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<h2 class="hd hd-2 unit-title">Maxwell's equation in Differential form</h2>
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<p><center><b>Maxwell's Equations - Differential Form </b></center></p><p/><p> So far, we have been using the integral representation of Maxwell's equations:</p><table width="200"><tr><th colspan="2" align="center">Integral Form</th></tr><tr><td>\(\underset{S}{\Large\unicode{x222F}}\;\mathbf{\vec{E}} \cdot d\mathbf{\vec{A}
}=\dfrac{1}{\epsilon_0}\underset{V}{\iiint}\rho dV\)</td><td>Gauss' Law</td></tr><tr><td>\(\underset{S}{\Large\unicode{x222F}}\;\mathbf{\vec{B}} \cdot \mathbf{\hat{n}}\;dA=0\)</td><td>Magnetic Gauss' Law</td></tr><tr><td>\(\underset{C}{\oint}\mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} =-\dfrac{\partial }{\partial t}\underset{S}{\iint}\mathbf{\vec{B}}\cdot \mathbf{\hat{n}}\;dA \)</td><td>Faraday's Law</td></tr><tr><td>\( \underset{C}{\oint}\;\mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} =\mu_0\underset{S}{\iint}\mathbf{\vec{J}}\cdot \mathbf{\hat{n}}\;dA + \mu_0\epsilon_0\dfrac{\partial }{\partial t}\underset{S}{\iint}\mathbf{\vec{E}}\cdot \mathbf{\hat{n}}\;dA \)</td><td>Ampere-Maxwell's Law</td></tr></table><p>In this lesson, we will use two theorems of vector calculus (<a href="https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/4.-triple-integrals-and-surface-integrals-in-3-space/part-b-flux-and-the-divergence-theorem/session-84-divergence-theorem/"><b>Divergence</b></a> and <a href="https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/4.-triple-integrals-and-surface-integrals-in-3-space/part-c-line-integrals-and-stokes-theorem/session-95-stokes-theorem-and-surface-independence/"><b>Stokes</b></a>) to express Maxwell's equations in terms of derivatives. We will call this representation the differential form of Maxwell's equations. For any vector \(\mathbf{\vec{F}}\), the <b>Divergence</b> and <b>Stokes</b> Theorems are the following: </p><table><tr><td>\[\underset{\text{V}}{\iiint}\vec{\nabla}\cdot\mathbf{\vec{F}} \;dV = \underset{\text{S}}{\Large\unicode{x222F}}\;\mathbf{\vec{F}} \cdot \mathbf{\hat{n}}\;dA\]
<p>where \(V\) is the volume enclosed by the closed surface \(S\)</p></td><td><p/><p/><p><b>Divergence Theorem</b></p></td></tr><tr><td>\[\underset{\text{S}}{\iint} (\vec{\nabla}\times\mathbf{\vec{F}})\cdot \mathbf{\hat{n}}\;dA = \underset{\text{C}}{\oint}\mathbf{\vec{F}} \cdot d\mathbf{\vec{s}}\]
<p>where \(C\) is a closed path and \(S\) is a surface with \(C\) as its boundary.</p></td><td><p/><p/><p><b>Stokes' Theorem</b></p></td></tr></table>
You will show that after applying these two theorems the intergral forms of Maxwell's Equations are the following:
<table><tr><th colspan="2">Differential Form</th></tr><tr><td>\(\mathbf{\vec{\nabla}}\cdot \mathbf{\vec{E}}= \dfrac{\rho}{\epsilon_0}\)</td><td>Gauss' Law</td></tr><tr><td>\(\mathbf{\vec{\nabla}}\cdot \mathbf{\vec{B}}=0\)</td><td>Magnetic Gauss' Law</td></tr><tr><td>\(\mathbf{\vec{\nabla}}\times\mathbf{\vec{E}}=-\dfrac{\partial }{\partial t}\mathbf{\vec{B}} \)</td><td>Faraday's Law</td></tr><tr><td>\(\mathbf{\vec{\nabla}}\times\mathbf{\vec{B}}=\mu_0\mathbf{\vec{J}} + \mu_0\epsilon_0\dfrac{\partial }{\partial t}\mathbf{\vec{E}} \)</td><td>Ampere-Maxwell's Law</td></tr></table>
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<h2 class="hd hd-2 unit-title">L35v1: Maxwell's Equations and the Divergence Theorem</h2>
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Maxwell&#39;s Equations and the Divergence Theorem
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<p><b class="bfseries">(Part a)</b> Consider a closed surface [mathjaxinline]S[/mathjaxinline] and the volume [mathjaxinline]V[/mathjaxinline] enclosed by the surface. Which of the following statements about the flux of the electric field [mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline] through [mathjaxinline]S[/mathjaxinline] are true? In the equations, [mathjaxinline]\rho[/mathjaxinline] indicates charge per unit volume, [mathjaxinline]\sigma[/mathjaxinline] indicated charge per unit area on a surface, and [mathjaxinline]\mathbf{\vec{J}}[/mathjaxinline] indicates current per unit area through a surface. Also note that [mathjaxinline]d\mathbf{\vec{A}}[/mathjaxinline] and [mathjaxinline]\mathbf{\hat{n}}\; dA[/mathjaxinline] are two ways of writing exactly the same thing. </p>
