以下电磁场知识点整理及题目均基于 马西奎 工程电磁场导论 北京 高等教育出版社, 2000. Print. 面向21世纪课程教材 Mian Xiang 21shi Ji Ke Cheng Jiao Cai = Textbook Series for 21st Century Eng. （第一版） 旨在为四川大学电气工程及其自动化专业的电磁场课程提供一个知识纲要性质的文本，以供课程平时学习及期末复习使用。考点及考试范围主要参考王仲老师的讲课内容及2024-2025学年期末考试题目，由于历年考试范围不同，其覆盖范围可能与实际考试范围有所出入。
第四章 时变电磁场
一、知识归纳
1. 电磁感应定律
闭合回路内感应电动势 E \mathcal{E} E d Φ m d t \frac{d\Phi_m}{dt} d t d Φ m
E = − d Φ m d t = − d d t ∫ S B ⃗ ⋅ d S ⃗ \mathcal{E} = -\frac{d\Phi_m}{dt} = -\frac{d}{dt}\int_S \vec{B} \cdot d\vec{S}
E = − d t d Φ m = − d t d ∫ S B ⋅ d S
一般感应电动势为感生与动生电动势叠加，有
E = − ∫ S d B d t ⋅ d S ⃗ + ∮ l ( v ⃗ × B ⃗ ) ⋅ d l ⃗ \mathcal{E} = -\int_S \frac{dB}{dt} \cdot d\vec{S} + \oint_l (\vec{v} \times \vec{B}) \cdot d\vec{l}
E = − ∫ S d t d B ⋅ d S + ∮ l ( v × B ) ⋅ d l
2.
对给定导体回路，感应电动势I 感 ∝ E 感 I_感 \propto \mathcal{E}_感 I 感 ∝ E 感 d Φ m d t \frac{d\Phi_m}{dt} d t d Φ m
3.
由变化的磁场产生的电场为感应电场，与空间有无回路、导体何种媒介中无关（区别于法拉第电磁感应对回路而言），且有
∮ l E ⃗ 感 ⋅ d l ⃗ = − ∫ S d B d t ⋅ d S ⃗ + ∮ l ( v ⃗ × B ⃗ ) ⋅ d l ⃗ \oint_l \vec{E}_感 \cdot d\vec{l} = -\int_S \frac{dB}{dt} \cdot d\vec{S} + \oint_l (\vec{v} \times \vec{B}) \cdot d\vec{l}
∮ l E 感 ⋅ d l = − ∫ S d t d B ⋅ d S + ∮ l ( v × B ) ⋅ d l
由于 ∮ l E ⃗ 源 ⋅ d l ⃗ = 0 \oint_l \vec{E}_源 \cdot d\vec{l} = 0 ∮ l E 源 ⋅ d l = 0
∮ l E ⃗ ⋅ d l ⃗ = − ∫ S d B d t ⋅ d S ⃗ + ∮ l ( v ⃗ × B ⃗ ) ⋅ d l ⃗ \oint_l \vec{E} \cdot d\vec{l} = -\int_S \frac{dB}{dt} \cdot d\vec{S} + \oint_l (\vec{v} \times \vec{B}) \cdot d\vec{l}
∮ l E ⋅ d l = − ∫ S d t d B ⋅ d S + ∮ l ( v × B ) ⋅ d l
取其微分形式为
∇ × E ⃗ = − ∂ B ∂ t + ∇ × ( v ⃗ × B ⃗ ) \nabla \times \vec{E} = -\frac{\partial B}{\partial t} + \nabla \times (\vec{v} \times \vec{B})
∇ × E = − ∂ t ∂ B + ∇ × ( v × B )
4.
Maxwell 将电位移矢量D ⃗ \vec{D} D
∮ S ( J ⃗ + J d ⃗ ) ⋅ d S ⃗ = 0 \oint_S (\vec{J}+\vec{J_d}) \cdot d\vec{S} = 0
∮ S ( J + J d ) ⋅ d S = 0
从而有
∮ S J d ⃗ ⋅ d S ⃗ = − ∮ S J ⃗ ⋅ d S ⃗ = d q d t = d d t ∮ S D ⃗ ⋅ d S ⃗ = ∮ S d D ⃗ d t ⋅ d S ⃗ \oint_S \vec{J_d} \cdot d\vec{S} = -\oint_S \vec{J} \cdot d\vec{S} = \frac{dq}{dt} = \frac{d}{dt} \oint_S \vec{D} \cdot d\vec{S}= \oint_S \frac{d\vec{D}}{dt} \cdot d\vec{S}
∮ S J d ⋅ d S = − ∮ S J ⋅ d S = d t d q = d t d ∮ S D ⋅ d S = ∮ S d t d D ⋅ d S
进而有
J d ⃗ = d D ⃗ d t \vec{J_d} = \frac{d\vec{D}}{dt}
J d = d t d D
从而对于非恒定电流，有
∮ l H ⃗ ⋅ d l ⃗ = ∫ S ( J ⃗ + ∂ D ⃗ ∂ t ) ⋅ d S ⃗ \oint_l \vec{H} \cdot d\vec{l} = \int_S (\vec{J}+\frac{\partial \vec{D}}{\partial t}) \cdot d\vec{S}
∮ l H ⋅ d l = ∫ S ( J + ∂ t ∂ D ) ⋅ d S
其微分形式为
∇ × H ⃗ = J ⃗ + J d ⃗ \nabla \times \vec{H} = \vec{J} + \vec{J_d}
∇ × H = J + J d
称为全电流定律。
5.
位移电流虽与传导电流一样能激发磁场，但位移电流不伴随电荷运动（只表示电场变化），同时不产生焦耳热。虽然在电介质中由于 ∂ P ⃗ ∂ t \frac{\partial \vec{P}}{\partial t} ∂ t ∂ P
6. 电磁场基本方程组（Maxwell方程组）
积分形式：
{ ∮ ∂ V D ⋅ d S = Q ∮ ∂ V B ⋅ d S = 0 ∮ ∂ S E ⋅ d l = − d d t ∫ S B ⋅ d S ∮ ∂ S H ⋅ d l = ∫ S J ⋅ d S + d d t ∫ S D ⋅ d S \begin{cases}
\displaystyle \oint_{\partial V} \mathbf{D} \cdot d\mathbf{S} = Q \\[2ex]
\displaystyle \oint_{\partial V} \mathbf{B} \cdot d\mathbf{S} = 0 \\[2ex]
\displaystyle \oint_{\partial S} \mathbf{E} \cdot d\mathbf{l} = -\dfrac{d}{dt} \int_S \mathbf{B} \cdot d\mathbf{S} \\[2ex]
\displaystyle \oint_{\partial S} \mathbf{H} \cdot d\mathbf{l} = \int_S \mathbf{J} \cdot d\mathbf{S} + \frac{d}{dt} \int_S \mathbf{D} \cdot d\mathbf{S}
\end{cases}
⎩ ⎨ ⎧ ∮ ∂ V D ⋅ d S = Q ∮ ∂ V B ⋅ d S = 0 ∮ ∂ S E ⋅ d l = − d t d ∫ S B ⋅ d S ∮ ∂ S H ⋅ d l = ∫ S J ⋅ d S + d t d ∫ S D ⋅ d S
微分形式：
{ ∇ ⋅ D = ρ ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × H = J + ∂ D ∂ t \begin{cases}
\nabla \cdot \mathbf{D} = \rho \\
\nabla \cdot \mathbf{B} = 0 \\
\nabla \times \mathbf{E} = -\dfrac{\partial \mathbf{B}}{\partial t} \\
\nabla \times \mathbf{H} = \mathbf{J} + \dfrac{\partial \mathbf{D}}{\partial t}
\end{cases}
⎩ ⎨ ⎧ ∇ ⋅ D = ρ ∇ ⋅ B = 0 ∇ × E = − ∂ t ∂ B ∇ × H = J + ∂ t ∂ D
及本构方程
{ D ⃗ = ε E ⃗ J ⃗ = γ E ⃗ B ⃗ = μ H ⃗ \begin{cases}
\vec{D}=\varepsilon \vec{E} \\
\vec{J} = \gamma \vec{E} \\
\vec{B} = \mu \vec{H}
\end{cases}
⎩ ⎨ ⎧ D = ε E J = γ E B = μ H
7. 电磁场边界衔接条件
与电场+磁场一致，即
{ D 2 n − D 1 n = σ E 1 t = E 2 t \begin{cases}
D_{2n} - D_{1n} = \sigma \\
E_{1t} = E_{2t}
\end{cases}
{ D 2 n − D 1 n = σ E 1 t = E 2 t
及
{ H 1 t − H 2 t = K B 1 n = B 2 n \begin{cases}
H_{1t} - H_{2t} = K \\
B_{1n}=B_{2n}
\end{cases}
{ H 1 t − H 2 t = K B 1 n = B 2 n
同理当 σ = 0 \sigma=0 σ = 0 K = 0 K=0 K = 0
8.
