38 lines
1.4 KiB
TeX
38 lines
1.4 KiB
TeX
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\chapter{Maxwell Equations}
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因为在柱坐标系下,$\overline{\overline\mu}$是对角的,所以Maxwell方程组中电场$\bf
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E$的旋度
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所以$\bf H$的各个分量可以写为:
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\begin{subequations}
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\begin{eqnarray}
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H_r=\frac{1}{\mathbf{i}\omega\mu_r}\frac{1}{r}\frac{\partial
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E_z}{\partial\theta } \\
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H_\theta=-\frac{1}{\mathbf{i}\omega\mu_\theta}\frac{\partial E_z}{\partial r}
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\end{eqnarray}
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\end{subequations}
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同样地,在柱坐标系下,$\overline{\overline\epsilon}$是对角的,所以Maxwell方程组中磁场$\bf
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H$的旋度
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\begin{subequations}
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\begin{eqnarray}
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&&\nabla\times{\bf H}=-\mathbf{i}\omega{\bf D}\\
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&&\left[\frac{1}{r}\frac{\partial}{\partial
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r}(rH_\theta)-\frac{1}{r}\frac{\partial
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H_r}{\partial\theta}\right]{\hat{\bf
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z}}=-\mathbf{i}\omega{\overline{\overline\epsilon}}{\bf
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E}=-\mathbf{i}\omega\epsilon_zE_z{\hat{\bf z}} \\
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&&\frac{1}{r}\frac{\partial}{\partial
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r}(rH_\theta)-\frac{1}{r}\frac{\partial
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H_r}{\partial\theta}=-\mathbf{i}\omega\epsilon_zE_z
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\end{eqnarray}
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\end{subequations}
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由此我们可以得到关于$E_z$的波函数方程:
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\begin{eqnarray}
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\frac{1}{\mu_\theta\epsilon_z}\frac{1}{r}\frac{\partial}{\partial r}
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\left(r\frac{\partial E_z}{\partial r}\right)+
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\frac{1}{\mu_r\epsilon_z}\frac{1}{r^2}\frac{\partial^2E_z}{\partial\theta^2}
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+\omega^2 E_z=0
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\end{eqnarray}
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