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<text>1. [mathjaxinline]\underset {S}{\Large \unicode {x222F}}\; \mathbf{\vec{E}} \cdot \mathbf{\hat{n}}\; dA = \underset {\text {S}}{\iint } (\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}})\cdot \; \mathbf{\hat{n}}\; dA[/mathjaxinline]</text>
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<text>2. [mathjaxinline]\underset {S}{\Large \unicode {x222F}}\; \mathbf{\vec{E}} \cdot d\mathbf{\vec{A} }=\dfrac {1}{\epsilon _0}\underset {V}{\iiint }\rho dV[/mathjaxinline]</text>
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<text>3. [mathjaxinline]\underset {S}{\Large \unicode {x222F}} \; \mathbf{\vec{E}} \cdot \mathbf{\hat{n}}\; dA = \underset {\text {V}}{\iiint } \mathbf{\vec{\nabla }}\cdot \mathbf{\vec{E}} \; dV[/mathjaxinline]</text>
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<text>4. [mathjaxinline]\underset {S}{\Large \unicode {x222F}}\; \mathbf{\vec{E}} \cdot \mathbf{\hat{n}}\; dA = \underset {\text { S}}{\iint }\sigma \mathbf{\vec{J}}\cdot \mathbf{\hat{n}}\; dA[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> Using the result of part (a), which one of the following statements is true? </p>
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<text> 1. [mathjaxinline]\underset {\text {S}}{\iint } (\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}})\cdot \mathbf{\hat{n}}\; dA=\dfrac {1}{\epsilon _0}\underset {V}{\iiint }\rho dV[/mathjaxinline]</text>
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<text> 2. [mathjaxinline]\underset {\text {V}}{\iiint } \mathbf{\vec{\nabla }}\cdot \mathbf{\vec{E}} \; dV=\dfrac {1}{\epsilon _0}\underset {V}{\iiint }\rho dV[/mathjaxinline]</text>
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<text> 3. [mathjaxinline]\underset {\text { S}}{\iint }\dfrac {\mathbf{\vec{E}}}{\sigma }\cdot \mathbf{\hat{n}}\; dA=\dfrac {1}{\epsilon _0}\underset {V}{\iiint }\rho dV[/mathjaxinline]</text>
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<p><b class="bfseries">(Part c)</b> Consider the scalar functions [mathjaxinline]f(x,y,z)[/mathjaxinline] and [mathjaxinline]g(x,y,z)[/mathjaxinline], and let [mathjaxinline]V[/mathjaxinline] be any arbitrary volume in space, then </p>
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[mathjaxinline]\displaystyle \text {If}\; \; \underset {V}{\iiint } f \; dV=\underset {V}{\iiint } g\; dV \Longrightarrow f(x,y,z) = g(x,y,z) \; \; \text {for any arbitrary point } \; (x,y,z)[/mathjaxinline]
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Note that this result is only true for all points [mathjaxinline](x,y,z)[/mathjaxinline] if the two integrals are equal for any volume [mathjaxinline]V[/mathjaxinline] that you chose. </p>
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Use the result of the previous part and the statement above to obtain an expression for [mathjaxinline]\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{E}}[/mathjaxinline], the divergence of the electric field. Expres your answer in terms of rho for [mathjaxinline]\rho[/mathjaxinline] and epsilon_0 for [mathjaxinline]\epsilon _0[/mathjaxinline] </p>
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<p style="display:inline">[mathjaxinline]\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{E}} =[/mathjaxinline]</p>
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<p><b class="bfseries">(Part d)</b> Calculate [mathjaxinline]\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{B}}[/mathjaxinline], the divergence of a given magnetic field [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline]. </p>
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<p style="display:inline">[mathjaxinline]\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{B}} =[/mathjaxinline]</p>
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<h3 class="hd hd-2">L35v1: Maxwell's Equations and the Divergence Theorem</h3>