在理想导体内，由于 γ → ∞ \gamma \to \infty γ → ∞ E ⃗ = 0 \vec{E}=0 E = 0 J ⃗ = 0 \vec{J}=0 J = 0 H ⃗ = 0 \vec{H}=0 H = 0
{ H 2 t = K B 2 n = B 1 n = 0 \begin{cases}
H_{2t} = K \\
B_{2n} = B_{1n} = 0
\end{cases}
{ H 2 t = K B 2 n = B 1 n = 0
{ D 2 n = σ E 1 t = E 2 t = 0 \begin{cases}
D_{2n}=\sigma \\
E_{1t}=E_{2t}=0
\end{cases}
{ D 2 n = σ E 1 t = E 2 t = 0
这表明理想导体外侧介质中，磁力线平行于分界面而电力线垂直于分界面。
9.
引入矢量A ⃗ \vec{A} A B ⃗ = ∇ × A ⃗ \vec{B} = \nabla \times \vec{A} B = ∇ × A ∂ B ⃗ ∂ t = − ∇ × E ⃗ \frac{\partial \vec{B}}{\partial t} = - \nabla \times \vec{E} ∂ t ∂ B = − ∇ × E
∇ × ( E ⃗ + ∂ A ⃗ ∂ t ) = 0 \nabla \times (\vec{E} + \frac{\partial \vec{A}}{\partial t}) = 0
∇ × ( E + ∂ t ∂ A ) = 0
故而存在一个标量函数 φ \varphi φ E ⃗ + ∂ A ⃗ ∂ t = − ∇ φ \vec{E} + \frac{\partial \vec{A}}{\partial t} = -\nabla \varphi E + ∂ t ∂ A = − ∇ φ
E ⃗ = − ∂ A ⃗ ∂ t − ∇ φ \vec{E} = -\frac{\partial \vec{A}}{\partial t} - \nabla \varphi
E = − ∂ t ∂ A − ∇ φ
其中 A ⃗ \vec{A} A φ \varphi φ
10.
将动态位函数及本构方程代入∇ × H ⃗ = J ⃗ + J d ⃗ \nabla \times \vec{H} = \vec{J} + \vec{J_d} ∇ × H = J + J d
∇ 2 A ⃗ − μ ε ∂ 2 A ⃗ ∂ t 2 = − μ J ⃗ + ∇ ( ∇ ⋅ A ⃗ + μ ε ∂ 2 φ ∂ t 2 ) \nabla^2 \vec{A} - \mu \varepsilon \frac{\partial ^2 \vec{A}}{\partial t^2} = -\mu \vec{J} + \nabla(\nabla \cdot \vec{A} + \mu \varepsilon \frac{\partial ^2 \varphi}{\partial t^2})
∇ 2 A − μ ε ∂ t 2 ∂ 2 A = − μ J + ∇ ( ∇ ⋅ A + μ ε ∂ t 2 ∂ 2 φ )
∇ 2 φ + ∂ ∂ t ( ∇ ⋅ A ⃗ ) = − ρ ε \nabla^2 \varphi + \frac{\partial }{\partial t}(\nabla \cdot \vec{A})=-\frac{\rho}{\varepsilon}
∇ 2 φ + ∂ t ∂ ( ∇ ⋅ A ) = − ε ρ
规定
∇ ⋅ A ⃗ = − μ ε ∂ φ ∂ t \nabla \cdot \vec{A} = -\mu \varepsilon \frac{\partial \varphi}{\partial t}
∇ ⋅ A = − μ ε ∂ t ∂ φ
上式化简为方程组
{ ∇ 2 A ⃗ − μ ε ∂ 2 A ⃗ ∂ t 2 = − μ J ⃗ ∇ 2 φ − μ ε ∂ 2 φ ∂ t 2 = − ρ ε \begin{cases}
\nabla^2 \vec{A} - \mu \varepsilon \frac{\partial ^2 \vec{A}}{\partial t^2} = -\mu \vec{J} \\
\nabla^2 \varphi - \mu \varepsilon \frac{\partial^2 \varphi}{\partial t^2}=-\frac{\rho}{\varepsilon}
\end{cases}
{ ∇ 2 A − μ ε ∂ t 2 ∂ 2 A = − μ J ∇ 2 φ − μ ε ∂ t 2 ∂ 2 φ = − ε ρ
称为 D’Alembert 方程。
11.
