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<h2 class="hd hd-2 unit-title">Introduction to the Poynting Vector</h2>
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<p>Last week, we discovered how Maxwell's Equations predict the existence of electromagnetic waves that can propagate in free space.</p><p> Waves can transport energy. Ocean waves generated by offshore winds erode beaches, sound waves produced by the string of a guitar set up the vibration of our ear drums, and electromagnetic waves generated by the sun travel to the earth and warm it up.</p><p> This week, we will look at the flow and transferal of energy by electromagnetic waves. Starting with conservation of energy in electromagnetism, we will define the <b>Poynting Vector</b>, which describes the rate at which electromagnetic energy flows per unit area and whose direction is the direction of wave propagation.</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 13.6</a></li>
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<h2 class="hd hd-2 unit-title">L36v1: Conservation Laws</h2>
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<h2 class="hd hd-2 unit-title">L36Q1: Conservation of Energy</h2>
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Conservation of Energy
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Consider a region in space in which there are electric [mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline] and magnetic [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline] fields as well as a certain charge distribution. A surface encloses a portion of that region of volume [mathjaxinline]V[/mathjaxinline]. The energy [mathjaxinline]U[/mathjaxinline] in the volume will change if the electric or magnetic energies change, or if there is work done on the charges. If the volume is determined by a closed surface with an outwards unit normal [mathjaxinline]\mathbf{\hat{n}}_{\text {out}}[/mathjaxinline], the mathematical representation of the conservation of energy is: </p>
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[mathjaxinline]\displaystyle Power_{out} =-\dfrac {d}{dt}\underbrace{ \underset {V}{\iiint }(u_ E+u_ B)\; dV }_{\text {Electric+Magnetic Energy}}-\underbrace{\underset {V}{\iiint }\mathbf{\vec{J}}\cdot \mathbf{\vec{E}} \; dV}_{\text {Work on charges}}[/mathjaxinline]
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<span>(<span>1</span>)</span>
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In the next video, we will define a vector that represents the flow of energy per unit area, called the Poynting vector, [mathjaxinline]\mathbf{\vec{S}}[/mathjaxinline], such that: </p>
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[mathjaxinline]\displaystyle Power_{out} = \underset {\text {surface}}{\Large \unicode {x222F}}\; \mathbf{\vec{S}} \cdot \mathbf{\hat{n}}_{\text {out}}\; da[/mathjaxinline]
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<span>(<span>2</span>)</span>
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where [mathjaxinline]\mathbf{\hat{n}}_{\text {out}}[/mathjaxinline] is the outward unit normal to the surface. </p>
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<b class="bfseries">Conservation of Energy in Free Space</b>
</p>
<p>
In the particular case when there are no charges in the volume, the second term in the energy conservation equation is zero. </p>
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[mathjaxinline]\displaystyle Power_{out} =-\dfrac {d}{dt}\underbrace{\underset {V}{\iiint }(u_ E+u_ B)\; dV }_{\text {Electric+Magnetic Energy}}[/mathjaxinline]
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<span>(<span>3</span>)</span>
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<p>
In a given volume, [mathjaxinline]V[/mathjaxinline], of space which has electric and magnetic fields, but no charges, which of the following statements are true about the relationship between the [mathjaxinline]Power_{out}[/mathjaxinline] through a certain surface and the energy stored in the electric and magnetic fields in the volume enclosed by that surface? </p>
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<text>1. If [mathjaxinline]Power_{out}&gt;0[/mathjaxinline] energy is flowing into the surface and the EM energy is increasing inside [mathjaxinline]V[/mathjaxinline].</text>