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<h2 class="hd hd-2 unit-title">L35v2: Maxwell's Equations and Stokes' Theorem</h2>
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Maxwell&#39;s Equations and Stokes&#39; Theorem
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<p><b class="bfseries">(Part a)</b> Consider a closed path [mathjaxinline]C[/mathjaxinline] and an open surface [mathjaxinline]S[/mathjaxinline] with [mathjaxinline]C[/mathjaxinline] as its boundary. Which of the following statements about [mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline] and [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline] are true? </p>
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<text>1. [mathjaxinline]\underset {C}{\oint }\mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} =\dfrac {\partial }{\partial t}\underset {S}{\iint }\mathbf{\vec{E}}\cdot \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<text>2. [mathjaxinline]\underset {C}{\oint }\mathbf{\vec{E}} \cdot d\mathbf{\vec{s}} =-\dfrac {\partial }{\partial t}\underset {S}{\iint }\mathbf{\vec{B}}\cdot \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<text>3. [mathjaxinline]\underset {C}{\oint }\mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} =\underset {V}{\iiint }(\mathbf{\nabla }\cdot \mathbf{\vec{E}})\; \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<text>4. [mathjaxinline]\underset {\text {C}}{\oint }\mathbf{\vec{E}} \cdot d\mathbf{\vec{s}}=\underset {\text {S}}{\iint } (\vec{\nabla }\times \mathbf{\vec{E}})\cdot \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> Using the result of part (a), which one of the following statements is true? </p>
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<text> 1. [mathjaxinline]\underset {\text {S}}{\iint } (\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}})\cdot \mathbf{\hat{n}}dA=\underset {S}{\iint }(\mathbf{\nabla }\cdot \mathbf{\vec{E}})\cdot \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<text> 2. [mathjaxinline]\underset {\text {V}}{\iiint } \mathbf{\vec{\nabla }}\cdot \mathbf{\vec{E}} \; dV=\dfrac {\partial }{\partial t}\underset {S}{\iint }\mathbf{\vec{B}}\cdot \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<text> 3. [mathjaxinline]\underset {\text {S}}{\iint } (\vec{\nabla }\times \mathbf{\vec{E}})\cdot \mathbf{\hat{n}}dA=-\dfrac {\partial }{\partial t}\underset {S}{\iint }\mathbf{\vec{B}}\cdot \mathbf{\hat{n}}dA[/mathjaxinline]</text>
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<p><b class="bfseries">(Part c)</b> Consider the vector functions [mathjaxinline]\mathbf{\vec{f}}(x,y,z)[/mathjaxinline] and [mathjaxinline]\mathbf{\vec{g}}(x,y,z)[/mathjaxinline], and [mathjaxinline]S[/mathjaxinline] to be any arbitrary open surface in space, then </p>
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[mathjaxinline]\displaystyle \text {If}\; \; \underset {S}{\iint } \mathbf{\vec{f}}\cdot \mathbf{\hat{n}}dA=\underset {S}{\iint } \mathbf{\vec{g}}\cdot \mathbf{\hat{n}}dA \Longrightarrow \mathbf{\vec{f}}(x,y,z) = \mathbf{\vec{g}}(x,y,z) \; \; \text {for any arbitrary point } \; (x,y,z)[/mathjaxinline]
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Note that this result is only true for all points [mathjaxinline](x,y,z)[/mathjaxinline] if the two integrals are equal for any surface [mathjaxinline]S[/mathjaxinline] that you chose. </p>
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Using the result of the previous part and the statement above, indicate which one of the following is true? </p>
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<text> 1. [mathjaxinline]\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}}=\dfrac {\partial }{\partial t}\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}}[/mathjaxinline]</text>
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<text> 2. [mathjaxinline]\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}} =-\dfrac {\partial \mathbf{\vec{B}}}{\partial t}[/mathjaxinline]</text>
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<text> 3. [mathjaxinline]\vec{\nabla }\cdot \mathbf{\vec{E}}=\dfrac {\partial \mathbf{\vec{B}}}{\partial t}[/mathjaxinline]</text>
