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<h2 class="hd hd-2 unit-title">Introduction to Faraday's Law</h2>
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<p><b>Electromagnetic Induction.</b></p><p> In 1831 Michael Faraday in England and independently Joseph Henry in the United States found experimentally that a change of magnetic field through a coil wire induces a current in the wire. The results of these experiments were the foundation of a new law of nature called Faraday's law of induction.</p><p> The phenomenon of electromagnetic induction has important technological applications encountered in our every day life (electric generators, computer hardware, cellphones, to name a few). In addition, electromagnetic induction is the phenomenon that provides the link between electric and magnetic fields needed to explain the existence of electromagnetic waves.</p><p>To discuss Faraday's law and its implications we need to review two concepts:</p><p><ul><li> Magnetic Flux.</li><li> Electromotive Force (emf).</li></ul></p>
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<p>Textbook Links</p>
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<li><a href="https://openlearninglibrary.mit.edu/courses/course-v1:MITx+8.02.3x+1T2019/pdfbook/0/#viewer-frame" target="[object Object]">Chapter 10.1-10.2: Faraday's Law and Motional EMF</a></li>
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<h2 class="hd hd-2 unit-title">L23Q1: Magnetic Flux - Review</h2>
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<p><b>Magnetic Flux</b></p><p>Consider a given surface \(S\) in a region in space where there is a magnetic field \(\mathbf{\vec{B}}\) as shown.</p><center><img src="/assets/courseware/v1/c8440da6868d946de89b5ee8aa98be6b/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_html_lesson_23_01c.svg" width="400"/></center><p>The magnetic flux through the surface is defined as: </p><p>
\[\Phi_{B} = \iint_{S} \mathbf{\vec{B}}\cdot \mathbf{d\vec{A}} \]
</p><p> where \(\mathbf{d\vec{A}}\) is given by \(d\mathbf{\vec{A}}=dA\mathbf{\hat{n}}\), with \(dA\) being the element of area of the surface and \(\mathbf{\hat{n}}\) the unit vector perpendicular to the surface.</p><p>Note: In the definition above we are calculating the magnetic flux over an <b>open</b> surface.
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CQ: Magnetic Flux
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<p><b class="bfseries">(Part a)</b> Consider the circular loop perpendicular to the axis of the bar magnet in figure a). The south pole points along the [mathjaxinline]+\mathbf{\hat{k}}[/mathjaxinline] - direction, the direction of [mathjaxinline]\mathbf{\hat{n}}[/mathjaxinline]. The flux through the open surface enclosed by the circle is </p>
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<p><b class="bfseries">(Part b)</b> Consider a rectangular loop parallel to the infinite current wire shown in figure b). The unit normal to the surface enclosed by the loop, [mathjaxinline]\mathbf{\hat{n}}[/mathjaxinline], points out of the screen, in the [mathjaxinline]-\mathbf{\hat{\theta }}[/mathjaxinline] direction. The flux through the open surface enclosed by the loop is </p>
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Consider now a sphere in front of the bar magnet. Consider [mathjaxinline]\mathbf{\hat{n}}[/mathjaxinline] to be the normal to the sphere's surface pointing outward. The flux through the closed spherical surface is </p>
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Given a force [mathjaxinline]\mathbf{\vec{F}}[/mathjaxinline] acting on a charge [mathjaxinline]q[/mathjaxinline], the electromotive force or emf is defined as the line integral over a closed loop [mathjaxinline]C[/mathjaxinline]: </p>
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[mathjaxinline]\displaystyle \mathcal{E}=\oint _{C}\frac{\mathbf{\vec{F}}}{q}\cdot d\mathbf{\vec{s}}[/mathjaxinline]
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Which of the following statements about the electromotive force are true? (Check all that apply) </p>
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<text>The electromotive force is not a force, it is the work done by the force [mathjaxinline]\mathbf{\vec{F}}[/mathjaxinline] per unit charge to move the charges through a closed loop.</text>
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<text>The force [mathjaxinline]\mathbf{\vec{F}}[/mathjaxinline] is a conservative force.</text>
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<text>The closed loop in the definition of the emf must be a conducting wire.</text>
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<text>If the closed loop in the definition of the emf is a conducting wire of resistance [mathjaxinline]R[/mathjaxinline] a current [mathjaxinline]I = \mathcal{E}/R[/mathjaxinline] flows in the wire.</text>
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<h2 class="hd hd-2 unit-title">L23v2: Motional EMF</h2>
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Motional EMF
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We have already seen the concept of emf in the context of DC circuits, where the force moving the charges is the chemical force due to the chemical reactions inside the battery. Other examples of emf are the work done per unit charge by the motor moving the belt of a Van der Graaff generator, or the solar energy moving charges in a solar cell. </p>
<p>