D’Alembert 方程的通解为
φ = f 1 ( t − r v ) v + f 2 ( t + r v ) v v = 1 ε 0 μ 0 \varphi = \frac{f_1(t - \frac{r}{v})}{v} + \frac{f_2(t + \frac{r}{v})}{v} \quad v=\frac{1}{\sqrt{\varepsilon_0 \mu_0}}
φ = v f 1 ( t − v r ) + v f 2 ( t + v r ) v = ε 0 μ 0 1
其中 f 1 ( t − r v ) f_1(t - \frac{r}{v}) f 1 ( t − v r ) f 2 ( t + r v ) f_2(t + \frac{r}{v}) f 2 ( t + v r )
12. 动态位的积分形式解
φ ( r ⃗ , t ) = 1 4 π ε ∫ V ρ ( r ⃗ ′ , t ′ ) d V ′ ( r − r ′ ) 2 \varphi(\vec{r}, t) = \frac{1}{4\pi \varepsilon} \int_V \frac{\rho(\vec{r}', t') dV'}{\sqrt{(r - r')^2}}
φ ( r , t ) = 4 π ε 1 ∫ V ( r − r ′ ) 2 ρ ( r ′ , t ′ ) d V ′
A ⃗ ( r ⃗ , t ) = μ 4 π ∫ V J ⃗ ( r ⃗ ′ , t ′ ) d V ′ ( r − r ′ ) 2 \vec{A}(\vec{r}, t) = \frac{\mu}{4\pi} \int_V \frac{\vec{J}(\vec{r}', t') dV'}{\sqrt{(r - r')^2}}
A ( r , t ) = 4 π μ ∫ V ( r − r ′ ) 2 J ( r ′ , t ′ ) d V ′
说明在时刻，场中某点 r r r t = t 0 t = t_0 t = t 0 v = 1 ε 0 μ 0 v = \frac{1}{\sqrt{\varepsilon_0 \mu_0}} v = ε 0 μ 0 1 3 × 10 8 3 \times 10^8 3 × 1 0 8
13. 电磁场能量
电磁场能量定域于场中，其体密度为电场与磁场能量密度之和。从而有
w = 1 2 ( E ⋅ D + B ⋅ H ) w = \frac{1}{2}(\mathbf{E}\cdot\mathbf{D} + \mathbf{B}\cdot\mathbf{H})
w = 2 1 ( E ⋅ D + B ⋅ H )
W = ∫ V w d V = ∫ V 1 2 ( E ⋅ D + B ⋅ H ) d V W = \int_V w \, dV = \int_V \frac{1}{2}(\mathbf{E}\cdot\mathbf{D} + \mathbf{B}\cdot\mathbf{H}) \, dV
W = ∫ V w d V = ∫ V 2 1 ( E ⋅ D + B ⋅ H ) d V
考虑能量随时间的变化率，利用麦克斯韦方程组及矢量恒等式∇ ⋅ ( E × H ) = H ⋅ ( ∇ × E ) − E ⋅ ( ∇ × H ) \nabla\cdot(\mathbf{E}\times\mathbf{H}) = \mathbf{H}\cdot(\nabla\times\mathbf{E}) - \mathbf{E}\cdot(\nabla\times\mathbf{H}) ∇ ⋅ ( E × H ) = H ⋅ ( ∇ × E ) − E ⋅ ( ∇ × H )
− ∂ W ∂ t = ∮ A S ⋅ d A + ∫ V E ⋅ J d V -\frac{\partial W}{\partial t} = \oint_A \mathbf{S}\cdot d\mathbf{A} + \int_V \mathbf{E}\cdot\mathbf{J} \, dV
− ∂ t ∂ W = ∮ A S ⋅ d A + ∫ V E ⋅ J d V
其中 S = E × H \mathbf{S} = \mathbf{E}\times\mathbf{H} S = E × H ∫ V E ⋅ J d V \int_V \mathbf{E}\cdot\mathbf{J} \, dV ∫ V E ⋅ J d V V V V
若存在电源局外场强 E e \mathbf{E}_e E e E + E e \mathbf{E} + \mathbf{E}_e E + E e J \mathbf{J} J J = σ ( E + E e ) \mathbf{J} = \sigma(\mathbf{E}+\mathbf{E}_e) J = σ ( E + E e )
− ∂ W ∂ t = ∮ A S ⋅ d A + ∫ V E ⋅ J d V − ∫ V E e ⋅ J d V -\frac{\partial W}{\partial t} = \oint_A \mathbf{S}\cdot d\mathbf{A} + \int_V \mathbf{E}\cdot\mathbf{J} \, dV - \int_V \mathbf{E}_e\cdot\mathbf{J} \, dV
− ∂ t ∂ W = ∮ A S ⋅ d A + ∫ V E ⋅ J d V − ∫ V E e ⋅ J d V
或等价地
∮ A S ⋅ d A = − ∂ W ∂ t − ∫ V E ⋅ J d V + ∫ V E e ⋅ J d V \oint_A \mathbf{S}\cdot d\mathbf{A} = -\frac{\partial W}{\partial t} - \int_V \mathbf{E}\cdot\mathbf{J} \, dV + \int_V \mathbf{E}_e\cdot\mathbf{J} \, dV
∮ A S ⋅ d A = − ∂ t ∂ W − ∫ V E ⋅ J d V + ∫ V E e ⋅ J d V
此即电磁场中的能量守恒及转化定律（电磁能流定理/坡印廷定理）。其中 ∫ V E ⋅ J d V \int_V \mathbf{E}\cdot\mathbf{J} \, dV ∫ V E ⋅ J d V V V V ∫ V E e ⋅ J d V \int_V \mathbf{E}_e\cdot\mathbf{J} \, dV ∫ V E e ⋅ J d V ∮ A S ⋅ d A \oint_A \mathbf{S}\cdot d\mathbf{A} ∮ A S ⋅ d A A A A
坡印廷矢量 S = E × H \mathbf{S} = \mathbf{E}\times\mathbf{H} S = E × H
对恒定场，由于场不随时间变化，∂ W ∂ t = 0 \frac{\partial W}{\partial t}=0 ∂ t ∂ W = 0 E e = 0 \mathbf{E}_e = 0 E e = 0
∮ A S ⋅ d A = − ∫ V E ⋅ J d V \oint_A \mathbf{S}\cdot d\mathbf{A} = -\int_V \mathbf{E}\cdot\mathbf{J} \, dV
∮ A S ⋅ d A = − ∫ V E ⋅ J d V
负号表示流入闭合面的能流等于体积内的焦耳热损耗，说明导电媒质内的焦耳热是由外部流入的电磁能供给的。
14.
电磁能量的储存与传递都由电磁场完成，导体仅起引导电磁能流的作用，故而称为导波系统。
15. 正弦时变电磁场的复数形式
对于正弦时变电磁场，可将其写作相量形式（具体方式参见电路原理），由此可得到复数形式的Maxwell方程组。在正弦稳态下，场量随时间做简谐变化，可用复矢量表示，时间导数 ∂ ∂ t \frac{\partial}{\partial t} ∂ t ∂ j ω j\omega jω
复数形式的Maxwell方程组为：
∇ × H ~ = J ~ + j ω D ~ \nabla \times \mathbf{\tilde{H}} = \mathbf{\tilde{J}} + j\omega \mathbf{\tilde{D}}
∇ × H ~ = J ~ + jω D ~
∇ × E ~ = − j ω B ~ \nabla \times \mathbf{\tilde{E}} = -j\omega \mathbf{\tilde{B}}
∇ × E ~ = − jω B ~
∇ ⋅ D ~ = ρ ~ \nabla \cdot \mathbf{\tilde{D}} = \tilde{\rho}
∇ ⋅ D ~ = ρ ~
∇ ⋅ B ~ = 0 \nabla \cdot \mathbf{\tilde{B}} = 0
∇ ⋅ B ~ = 0
对于线性各向同性媒质，本构关系为：
D ~ = ε E ~ \mathbf{\tilde{D}} = \varepsilon \mathbf{\tilde{E}}
