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试推导矩形波导中TE10模的磁力线方程。

大学 2022-03-06 08:19:37 波导,磁力线,矩形已帮助人

试推导矩形波导中TE₁₀模的磁力线方程。

热搜：流体力学三大方程推导所用的定理

答案解析

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当矩形波导中传输TE10模时，磁场的复矢量为
$latex.action?latex=XG92ZXJsaW5le0h9PVxsZWZ0KCBcaGF0e3h9alxiZXRhIF97MTB9XGZyYWN7YX17XHBpIH1IXzBz%0D%0AaW5cZnJhY3tccGl4fXthfStcaGF0e3p9SF8wY29zXGZyYWN7XHBpIHh9e2F9IFxyaWdodCllXnst%0D%0AalxiZXRhIF97MTB9en0%3D$
磁场的瞬时值为
$latex.action?latex=XG92ZXJsaW5le0h9KFxvdmVybGluZXtyfe%2B8jHQpPS1caGF0e3h9XGJldGEgX3sxMH1cZnJhY3th%0D%0AfXtccGkgfUhfMHNpblxmcmFje1xwaSB4fXthfXNpbihcb21lZ2EgdC1cYmV0YSBfezEwfXopK1xo%0D%0AYXR7en1IXzBjb3NcZnJhY3tccGkgeH17YX1jb3MoXG9tZWdhICB0LVxiZXRhIF97MTB9eik%3D$
于是所求的曲线应满足方程
$latex.action?latex=XGZyYWN7ZHh9e0hfeH09XGZyYWN7ZHl9e0hfeX0%3D$
即 $latex.action?latex=SF8wY29zXGZyYWN7XHBpeH17YX1jb3MoXG9tZWdhIHQtXGJldGEgX3sxMH16KWR4PS1cYmV0YSBf%0D%0AezEwfVxmcmFje2F9e1xwaSB9SF8wc2luXGZyYWN7XHBpIHh9e2F9c2luKFxvbWVnYSB0LVxiZXRh%0D%0AIF97MTB9eilkeg%3D%3D$
变形为 $latex.action?latex=XGZyYWN7XHBpIH17YX1jb3RcZnJhY3tccGkgeH17YX1keD0tXGJldGEgX3sxMH10YW4oXG9tZWdh%0D%0AIHQtXGJldGEgX3sxMH16KWR6$
两边同时积分得 $latex.action?latex=bG5cbGVmdCggc2luXGZyYWN7XHBpIHh9e2F9IFxyaWdodCk9LWxuW2Nvcyhcb21lZ2EgdC1cYmV0%0D%0AYSBfezEwfXopXStsbkM%3D$
所以 $latex.action?latex=c2luXGZyYWN7XHBpIHh9e2F9XGNkb3QgY29zKFxvbWVnYSB0LVxiZXRhIF97MTB9eik9Qw%3D%3D$
令t=0，得到TE10模磁力线满足的方程为
$latex.action?latex=c2luXGZyYWN7XHBpIHh9e2F9XGNkb3QgY29zXGJldGEgX3sxMH16PUM%3D$