In this question and in the next video, we will consider the magnetic force on a charge inside a conductor moving in a magnetic field. This magnetic force will induce a current in the conductor. Because the resulting emf is due to the motion of the conductor with respect to the magnetic field we will call it "Motional emf". </p>
<p>
Consider a rectangular loop of length [mathjaxinline]l[/mathjaxinline] and width [mathjaxinline]w[/mathjaxinline] moving with speed [mathjaxinline]v[/mathjaxinline] along the [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline] direction. The loop is contained in the [mathjaxinline](x,y)[/mathjaxinline] plane perpendicular to a uniform magnetic field of magnitude [mathjaxinline]B[/mathjaxinline] and direction [mathjaxinline]+\mathbf{\hat{k}}[/mathjaxinline] (out of the screen). </p>
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At the instant shown in the figure above, the loop is moving out of the magnetic field region, therefore only a part of it is submerged in the magnetic field. We will assume that the charges that are free to move in the conductor are the positive charges (adopt the standard convention for current). </p>
<p><b class="bfseries">(Part a)</b> Calculate the force on the charges labeled as [mathjaxinline]q_1[/mathjaxinline], [mathjaxinline]q_2[/mathjaxinline], [mathjaxinline]q_3[/mathjaxinline], and [mathjaxinline]q_4[/mathjaxinline] in the figure, where [mathjaxinline]q_1=q_2=q_3=q_4=q&gt;0[/mathjaxinline]. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]q[/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>
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<p style="display:inline">On charge 1, [mathjaxinline]\mathbf{\vec{F}}_{mag}=[/mathjaxinline] </p>
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<p style="display:inline">On charge 2, [mathjaxinline]\mathbf{\vec{F}}_{mag}=[/mathjaxinline] </p>
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<p style="display:inline">On charge 3, [mathjaxinline]\mathbf{\vec{F}}_{mag}=[/mathjaxinline] </p>
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<p style="display:inline">On charge 4, [mathjaxinline]\mathbf{\vec{F}}_{mag}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> The direction of the resulting (conventional) current is: </p>
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<text> clockwise</text>
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<text> counterclockwise</text>
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<p><b class="bfseries">(Part c)</b> Calculate the motional emf by evaluating the line integral of [mathjaxinline]\displaystyle \frac{\mathbf{\vec{F}}_{mag}}{q}[/mathjaxinline] over the rectangle. </p>
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[mathjaxinline]\displaystyle \mathcal{E} = \oint _{rectangle}\frac{\mathbf{\vec{F}}_{mag}}{q}\cdot d\mathbf{\vec{s}}[/mathjaxinline]
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<p>
Circulate along the loop in the counterclockwise direction. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]v[/mathjaxinline], [mathjaxinline]q[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], and [mathjaxinline]w[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]\mathcal{E}=[/mathjaxinline] </p>
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<h3 class="hd hd-2">L23v2: Motional EMF</h3>
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<h2 class="hd hd-2 unit-title">L23Q3: Motional EMF on a Wire</h2>
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Moving away from a wire
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A thin conducting rod of length [mathjaxinline]L[/mathjaxinline] is moving away from an infinite wire with a constant velocity [mathjaxinline]\mathbf{\vec{v}} = v_0\mathbf{\hat{r}}[/mathjaxinline] and its length parallel to the wire. The distance between the wire and the rod is [mathjaxinline]r[/mathjaxinline]. A current [mathjaxinline]I[/mathjaxinline] is flowing along the wire as shown. </p>
<p><b class="bfseries">(Part a)</b> Consider the direction of [mathjaxinline]\mathbf{\vec{F}}[/mathjaxinline], the force exerted on a charge [mathjaxinline]q&gt;0[/mathjaxinline] in the conductor. </p>
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<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]\mathcal{E} = \int _{-}^{+}\mathbf{\vec{v}}\times \mathbf{\vec{B}} \cdot \mathbf{d\vec{s}}[/mathjaxinline], the motional emf as the line integral of [mathjaxinline]\mathbf{\vec{v}}\times \mathbf{\vec{B}}[/mathjaxinline] between the negative and positive end of the rod. Express your answer in terms of [mathjaxinline]I[/mathjaxinline], [mathjaxinline]L[/mathjaxinline], [mathjaxinline]r[/mathjaxinline], v_0 for [mathjaxinline]v_0[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\mathcal{E}=[/mathjaxinline] </p>
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<h2 class="hd hd-2 unit-title">L23v3: Motional EMF from Magnetic Flux</h2>
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Changing Magnetic Flux
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Consider a rectangular loop of length [mathjaxinline]l[/mathjaxinline] and width [mathjaxinline]w[/mathjaxinline] moving with speed [mathjaxinline]v[/mathjaxinline] along the [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline] direction. The loop is contained in the [mathjaxinline](x,y)[/mathjaxinline] plane perpendicular to a uniform magnetic field of magnitude [mathjaxinline]B[/mathjaxinline] and out of the screen ([mathjaxinline]+\mathbf{\hat{k}}[/mathjaxinline]). </p>