D ~ = ε E ~
B ~ = μ H ~ \mathbf{\tilde{B}} = \mu \mathbf{\tilde{H}}
B ~ = μ H ~
J ~ = σ E ~ \mathbf{\tilde{J}} = \sigma \mathbf{\tilde{E}}
J ~ = σ E ~
16. Poynting 矢量复数形式
对于正弦时变电磁场，引入复矢量后，可定义复数形式的Poynting矢量：
S ~ = 1 2 E ~ × H ~ ∗ \tilde{\mathbf{S}} = \frac{1}{2} \tilde{\mathbf{E}} \times \tilde{\mathbf{H}}^*
S ~ = 2 1 E ~ × H ~ ∗
其中 E ~ \tilde{\mathbf{E}} E ~ H ~ \tilde{\mathbf{H}} H ~ H ~ ∗ \tilde{\mathbf{H}}^* H ~ ∗ H ~ \tilde{\mathbf{H}} H ~
平均功率流密度（时间平均Poynting矢量）为：
S a v = Re [ S ~ ] = 1 2 Re [ E ~ × H ~ ∗ ] \mathbf{S}_{av} = \text{Re}[\tilde{\mathbf{S}}] = \frac{1}{2} \text{Re}[\tilde{\mathbf{E}} \times \tilde{\mathbf{H}}^*]
S a v = Re [ S ~ ] = 2 1 Re [ E ~ × H ~ ∗ ]
由此可导出复数形式的Poynting定理（复功率定理）。对于包含线性媒质的区域 V V V A A A
− ∮ A S ~ ⋅ d A = ∫ V 1 2 J ~ ∗ ⋅ E ~ d V + j ω ∫ V ( 1 2 μ ∣ H ~ ∣ 2 − 1 2 ε ∣ E ~ ∣ 2 ) d V -\oint_A \tilde{\mathbf{S}} \cdot d\mathbf{A} = \int_V \frac{1}{2} \mathbf{\tilde{J}}^* \cdot \tilde{\mathbf{E}} \, dV + j\omega \int_V \left( \frac{1}{2} \mu |\tilde{\mathbf{H}}|^2 - \frac{1}{2} \varepsilon |\tilde{\mathbf{E}}|^2 \right) dV
− ∮ A S ~ ⋅ d A = ∫ V 2 1 J ~ ∗ ⋅ E ~ d V + jω ∫ V ( 2 1 μ ∣ H ~ ∣ 2 − 2 1 ε ∣ E ~ ∣ 2 ) d V
其中 J ~ \tilde{\mathbf{J}} J ~ μ \mu μ ε \varepsilon ε
在无源区域（J ~ = 0 \tilde{\mathbf{J}} = 0 J ~ = 0
− ∮ A S ~ ⋅ d A = P + j Q -\oint_A \tilde{\mathbf{S}} \cdot d\mathbf{A} = P + jQ
− ∮ A S ~ ⋅ d A = P + j Q
其中实部 P P P A A A V V V P = I 2 R P = I^2 R P = I 2 R Q Q Q Q = I 2 X Q = I^2 X Q = I 2 X R R R X X X
17. D’Alembert方程的复数形式及其解
对于正弦时变电磁场，在洛伦兹规范条件下，动态势（矢量磁位 A ~ \tilde{\mathbf{A}} A ~ ψ ~ \tilde{\psi} ψ ~
∇ 2 A ~ + β 2 A ~ = − μ J ~ \nabla^2 \tilde{\mathbf{A}} + \beta^2 \tilde{\mathbf{A}} = -\mu \tilde{\mathbf{J}}
∇ 2 A ~ + β 2 A ~ = − μ J ~
∇ 2 ψ ~ + β 2 ψ ~ = − ρ ~ ε \nabla^2 \tilde{\psi} + \beta^2 \tilde{\psi} = -\frac{\tilde{\rho}}{\varepsilon}
∇ 2 ψ ~ + β 2 ψ ~ = − ε ρ ~
其中 β = ω / c = ω μ ε \beta = \omega / c = \omega \sqrt{\mu\varepsilon} β = ω / c = ω μ ε J ~ \tilde{\mathbf{J}} J ~ ρ ~ \tilde{\rho} ρ ~
上述方程的解为推迟势（retarded potentials），考虑滞后效应后其复数形式为 ：
ψ ~ ( r ) = 1 4 π ε ∫ V ρ ~ ( r ′ ) e − j β R R d V ′ \tilde{\psi}(\mathbf{r}) = \frac{1}{4\pi\varepsilon} \int_V \frac{\tilde{\rho}(\mathbf{r}') e^{-j\beta R}}{R} dV'
ψ ~ ( r ) = 4 π ε 1 ∫ V R ρ ~ ( r ′ ) e − j βR d V ′
A ~ ( r ) = μ 4 π ∫ V J ~ ( r ′ ) e − j β R R d V ′ \tilde{\mathbf{A}}(\mathbf{r}) = \frac{\mu}{4\pi} \int_V \frac{\tilde{\mathbf{J}}(\mathbf{r}') e^{-j\beta R}}{R} dV'
A ~ ( r ) = 4 π μ ∫ V R J ~ ( r ′ ) e − j βR d V ′
其中 R = ∣ r − r ′ ∣ R = |\mathbf{r} - \mathbf{r}'| R = ∣ r − r ′ ∣ e − j β R e^{-j\beta R} e − j βR
求得势函数后，电磁场量可由下式计算：
B ~ = ∇ × A ~ \tilde{\mathbf{B}} = \nabla \times \tilde{\mathbf{A}}
B ~ = ∇ × A ~
E ~ = − j ω A ~ − ∇ ψ ~ \tilde{\mathbf{E}} = -j\omega \tilde{\mathbf{A}} - \nabla \tilde{\psi}
E ~ = − jω A ~ − ∇ ψ ~
当 β R ≪ 1 \beta R \ll 1 βR ≪ 1 R ≪ λ R \ll \lambda R ≪ λ λ \lambda λ e − j β R ≈ 1 e^{-j\beta R} \approx 1 e − j βR ≈ 1 β R ≪ 1 \beta R \ll 1 βR ≪ 1
二、题型总结
第四章的题目以概念直接应用为主，下面以此举例
1、电磁感应定律
例 4.1.1 （P160 T4-1-1）长直导线载有电流 i = I m sin ω t i = I_m \sin\omega t i = I m sin ω t v v v
由毕-萨定理/由Ampere环路定律，有 B ⋅ 2 π r = μ 0 I m sin ω t B \cdot 2\pi r = \mu_0 I_m \sin\omega t B ⋅ 2 π r = μ 0 I m sin ω t
⇒ B = μ 0 i 2 π r \Rightarrow B = \frac{\mu_0 i}{2\pi r} ⇒ B = 2 π r μ 0 i Φ m = ∫ c c + a μ 0 i 2 π r ⋅ b ⋅ d r = μ 0 b i 2 π ln ( 1 + a c ) \Phi_m = \int_c^{c+a} \frac{\mu_0 i}{2\pi r} \cdot b \cdot dr = \frac{\mu_0 b i}{2\pi} \ln\left(1+\frac{a}{c}\right) Φ m = ∫ c c + a 2 π r μ 0 i ⋅ b ⋅ d r = 2 π μ 0 bi ln ( 1 + c a )
∂ Φ m ∂ t = μ 0 b 2 π ln ( 1 + a c ) I m ω cos ω t \frac{\partial \Phi_m}{\partial t} = \frac{\mu_0 b}{2\pi} \ln\left(1+\frac{a}{c}\right) I_m \omega \cos\omega t
∂ t ∂ Φ m = 2 π μ 0 b ln ( 1 + c a ) I m ω cos ω t
