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<p>This week, we find an easier way to calculate magnetic fields in certain situations. Just like Gauss's Law for electric fields, Ampere's Law for magnetic fields is true in all cases but can only be used to find an easy analytical solution for certain geometries.</p>
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<h2 class="hd hd-2 unit-title">L21v1: Ampere's Law</h2>
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The integral expression </p>
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[mathjaxinline]\displaystyle \oint _{\text {closed path}}\vec{B}\cdot d\vec{s}[/mathjaxinline]
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<text>is equal to the magnetic work done around a closed path.</text>
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<text>is an infinite sum of the product of the tangent component of the magnetic field along a small element of the closed path with the length of that small element up to a choice of plus or minus sign.</text>
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<text>is equal to the magnetic potential energy between two points.</text>
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Consider the loop shown in the figure below. We choose the unit normal to the surface enclosed by the loop [mathjaxinline]\hat{n}[/mathjaxinline] to be directed into the screen. The circles are the cross sectional area of [mathjaxinline]8[/mathjaxinline] parallel wires. Each wire carries a current of magnitude [mathjaxinline]I[/mathjaxinline] pointing either into (for blue circles with X's) or out of (for red circles with dots) the screen. Integrating [mathjaxinline]B[/mathjaxinline] around the loop yields </p>
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<h2 class="hd hd-2 unit-title">Ampere's Law - Sign Convention and the Right Hand Rule </h2>
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<p><b>Ampere's Law. Terminology, sign convention and the Right Hand Rule</b></p><p><b> Amperian loop </b></p>
Ampere's law relates a line integral of the B-field and current:
\[ \oint_{\text{closed path}} \vec{B}\cdot d\vec{s} = \mu_o I_{enc} \]
<p> where the line integral is over a closed path. The closed path is often called the <i>Amperian loop</i> or simply <i>loop</i>.</p>
<p><b> Current enclosed.</b></p>
<p> As an abbreviation, we will use the term "current enclosed" to refer to the <i>current passing through the surface enclosed by the loop</i>, the term \(I_{enc}\) in the right hand side of Ampere's law.</p>
<p><b> Current enclosed and the flux of the current density.</b></p>
<p> The current \(I_{enc}\) is the flux over the surface enclosed by the loop of the current density vector \(\vec{J}\). Consider a region in space with a current density vector \(\vec{J}\) as shown in the figure below. </p>
<center><img src="/assets/courseware/v1/7af21b96bea220f2b683b073570fb283/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_html_lesson21_01.svg" width="400"/></center>
<p>If we pick the closed loop indicated in the figure, the current enclosed is given by: </p>
<p>
\[I_{enc} = \iint_{S} \vec{J}\cdot d\vec{A} \]
</p>
<p> where \(S\) is the surface enclosed by the loop</p>
<p><b> The Right Hand Rule and the sign of the current.</b></p>
<p>To calculate the flux of the current density vector we need to define the element of surface \(d\vec{A}= dA\hat{n}\), where \(\hat{n}\) is the unit vector perpendicular to the surface. The direction of the unit vector is related to the direction of circulation in the loop and is defined by the right hand rule. Below, the loop is contained in the plane of the screen. The two possible choices of direction of circulation along the loop, counter-clockwise or clockwise, are shown.</p>
<center><img src="/assets/courseware/v1/a4c09820f2be6e9334ff405376e624e9/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_html_lesson21_01b.svg" width="800"/></center>
<p> If we pick the direction of circulation to be <b>counter-clockwise</b>, left figure above, the right hand rule states that the direction of the unit normal \(\hat{n}\) is <b>out of</b> the screen, therefore \(d\vec{A}\) is out of the screen. As a result, \(\vec{J}\cdot d\vec{A}\) is positive or negative depending on the direction of \(\vec{J}\), out of or into the screen.</p>
