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<h2 class="hd hd-2 unit-title">Faraday's Law, Motional EMF, and Sign Convention</h2>
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Faraday&#39;s Law, Motional EMF, and Sign Convention
</h3>
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<img src="/assets/courseware/v1/198eb62c45c0ae311f211f30c1645a3c/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w10_33.svg" width="660"/>
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<p><b class="bfseries">(Part a)</b> In the left figure above a rectangular conducting loop is being pull into a region of magnetic field at a constant velocity [mathjaxinline]\mathbf{\vec{v}}[/mathjaxinline], in the right figure it is being pulled out from the region of magnetic field. The magnetic field points out of the screen. Use Lenz's law to determine the direction of the induced current in the two situations: </p>
<p>
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<text> 1. [mathjaxinline]I_{ind}[/mathjaxinline] is clockwise in the left figure and counterclockwise in the right one.</text>
</label>
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<div class="field">
<input type="radio" name="input_checkpoint_w10_33_2_1" id="input_checkpoint_w10_33_2_1_choice_2" class="field-input input-radio" value="choice_2"/><label id="checkpoint_w10_33_2_1-choice_2-label" for="input_checkpoint_w10_33_2_1_choice_2" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w10_33_2_1">
<text> 2. [mathjaxinline]I_{ind}[/mathjaxinline] is clockwise in the right figure and counterclockwise in the left one.</text>
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<input type="radio" name="input_checkpoint_w10_33_2_1" id="input_checkpoint_w10_33_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="checkpoint_w10_33_2_1-choice_3-label" for="input_checkpoint_w10_33_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w10_33_2_1">
<text> 3. [mathjaxinline]I_{ind}[/mathjaxinline] is clockwise in both situations.</text>
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<input type="radio" name="input_checkpoint_w10_33_2_1" id="input_checkpoint_w10_33_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="checkpoint_w10_33_2_1-choice_4-label" for="input_checkpoint_w10_33_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w10_33_2_1">
<text> 4. [mathjaxinline]I_{ind}[/mathjaxinline] is counterclockwise in both situations.</text>
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<p><b class="bfseries">(Part b)</b> We will use this example to review the sign conventions in the expression of Faraday's law for the case of motional emf. </p>
<p>
For a conductor moving with velocity [mathjaxinline]\mathbf{\vec{v}}[/mathjaxinline] in a magnetic field [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline], the motional emf is defined as the line integral over the loop of the work done by the magnetic force per unit charge and is related to the rate of change of the magnetic flux through the area enclosed by the loop as: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \oint _{loop}(\mathbf{\vec{v}}\times \mathbf{\vec{B}})\cdot \mathbf{d\vec{s}}=-\dfrac {d}{dt}\iint _{surface}\mathbf{\vec{B}}\cdot \mathbf{d\vec{A}}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
Consider the situation shown in the figure where the magnetic field [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline] is out of the screen, in the [mathjaxinline]+\mathbf{\hat{k}}[/mathjaxinline] - direction. In order to calculate the flux through the surface enclosed by the rectangle, the vector [mathjaxinline]\mathbf{d\vec{A}}[/mathjaxinline] must be perpendicular to the rectangle, i. e. along the [mathjaxinline]\pm \mathbf{\hat{k}}[/mathjaxinline]. Following the right hand rule, if we choose the unit normal out of the screen, [mathjaxinline]\mathbf{\hat{n}}=+ \mathbf{\hat{k}}[/mathjaxinline], the direction of circulation along the loop, [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline], must be counterclockwise as indicated in the left figure below. Otherwise, if [mathjaxinline]\mathbf{\hat{n}}=- \mathbf{\hat{k}}[/mathjaxinline] is chosen, the circulation is clockwise as in the right figure. </p>
<center>
<img src="/assets/courseware/v1/ed1470dbeee0c8f876cf83587854ba8c/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w10_33b.svg" width="770"/>
</center>
<p>
When the loop is <b class="bfseries">entering</b> the magnetic field, and with the choice of [mathjaxinline]\mathbf{\hat{n}}=+ \mathbf{\hat{k}}[/mathjaxinline] (left figure), </p>
<p>
<p style="display:inline"><b class="bfseries">1.</b> The magnetic flux [mathjaxinline]\Phi[/mathjaxinline] is </p>
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<option value="positive"> positive</option>
<option value="[mathjaxinline]0[/mathjaxinline]"> [mathjaxinline]0[/mathjaxinline]</option>
<option value="negative"> negative</option>
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<p>
<p style="display:inline"><b class="bfseries">2.</b> [mathjaxinline]\displaystyle \frac{d\Phi }{dt}[/mathjaxinline] is </p>