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<p><b class="bfseries">(Part d)</b> Applying Stokes' theorem to the magnetic field we obtain that for any closed path [mathjaxinline]C[/mathjaxinline] and any open surface [mathjaxinline]S[/mathjaxinline] with [mathjaxinline]C[/mathjaxinline] as a boundary: </p>
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[mathjaxinline]\displaystyle \underset {\text {C}}{\oint } \mathbf{\vec{B}}\cdot d\mathbf{\vec{s}} = \underset {\text {S}}{\iint }(\vec{\nabla }\times \mathbf{\vec{B}})\cdot \mathbf{\hat{n}}dA[/mathjaxinline]
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We also know that Ampere-Maxwell's law states that: </p>
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[mathjaxinline]\displaystyle \underset {C}{\oint }\; \mathbf{\vec{B}} \cdot d\mathbf{\vec{s}} =\mu _0\underset {S}{\iint }\mathbf{\vec{J}}\cdot \mathbf{\hat{n}}dA + \mu _0\epsilon _0\dfrac {\partial }{\partial t}\underset {S}{\iint }\mathbf{\vec{E}}\cdot \mathbf{\hat{n}}dA[/mathjaxinline]
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Using the same steps followed in parts (a) to (c), complete the line below: </p>
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<p style="display:inline">[mathjaxinline]\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}}=[/mathjaxinline]</p>
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<option value="[mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline]"> [mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline]</option>
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<p style="display:inline">[mathjaxinline]+[/mathjaxinline]</p>
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<option value="[mathjaxinline]\mu _{0}[/mathjaxinline]"> [mathjaxinline]\mu _{0}[/mathjaxinline]</option>
<option value="[mathjaxinline]\epsilon _{0}[/mathjaxinline]"> [mathjaxinline]\epsilon _{0}[/mathjaxinline]</option>
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<p style="display:inline">[mathjaxinline]\dfrac {\partial }{\partial t}[/mathjaxinline]</p>
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<h3 class="hd hd-2">L35v2: Maxwell's Equations and Stoke's Theorem</h3>
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<h2 class="hd hd-2 unit-title">Maxwell's Equation - Wave Equation.</h2>
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<p><center><b>Maxwell's Equations - Wave Equation </b></center></p><p><br/></p><p>
In the case when there is no charge or current, \(\rho = 0\) and \(\mathbf{\vec{J}} = 0\), Maxwell's equations in differential form become:
</p><table><tr><th colspan="2">Maxwell's equations in free space</th></tr><tr><td>\(\mathbf{\vec{\nabla}}\cdot \mathbf{\vec{E}}= 0\)</td><td>Gauss' Law</td></tr><tr><td>\(\mathbf{\vec{\nabla}}\cdot \mathbf{\vec{B}}=0\)</td><td>Magnetic Gauss' Law</td></tr><tr><td>\(\mathbf{\vec{\nabla}}\times\mathbf{\vec{E}}=-\dfrac{\partial }{\partial t}\mathbf{\vec{B}} \)</td><td>Faraday's Law</td></tr><tr><td>\(\mathbf{\vec{\nabla}}\times\mathbf{\vec{B}}= \mu_0\epsilon_0\dfrac{\partial }{\partial t}\mathbf{\vec{E}} \)</td><td>Ampere-Maxwell's Law</td></tr></table>
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Maxwell&#39;s Equations - Wave Equation
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<p>
You will use the differential form of Maxwell's equations in free space and the vector identity below, to derive the wave equation for [mathjaxinline]\mathbf{\vec{E}}[/mathjaxinline]. </p>
<table id="a0000000002" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \mathbf{\vec{\nabla }}\times (\mathbf{\vec{\nabla }}\times \mathbf{\vec{F}})=-\nabla ^2\mathbf{\vec{F}}+\mathbf{\vec{\nabla }}(\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{F}})[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
</table>
<p>
where [mathjaxinline]\mathbf{\vec{F}}[/mathjaxinline] is a any vector with well defined derivatives. </p>
<p><b class="bfseries">(Part a)</b> Starting from Faraday's law in differential form </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
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<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \mathbf{\vec{\nabla }}\times \mathbf{\vec{E}} = -\dfrac {\partial \mathbf{\vec{B}}}{\partial t}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
