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想要公式代码可以私信我
e = 1.6 × 1 0 − 19 C e=1.6\times 10^{-19}\rm C e=1.6×10−19C
F 21 = k q 1 q 2 r 21 2 e r 21 \boldsymbol F_{21}=k\frac{q_1q_2}{r_{21}^2} \boldsymbol {e}_{r21} F21=kr212q1q2er21
k = 9 × 1 0 9 N ⋅ m 2 / C 2 k=9\times 10^9\rm N\cdot m^2/C^2 k=9×109N⋅m2/C2
k = 1 4 π ε 0 k=\frac{1}{4\pi\varepsilon_0} k=4πε01
ε 0 = 8.85 × 1 0 − 12 C 2 / ( N ⋅ m 2 ) \varepsilon_0=8.85\times 10^{-12}\rm C^2/(N\cdot m^2) ε0=8.85×10−12C2/(N⋅m2)
1 V / m = 1 N / C 1\rm V/m=1\rm N/C 1V/m=1N/C
E = d Φ d S ⊥ E=\frac{\text d \Phi}{\text d S_\perp} E=dS⊥dΦ
电通量: 通过面元的电场线条数 E ⋅ d S \boldsymbol E\cdot \text d \boldsymbol S E⋅dS
∮ S E ⋅ d S = 1 ε 0 ∑ q i n \oint_S \boldsymbol E\cdot \text d \boldsymbol S=\frac{1}{\varepsilon_0}\sum q_{in} ∮SE⋅dS=ε01∑qin
高斯定律比库仑定律更普遍
球面内 E = 0 E=0 E=0
球面外 E = q 4 π ε 0 1 r 2 E=\frac{q}{4\pi \varepsilon_0}\frac{1}{r^2} E=4πε0qr21
球体内 E = q 4 π ε 0 r R 3 = ρ 3 ε 0 r E=\frac{q}{4\pi \varepsilon_0}\frac{r}{R^3}=\frac{\rho}{3\varepsilon_0} r E=4πε0qR3r=3ε0ρr
球体外 E = q 4 π ε 0 1 r 2 E=\frac{q}{4\pi \varepsilon_0}\frac{1}{r^2} E=4πε0qr21
长直导线 E = λ 2 π ε 0 r E=\frac{\lambda}{2\pi\varepsilon_0 r} E=2πε0rλ
平面 E = σ 2 ε 0 E=\frac{\sigma}{2\varepsilon_0} E=2ε0σ
圆盘周线上的场强 E = σ x 2 ε 0 [ 1 x 2 + R 1 2 − 1 x 2 + R 2 2 ] E=\frac{\sigma x}{2\varepsilon_0}[\frac{1}{\sqrt{x^2+R_1^2}}-\frac{1}{x^2+R_2^2}] E=2ε0σx[x2+R121−x2+R221]
沿 p ⃗ \vec p p 方向的场强 E = 2 p 4 π ε 0 r 3 \boldsymbol E=\frac{2\boldsymbol p}{4\pi \varepsilon_0 r^3} E=4πε0r32p
中垂线上的场强 E = − p 4 π ε 0 r 3 \boldsymbol E=-\frac{\boldsymbol p}{4\pi \varepsilon_0 r^3} E=−4πε0r3p
一般情况场强 E = 1 4 π ε 0 r 3 [ 3 ( r ⋅ p ) r r 2 − p ] \boldsymbol E=\frac{1}{4\pi \varepsilon_0 r^3}[\frac{3(\boldsymbol r\cdot \boldsymbol p)\boldsymbol r}{r^2}-\boldsymbol p] E=4πε0r31[r23(r⋅p)r−p]
力矩 M = p × E \boldsymbol M=\boldsymbol p\times \boldsymbol E M=p×E
∮ E ⋅ d r = 0 \oint\boldsymbol E\cdot \text d \boldsymbol r=0 ∮E⋅dr=0
U 12 = φ 1 − φ 2 U_{12}=\varphi_1-\varphi_2 U12=φ1−φ2
φ = ∫ P ∞ E ⋅ d r \varphi=\int_P^{\infty}\boldsymbol E\cdot \text d \boldsymbol r φ=∫P∞E⋅dr