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<option value="positive"> positive</option>
<option value="[mathjaxinline]0[/mathjaxinline]"> [mathjaxinline]0[/mathjaxinline]</option>
<option value="negative"> negative</option>
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<p>
<p style="display:inline"><b class="bfseries">3.</b> The induced emf [mathjaxinline]\mathcal{E}[/mathjaxinline] is </p>
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<option value="option_checkpoint_w10_33_5_1_dummy_default">Select an option</option>
<option value="positive"> positive</option>
<option value="[mathjaxinline]0[/mathjaxinline]"> [mathjaxinline]0[/mathjaxinline]</option>
<option value="negative"> negative</option>
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<p>
<p style="display:inline"><b class="bfseries">4.</b> The induced current [mathjaxinline]I_{ind}[/mathjaxinline] is in the </p>
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<option value="option_checkpoint_w10_33_6_1_dummy_default">Select an option</option>
<option value="same direction as [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]"> same direction as [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]</option>
<option value="opposite direction of [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]"> opposite direction of [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]</option>
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<p>
<p style="display:inline"><b class="bfseries">5.</b> [mathjaxinline](\mathbf{\vec{v}}\times \mathbf{\vec{B}})\cdot \mathbf{d\vec{s}}[/mathjaxinline] is equal to zero except on </p>
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<option value="leg 1"> leg 1</option>
<option value="leg 2"> leg 2</option>
<option value="leg 3"> leg 3</option>
<option value="leg 4"> leg 4</option>
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<option value="option_checkpoint_w10_33_8_1_dummy_default">Select an option</option>
<option value="and is positive"> and is positive</option>
<option value="and is negative"> and is negative</option>
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<p>
When the loop is <b class="bfseries">entering</b> the magnetic field, and with the choice of [mathjaxinline]\mathbf{\hat{n}}=- \mathbf{\hat{k}}[/mathjaxinline], </p>
<p>
<p style="display:inline"><b class="bfseries">1.</b> The magnetic flux [mathjaxinline]\Phi[/mathjaxinline] is </p>
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<option value="option_checkpoint_w10_33_9_1_dummy_default">Select an option</option>
<option value="positive"> positive</option>
<option value="[mathjaxinline]0[/mathjaxinline]"> [mathjaxinline]0[/mathjaxinline]</option>
<option value="negative"> negative</option>
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<p>
<p style="display:inline"><b class="bfseries">2.</b> [mathjaxinline]\displaystyle \frac{d\Phi }{dt}[/mathjaxinline] is </p>
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<option value="option_checkpoint_w10_33_10_1_dummy_default">Select an option</option>
<option value="positive"> positive</option>
<option value="[mathjaxinline]0[/mathjaxinline]"> [mathjaxinline]0[/mathjaxinline]</option>
<option value="negative"> negative</option>
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<p>
<p style="display:inline"><b class="bfseries">3.</b> The induced emf [mathjaxinline]\mathcal{E}[/mathjaxinline] is </p>
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<option value="option_checkpoint_w10_33_11_1_dummy_default">Select an option</option>
<option value="positive"> positive</option>
<option value="[mathjaxinline]0[/mathjaxinline]"> [mathjaxinline]0[/mathjaxinline]</option>
<option value="negative"> negative</option>
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</p>
<p>
<p style="display:inline"><b class="bfseries">4.</b> The induced current [mathjaxinline]I_{ind}[/mathjaxinline] is in the </p>
<div class="inline" tabindex="-1" aria-label="Question 11" role="group"><div class="inputtype option-input inline">
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<option value="option_checkpoint_w10_33_12_1_dummy_default">Select an option</option>
<option value="same direction as [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]"> same direction as [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]</option>
<option value="opposite direction of [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]"> opposite direction of [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline]</option>
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<p>
<p style="display:inline"><b class="bfseries">5.</b> [mathjaxinline](\mathbf{\vec{v}}\times \mathbf{\vec{B}})\cdot \mathbf{d\vec{s}}[/mathjaxinline] is equal to zero except on </p>
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<option value="option_checkpoint_w10_33_13_1_dummy_default">Select an option</option>
<option value="leg 1"> leg 1</option>
<option value="leg 2"> leg 2</option>
<option value="leg 3"> leg 3</option>
