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on 17-Nov-2023 (Fri)

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Divergence of gradient

Divergence of gradient: \(\begin{align}\nabla\cdot(\nabla T)=\frac{\partial^{2} T}{\partial x^{2}} +\frac{\partial^{2} T}{\partial y^{2}} +\frac{\partial^{2} T}{\partial z^{2}}\end{align}\)

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Flashcard 7597278498060

Question

Divergence of gradient: [...]

Answer

\(\begin{align}\nabla\cdot(\nabla T)=\frac{\partial^{2} T}{\partial x^{2}} +\frac{\partial^{2} T}{\partial y^{2}} +\frac{\partial^{2} T}{\partial z^{2}}\end{align}\)


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Divergence of gradient
Divergence of gradient: \(\begin{align}\nabla\cdot(\nabla T)=\frac{\partial^{2} T}{\partial x^{2}} +\frac{\partial^{2} T}{\partial y^{2}} +\frac{\partial^{2} T}{\partial z^{2}}\end{align}\)







Curl of gradient

Curl of gradient: The curl of a gradient is always zero:\(\nabla\times(\nabla T)=\mathbf{0}\)

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Flashcard 7597282954508

Question

Curl of gradient: [...]

Answer
The curl of a gradient is always zero:\(\nabla\times(\nabla T)=\mathbf{0}\)

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Curl of gradient
Curl of gradient: The curl of a gradient is always zero:\(\nabla\times(\nabla T)=\mathbf{0}\)







Divergence of curl
Divergence of curl: The divergence of a curl, like the curl of a gradient, is always zero: \(\nabla\cdot(\nabla\times T)=\mathbf{0}\)
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Flashcard 7597287148812

Question
Divergence of curl: [...]
Answer
The divergence of a curl, like the curl of a gradient, is always zero: \(\nabla\cdot(\nabla\times T)=\mathbf{0}\)

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Divergence of curl
Divergence of curl: The divergence of a curl, like the curl of a gradient, is always zero: \(\nabla\cdot(\nabla\times T)=\mathbf{0}\)







Curl of curl

Curl of curl: Curl-of-curl gives nothing new: \(\nabla\times(\nabla\times T)=\nabla(\nabla\cdot\mathbf{v})-\nabla^{2}\mathbf{v}\)

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Flashcard 7597290032396

Question

Curl of curl: Curl-of-curl gives nothing new: [...]

Answer
\(\nabla\times(\nabla\times T)=\nabla(\nabla\cdot\mathbf{v})-\nabla^{2}\mathbf{v}\)

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Curl of curl
Curl of curl: Curl-of-curl gives nothing new: \(\nabla\times(\nabla\times T)=\nabla(\nabla\cdot\mathbf{v})-\nabla^{2}\mathbf{v}\)







Differential Form of Maxwell's Equations
Differential Form of Maxwell's Equations: \(\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)
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Flashcard 7597297634572

Question
Differential Form of Maxwell's Equations: [...]
Answer
\(\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)

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Differential Form of Maxwell's Equations
Differential Form of Maxwell's Equations: \(\begin{array}{rrl}\\ (\text{i})&\qquad\nabla\cdot\mathbf{E}&=&\displaystyle\frac{1}{\epsilon_{0}}\rho\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_{0}\mathbf{J}+\mu_0 \epsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)







Integrational Form of Maxwell's Equations
Integrational Form of Maxwell's Equations: \(\begin{array}{rrl}\\ (\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&\displaystyle\frac{1}{\varepsilon_{0}}\int_{V}\rho\ \mathrm{d}\tau\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_{0}\int_{S}\mathbf{J}\cdot\mathrm{d}\mathbf{a}+\mu_0 \epsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\ \end{array}\)
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Flashcard 7597300518156

Question
Integrational Form of Maxwell's Equations: [...]
Answer
\(\begin{array}{rrl}\\ (\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&\displaystyle\frac{1}{\varepsilon_{0}}\int_{V}\rho\ \mathrm{d}\tau\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_{0}\int_{S}\mathbf{J}\cdot\mathrm{d}\mathbf{a}+\mu_0 \epsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\ \end{array}\)

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Integrational Form of Maxwell's Equations
Integrational Form of Maxwell's Equations: \(\begin{array}{rrl}\\ (\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&\displaystyle\frac{1}{\varepsilon_{0}}\int_{V}\rho\ \mathrm{d}\tau\qquad &\text{(Gauss' law)}\\\\ (\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\ (\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\ (\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_{0}\int_{S}\mathbf{J}\cdot\mathrm{d}\mathbf{a}+\mu_0 \epsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\ \end{array}\)







