Waveguides and resonating cavities Equation Formula Latex Code and Summary

Waveguides and resonating cavities

Tags: #physics #waveguides

Equation

$$\begin{array}{ll} \vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma~~&~~\vec{n}\times(\vec{E}_2-\vec{E}_1)=0\\ \vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0~~&~~\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K} \end{array} \\ \vec{E}(\vec{x},t)=\vec{\cal E}(x,y){e}^{i(kz-\omega t)} \\ \vec{B}(\vec{x},t)=\vec{\cal{B}}(x,y){e}^{i(kz-\omega t)}$$

Latex Code

                                 \begin{array}{ll} \vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma~~&~~\vec{n}\times(\vec{E}_2-\vec{E}_1)=0\\ \vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0~~&~~\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K} \end{array} \\ \vec{E}(\vec{x},t)=\vec{\cal E}(x,y){e}^{i(kz-\omega t)} \\ \vec{B}(\vec{x},t)=\vec{\cal{B}}(x,y){e}^{i(kz-\omega t)}

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Introduction

Equation

$\begin{array}{ll} \vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma~~&~~\vec{n}\times(\vec{E}_2-\vec{E}_1)=0\\ \vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0~~&~~\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K} \end{array} \\ \vec{E}(\vec{x},t)=\vec{\cal E}(x,y){e}^{i(kz-\omega t)} \\ \vec{B}(\vec{x},t)=\vec{\cal{B}}(x,y){e}^{i(kz-\omega t)}$

Latex Code

            \begin{array}{ll} \vec{n}\cdot(\vec{D}_2-\vec{D}_1)=\sigma~~&~~\vec{n}\times(\vec{E}_2-\vec{E}_1)=0\\ \vec{n}\cdot(\vec{B}_2-\vec{B}_1)=0~~&~~\vec{n}\times(\vec{H}_2-\vec{H}_1)=\vec{K} \end{array} \\
            \vec{E}(\vec{x},t)=\vec{\cal E}(x,y){e}^{i(kz-\omega t)} \\ 
            \vec{B}(\vec{x},t)=\vec{\cal{B}}(x,y){e}^{i(kz-\omega t)} \\
            \begin{array}{ll} \displaystyle {\cal B}_x=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal B}_z}{\partial x}-\varepsilon\mu\omega\frac{\partial {\cal E}_z}{\partial y}\right)~~&~~ \displaystyle {\cal B}_y=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal B}_z}{\partial y}+\varepsilon\mu\omega\frac{\partial {\cal E}_z}{\partial x}\right)\\ \displaystyle {\cal E}_x=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal E}_z}{\partial x}+\varepsilon\mu\omega\frac{\partial {\cal B}_z}{\partial y}\right)~~&~~ \displaystyle {\cal E}_y=\frac{i}{\varepsilon\mu\omega^2-k^2}\left(k\frac{\partial {\cal E}_z}{\partial y}-\varepsilon\mu\omega\frac{\partial {\cal B}_z}{\partial x}\right) \end{array}

Explanation

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Jacqueline Snyder

Success on this exam is all I've been thinking about.

2023-09-01 00:00
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Best Wishes.

2023-09-10 00:00:00.0
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Mildred Turner

Would be great if I could pass this test.

2023-11-24 00:00
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Noel Bowers reply to Mildred Turner

Nice~

2023-12-11 00:00:00.0
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Andrea Scott

It's do or die with this test.

2024-03-21 00:00
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Gregory Edwards reply to Andrea Scott

Best Wishes.

2024-04-06 00:00:00.0
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