<p> If we pick the direction of circulation to be <b>clockwise</b>, right figure above, the right hand rule states that the direction of the unit normal \(\hat{n}\) is <b>into</b> the screen, therefore \(d\vec{A}\) is into the screen. As a result, \(\vec{J}\cdot d\vec{A}\) is positive or negative depending on the direction of \(\vec{J}\) into or out of the screen.</p>
<p><b> Review of the Right hand rule.</b></p>
<p> The right hand rule used to determine the direction of the unit normal given the direction of circulation is as follows: <i>"Curl the fingers of your right hand in the direction of the circulation, your thumb points in the direction of \(\hat{n}\)."</i></p>
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<h2 class="hd hd-2 unit-title">L21Q3: A Cylinder and a Cylindrical Shell</h2>
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A cylinder and a cylindrical shell
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<p>
Consider two conductors. An inner cylinder of radius [mathjaxinline]a[/mathjaxinline] and a concentric cylinderical shell of inner radius [mathjaxinline]3a[/mathjaxinline] and outer radius [mathjaxinline]b[/mathjaxinline]. Vectors [mathjaxinline]\vec{J}_1[/mathjaxinline] and [mathjaxinline]\vec{J}_2[/mathjaxinline] are the current density vectors in the inner cylinder and the outer shell, respectively. They are uniform, have the same magnitude [mathjaxinline]J_0[/mathjaxinline], but are in opposite directions. </p>
<p>
A circular Amperian loop of radius [mathjaxinline]r&gt;b[/mathjaxinline], concentric to the cylinders, and with a clockwise circulation is used to calculate the line integral of the resulting magnetic field. </p>
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<p>
It is found that [mathjaxinline]\oint _{loop}\vec{B}\cdot d\vec{s} &lt;0[/mathjaxinline]. Which of the following options for the ratio [mathjaxinline]b^2/a^2[/mathjaxinline] could be correct? </p>
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<text> [mathjaxinline]\displaystyle \frac{b^2}{a^2} = 9.3[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{b^2}{a^2} = 9.5[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{b^2}{a^2} = 9.7[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{b^2}{a^2} = 10[/mathjaxinline]</text>
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<text> [mathjaxinline]\displaystyle \frac{b^2}{a^2} = 10.3[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L21Q4: Three Currents</h2>
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Three Currents
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<img src="/assets/courseware/v1/1c65e42c15b03537ef2c628cd6d5e9fa/asset-v1:MITx+8.02.2x+2T2018+type@asset+block/images_checkpoint_w9_2.svg" width="220"/>
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<p>
The figure shows three wires carrying current [mathjaxinline]I_1[/mathjaxinline], [mathjaxinline]I_2[/mathjaxinline], and [mathjaxinline]I_3[/mathjaxinline], with an Amperian circular loop around [mathjaxinline]I_1[/mathjaxinline] and [mathjaxinline]I_2[/mathjaxinline]. The wires are all perpendicular to the screen. </p>
<p>
Which currents produce a magnetic field at the point [mathjaxinline]P[/mathjaxinline] shown in the figure? </p>
<p>
<div class="wrapper-problem-response" tabindex="-1" aria-label="Question 1" role="group"><div class="choicegroup capa_inputtype" id="inputtype_checkpoint_w9_2_2_1">
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<text> [mathjaxinline]I_1[/mathjaxinline] only.</text>
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<input type="radio" name="input_checkpoint_w9_2_2_1" id="input_checkpoint_w9_2_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="checkpoint_w9_2_2_1-choice_2-label" for="input_checkpoint_w9_2_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w9_2_2_1">
<text> [mathjaxinline]I_1[/mathjaxinline] and [mathjaxinline]I_2[/mathjaxinline].</text>
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<text> [mathjaxinline]I_1[/mathjaxinline], [mathjaxinline]I_2[/mathjaxinline] and [mathjaxinline]I_3[/mathjaxinline].</text>
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<h2 class="hd hd-2 unit-title">L21DD1: Deep Dive: Proof of Ampere's Law</h2>
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<h3 class="hd hd-2">L21DD01: Deep Dive: Proof of Ampere's Law</h3>
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<h4 class="hd hd-5">Transcripts</h4>
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