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<h2 class="hd hd-2 unit-title">Introduction to Radiation Pressure, Momentum, and Power</h2>
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<p>We have already seen that the Poynting Vector relates the electric and magnetic fields of light to the rate at which that light is transmitting energy. In this lesson, we will use the fact that light also carries momentum to think about how light being absorbed on or reflecting off of a surface will also transfer momentum to that surface, creating a pressure on that surface known as radiation pressure.</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 13.7</a></li>
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<h2 class="hd hd-2 unit-title">L37v1: Radiation Pressure</h2>
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<h3 class="hd hd-2">L37v1: Radiation Pressure</h3>
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<h2 class="hd hd-2 unit-title">L37v2: Momentum Carried by Electromagnetic Radiation</h2>
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<h2 class="hd hd-2 unit-title">L37Q1: Momentum of a Pulse</h2>
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Momentum Transfer by a Pulse
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<p>
To have a qualitative understanding on how electromagnetic waves can transfer momentum to an object, let's consider the simple case of an electric and magnetic "pulse" interacting with a single positive charge [mathjaxinline]+q[/mathjaxinline] hanging from a non-conducting spring. </p>
<center>
<img src="/assets/courseware/v1/59610e906220f0282a0af3578587aa59/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w14_17.svg" width="550"/>
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<p>
To imagine the electromagnetic pulse, you need to consider a rectangular "slab" in space of thickness [mathjaxinline]d[/mathjaxinline], infinite height and infinite depth, filled with a uniform electric field [mathjaxinline]\mathbf{\vec{E}}=E_0\mathbf{\hat{j}}[/mathjaxinline] and uniform magnetic field [mathjaxinline]\mathbf{\vec{B}}=B_0\mathbf{\hat{k}}[/mathjaxinline]. There are no fields outside the slab. The slab moves along the [mathjaxinline]+x[/mathjaxinline] axis with speed [mathjaxinline]c[/mathjaxinline]. A plot of the E-field as a function of time at a fixed location of space is shown on the left figure, and a portion of the slab with the fields inside is shown on the right figure. </p>
<p>
Answer the following questions in terms of hati for [mathjaxinline]\mathbf{\hat{i}}[/mathjaxinline], hatj for [mathjaxinline]\mathbf{\hat{j}}[/mathjaxinline], and hatk for [mathjaxinline]\mathbf{\hat{k}}[/mathjaxinline], and use a minus to indicate a negative direction as needed. </p>
<p><b class="bfseries">(Part a)</b> What is the direction of the force exerted by the E-field on the charge when the pulse passes through the location of the charge? </p>
<p>
<div class="inline" tabindex="-1" aria-label="Question 1" role="group"><div id="inputtype_checkpoint_w14_17_2_1" class="text-input-dynamath capa_inputtype inline textline">
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<p><b class="bfseries">(Part b)</b> Because the electric field causes the charge to accelerate, it will begin moving and therefore the magnetic field also exerts a force on the charge. What is the direction of the force exerted by the B-field on the charge? </p>
<p>
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<p><div class="solution-span">
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</div><b class="bfseries">(Part c)</b> What is the direction of the force exerted by the B-field on a negative charge? </p>
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<p><b class="bfseries">(Part d)</b> Imagine now that a sinusoidal planar electromagnetic wave passes through the location of the positively charged particle as shown. </p>
<center>
<img src="/assets/courseware/v1/9aa30fc7e878b5f2cae75eed504332f2/asset-v1:MITx+8.02.3x+1T2019+type@asset+block/images_checkpoint_w14_17b.svg" width="330"/>
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<p>
Which of the following statements are true? </p>
<p>
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<text>1. The time average electric force on the particle is along the [mathjaxinline]+\mathbf{\hat{j}}[/mathjaxinline]-direction.</text>
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<input type="checkbox" name="input_checkpoint_w14_17_5_1[]" id="input_checkpoint_w14_17_5_1_choice_1" class="field-input input-checkbox" value="choice_1"/><label id="checkpoint_w14_17_5_1-choice_1-label" for="input_checkpoint_w14_17_5_1_choice_1" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w14_17_5_1">