<option value="leg 4"> leg 4</option>
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<option value="option_checkpoint_w10_33_14_1_dummy_default">Select an option</option>
<option value="and is positive"> and is positive</option>
<option value="and is negative"> and is negative</option>
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<p>
<div class="solution-span">
<span id="solution_checkpoint_w10_33_solution_3"/>
</div></p>
<p>
<b class="bfseries">Summary</b>
<p style="display:inline">If the induced emf [mathjaxinline]\mathcal{E} &lt;0[/mathjaxinline] the direction of [mathjaxinline]I_{ind}[/mathjaxinline] is </p>
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<option value="option_checkpoint_w10_33_15_1_dummy_default">Select an option</option>
<option value="in the same direction of the circulation"> in the same direction of the circulation</option>
<option value="opposite to the direction of the circulation"> opposite to the direction of the circulation</option>
</select>
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<span class="status unanswered" id="status_checkpoint_w10_33_15_1" data-tooltip="Not yet answered.">
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<p>
Note that the choice of the direction of [mathjaxinline]\mathbf{\hat{n}}[/mathjaxinline] is completely <b class="bf">arbitrary</b>. However, once that choice is made, the direction of [mathjaxinline]\mathbf{d\vec{s}}[/mathjaxinline] is fixed. The signs of some quantities in the equations will change, <b class="bf">but</b> the physics will <b class="bf">not</b> change. You will still find the same currents and the same forces in the same directions in any part of the circuit. </p>
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<h2 class="hd hd-2 unit-title">Pulling a Loop out of a Magnetic Field</h2>
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Pulling a loop out of a field
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A rectangular conducting loop of length [mathjaxinline]l[/mathjaxinline] (along the [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline] direction), height [mathjaxinline]h[/mathjaxinline] (along the [mathjaxinline]+\mathbf{\hat{j}}[/mathjaxinline] direction), and resistance [mathjaxinline]R[/mathjaxinline], is being pull out from a region of uniform magnetic field [mathjaxinline]\mathbf{\vec{B}}=B\mathbf{\hat{k}}[/mathjaxinline] at a constant velocity [mathjaxinline]\mathbf{\vec{v}}=v\mathbf{\hat{i}}[/mathjaxinline]. Solutions to the following questions can be found in the video at the end of the problem. </p>
<p><b class="bfseries">(Part a)</b> Consider the unit normal of the surface enclosed by the rectangular loop to be out of the screen, along the [mathjaxinline]+\mathbf{\hat{k}}[/mathjaxinline] direction. Calculate [mathjaxinline]\mathcal{E}[/mathjaxinline], the induced emf. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and [mathjaxinline]h[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\mathcal{E}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]I_{ind}[/mathjaxinline], the induced current in the loop. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and [mathjaxinline]h[/mathjaxinline] as needed. Consistent with the choice of unit normal, the current is positive if it circulates counterclockwise, and negative otherwise. <p style="display:inline">[mathjaxinline]I_{ind}=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 2" role="group"><div id="inputtype_checkpoint_w10_34_3_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part c)</b> Calculate [mathjaxinline]\mathbf{\vec{F}}^{ind}[/mathjaxinline], the force exerted by the magnetic field on the induced current [mathjaxinline]I_{ind}[/mathjaxinline] calculated in part (b). Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], hati for [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline], hatj for [mathjaxinline]\mathbf{\hat{j}}[/mathjaxinline], and hatk for [mathjaxinline]\mathbf{\hat{k}}[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]\mathbf{\vec{F}}^{ind}=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 3" role="group"><div id="inputtype_checkpoint_w10_34_4_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part d)</b> Calculate [mathjaxinline]\mathbf{\vec{F}}^{P}[/mathjaxinline], the external force which must be exerted on the loop in order to pull it out from the region of magnetic field at a constant velocity. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], hati for [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline], hatj for [mathjaxinline]\mathbf{\hat{j}}[/mathjaxinline], and hatk for [mathjaxinline]\mathbf{\hat{k}}[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]\mathbf{\vec{F}}^{P}=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 4" role="group"><div id="inputtype_checkpoint_w10_34_5_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part d)</b> Calculate [mathjaxinline]P[/mathjaxinline], the power exerted by the external pulling force on the loop. