\) find the associated current \({\underline I}\).</p><p><img src="@@PLUGINFILE@@/magfield.svg" alt="Magnetic field profile" width="196" height="169" class="img-fluid atto_image_button_text-bottom"><br></p>
 
 <p><strong>Mathematical statement of problem</strong><br> We need to evaluate the curl of \(\underline B\). <br> <strong>Mathematical analysis</strong><br>\[ { \underline \nabla}\times {\underline B}  =  \left|\begin{matrix} {\underline i} & {\underline j} & {\underline k}\\\\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y}  &\dfrac{\partial}{\partial z}\\\\ 0 & 0 & B_0x\end{matrix}\right|\\\\   =  0{\underline i}-B_0{\underline j}+0{\underline k}\\   =  -B_0{\underline j}\] and so \({\underline I}=-\dfrac{B_0}{\mu_0}{\underline j}\).</p>
<p> <strong>Interpretation</strong> <br> The current is perpendicular to the field and to the direction of variation of the field.</p>]]></text>
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<file name="magfield.svg" path="/" encoding="base64">[encoded data, 11400 characters]</file>
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