<text>2. The time average electric force on the particle is zero.</text>
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<text>3. The time average magnetic force is along [mathjaxinline]+\mathbf{\hat{i}}[/mathjaxinline]-direction.</text>
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<input type="checkbox" name="input_checkpoint_w14_17_5_1[]" id="input_checkpoint_w14_17_5_1_choice_3" class="field-input input-checkbox" value="choice_3"/><label id="checkpoint_w14_17_5_1-choice_3-label" for="input_checkpoint_w14_17_5_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w14_17_5_1">
<text>4. The time average magnetic force is zero.</text>
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<input type="checkbox" name="input_checkpoint_w14_17_5_1[]" id="input_checkpoint_w14_17_5_1_choice_4" class="field-input input-checkbox" value="choice_4"/><label id="checkpoint_w14_17_5_1-choice_4-label" for="input_checkpoint_w14_17_5_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w14_17_5_1">
<text>5. Both the [mathjaxinline]x[/mathjaxinline] and [mathjaxinline]y[/mathjaxinline] components of the momentum of the particle change</text>
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<input type="checkbox" name="input_checkpoint_w14_17_5_1[]" id="input_checkpoint_w14_17_5_1_choice_5" class="field-input input-checkbox" value="choice_5"/><label id="checkpoint_w14_17_5_1-choice_5-label" for="input_checkpoint_w14_17_5_1_choice_5" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w14_17_5_1">
<text>6. Only the [mathjaxinline]x[/mathjaxinline]-component of the momentum of the particle changes.</text>
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<span id="answer_checkpoint_w14_17_5_1"/>
</fieldset>
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<span class="status unanswered" id="status_checkpoint_w14_17_5_1" data-tooltip="Not yet answered.">
<span class="sr">unanswered</span><span class="status-icon" aria-hidden="true"/>
</span>
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<h2 class="hd hd-2 unit-title">L37Q2: Radiation Pressure</h2>
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Laser Pointer
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<p>
Let's come back to our laser pointer. </p>
<p><b class="bfseries">(Part a)</b> Calculate [mathjaxinline]I[/mathjaxinline], the intensity of the beam of a 3.0 mW pointer that creates a spot on a screen of 2.00 mm diameter? (Note: The power 3.0 mW is a time averaged value.) </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_18_2_1">[mathjaxinline]\mathrm{(W\cdot m^{-2} ) }[/mathjaxinline]</span>
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<p><b class="bfseries">(Part b)</b> Calculate [mathjaxinline]\mathcal{P}_{rad}[/mathjaxinline], the radiation pressure exerted by the laser light on the screen if it reflects [mathjaxinline]70 \%[/mathjaxinline] of the light that strikes it. </p>
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<p style="display:inline">[mathjaxinline]\mathcal{P}_{rad}=[/mathjaxinline]</p>
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<span class="trailing_text" id="trailing_text_checkpoint_w14_18_3_1">[mathjaxinline]\mathrm{(Pa) }[/mathjaxinline]</span>
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<h2 class="hd hd-2 unit-title">Review of Poynting Vector Relationships</h2>
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A plane electromagnetic wave transports energy in the direction of propagation of the wave. The power per unit area is given by the Poynting vector </p><table id="a0000000002" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\vec{\mathbf{S}}=\frac{\vec{\mathbf{E}}\times \vec{\mathbf{B}}}{\mu _0}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The power that flows into a rectangular volume of cross-sectional area [mathjaxinline]A[/mathjaxinline] perpendicular to the propagation direction and length [mathjaxinline]c\Delta t[/mathjaxinline] parallel to the direction of the wave (where [mathjaxinline]c[/mathjaxinline] is the speed of light) is equal to the rate of change of the energy stored in the fields inside that volume </p><table id="a0000000003" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]P=\langle |\vec{\mathbf{S}}|\rangle A=\dfrac{d}{dt}\langle U\rangle \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The electromagnetic wave also transports momentum, and hence can exert a <i class="it">radiation pressure</i> on a surface due to the absorption or reflection of the light. </p><p>