而 ∫ l ( v ⃗ × B ⃗ ) ⋅ d l ⃗ = b v × μ 0 I m sin ω t 2 π ( 1 c − 1 c + a ) = μ 0 a b v I m sin ω t 2 π c ( c + a ) \int_l (\vec{v} \times \vec{B}) \cdot d\vec{l} = bv \times \frac{\mu_0 I_m \sin\omega t}{2\pi} \left(\frac{1}{c} - \frac{1}{c+a}\right) = \frac{\mu_0 abv I_m \sin\omega t}{2\pi c(c+a)} ∫ l ( v × B ) ⋅ d l = b v × 2 π μ 0 I m s i n ω t ( c 1 − c + a 1 ) = 2 π c ( c + a ) μ 0 ab v I m s i n ω t
∴ E = − μ 0 b 2 π ln ( 1 + a c ) I m ω cos ω t + μ 0 a b v 2 π c ( c + a ) I m sin ω t ，方向沿顺时针 \therefore \mathcal{E} = -\frac{\mu_0 b}{2\pi} \ln\left(1+\frac{a}{c}\right) I_m \omega \cos\omega t + \frac{\mu_0 abv}{2\pi c(c+a)} I_m \sin\omega t，方向沿顺时针
∴ E = − 2 π μ 0 b ln ( 1 + c a ) I m ω cos ω t + 2 π c ( c + a ) μ 0 ab v I m sin ω t ，方向沿顺时针
2、传导电流与位移电流
例 4.2.1 （P151 T4-1-2）电场强度 E ( t ) = E m cos ω t E(t) = E_m \cos\omega t E ( t ) = E m cos ω t ω = 10 3 \omega = 10^3 ω = 1 0 3 γ = 5.8 × 10 7 \gamma = 5.8 \times 10^7 γ = 5.8 × 1 0 7 ε r = 1 \varepsilon_r = 1 ε r = 1 γ = 10 − 16 \gamma = 10^{-16} γ = 1 0 − 16 ε r = 2.53 \varepsilon_r = 2.53 ε r = 2.53
（1）J m 传 = γ E m J_{m\text{传}} = \gamma E_m J m 传 = γ E m J m 位 = ε E m ω J_{m\text{位}} = \varepsilon E_m \omega J m 位 = ε E m ω J m 传 J m 位 = γ E m ε r E m ω = γ ε r ω = 5.8 × 10 4 \frac{J_{m\text{传}}}{J_{m\text{位}}} = \frac{\gamma E_m}{\varepsilon_r E_m \omega} = \frac{\gamma}{\varepsilon_r \omega} = 5.8 \times 10^4 J m 位 J m 传 = ε r E m ω γ E m = ε r ω γ = 5.8 × 1 0 4
（2）同理有 J m 传 J m 位 = γ ε r ω = 10 − 16 2.53 × 10 3 = 3.95 × 10 − 20 \frac{J_{m\text{传}}}{J_{m\text{位}}} = \frac{\gamma}{\varepsilon_r \omega} = \frac{10^{-16}}{2.53 \times 10^3} = 3.95 \times 10^{-20} J m 位 J m 传 = ε r ω γ = 2.53 × 1 0 3 1 0 − 16 = 3.95 × 1 0 − 20
3、全电流定律
例 4.3.1 （P184 T4-3）由圆形极板构成平行板电容器，板间距离为 d d d γ \gamma γ ε \varepsilon ε μ 0 \mu_0 μ 0 u = U m sin ω t u = U_m \sin\omega t u = U m sin ω t
∵ u = U m sin ω t ⇒ E = u d = U m d sin ω t ， D = ε E = ε U m d sin ω t \because u = U_m \sin\omega t \Rightarrow E = \frac{u}{d} = \frac{U_m}{d} \sin\omega t，D = \varepsilon E = \frac{\varepsilon U_m}{d} \sin\omega t
∵ u = U m sin ω t ⇒ E = d u = d U m sin ω t ， D = εE = d ε U m sin ω t
∴ J d = ∂ D ∂ t = ε U m ω d cos ω t ，方向从正极板指向负极板 \therefore J_d = \frac{\partial D}{\partial t} = \frac{\varepsilon U_m \omega}{d} \cos\omega t，方向从正极板指向负极板
∴ J d = ∂ t ∂ D = d ε U m ω cos ω t ，方向从正极板指向负极板
由全电流定律，设某点到轴线距离为 r r r H ⋅ 2 π r = π r 2 ⋅ ε U m ω d cos ω t + γ U m d π r 2 sin ω t H \cdot 2\pi r = \pi r^2 \cdot \frac{\varepsilon U_m \omega}{d} \cos\omega t + \frac{\gamma U_m}{d} \pi r^2 \sin\omega t H ⋅ 2 π r = π r 2 ⋅ d ε U m ω cos ω t + d γ U m π r 2 sin ω t
⇒ H = ε U m ω r 2 d cos ω t + γ U m r 2 d sin ω t \Rightarrow H = \frac{\varepsilon U_m \omega r}{2d} \cos\omega t + \frac{\gamma U_m r}{2d} \sin\omega t
⇒ H = 2 d ε U m ω r cos ω t + 2 d γ U m r sin ω t
∴ B = μ 0 H = μ 0 U m r 2 d ( ε ω cos ω t + γ sin ω t ) \therefore B = \mu_0 H = \frac{\mu_0 U_m r}{2d} (\varepsilon \omega \cos\omega t + \gamma \sin\omega t)
∴ B = μ 0 H = 2 d μ 0 U m r ( ε ω cos ω t + γ sin ω t )
4、Maxwell方程组及衔接条件（理想介质）
例 4.4.1 （P155 T4-2-1）证明下述场量满足Maxwell方程组：E ⃗ = cos ( y − c t ) e ⃗ z \vec{E} = \cos(y-ct)\vec{e}_z E = cos ( y − c t ) e z B ⃗ = cos ( y − c t ) c e ⃗ x \vec{B} = \frac{\cos(y-ct)}{c}\vec{e}_x B = c c o s ( y − c t ) e x
∵ ∇ × E ⃗ = − sin ( y − c t ) e ⃗ x ， ∂ B ⃗ ∂ t = sin ( y − c t ) e ⃗ x ， ∇ × E ⃗ = − ∂ B ⃗ ∂ t \because \nabla \times \vec{E} = -\sin(y-ct)\vec{e}_x，\frac{\partial \vec{B}}{\partial t} = \sin(y-ct)\vec{e}_x，\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}
∵ ∇ × E = − sin ( y − c t ) e x ， ∂ t ∂ B = sin ( y − c t ) e x ， ∇ × E = − ∂ t ∂ B
∇ ⋅ B ⃗ = ∂ ( cos ( y − c t ) ) ∂ x = 0 ， ∇ ⋅ D ⃗ = ε ∇ ⋅ E ⃗ = 0 \nabla \cdot \vec{B} = \frac{\partial(\cos(y-ct))}{\partial x} = 0，\nabla \cdot \vec{D} = \varepsilon \nabla \cdot \vec{E} = 0
∇ ⋅ B = ∂ x ∂ ( cos ( y − c t )) = 0 ， ∇ ⋅ D = ε ∇ ⋅ E = 0