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<p>
Take the curl in both sides of the equation to obtain: </p>
<table id="a0000000006" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000007">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \mathbf{\vec{\nabla }} \times (\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}}) =-\mathbf{\vec{\nabla }}\times \dfrac {\partial }{\partial t} \mathbf{\vec{B}}= -\dfrac {\partial }{\partial t} \mathbf{\vec{\nabla }}\times \mathbf{\vec{B}}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
where we have use the fact that the order of the spacial derivative and time derivative can be interchanged, [mathjaxinline]\dfrac {\partial }{\partial x}\dfrac {\partial }{\partial t}=\dfrac {\partial }{\partial t}\dfrac {\partial }{\partial t}[/mathjaxinline], etc. </p>
<p>
Applying eq. (1) and Gauss's law in the expression above you will obtain which of the following equations? </p>
<p>
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<text> 1. [mathjaxinline]\mathbf{\vec{\nabla }}(\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{E}}) =-\dfrac {\partial }{\partial t}(\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}})[/mathjaxinline]</text>
</label>
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<text> 2. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =-\dfrac {\partial }{\partial t}(\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}})[/mathjaxinline]</text>
</label>
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<text> 3. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\dfrac {\partial }{\partial t}(\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}})[/mathjaxinline]</text>
</label>
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<text> 4. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\dfrac {\partial }{\partial t}\nabla ^2\mathbf{\vec{B}}[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> Use Ampere-Maxwell's law in free space, [mathjaxinline]\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}}= \mu _0\epsilon _0\dfrac {\partial }{\partial t}\mathbf{\vec{E}}[/mathjaxinline], in the right hand side of the equation you obtained in part (a), to find out which of the following is true? </p>
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<text> 1. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\mu _0\epsilon _0\dfrac {\partial \mathbf{\vec{E}}}{\partial t}[/mathjaxinline]</text>
</label>
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<text> 2. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\mu _0\epsilon _0\dfrac {\partial ^2 \mathbf{\vec{E}}}{\partial t^2}[/mathjaxinline]</text>
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<input type="radio" name="input_checkpoint_w13_36_3_1" id="input_checkpoint_w13_36_3_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="checkpoint_w13_36_3_1-choice_3-label" for="input_checkpoint_w13_36_3_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w13_36_3_1">
<text> 3. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =-\mu _0\epsilon _0\dfrac {\partial ^2 \mathbf{\vec{E}}}{\partial t^2}[/mathjaxinline]</text>
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<text> 4. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\dfrac {\partial }{\partial t}\nabla ^2\mathbf{\vec{B}}[/mathjaxinline]</text>
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Maxwell&#39;s Equations - Wave Equation - Part 2
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<p>
You will use the differential form of Maxwell's equations in free space and the vector identity below to derive the wave equation for [mathjaxinline]\mathbf{\vec{B}}[/mathjaxinline]. </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 \mathbf{\vec{\nabla }}\times (\mathbf{\vec{\nabla }}\times \mathbf{\vec{F}})=-\nabla ^2\mathbf{\vec{F}}+\mathbf{\vec{\nabla }}(\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{F}})[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">
<span>(<span>1</span>)</span>
</td>
</tr>
</table>
<p>