The momentum carried by an electromagnetic wave is related to the energy of the wave according to [mathjaxinline]U=c|\vec{\mathbf{p}}|[/mathjaxinline]. If the plane electromagnetic wave is completely absorbed by a surface of cross-sectional area [mathjaxinline]A[/mathjaxinline], then the momentum [mathjaxinline]|\Delta \vec{\mathbf{p}}|[/mathjaxinline] delivered to the surface in a time [mathjaxinline]\Delta t[/mathjaxinline] is given by [mathjaxinline]|\Delta \vec{\mathbf{p}}|=\Delta U/c[/mathjaxinline]. The force that the wave exerts on the surface is then the rate of change of the momentum in time </p><table id="a0000000004" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]F=\lim _{\Delta t\to 0}\frac{|\Delta \vec{\mathbf{p}}|}{\Delta t}=\lim _{\Delta t\to 0}\frac{1}{c}\frac{\Delta U}{\Delta t}=\frac{1}{c}\frac{dU}{dt}\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
Since the rate of change of energy is related to the power flowing across the surface, the force is </p><table id="a0000000005" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]F=\frac{1}{c}\frac{dU}{dt}=\frac{1}{c}\langle |\vec{\mathbf{S}}|\rangle A\, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
The radiation pressure [mathjaxinline]\mathcal{P}_{rad}[/mathjaxinline] is then defined to be the force per area that the wave exerts on the surface </p><table id="a0000000006" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\mathcal{P}_{rad}^{\text {abs}}\equiv \frac{F}{A}=\frac{1}{c}\langle |\vec{\mathbf{S}}|\rangle[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table><p>
if the material is perfectly absorbing. When the surface completely reflects the wave, then the change in momentum is twice the absorbing case since the wave completely reverses direction [mathjaxinline]|\Delta \vec{\mathbf{p}}|=2\Delta U/c[/mathjaxinline]. Therefore, the radiation pressure of a wave on a perfectly reflecting surface is </p><table id="a0000000007" class="equation" width="100%" cellspacing="0" cellpadding="7" style="table-layout:auto"><tr><td class="equation" style="width:80%; border:none">[mathjax]\mathcal{P}_{rad}^{\text {ref}}\equiv \frac{F}{A}=\frac{2}{c}\langle |\vec{\mathbf{S}}|\rangle \, .[/mathjax]</td><td class="eqnnum" style="width:20%; border:none"> </td></tr></table>
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<h2 class="hd hd-2 unit-title">L37Q3: Force from Sunlight</h2>
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Force of Sunlight
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<p>
As you lie on the beach in the bright midday sun, approximately what force does the light exert on you? (Assume light is a plane wave and ignore effects of latitude.) This will be an order of magnitude estimate that we will break down and think about step-by-step. </p>
<p>
Let's start with an estimate of the intensity of light that reaches you while you lie on the beach. You might find the following helpful: </p>
<ul class="itemize">
<li>
<p>
Distance from the Sun to the Earth, [mathjaxinline]r_{ES}= 1.5\times 10^{11}~ \mathrm{m}[/mathjaxinline] </p>
</li>
<li>
<p>
Power output from the Sun, [mathjaxinline]P = 4 \times 10^{26}~ \mathrm{W}[/mathjaxinline] </p>
</li>
<li>
<p>
Earth's atmosphere reflects around [mathjaxinline]29\%[/mathjaxinline] of the incoming solar radiation </p>
</li>
<li>
<p>
Earth's atmosphere absorbs around [mathjaxinline]23\%[/mathjaxinline] of the incoming solar radiation </p>
</li>
</ul>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle \langle | \vec{S} | \rangle =[/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline]\mathrm{W\, m^{-2}}[/mathjaxinline])</p>
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What is the approximate area (your surface area) that the sunlight is hitting? </p>
<p>
<p style="display:inline">[mathjaxinline]\displaystyle A=[/mathjaxinline] </p>
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<p style="display:inline"> (in [mathjaxinline]\mathrm{m^2}[/mathjaxinline])</p>
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Is the light mainly reflected or absorbed? Put in a value, [mathjaxinline]a[/mathjaxinline], between 1 (totally absorbed) and 2 (totally reflected). </p>
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<p style="display:inline">[mathjaxinline]\displaystyle a=[/mathjaxinline] </p>
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Now, what is the order of magnitude of the force from the sunlight on you? (Do you expect to be able to feel it?) We are interested in the order of magnitude, so force is [mathjaxinline]F \approx 10^{x}[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\displaystyle x=[/mathjaxinline] </p>
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