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<text>3. If [mathjaxinline]Power_{out}&lt;0[/mathjaxinline] energy is flowing into the surface and the EM energy is increasing inside [mathjaxinline]V[/mathjaxinline].</text>
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<text>4. If [mathjaxinline]Power_{out}&lt;0[/mathjaxinline] energy is flowing out of the surface and the EM energy is decreasing inside [mathjaxinline]V[/mathjaxinline].</text>
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<h2 class="hd hd-2 unit-title">L36Q2: Charging Capacitor Poynting Vector</h2>
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Direction of the Poynting Vector
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<p>Answer the following questions regarding the circuit element shown in the figure. (Note the indicated direction of current and polarity of the charge, and assume \(I(t)\ is constant. )</p>
<p>(a) Is the electromagnetic energy within the circuit element increasing or decreasing?</p>
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</div><p>(b) What is the direction of the Electric field vector at point \(P\)?</p>
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</div><p>(c) What is the direction of the Magnetic field vector at point \(P\)?</p>
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</div><p>(d) What is the direction of the Poynting vector at point \(P\)?</p>
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<h2 class="hd hd-2 unit-title">L36v3: Direction of the Poynting Vector</h2>
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<h2 class="hd hd-2 unit-title">L36Q3: Poynting Vector</h2>
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Poynting Vector of a planar EM wave
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<p>
In a certain region of space, the electric and magnetic fields are given by [mathjaxinline]\mathbf{\vec{E}}(y,t)=-E_{0} \sin {(ky + \omega t)}\mathbf{\hat{i}}[/mathjaxinline] and [mathjaxinline]\mathbf{\vec{B}}(y,t)=-B_{0} \sin {(ky + \omega t)}\mathbf{\hat{k}}[/mathjaxinline], respectively. </p>
<p><b class="bfseries">(Part a)</b> Find the Poynting vector, [mathjaxinline]\mathbf{\vec{S}}[/mathjaxinline] in this region of space. </p>
<p>
Write your answer in terms of E_0 for [mathjaxinline]E_0[/mathjaxinline], B_0 for [mathjaxinline]B_0[/mathjaxinline], [mathjaxinline]k[/mathjaxinline], [mathjaxinline]y[/mathjaxinline], omega for [mathjaxinline]\omega[/mathjaxinline], [mathjaxinline]t[/mathjaxinline], hati for [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline], hatj for [mathjaxinline]\mathbf{\hat{j}}[/mathjaxinline], hatk for [mathjaxinline]\mathbf{\hat{k}}[/mathjaxinline], epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline] as needed. </p>
<p>
<p style="display:inline">[mathjaxinline]\vec{S}(t)=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> The direction of wave propagation is: </p>
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<text> [mathjaxinline]+\mathbf{\hat{i}}[/mathjaxinline]</text>
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<h2 class="hd hd-2 unit-title">L36v4: Rate of Energy Flux in an Electromagnetic Wave</h2>
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Energy Flow in Electromagnetic Waves
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Consider a region in space where there are electric [mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline] and magnetic [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline] fields but no electric charge. In a closed volume [mathjaxinline]V[/mathjaxinline] of that region, the power flowing out of the surface, [mathjaxinline]Power_{out}[/mathjaxinline], is given by the change of electric and magnetic energy in that volume: </p>
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[mathjaxinline]\displaystyle Power_{out} = -\dfrac {d}{dt}\underset {V}{\iiint }(u_ E+u_ B)\; dV[/mathjaxinline]
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In addition, we will assume that the electric and magnetic fields are part of an electromagnetic wave propagating along the [mathjaxinline]+\mathbf{\hat{i}}[/mathjaxinline] direction. We choose the volume to be a cylinder of cross sectional [mathjaxinline]A[/mathjaxinline] with its axis parallel to the [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline] direction and of length [mathjaxinline]c dt[/mathjaxinline] which is the distance traveled by the wave during a time interval [mathjaxinline]dt[/mathjaxinline]. </p>