而 ∇ × H ⃗ = 1 μ ∇ × B ⃗ = 1 μ sin ( y − c t ) c e ⃗ z \nabla \times \vec{H} = \frac{1}{\mu} \nabla \times \vec{B} = \frac{1}{\mu} \frac{\sin(y-ct)}{c}\vec{e}_z ∇ × H = μ 1 ∇ × B = μ 1 c s i n ( y − c t ) e z ∂ D ⃗ ∂ t = ε ∂ E ⃗ ∂ t = c ε sin ( y − c t ) e ⃗ z \frac{\partial \vec{D}}{\partial t} = \varepsilon \frac{\partial \vec{E}}{\partial t} = c\varepsilon \sin(y-ct)\vec{e}_z ∂ t ∂ D = ε ∂ t ∂ E = c ε sin ( y − c t ) e z
又 ∵ \because ∵ γ = 0 \gamma = 0 γ = 0 J ⃗ = 0 \vec{J} = 0 J = 0 c 2 = 1 μ ε c^2 = \frac{1}{\mu\varepsilon} c 2 = μ ε 1 ∴ ∇ × H ⃗ = ∂ D ⃗ ∂ t \therefore \nabla \times \vec{H} = \frac{\partial \vec{D}}{\partial t} ∴ ∇ × H = ∂ t ∂ D
从而Maxwell第一、二方程得证，亦即符合Maxwell方程组
P.S. Maxwell方程组中事实上只有Maxwell第一、二方程是独立的，若要证明场量满足Maxwell方程组，仅需验证前两个。
例 4.4.2 （P156 T4-2-3）设 z = 0 z = 0 z = 0 z < 0 z \lt 0 z < 0 H ⃗ ( x , y , z , t ) = H 0 sin β x cos ( ω t − β y ) e ⃗ x \vec{H}(x,y,z,t) = H_0 \sin\beta x \cos(\omega t - \beta y)\vec{e}_x H ( x , y , z , t ) = H 0 sin β x cos ( ω t − β y ) e x E ⃗ \vec{E} E
由题意，∇ × H ⃗ = ∂ D ⃗ ∂ t = − β H 0 sin β x sin ( ω t − β y ) e ⃗ z = ε 0 ∂ E ⃗ ∂ t \nabla \times \vec{H} = \frac{\partial \vec{D}}{\partial t} = -\beta H_0 \sin\beta x \sin(\omega t - \beta y)\vec{e}_z = \varepsilon_0 \frac{\partial \vec{E}}{\partial t} ∇ × H = ∂ t ∂ D = − β H 0 sin β x sin ( ω t − β y ) e z = ε 0 ∂ t ∂ E
从而 E ⃗ = β H 0 ε 0 ω sin β x cos ( ω t − β y ) e ⃗ z \vec{E} = \frac{\beta H_0}{\varepsilon_0 \omega} \sin\beta x \cos(\omega t - \beta y)\vec{e}_z E = ε 0 ω β H 0 sin β x cos ( ω t − β y ) e z E t = 0 E_t = 0 E t = 0
又 K ⃗ = n ⃗ × H ⃗ \vec{K} = \vec{n} \times \vec{H} K = n × H K = H z t = H 0 sin β x cos ( ω t − β y ) K = H_{zt} = H_0 \sin\beta x \cos(\omega t - \beta y) K = H z t = H 0 sin β x cos ( ω t − β y ) K ⃗ = n ⃗ × H ⃗ = K e ⃗ y \vec{K} = \vec{n} \times \vec{H} = K\vec{e}_y K = n × H = K e y
P.S. 注意理想导体分界面条件，依次带入衔接条件计算即可。
5、动态位及D’Alembert方程
例 4.5.1 （P160 T4-3-1）在时变电磁场中，已知矢量位函数 A ⃗ = A m sin ( ω t − β z ) e ⃗ x \vec{A} = A_m \sin(\omega t - \beta z)\vec{e}_x A = A m sin ( ω t − β z ) e x A m A_m A m β \beta β E ⃗ \vec{E} E H ⃗ \vec{H} H
由动态位的D’Alembert方程有 ∇ 2 A ⃗ − μ ε ∂ 2 A ⃗ ∂ t 2 = [ β 2 A m sin ( ω t − β z ) + μ ε A m ω 2 sin ( ω t − β z ) ] e ⃗ x \nabla^2 \vec{A} - \mu\varepsilon \frac{\partial^2 \vec{A}}{\partial t^2} = \left[\beta^2 A_m \sin(\omega t - \beta z) + \mu\varepsilon A_m \omega^2 \sin(\omega t - \beta z)\right]\vec{e}_x ∇ 2 A − μ ε ∂ t 2 ∂ 2 A = [ β 2 A m sin ( ω t − β z ) + μ ε A m ω 2 sin ( ω t − β z ) ] e x
J ⃗ = [ − ε A m ω 2 sin ( ω t − β z ) + β 2 μ A m sin ( ω t − β z ) ] e ⃗ x \vec{J} = \left[-\varepsilon A_m \omega^2 \sin(\omega t - \beta z) + \frac{\beta^2}{\mu} A_m \sin(\omega t - \beta z)\right]\vec{e}_x
J = [ − ε A m ω 2 sin ( ω t − β z ) + μ β 2 A m sin ( ω t − β z ) ] e x
∵ ∇ ⋅ A ⃗ = 0 \because \nabla \cdot \vec{A} = 0 ∵ ∇ ⋅ A = 0 μ ε ∂ φ ∂ t = 0 \mu\varepsilon \frac{\partial \varphi}{\partial t} = 0 μ ε ∂ t ∂ φ = 0 φ = C ( x , y , z ) \varphi = C(x,y,z) φ = C ( x , y , z )
在时变电磁场中，不考虑静电场存在，从而有 ∇ φ = 0 \nabla \varphi = 0 ∇ φ = 0
E ⃗ = − ∂ A ⃗ ∂ t = − ω A m cos ( ω t − β z ) e ⃗ x \vec{E} = -\frac{\partial \vec{A}}{\partial t} = -\omega A_m \cos(\omega t - \beta z)\vec{e}_x
E = − ∂ t ∂ A = − ω A m cos ( ω t − β z ) e x
而 B ⃗ = ∇ × A ⃗ = − β A m sin ( ω t − β z ) e ⃗ y \vec{B} = \nabla \times \vec{A} = -\beta A_m \sin(\omega t - \beta z)\vec{e}_y B = ∇ × A = − β A m sin ( ω t − β z ) e y ∴ H ⃗ = B ⃗ μ = − β A m μ sin ( ω t − β z ) e ⃗ y \therefore \vec{H} = \frac{\vec{B}}{\mu} = -\frac{\beta A_m}{\mu} \sin(\omega t - \beta z)\vec{e}_y ∴ H = μ B = − μ β A m sin ( ω t − β z ) e y
P.S. 本题特殊之处在于"在时变电磁场中，暂不考虑静电场存在"，从而有 ∇ φ = 0 \nabla \varphi = 0 ∇ φ = 0
6、电磁能传输计算
例 4.6.1 （P164 T4-4-2）圆柱形导线长为 l l l R R R I I I − ∮ A ( E ⃗ × H ⃗ ) ⋅ d A ⃗ -\oint_A (\vec{E} \times \vec{H}) \cdot d\vec{A} − ∮ A ( E × H ) ⋅ d A I 2 R I^2 R I 2 R
依题意，取单位长导线，设其截面积为 S S S R = 1 γ ⋅ l S ⇒ γ = l S R R = \frac{1}{\gamma} \cdot \frac{l}{S} \Rightarrow \gamma = \frac{l}{SR} R = γ 1 ⋅ S l ⇒ γ = SR l