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline],[mathjaxinline]R[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and [mathjaxinline]h[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]P=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 5" role="group"><div id="inputtype_checkpoint_w10_34_6_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part e)</b> Calculate [mathjaxinline]P_ R[/mathjaxinline], the power dissipated by the current [mathjaxinline]I_{ind}[/mathjaxinline] while it flows through the loop's resistance [mathjaxinline]R[/mathjaxinline]. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and [mathjaxinline]h[/mathjaxinline] as needed. <p style="display:inline">[mathjaxinline]P_ R=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 6" role="group"><div id="inputtype_checkpoint_w10_34_7_1" class="text-input-dynamath capa_inputtype inline textline">
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<h3 class="hd hd-2">W10PS1: Worked Example - Pulling a Loop out of a Magnetic Field</h3>
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<h2 class="hd hd-2 unit-title">Forces on a Moving Loop</h2>
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Forces on a moving loop
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Two identical conducting loops are moving in a region of magnetic field with the same constant velocity [mathjaxinline]\mathbf{\vec{v}}[/mathjaxinline]. In both cases the magnetic field is perpendicular to the plane of the loop, in the [mathjaxinline]+\mathbf{\hat{k}}[/mathjaxinline] direction. In Figure (a), the magnetic field is uniform while in Figure (b), the magnitude of the field increases as the loop moves to the right. </p>
<p>
We will assume that the moving charges are positive in agreement with the charge carriers in the conventional current. Consider [mathjaxinline]q_ R[/mathjaxinline] to be a charge on the right vertical leg of the loop, and [mathjaxinline]q_ L[/mathjaxinline] to be a charge on the left vertical leg of the loop, where [mathjaxinline]q_ R=q_ L=q&gt;0[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> Consider the moving loop in Figure (a). Which of the following statements about [mathjaxinline]\mathbf{\vec{F}_ R}[/mathjaxinline] and [mathjaxinline]\mathbf{\vec{F}_ L}[/mathjaxinline], the forces on [mathjaxinline]q_ R[/mathjaxinline] and [mathjaxinline]q_ L[/mathjaxinline], respectively, are true? </p>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|=|\mathbf{\vec{F}_ R}|[/mathjaxinline], and both are along the [mathjaxinline]-\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|=|\mathbf{\vec{F}_ R}|[/mathjaxinline], and both are along the [mathjaxinline]+\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|\neq |\mathbf{\vec{F}_ R}|[/mathjaxinline] and both are along [mathjaxinline]-\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|\neq |\mathbf{\vec{F}_ R}|[/mathjaxinline], and both are along the [mathjaxinline]+\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<p><b class="bfseries">(Part b)</b> Consider the moving loop in Figure (b). Which of the following statements about [mathjaxinline]\mathbf{\vec{F}_ R}[/mathjaxinline] and [mathjaxinline]\mathbf{\vec{F}_ L}[/mathjaxinline], the forces on [mathjaxinline]q_ R[/mathjaxinline] and [mathjaxinline]q_ L[/mathjaxinline], respectively, are true? </p>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|=|\mathbf{\vec{F}_ R}|[/mathjaxinline], and both are along the [mathjaxinline]-\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|=|\mathbf{\vec{F}_ R}|[/mathjaxinline], and both are along the [mathjaxinline]+\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|\neq |\mathbf{\vec{F}_ R}|[/mathjaxinline] and both are along [mathjaxinline]-\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<text> [mathjaxinline]|\mathbf{\vec{F}_ L}|\neq |\mathbf{\vec{F}_ R}|[/mathjaxinline], and both are along the [mathjaxinline]+\mathbf{\hat{j}}[/mathjaxinline] direction</text>
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<p><b class="bfseries">(Part c)</b> As a consequence of the forces acting on the charges in the loop of Figure (a): </p>
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<text> The same amount of positive charges move upwards along the left and right legs of the loop. As a result there is no induced current.</text>
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<text> The same amount of positive charges move downwards along the left and right leg of the loop. As a result there is no induced current.</text>
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<text> An induced clockwise current is established.</text>
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<text> An induced counterclockwise current is established.</text>
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<p><b class="bfseries">(Part d)</b> As a consequence of the forces acting on the charges in the loop of Figure (b): </p>
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<text> The same amount of positive charges move upwards along the left and right legs of the loop. As a result there is no induced current.</text>