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<p><b class="bfseries">(Part a)</b> If [mathjaxinline]u_ E[/mathjaxinline] and [mathjaxinline]u_ B[/mathjaxinline] are the electric and magnetic energy densities (energy per unit volume) calculate [mathjaxinline]dU[/mathjaxinline], the amount of electric and magnetic energy inside the cylinder. Assume that [mathjaxinline]u_ E[/mathjaxinline] and [mathjaxinline]u_ B[/mathjaxinline] can be approximated as being uniform over the volume (that would be true, for example, if [mathjaxinline]dt[/mathjaxinline] was very small). Express your answer in terms of [mathjaxinline]c[/mathjaxinline], [mathjaxinline]A[/mathjaxinline], [mathjaxinline]dt[/mathjaxinline], u_E for [mathjaxinline]u_ E[/mathjaxinline], and u_B for [mathjaxinline]u_ B[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]dU=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part b)</b> Consider [mathjaxinline]E[/mathjaxinline] and [mathjaxinline]B[/mathjaxinline] to be the amplitudes of the electric and magnetic fields, respectively. What are [mathjaxinline]u_ E[/mathjaxinline] and [mathjaxinline]u_ B[/mathjaxinline], the electric and magnetic energy densities. Express your answer in terms of [mathjaxinline]E[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], and epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]u_ E=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]u_ B=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part c)</b> Because the electric and magnetic fields are part of an electromagnetic wave, which must be a solution of Maxwell's equations, their amplitudes are such that [mathjaxinline]B=E/c[/mathjaxinline]. Use this relationship to express the electric and magnetic energy densities [mathjaxinline]u_ E[/mathjaxinline] and [mathjaxinline]u_ B[/mathjaxinline] in terms of the product [mathjaxinline]EB[/mathjaxinline]. Express your answer in terms of [mathjaxinline]E[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline], and [mathjaxinline]c[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]u_ E=[/mathjaxinline]</p>
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<p style="display:inline">[mathjaxinline]u_ B=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part d)</b> Use the results in parts (a) and (c) to calculate [mathjaxinline]Power_{out}[/mathjaxinline], the rate of change of energy through the cylinder in terms of the product [mathjaxinline]EB[/mathjaxinline]. Express your answer in terms of [mathjaxinline]E[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], and [mathjaxinline]A[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]Power_{out}=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part e)</b> Calculate the rate of energy per unit area [mathjaxinline](Power)_{out}/A[/mathjaxinline] flowing through the end of the cylinder. Express your answer in terms of [mathjaxinline]E[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline](Power)_{out}/A=[/mathjaxinline]</p>
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<p><b class="bfseries">(Part f)</b> What is [mathjaxinline]|\mathbf{\vec{S}}|[/mathjaxinline], the magnitude of the Poynting vector. Express your answer in terms of [mathjaxinline]E[/mathjaxinline], [mathjaxinline]B[/mathjaxinline], and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]|\mathbf{\vec{S}}|=[/mathjaxinline]</p>
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<h3 class="hd hd-2">L36v4: Rate of Energy Flux in an Electromagnetic Wave</h3>
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<h2 class="hd hd-2 unit-title">L36v5: Intensity of Light</h2>
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Energy in an EM Wave
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A plane sinusoidal electromagnetic wave is moving in the [mathjaxinline]+x[/mathjaxinline]-direction. The electric and magnetic fields are given by: </p>
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[mathjaxinline]\displaystyle \mathbf{\vec{E}(x,t)}[/mathjaxinline]
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[mathjaxinline]\displaystyle =[/mathjaxinline]
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<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle E_0 \sin (kx-\omega t ) \mathbf{\hat{j}}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle \mathbf{\vec{B}(x,t)}[/mathjaxinline]
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<td style="vertical-align:middle; text-align:center; border:none">