而 J ⃗ = I S e ⃗ z \vec{J} = \frac{I}{S}\vec{e}_z J = S I e z E ⃗ = 1 γ J ⃗ = S R l J ⃗ \vec{E} = \frac{1}{\gamma}\vec{J} = \frac{SR}{l}\vec{J} E = γ 1 J = l SR J H ⋅ 2 π r = I S ⋅ π r 2 ⇒ H ⃗ = I r 2 S e ⃗ ϕ H \cdot 2\pi r = \frac{I}{S} \cdot \pi r^2 \Rightarrow \vec{H} = \frac{Ir}{2S}\vec{e}_\phi H ⋅ 2 π r = S I ⋅ π r 2 ⇒ H = 2 S I r e ϕ
∴ S ⃗ = E ⃗ × H ⃗ = − R I l ⋅ I r 2 S e ⃗ ρ = − I 2 R r 2 S l e ⃗ ρ \therefore \vec{S} = \vec{E} \times \vec{H} = -\frac{RI}{l} \cdot \frac{Ir}{2S}\vec{e}_\rho = -\frac{I^2 R r}{2Sl}\vec{e}_\rho
∴ S = E × H = − l R I ⋅ 2 S I r e ρ = − 2 Sl I 2 R r e ρ
从而在半径为 a a a S = π a 2 S = \pi a^2 S = π a 2 − ∮ A S ⃗ ⋅ d A ⃗ = I 2 R a 2 S l ⋅ 2 π a ⋅ l = I 2 R -\oint_A \vec{S} \cdot d\vec{A} = \frac{I^2 R a}{2Sl} \cdot 2\pi a \cdot l = I^2 R − ∮ A S ⋅ d A = 2 Sl I 2 R a ⋅ 2 πa ⋅ l = I 2 R
例 4.6.2 （P164 T4-6）已知自由空间中电磁波两个场分量 E x = 1000 cos ( ω t − β z ) E_x = 1000\cos(\omega t - \beta z) E x = 1000 cos ( ω t − β z ) H y = 2.65 cos ( ω t − β z ) H_y = 2.65\cos(\omega t - \beta z) H y = 2.65 cos ( ω t − β z ) f = 20 f = 20 f = 20 β = ω μ 0 ε 0 = 0.42 \beta = \omega\sqrt{\mu_0\varepsilon_0} = 0.42 β = ω μ 0 ε 0 = 0.42
（1）瞬时Poynting矢量 （2）平均Poynting矢量 （3）流入图中平行六面体的瞬时功率
（1）S ⃗ = E ⃗ × H ⃗ = 2650 cos 2 ( ω t − β z ) e ⃗ z \vec{S} = \vec{E} \times \vec{H} = 2650\cos^2(\omega t - \beta z)\vec{e}_z S = E × H = 2650 cos 2 ( ω t − β z ) e z 2 ^2 2
（2）∵ E ⃗ ˙ = 1000 2 ∠ ( − β z ) e ⃗ x \because \dot{\vec{E}} = \frac{1000}{\sqrt{2}}\angle(-\beta z)\vec{e}_x ∵ E ˙ = 2 1000 ∠ ( − β z ) e x H ⃗ ˙ = 2.65 2 ∠ ( − β z ) e ⃗ y \dot{\vec{H}} = \frac{2.65}{\sqrt{2}}\angle(-\beta z)\vec{e}_y H ˙ = 2 2.65 ∠ ( − β z ) e y
S ⃗ ˙ = E ⃗ ˙ × H ⃗ ˙ ∗ = 2650 2 ∠ 0 e ⃗ z = 1325 e ⃗ z \dot{\vec{S}} = \dot{\vec{E}} \times \dot{\vec{H}}^* = \frac{2650}{2}\angle 0 \vec{e}_z = 1325\vec{e}_z
S ˙ = E ˙ × H ˙ ∗ = 2 2650 ∠0 e z = 1325 e z
∴ S a v = Re [ S ⃗ ˙ ] = 1325 e ⃗ z W/m 2 \therefore S_{av} = \text{Re}[\dot{\vec{S}}] = 1325\vec{e}_z \text{ W/m}^2
∴ S a v = Re [ S ˙ ] = 1325 e z W/m 2
（3）P = ( S ⃗ ∣ z = 0 − S ⃗ ∣ z = 1 ) ⋅ S 底 = 2650 [ cos 2 ( ω t ) − cos 2 ( ω t − β ) ] × 0.25 = 662.5 [ cos 2 ( ω t ) − cos 2 ( ω t − β ) ] P = (\vec{S}|_{z=0} - \vec{S}|_{z=1}) \cdot S_{\text{底}} = 2650[\cos^2(\omega t) - \cos^2(\omega t - \beta)] \times 0.25 = 662.5[\cos^2(\omega t) - \cos^2(\omega t - \beta)] P = ( S ∣ z = 0 − S ∣ z = 1 ) ⋅ S 底 = 2650 [ cos 2 ( ω t ) − cos 2 ( ω t − β )] × 0.25 = 662.5 [ cos 2 ( ω t ) − cos 2 ( ω t − β )]
P.S. 注意区分：瞬时功率 S ⃗ = E ⃗ × H ⃗ \vec{S} = \vec{E} \times \vec{H} S = E × H S ⃗ a v = Re ( S ⃗ ˙ ) = Re ( E ⃗ ˙ × H ⃗ ˙ ∗ ) \vec{S}_{av} = \text{Re}(\dot{\vec{S}}) = \text{Re}(\dot{\vec{E}} \times \dot{\vec{H}}^*) S a v = Re ( S ˙ ) = Re ( E ˙ × H ˙ ∗ )
电阻 R = − 1 I 2 Re [ ∮ A ( E ⃗ × H ⃗ ∗ ) ⋅ d A ⃗ ] R = -\frac{1}{I^2}\text{Re}\left[\oint_A (\vec{E} \times \vec{H}^*) \cdot d\vec{A}\right] R = − I 2 1 Re [ ∮ A ( E × H ∗ ) ⋅ d A ] X = − 1 I 2 Im [ ∮ A ( E ⃗ × H ⃗ ∗ ) ⋅ d A ⃗ ] X = -\frac{1}{I^2}\text{Im}\left[\oint_A (\vec{E} \times \vec{H}^*) \cdot d\vec{A}\right] X = − I 2 1 Im [ ∮ A ( E × H ∗ ) ⋅ d A ]
7、综合题
例 4.7.1 （P164 T4-8）同轴电缆内导体半径为 a = 1 a = 1 a = 1 b = 4 b = 4 b = 4 ε r = 2.25 \varepsilon_r = 2.25 ε r = 2.25 μ r = 1 \mu_r = 1 μ r = 1 γ = 0 \gamma = 0 γ = 0 E ⃗ = 100 ρ cos ( 10 8 t − β z ) e ⃗ ρ \vec{E} = \frac{100}{\rho}\cos(10^8 t - \beta z)\vec{e}_\rho E = ρ 100 cos ( 1 0 8 t − β z ) e ρ
（1）说明 E ⃗ \vec{E} E β \beta β H ⃗ \vec{H} H 0 ≤ z ≤ 1 0 \leq z \leq 1 0 ≤ z ≤ 1
（1）使式中 10 8 t − β z 10^8 t - \beta z 1 0 8 t − β z E ⃗ \vec{E} E 10 8 β = v \frac{10^8}{\beta} = v β 1 0 8 = v z z z
（2）∵ β 10 8 = 1 c = μ ε = 1.5 μ 0 ε 0 1 = 1.5 3 × 10 8 \because \frac{\beta}{10^8} = \frac{1}{c} = \sqrt{\mu\varepsilon} = \frac{1.5\sqrt{\mu_0\varepsilon_0}}{1} = \frac{1.5}{3 \times 10^8} ∵ 1 0 8 β = c 1 = μ ε = 1 1.5 μ 0 ε 0 = 3 × 1 0 8 1.5 ∴ β = 0.5 \therefore \beta = 0.5 ∴ β = 0.5