Differential Form of Maxwell's Equations in vacuum
Differential Form of Maxwell's Equations in Vacuum: \(\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)
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Flashcard 7597304450316

Question
Differential Form of Maxwell's Equations in Vacuum: [...]
Answer
\(\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)

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Differential Form of Maxwell's Equations in vacuum
Differential Form of Maxwell's Equations in Vacuum: \(\begin{array}{rrl}\\(\text{i})&\qquad\nabla\cdot\mathbf{E}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\nabla\cdot\mathbf{B}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\nabla\times\mathbf{E}&=&\displaystyle-\frac{\partial\mathbf{B}}{\partial t}\qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\nabla\times\mathbf{B}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}\qquad &\text{(Ampere's law with Maxwell's correction)}\end{array}\)







Integrational Form of Maxwell's Equations in Vacuum

Integrational Form of Maxwell's Equations in Vacuum:

\(\begin{array}{rrl}\\(\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\\end{array}\)

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Flashcard 7597309955340

Question
Integrational Form of Maxwell's Equations in Vacuum: [...]
Answer

\(\begin{array}{rrl}\\(\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\\end{array}\)


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Integrational Form of Maxwell's Equations in Vacuum
Integrational Form of Maxwell's Equations in Vacuum: \(\begin{array}{rrl}\\(\text{i})&\qquad\displaystyle\oint_{S}\mathbf{E}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss' law)}\\\\(\text{ii})&\qquad\displaystyle\oint_{S}\mathbf{B}\cdot \mathrm{d}\mathbf{a}&=&0\qquad &\text{(Gauss's law for magnetism)}\\\\(\text{iii})&\qquad\displaystyle\oint_{L}\ \mathbf{E} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle - \frac {\partial}{\partial t}\int_{S}\mathbf{B}\cdot\mathrm{d}\mathbf{a} \qquad &\text{(Faraday's law)}\\\\(\text{iv})&\qquad\displaystyle\oint_{L}\ \mathbf{B} \cdot \mathrm{d}\boldsymbol{\mathbf{r}}&=&\displaystyle\mu_0 \varepsilon_0 \frac{\partial}{\partial t}\int_{S}\mathbf{E}\cdot\mathrm{d}\mathbf{a}\qquad &\text{(Ampere's law with Maxwell's correction)}\\\end{array}\)







Plane wave form of Electromagnatic wave

The Plane wave form of Electromagnetic wave is

\(\begin{array}{rl}\vec{E}&=&E_{0}\exp[i(kz-\omega t)]\cdot\hat{x}\\\vec{B}&=&B_{0}\exp[i(kz-\omega t)]\cdot\hat{y}\end{array}\)

where
- \(E_{0}\) is the amplitude of the electric field.
- \(B_{0}\) is the amplitude of the magnetic field.

- We also have \(B_{0}=\frac{k}{\omega}E_{0}=\frac{1}{c}E_{0}=\sqrt{\mu_{0}\varepsilon_{0}}E_{0}\)
- \(k\) is the wave number, or wave vector, representing the propagation direction.
- \(\omega\) is the angular frequency.
- \(\hat{x},\hat{y}\) is the polarization vector, representing the polarization direction.

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Flashcard 7597319130380

Question
The plane wave form of Electromagnetic wave is [...]
Answer

\(\begin{array}{rl}\vec{E}&=&E_{0}\exp[i(kz-\omega t)]\cdot\hat{x}\\\vec{B}&=&B_{0}\exp[i(kz-\omega t)]\cdot\hat{y}\end{array}\)

where
- \(E_{0}\) is the amplitude of the electric field.
- \(B_{0}\) is the amplitude of the magnetic field.

- We also have \(B_{0}=\frac{k}{\omega}E_{0}=\frac{1}{c}E_{0}=\sqrt{\mu_{0}\varepsilon_{0}}E_{0}\)
- \(k\) is the wave number, or wave vector, representing the propagation direction.
- \(\omega\) is the angular frequency.
- \(\hat{x},\hat{y}\) is the polarization vector, representing the polarization direction.


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Plane wave form of Electromagnatic wave
Plane wave form of Electromagnatic wave is \(\begin{array}{rl}\vec{E}&=&E_{0}\exp[i(kz-\omega t)]\cdot\hat{x}\\\vec{B}&=&B_{0}\exp[i(kz-\omega t)]\cdot\hat{y}\end{array}\) where - \(E_{0}\) is the amplitude of the electric field. - \(B_{0}\) is the amplitude of the magnatic field. - \(k\) is the wave number, or wave vector, represent the propagation direction. - \(\omega\) is the angular frequency. - \(\hat{x},\hat{y}\) is the polarization vector, represent the polarization direction.