[mathjaxinline]\displaystyle =[/mathjaxinline]
</td>
<td style="vertical-align:middle; text-align:left; border:none">
[mathjaxinline]\displaystyle B_0 \sin (kx-\omega t ) \mathbf{\hat{k}}[/mathjaxinline]
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<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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<p>
Recall that the time average of a function [mathjaxinline]f(t)[/mathjaxinline] over a time interval [mathjaxinline]0&lt;t&lt;T[/mathjaxinline] is defined as: </p>
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<td style="width:40%; border:none">&#160;</td>
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[mathjaxinline]\displaystyle &lt;f(t)&gt; =\dfrac {1}{T}\int _{t=0}^{t=T}f(t)\; dt[/mathjaxinline]
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In particular, if [mathjaxinline]f(t) = b\sin ^2(kx-\omega t)[/mathjaxinline], where the period [mathjaxinline]T=2\pi /\omega[/mathjaxinline]: </p>
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[mathjaxinline]\displaystyle &lt;f(t)&gt; =\dfrac {1}{T}\int _{t=0}^{t=T}b\sin ^2(kx-\omega t)\; dt = \dfrac {b}{2}[/mathjaxinline]
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<p><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]&lt;u_ E&gt;[/mathjaxinline], the time average of the electric energy density. Express your answer in terms of E_0 for [mathjaxinline]E_0[/mathjaxinline] and epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]&lt;u_ E&gt;=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]&lt;u_ B&gt;[/mathjaxinline], the time average of the magnetic energy density. Express your answer in terms of B_0 for [mathjaxinline]B_0[/mathjaxinline] and mu_0 for [mathjaxinline]\mu _0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]&lt;u_ B&gt;=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part c)</b> Because the electric and magnetic fields are part of an electromagnetic wave which is a solution of Maxwell's equations, their components are such that [mathjaxinline]B_ z = E_ y/c[/mathjaxinline], therefore their amplitudes satisfy that [mathjaxinline]B_0=E_0/c[/mathjaxinline]. Using this relationship and the results of parts (a) and (b), express [mathjaxinline]&lt;u_E&gt;[/mathjaxinline] and [mathjaxinline]&lt;u_B&gt;[/mathjaxinline] in terms of the product of the amplitudes, [mathjaxinline]E_0 B_0[/mathjaxinline]. Use in your answers E_0 for [mathjaxinline]E_0[/mathjaxinline], B_0 for [mathjaxinline]B_0[/mathjaxinline], epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], and [mathjaxinline]c[/mathjaxinline] as needed. </p>
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<p style="display:inline">[mathjaxinline]&lt;u_ E&gt;=[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]&lt;u_ B&gt;=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part d)</b> Consider an imaginary cylindrical surface of cross sectional area [mathjaxinline]A[/mathjaxinline], length [mathjaxinline]c dt[/mathjaxinline], and its axis parallel to the [mathjaxinline]x[/mathjaxinline]-axis as shown. </p>
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Use the answer in part (c) to calculate the magnitude of the time-average rate of change of the electromagnetic energy, [mathjaxinline]&lt;Power&gt;[/mathjaxinline], the power flowing through the cylindrical surface in terms of the product of the amplitudes [mathjaxinline]E_0 B_0[/mathjaxinline]. Express your answer in terms of [mathjaxinline]A[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], E_0 for [mathjaxinline]E_0[/mathjaxinline], and B_0 for [mathjaxinline]B_0[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]&lt;Power&gt;=[/mathjaxinline] </p>
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<p><b class="bfseries">(Part e)</b> What is [mathjaxinline]&lt;\mathbf{\vec{S}}&gt;[/mathjaxinline], the time averaged Poynting vector for this wave? Express your answer in terms of [mathjaxinline]A[/mathjaxinline], mu_0 for [mathjaxinline]\mu _0[/mathjaxinline], E_0 for [mathjaxinline]E_0[/mathjaxinline], and B_0 for [mathjaxinline]B_0[/mathjaxinline], as well as a unit vector hati for [mathjaxinline]\hat{i}[/mathjaxinline], hatj for [mathjaxinline]\hat{j}[/mathjaxinline], and hatk for [mathjaxinline]\hat{k}[/mathjaxinline],. </p>
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<p style="display:inline">[mathjaxinline]&lt;\mathbf{\vec{S}}&gt;=[/mathjaxinline] </p>