（续例4.7.1）
（3）∵ ∇ × E ⃗ = 100 β ρ sin ( 10 8 t − β z ) e ⃗ ϕ = − ∂ B ⃗ ∂ t \because \nabla \times \vec{E} = \frac{100\beta}{\rho}\sin(10^8 t - \beta z)\vec{e}_\phi = -\frac{\partial \vec{B}}{\partial t} ∵ ∇ × E = ρ 100 β sin ( 1 0 8 t − β z ) e ϕ = − ∂ t ∂ B ∴ B ⃗ = 100 β 10 8 ρ cos ( 10 8 t − β z ) e ⃗ ϕ = 5 × 10 − 7 ρ cos ( 10 8 t − 0.5 z ) e ⃗ ϕ \therefore \vec{B} = \frac{100\beta}{10^8 \rho}\cos(10^8 t - \beta z)\vec{e}_\phi = \frac{5 \times 10^{-7}}{\rho}\cos(10^8 t - 0.5z)\vec{e}_\phi ∴ B = 1 0 8 ρ 100 β cos ( 1 0 8 t − β z ) e ϕ = ρ 5 × 1 0 − 7 cos ( 1 0 8 t − 0.5 z ) e ϕ
∴ H ⃗ = B ⃗ μ 0 = 5 × 10 − 7 ρ μ 0 cos ( 10 8 t − 0.5 z ) e ⃗ ϕ \therefore \vec{H} = \frac{\vec{B}}{\mu_0} = \frac{5 \times 10^{-7}}{\rho\mu_0}\cos(10^8 t - 0.5z)\vec{e}_\phi
∴ H = μ 0 B = ρ μ 0 5 × 1 0 − 7 cos ( 1 0 8 t − 0.5 z ) e ϕ
（4）由 K ⃗ = n ⃗ × H ⃗ ∣ ρ = R 1 ρ = R 2 \vec{K} = \vec{n} \times \vec{H}|_{\rho=R_1}^{\rho=R_2} K = n × H ∣ ρ = R 1 ρ = R 2 K ⃗ = 5 × 10 − 4 μ 0 cos ( 10 8 t − 0.5 z ) e ⃗ z \vec{K} = \frac{5 \times 10^{-4}}{\mu_0}\cos(10^8 t - 0.5z)\vec{e}_z K = μ 0 5 × 1 0 − 4 cos ( 1 0 8 t − 0.5 z ) e z
（5）J ⃗ D = ∂ D ⃗ ∂ t = ε ∂ E ⃗ ∂ t = − ε × 10 10 ρ sin ( 10 8 t − β z ) e ⃗ ρ \vec{J}_D = \frac{\partial \vec{D}}{\partial t} = \varepsilon \frac{\partial \vec{E}}{\partial t} = -\frac{\varepsilon \times 10^{10}}{\rho}\sin(10^8 t - \beta z)\vec{e}_\rho J D = ∂ t ∂ D = ε ∂ t ∂ E = − ρ ε × 1 0 10 sin ( 1 0 8 t − β z ) e ρ 2 ^2 2
从而 i D = ∫ S J ⃗ D ⋅ d S ⃗ = ∫ 0 1 − ε × 10 10 ρ sin ( 10 8 t − β z ) ⋅ 2 π ρ d z = − 2 π ε × 10 10 β [ cos ( 10 8 − β ) − cos ( 10 8 t ) ] i_D = \int_S \vec{J}_D \cdot d\vec{S} = \int_0^1 -\frac{\varepsilon \times 10^{10}}{\rho}\sin(10^8 t - \beta z) \cdot 2\pi\rho dz = \frac{-2\pi\varepsilon \times 10^{10}}{\beta}[\cos(10^8 - \beta) - \cos(10^8 t)] i D = ∫ S J D ⋅ d S = ∫ 0 1 − ρ ε × 1 0 10 sin ( 1 0 8 t − β z ) ⋅ 2 π ρ d z = β − 2 π ε × 1 0 10 [ cos ( 1 0 8 − β ) − cos ( 1 0 8 t )]
例 4.7.2 （P185 T4-13）在均匀非导电媒质（γ = 0 \gamma = 0 γ = 0
E ⃗ = 30 π cos ( ω t − 4 3 y ) e ⃗ z \vec{E} = 30\pi\cos\left(\omega t - \frac{4}{3}y\right)\vec{e}_z E = 30 π cos ( ω t − 3 4 y ) e z H ⃗ = 10 cos ( ω t − 4 3 y ) e ⃗ x \vec{H} = 10\cos\left(\omega t - \frac{4}{3}y\right)\vec{e}_x H = 10 cos ( ω t − 3 4 y ) e x μ r = 1 \mu_r = 1 μ r = 1 ω \omega ω ε r \varepsilon_r ε r
∵ γ = 0 \because \gamma = 0 ∵ γ = 0 ∴ J ⃗ = γ E ⃗ = 0 \therefore \vec{J} = \gamma\vec{E} = 0 ∴ J = γ E = 0 ∴ ∇ × H ⃗ = 40 3 sin ( ω t − 4 3 y ) e ⃗ z = ∂ D ⃗ ∂ t = ε 0 ε r ∂ E ⃗ ∂ t \therefore \nabla \times \vec{H} = \frac{40}{3}\sin\left(\omega t - \frac{4}{3}y\right)\vec{e}_z = \frac{\partial \vec{D}}{\partial t} = \varepsilon_0\varepsilon_r \frac{\partial \vec{E}}{\partial t} ∴ ∇ × H = 3 40 sin ( ω t − 3 4 y ) e z = ∂ t ∂ D = ε 0 ε r ∂ t ∂ E
= − 30 π ε 0 ε r ω sin ( ω t − 4 3 y ) = -30\pi\varepsilon_0\varepsilon_r\omega\sin\left(\omega t - \frac{4}{3}y\right) = − 30 π ε 0 ε r ω sin ( ω t − 3 4 y )
∴ 30 π ε 0 ε r ω = 40 3 \therefore 30\pi\varepsilon_0\varepsilon_r\omega = \frac{40}{3}
∴ 30 π ε 0 ε r ω = 3 40
又 ∵ ∇ × E ⃗ = 40 π sin ( ω t − 4 3 y ) e ⃗ x = − ∂ B ⃗ ∂ t = − μ 0 ∂ H ⃗ ∂ t \because \nabla \times \vec{E} = 40\pi\sin\left(\omega t - \frac{4}{3}y\right)\vec{e}_x = -\frac{\partial \vec{B}}{\partial t} = -\mu_0 \frac{\partial \vec{H}}{\partial t} ∵ ∇ × E = 40 π sin ( ω t − 3 4 y ) e x = − ∂ t ∂ B = − μ 0 ∂ t ∂ H
= − μ 0 μ r ⋅ 10 ω sin ( ω t − 4 3 y ) e ⃗ x = -\mu_0\mu_r \cdot 10\omega\sin\left(\omega t - \frac{4}{3}y\right)\vec{e}_x = − μ 0 μ r ⋅ 10 ω sin ( ω t − 3 4 y ) e x ∴ 10 ω μ 0 μ r = 40 π \therefore 10\omega\mu_0\mu_r = 40\pi ∴ 10 ω μ 0 μ r = 40 π ω = 4 π μ 0 \omega = \frac{4\pi}{\mu_0} ω = μ 0 4 π
代入 30 π ε 0 ε r ω = 40 3 30\pi\varepsilon_0\varepsilon_r\omega = \frac{40}{3} 30 π ε 0 ε r ω = 3 40 ε r = μ 0 9 π 2 ε 0 \varepsilon_r = \frac{\mu_0}{9\pi^2\varepsilon_0} ε r = 9 π 2 ε 0 μ 0