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<text> The same amount of positive charges move downwards along the left and right legs of the loop. As a result there is no induced current.</text>
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<text> An induced clockwise current is established.</text>
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<text> An induced counterclockwise current is established.</text>
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<h2 class="hd hd-2 unit-title">Falling Loop</h2>
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Falling Loop, part (a)
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A rectangular loop of wire with mass [mathjaxinline]m[/mathjaxinline], width [mathjaxinline]w[/mathjaxinline], vertical length [mathjaxinline]l[/mathjaxinline], and resistance [mathjaxinline]R[/mathjaxinline] falls out of a magnetic field under the influence of gravity. The magnetic field is uniform and given by: </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000003">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \vec{B}_0 = \left\{ \begin{array}{ll} B\hat{k} &amp; \quad y \leq 0 \\ 0 &amp; \quad y &gt; 0 \end{array} \right.[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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where [mathjaxinline]B[/mathjaxinline] is a positive constant. Note that the [mathjaxinline]+\hat{j}[/mathjaxinline] (or [mathjaxinline]+\hat{y}[/mathjaxinline]) direction points downward. At the time shown in the figure, the loop is moving down and exiting the magnetic field with a velocity [mathjaxinline]\vec{v}(t)=v_ y(t)\hat{y}[/mathjaxinline], where [mathjaxinline]v_ y(t)&gt;0[/mathjaxinline]. At time [mathjaxinline]t[/mathjaxinline], the distance of the bottom of the loop measured from [mathjaxinline]y=0[/mathjaxinline] is [mathjaxinline]y(t)[/mathjaxinline] as shown. </p>
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If we define the area vector [mathjaxinline]\vec{A}[/mathjaxinline] to be out of the screen, in the same direction as [mathjaxinline]\vec{B}_0[/mathjaxinline], what is the magnetic flux [mathjaxinline]\Phi _ B[/mathjaxinline] through the loop at time [mathjaxinline]t[/mathjaxinline]? Express your answer using some or all of the following: [mathjaxinline]y[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\Phi _ B=[/mathjaxinline] </p>
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Falling Loop, part (b)
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<p>
Calculate [mathjaxinline]\dfrac {d\Phi _ B}{dt}[/mathjaxinline], the rate of change of the magnetic flux. Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \frac{d\Phi _ B}{dt}=[/mathjaxinline] </p>
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Falling Loop, part (c)
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If [mathjaxinline]d\Phi _ B/dt&lt;0[/mathjaxinline] then your induced emf (and current) will be right-handed with respect to [mathjaxinline]\vec{A}[/mathjaxinline], and vice versa. What is the direction of your induced current given your result in the previous part, clockwise or counterclockwise? </p>
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<text> Clockwise</text>
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Falling Loop, part (d)
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Calculate [mathjaxinline]I_{ind}[/mathjaxinline], the magnitude of the induced current flowing in the circuit at the time shown. Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline] and [mathjaxinline]g[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]I_{ind}=[/mathjaxinline] </p>
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Falling Loop, part (e)
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Calculate [mathjaxinline]\vec{F}_ B[/mathjaxinline], the force exerted by the magnetic field [mathjaxinline]\vec{B}_0[/mathjaxinline] on the induced current calculated in part (d). Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], [mathjaxinline]g[/mathjaxinline], v_y for [mathjaxinline]v_ y[/mathjaxinline], hati for [mathjaxinline]\hat{i}[/mathjaxinline], hatj for [mathjaxinline]\hat{j}[/mathjaxinline] and hatk for [mathjaxinline]\hat{k}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\vec{F}_{B}=[/mathjaxinline] </p>
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Falling Loop, part (f)
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Assume that the loop has reached &#8220;terminal velocity" &#8211; i. e. it is no longer accelerating. What is [mathjaxinline]v_{term}[/mathjaxinline], the magnitude of that terminal velocity? Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], and [mathjaxinline]g[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]v_{term}=[/mathjaxinline] </p>
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Falling Loop, part (g)
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While the loop is still falling at terminal velocity, what is [mathjaxinline]\dfrac {dW_ g}{dt}[/mathjaxinline], the rate at which gravity is doing work on the loop? Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], and [mathjaxinline]g[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\displaystyle \frac{dW_ g}{dt}=[/mathjaxinline] </p>