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<h3 class="hd hd-2">L36v5: Intensity of Light</h3>
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E-field of a Laser Beam
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When giving presentations, many people use a laser pointer to direct the attention of the audience to the information on a screen. </p>
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A small laser pointer produces a beam of red light of [mathjaxinline]d = 1~ \mathrm{mm}[/mathjaxinline] in diameter and has a power output of [mathjaxinline]2~ \mathrm{mW}[/mathjaxinline]. </p>
<p><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]I[/mathjaxinline], the intensity of the wave. <p style="display:inline">[mathjaxinline]I=[/mathjaxinline] </p> <div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_checkpoint_w14_12_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<span class="trailing_text" id="trailing_text_checkpoint_w14_12_2_1">[mathjaxinline]\, \mathrm{W/m^2}[/mathjaxinline]</span>
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<p><b class="bfseries">(Part b)</b> What is [mathjaxinline]E_0[/mathjaxinline], the amplitude of the electric field in the laser beam? </p>
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<p style="display:inline">[mathjaxinline]E_0=[/mathjaxinline] </p>
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<span class="trailing_text" id="trailing_text_checkpoint_w14_12_3_1">[mathjaxinline]\, \mathrm{V/m}[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">Planar versus Spherical Waves</h2>
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<p><center><b>Intensity of Electromagnetic Waves</b></center></p><p><b>Planar waves</b></p><p> Another way of visualizing a wave is the concept of <i>wavefront</i>, which is defined as the set of points in space where the wave has the same phase, for example, all the points in space where the wave has a maximum at the same time.</p><p>Plane sinusoidal waves are the type of waves we have been studying in the previous lessons. In these waves, the electric and magnetic fields are in phase at all time. Therefore, the wavefronts are parallel planes perpendicular to the direction of wave propagation.</p><center><img src="/assets/courseware/v1/e37f56e31e8d0f13dc058539fc87a970/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson36_01_c.svg" width="400"/></center><p><b>Intensity of Planar Waves</b></p><p>
In a previous exercise, we showed that the rate of change of energy is related to the Poynting vector by:
</p><p>
\[ Power = \dfrac{dU}{dt} = |\mathbf{\vec{S}}|A \]
</p><p>where \(A\) is the area perpendicular to the direction of wave propagation. Taking the time average of the expression above, </p><p>
\[ \langle Power \rangle=\langle \dfrac{dU}{dt} \rangle = IA \Longrightarrow I= \frac{\langle Power \rangle}{A} \]
</p><p>where \(I\) is the intensity. Because the amplitudes of a plane wave are assumed to be the same at all points on a wavefront plane, the time average intensity of a plane sinusoidal wave is independent of the position.</p><p><b>Spherical Waves</b></p><p> Consider the case of a point source emitting EM waves equally in all directions, radially away from the central emitting point. In this case, the wavefronts are concentric spheres with their centers coinciding with the source. </p><p>
The left figure below is a view from the \(+z\) axis, the black dot at the center is the emitting source. The concentric circles in the figure are the intersection of the spherical wavefronts with the \((x,y)\)-plane. For clarity, only a few wavefronts in the \((x,y)\)-plane are shown. The 2-dimensional representation of the wavefronts of the 3-dimensional spherical waves can be thought of as the water ripples formed when the free surface of water of a quiet lake is perturbed at a single point (see right figure below).</p><table style="width:100%"><tr><td><img src="/assets/courseware/v1/a08da708b998a495f699c1898f5f92d1/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson36_01_b.svg" width="400"/></td><td><center><img src="/assets/courseware/v1/f82512c0b471b8b3be098fa15e6b36b0/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_water-2634082_1920.jpg" width="300"/></center></td></tr><tr><td>Intersection of the spherical wavefronts with the \((x,y)\) plane.</td><td>Image by <a href="https://pixabay.com/users/BlazingFirebug-6008933/?utm_source=link-attribution&utm_medium=referral&utm_campaign=image&utm_content=2634082">Water Ripples by Brittney King</a> </td></tr></table><p><b>Intensity of Spherical Waves</b></p><p>Spherical waves also transport energy radially away from the source, equally in all directions. As a result of this spherical symmetry the Poynting vector at any point in space is given by:
</p><p>
\[\mathbf{\vec{S}} = S(r)\;\mathbf{\hat{r}}\]
</p><p>
where \( S(r)= |\mathbf{\vec{S}}|\) is the magnitude of the Poynting vector. The rate at which energy is radiated by the source is the flux of the Poynting vector through a closed sphere with its center coinciding with the source:</p><p>
\[Power_{out} = \underset{\text{Sphere}}{\Large\unicode{x222F}}\;\mathbf{\vec{S}} \cdot \mathbf{\hat{r}}\; da\]
</p><p>Using the spherical symmetry of the Poynting vector, we obtain: </p><p>
\[Power_{out}=\underset{\text{Sphere}}{\Large\unicode{x222F}}\;S(r)\mathbf{\hat{r}} \cdot \mathbf{\hat{r}}\; da = \underset{\text{Sphere}}{\Large\unicode{x222F}}\;S(r)\;da =S(r)\underset{\text{Sphere}}{\Large\unicode{x222F}}\;da=S(r)\;(4\pi r^2)
\]
</p><p>The time average of the equation above relates the intensity of the wave, \(I\), and the time average of the power:</p><p>
\[<Power > = < S(r) > \;(4\pi r^2) = I \;(4\pi r^2)\Longrightarrow I = \dfrac{< Power >}{4\pi r^2} \]
</p><p>The intensity of the ideal spherical EM waves is a function of the distance from the source.</p><p><b>Spherical Waves Far Away from the Source.</b></p><p> Far away from the source, a distance much larger than the wavelength of the waves, \(r>> \lambda\), the curvature of the wavefront becomes small and it can be approximated by a plane. Therefore, spherical waves can be approximated by sinusoidal planar waves at locations far away from the source.</p><center><img src="/assets/courseware/v1/4731a2029d0bfa98cef505d5040ba400/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_lesson36_01.svg" width="600"/></center><p> As a result, the intensity of the wave far from the source can be expressed in terms of the amplitude of the electric and magnetic fields,
\(E_0\) and \(B_0\),</p><p>
\[I = < |\mathbf{\vec{S}}| > =\dfrac{1}{2} \dfrac{E_0 B_0}{\mu_0} \]
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<h2 class="hd hd-2 unit-title">L37Q5: Light Bulb</h2>
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Power Emitted from a Light Bulb
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You are reading a book and decide to estimate the magnitude of the electric and magnetic fields of the light that is incident on the book from your desk lamp. You can treat the lamp as a point source that emits sinusoidal EM waves uniformly in all directions. The book is at [mathjaxinline]1~ \mathrm{m}[/mathjaxinline] from the 100 Watt light bulb. Assume that all the electric energy is transformed into visible light. </p>
<p><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]I[/mathjaxinline], the intensity of the light incident on your book? </p>
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<p style="display:inline">[mathjaxinline]I=[/mathjaxinline] </p>
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<span class="trailing_text" id="trailing_text_checkpoint_w14_16_2_1">[mathjaxinline]\mathrm{(W/m^{2} ) }[/mathjaxinline]</span>
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<p><b class="bfseries">(Part b)</b> Find [mathjaxinline]E_0[/mathjaxinline] and [mathjaxinline]B_0[/mathjaxinline], the amplitude of the electric and the magnetic fields of the EM waves emitted by the lamp. </p>
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<p style="display:inline">[mathjaxinline]E_0=[/mathjaxinline] </p>
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<span class="trailing_text" id="trailing_text_checkpoint_w14_16_3_1">[mathjaxinline]\, \mathrm{V/m}[/mathjaxinline]</span>
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<p style="display:inline">[mathjaxinline]B_0=[/mathjaxinline] </p>
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<span class="trailing_text" id="trailing_text_checkpoint_w14_16_4_1">[mathjaxinline]\, \mathrm{T}[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">L36DD1: Proof of Energy Conservation with the Poynting Vector</h2>
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<h3 class="hd hd-2">L36DD1: Proof of Energy Conservation with the Poynting Vector</h3>
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