where [mathjaxinline]\mathbf{\vec{F}}[/mathjaxinline] is a any vector with well defined derivatives. </p>
<p><b class="bfseries">(Part a)</b> Consider the differential form of Ampere - Maxwell's law </p>
<table id="a0000000004" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000005">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \mathbf{\vec{\nabla }}\times \mathbf{\vec{B}}= \mu _0\epsilon _0\dfrac {\partial }{\partial t}\mathbf{\vec{E}}[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
Take the curl in both sides of the equation </p>
<table id="a0000000006" cellpadding="7" width="100%" cellspacing="0" class="eqnarray" style="table-layout:auto">
<tr id="a0000000007">
<td style="width:40%; border:none">&#160;</td>
<td style="vertical-align:middle; text-align:right; border:none">
[mathjaxinline]\displaystyle \mathbf{\vec{\nabla }} \times (\mathbf{\vec{\nabla }}\times \mathbf{\vec{B}}) =\mu _0\epsilon _0\mathbf{\vec{\nabla }}\times (\dfrac {\partial }{\partial t} \mathbf{\vec{E}})= \mu _0\epsilon _0\dfrac {\partial }{\partial t} (\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}})[/mathjaxinline]
</td>
<td style="width:40%; border:none">&#160;</td>
<td style="width:20%; border:none" class="eqnnum">&#160;</td>
</tr>
</table>
<p>
Applying eq. (1) and the magnetic Gauss' law in the expression above you will obtain which of the following equations? </p>
<p>
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<text> 1. [mathjaxinline]\mathbf{\vec{\nabla }}(\mathbf{\vec{\nabla }}\cdot \mathbf{\vec{B}}) =\mu _0\epsilon _0\dfrac {\partial }{\partial t}(\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}})[/mathjaxinline]</text>
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<text> 2. [mathjaxinline]\nabla ^2\mathbf{\vec{B}} =-\mu _0\epsilon _0\dfrac {\partial }{\partial t}(\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}})[/mathjaxinline]</text>
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<input type="radio" name="input_checkpoint_w13_37_2_1" id="input_checkpoint_w13_37_2_1_choice_3" class="field-input input-radio" value="choice_3"/><label id="checkpoint_w13_37_2_1-choice_3-label" for="input_checkpoint_w13_37_2_1_choice_3" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w13_37_2_1">
<text> 3. [mathjaxinline]\nabla ^2\mathbf{\vec{B}} =+\mu _0\epsilon _0\dfrac {\partial }{\partial t}(\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}})[/mathjaxinline]</text>
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<input type="radio" name="input_checkpoint_w13_37_2_1" id="input_checkpoint_w13_37_2_1_choice_4" class="field-input input-radio" value="choice_4"/><label id="checkpoint_w13_37_2_1-choice_4-label" for="input_checkpoint_w13_37_2_1_choice_4" class="response-label field-label label-inline" aria-describedby="status_checkpoint_w13_37_2_1">
<text> 4. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\dfrac {1}{\mu _0\epsilon _0}\dfrac {\partial }{\partial t}\nabla ^2\mathbf{\vec{B}}[/mathjaxinline]</text>
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<p><b class="bfseries">(Part b)</b> Use the differential form of Faraday's law, [mathjaxinline]\mathbf{\vec{\nabla }}\times \mathbf{\vec{E}}=-\dfrac {\partial }{\partial t}\mathbf{\vec{B}}[/mathjaxinline] in the right hand side of the equation you obtained in part (a), and select the correct expression from the list below: </p>
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<text> 1. [mathjaxinline]\nabla ^2\mathbf{\vec{E}} =\mu _0\epsilon _0\dfrac {\partial \mathbf{\mathbf{\vec{\nabla }}\cdot \vec{E}}}{\partial t}[/mathjaxinline]</text>
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<text> 2. [mathjaxinline]\nabla ^2\mathbf{\vec{B}} =-\mu _0\epsilon _0\dfrac {\partial ^2 \mathbf{\vec{E}}}{\partial t^2}[/mathjaxinline]</text>
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<text> 3. [mathjaxinline]\nabla ^2\mathbf{\vec{B}} =\mu _0\epsilon _0\dfrac {\partial ^2 \mathbf{\vec{B}}}{\partial t^2}[/mathjaxinline]</text>
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<text> 4. [mathjaxinline]\nabla ^2\mathbf{\vec{B}} =\mu _0\epsilon _0\dfrac {\partial }{\partial t}\mathbf{\vec{B}}[/mathjaxinline]</text>
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