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Falling Loop, part (h)
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While the loop is still falling at terminal velocity, what is [mathjaxinline]P_ R[/mathjaxinline], the rate at which energy is being dissipated through the resistor? Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], and [mathjaxinline]g[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]P_ R=[/mathjaxinline] </p>
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Falling Loop, part (i)
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While the loop is still falling at terminal velocity, what is [mathjaxinline]\frac{dW_ B}{dt}[/mathjaxinline], the rate at which the magnetic force is doing work on the loop? Express your answer in terms of [mathjaxinline]l[/mathjaxinline], [mathjaxinline]m[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], [mathjaxinline]R[/mathjaxinline], and [mathjaxinline]g[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\displaystyle \frac{dW_ B}{dt}=[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">Self-Inductor Coaxial Cable</h2>
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Self-Inductor Coaxial Cable, part a)
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An inductor consists of two very thin conducting cylindrical shells, one of radius [mathjaxinline]a[/mathjaxinline] and one of radius [mathjaxinline]b[/mathjaxinline], both of length [mathjaxinline]h[/mathjaxinline]. Assume that the inner shell carries current [mathjaxinline]I[/mathjaxinline] in the [mathjaxinline]+\hat{k}[/mathjaxinline] direction out of the screen, and the outer shell carries current [mathjaxinline]I[/mathjaxinline] into the screen as shown in the right figure. The [mathjaxinline]z[/mathjaxinline]-axis is out of the page along the common axis of the shells and the [mathjaxinline]r[/mathjaxinline]-axis is the radial cylindrical axis perpendicular to the [mathjaxinline]z[/mathjaxinline]-axis. Assume that the current is uniformly distributed in both conductors and that [mathjaxinline]b-a &lt;&lt; h[/mathjaxinline] so we can neglect edge effects. </p>
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Calculate the magnetic field [mathjaxinline]\vec{\textbf{B}}[/mathjaxinline] between the cylindrical shells for [mathjaxinline]a &lt; r &lt; b[/mathjaxinline]. Express your answer using some or all of the following: [mathjaxinline]r[/mathjaxinline], [mathjaxinline]I[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], and hattheta for [mathjaxinline]\hat{\mathbf{\theta }}[/mathjaxinline]. </p>
<p>
<p style="display:inline">[mathjaxinline]\vec{\textbf{B}} (r) =[/mathjaxinline] </p>
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Self-Inductor Coaxial Cable, part b
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What is the magnetic energy density [mathjaxinline]u_ B[/mathjaxinline] for [mathjaxinline]a &lt; r &lt; b[/mathjaxinline]? Express your answer using some or all of the following: [mathjaxinline]r[/mathjaxinline], [mathjaxinline]I[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]u_ B(r) =[/mathjaxinline] </p>
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Self-Inductor Coaxial Cable, part c
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Calculate the inductance of this long inductor recalling that total magnetic energy [mathjaxinline]U_ B = \frac{1}{2} L I^2[/mathjaxinline], and using your results for the magnetic energy density in part (b). Express your answer using some or all of the following: [mathjaxinline]r[/mathjaxinline], [mathjaxinline]I[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]L =[/mathjaxinline] </p>
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Self-Inductor Coaxial Cable, part d
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Calculate the inductance of this long inductor by using the formula [mathjaxinline]\Phi = L \cdot I = \int _{open \, surface} \vec{\mathbf{B}} \cdot \mathbf{d} \vec{\mathbf{A}}[/mathjaxinline] and your results for the magnetic field in part (a). To do this, you must choose an appropriate open surface over which to evaluate the magnetic flux. Does your result calculated in this way agree with your result in part (c)? </p>
<p>
Express your answer using some or all of the following: [mathjaxinline]r[/mathjaxinline], [mathjaxinline]I[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], [mathjaxinline]a[/mathjaxinline], [mathjaxinline]b[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. <p style="display:inline">[mathjaxinline]L =[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_fridayw9_3d_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<a class="btn-link external-track" href="/assets/courseware/v1/87f30f5a82e8bc5a739c0f0b162eafdc/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/subs_W10PS02.srt">Download transcript</a>
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