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At the instant shown in the figure above only a part of the loop of length [mathjaxinline]s(t)[/mathjaxinline] is inside the region of magnetic field. Consider the element of area [mathjaxinline]d\mathbf{\vec{A}}[/mathjaxinline] to be out of the screen ([mathjaxinline]+\mathbf{\hat{k}})[/mathjaxinline] </p>
<p><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]\Phi _{mag}[/mathjaxinline], the magnetic flux through the area enclosed by the loop at the instant shown in the figure. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]s[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], [mathjaxinline]v[/mathjaxinline] as needed. </p>
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<p style="display:inline"> [mathjaxinline]\Phi _{mag}=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]\displaystyle \frac{d\Phi _ B}{dt}[/mathjaxinline], the rate of change of magnetic flux with respect to time. Express your answer in terms of [mathjaxinline]B[/mathjaxinline], [mathjaxinline]w[/mathjaxinline], [mathjaxinline]s[/mathjaxinline], [mathjaxinline]l[/mathjaxinline], [mathjaxinline]v[/mathjaxinline] as needed. </p>
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<h3 class="hd hd-2">L23v3: Motional EMF from Magnetic Flux</h3>
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<h3 class="hd hd-3 problem-header" id="checkpoint_w10_24-problem-title" aria-describedby="block-v1:MITx+8.02.3x+1T2019+type@problem+block@checkpoint_w10_24-problem-progress" tabindex="-1">
A rectangular loop
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<p>
A rectangular conducting coil of length [mathjaxinline]l[/mathjaxinline], height [mathjaxinline]h[/mathjaxinline], and resistance [mathjaxinline]R[/mathjaxinline] is moving parallel to the [mathjaxinline](x,y)[/mathjaxinline] plane at a constant velocity [mathjaxinline]\mathbf{\vec{v}}=v_0\mathbf{\hat{i}}[/mathjaxinline] as shown. It continues to move with this velocity through a region containing a uniform magnetic field of magnitude [mathjaxinline]B_0[/mathjaxinline]. The magnetic field is in the [mathjaxinline]-\mathbf{\hat{k}}[/mathjaxinline] direction and is confined in the region [mathjaxinline]0\le x \le 3l[/mathjaxinline]. Consider the unit normal to the surface enclosed by the loop to be in the same direction as the magnetic field, [mathjaxinline]\mathbf{\hat{n}} = -\mathbf{\hat{k}}[/mathjaxinline] </p>
<center>
<img src="/assets/courseware/v1/8f59f28a591e1b2d6ebf87bf8b778138/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w10_24.svg" width="660"/>
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<p><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]\mathcal{E}[/mathjaxinline] and [mathjaxinline]I[/mathjaxinline], the induced emf and the current in the loop, at the instant when the loop is entering the region of magnetic field (leftmost figure above). Express your answer in terms of [mathjaxinline]R[/mathjaxinline],[mathjaxinline]l[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], v_0 for [mathjaxinline]v_0[/mathjaxinline], and B_0 for [mathjaxinline]B_0[/mathjaxinline] as needed. A <b class="bfseries">clockwise</b> current is <b class="bfseries">positive</b>. A <b class="bfseries">counterclockwise</b> current is <b class="bfseries">negative</b>. </p>
<p>
<p style="display:inline">[mathjaxinline]\mathcal{E}=[/mathjaxinline] </p>
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<p>
<p style="display:inline">[mathjaxinline]I=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]\mathcal{E}[/mathjaxinline] and [mathjaxinline]I[/mathjaxinline], the induced emf and the current in the loop, at the instant when the loop is completely immersed in the magnetic field region (central figure above). Express your answer in terms of [mathjaxinline]R[/mathjaxinline],[mathjaxinline]l[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], v_0 for [mathjaxinline]v_0[/mathjaxinline], and B_0 for [mathjaxinline]B_0[/mathjaxinline] as needed. A <b class="bfseries">clockwise</b> current is <b class="bfseries">positive</b>. A <b class="bfseries">counterclockwise</b> current is <b class="bfseries">negative</b>. </p>
<p>
<p style="display:inline">[mathjaxinline]\mathcal{E}=[/mathjaxinline] </p>
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<p>
<p style="display:inline">[mathjaxinline]I=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Calculate [mathjaxinline]\mathcal{E}[/mathjaxinline] and [mathjaxinline]I[/mathjaxinline], the induced emf and the current in the loop, at the instant when the loop is exiting the region of magnetic field (rightmost figure). Express your answer in terms of [mathjaxinline]R[/mathjaxinline],[mathjaxinline]l[/mathjaxinline], [mathjaxinline]h[/mathjaxinline], v_0 for [mathjaxinline]v_0[/mathjaxinline], and B_0 for [mathjaxinline]B_0[/mathjaxinline] as needed. A <b class="bfseries">clockwise</b> current is <b class="bfseries">positive</b>. A <b class="bfseries">counterclockwise</b> current is <b class="bfseries">negative</b>. </p>
<p>
<p style="display:inline">[mathjaxinline]\mathcal{E}=[/mathjaxinline] </p>
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<span id="solution_checkpoint_w10_24_solution_5"/>
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<p style="display:inline">[mathjaxinline]I=[/mathjaxinline] </p>
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