1 V = 1 J / C 1\text V = 1\rm J/C 1V=1J/C
E = − ∇ φ \boldsymbol E=-\nabla \varphi E=−∇φ
一个电荷在电场中某点的电势能, 是属于该电荷与产生电场的电荷系所共有的, 是一种相互作用能
1 eV = 1.6 × 1 0 − 19 J 1\text e\text V=1.6\times 10^{-19}\rm J 1eV=1.6×10−19J
电荷系在原来状态的静电能: 将电荷分散到无穷远 电荷间静电力所做的功
W = 1 2 ∑ q i φ i = 1 2 ∫ q φ d q W=\frac{1}{2}\sum q_i \varphi_i=\frac{1}{2}\int_q \varphi \text d q W=21∑qiφi=21∫qφdq
静电学中上式与 W = ∫ V w e d V = ∫ V ε 0 E 2 2 d V W=\int_V w_e \text d V=\int_V \frac{\varepsilon_0 E^2}{2}\text d V W=∫VwedV=∫V2ε0E2dV 等价
电偶极子电势 φ = p cos θ 4 π ε 0 r 2 = p ⋅ e r 4 π ε 0 r 2 \varphi=\frac{p\cos \theta}{4\pi\varepsilon_0 r^2}=\frac{\boldsymbol p\cdot \boldsymbol e_r}{4\pi\varepsilon_0 r^2} φ=4πε0r2pcosθ=4πε0r2p⋅er
导体静电平衡的条件: E i n = 0 , E S ⊥ 表 面 \boldsymbol E_{in}=\boldsymbol{0}, \boldsymbol E_S \perp 表面 Ein=0,ES⊥表面
处于静电平衡的导体: σ = ε 0 E \sigma=\varepsilon_0 E σ=ε0E
有导体存在时静电场的计算: 静电场的基本规律, 电荷守恒, 导体静电平衡条件
静电屏蔽: 金属空壳的外表面上及壳外的电荷在壳内的合场强为 0, 因而对壳内无影响
唯一性定理:在给定条件下, 空间的电场分布和导体表面的电荷分布是唯一确定的
(1) 给定每个导体的总电量
(2) 给定每个导体的电势
(3) 给定一些导体的总电量和另一些导体的电势
可简述为: 给定边界条件后, 静电场的分布就唯一地确定了
镜像法求电场
将介质插入电容器 U = U 0 / ε r E = E 0 / ε r ε r > 1 U=U_0/\varepsilon_r\ E=E_0/\varepsilon_r\ \varepsilon_r>1 U=U0/εr E=E0/εr εr>1
电荷分布不对称的分子: 极性分子, 有固有电矩
正负电荷中心重合: 非极性分子, 无固有电矩. 外加电场会产生比固有电矩小得多的感生电矩
出现在电介质表面的电荷叫面束缚电荷 / 面极化电荷
分子电矩 p = q l \boldsymbol p=q\boldsymbol l p=ql
电极化强度: 单位体积内分子电矩矢量和 p = ∑ p i Δ V \boldsymbol p=\frac{\sum \boldsymbol p_i}{\Delta V} p=ΔV∑pi
P = n p \boldsymbol P=n\boldsymbol p P=np, n n n 为电介质单位体积内的分子数, 单位 C / m 2 \rm C/m^2 C/m2
电极化强度 P = ε 0 ( ε r − 1 ) E \boldsymbol P=\varepsilon_0(\varepsilon_r -1)\boldsymbol E P=ε0(εr−1)E
电极化率 χ = ε r − 1 \chi =\varepsilon_r -1 χ=εr−1
面束缚电荷 σ ′ = P ⋅ e n \sigma ' = \boldsymbol P\cdot\boldsymbol e_n σ′=P⋅en e n \boldsymbol e_n en 由介质指向真空
体束缚电荷 q i n ′ = − ∮ P ⋅ d S q_{in} '=-\oint \boldsymbol P\cdot\text d\boldsymbol S qin′=−∮P⋅dS
ρ ′ = − ∇ ⋅ P \rho'=-\nabla \cdot \boldsymbol P ρ′=−∇⋅P
电位移 D = ε 0 E + P \boldsymbol D=\varepsilon_0 \boldsymbol E+\boldsymbol P D=ε0E+P
∮ D ⋅ d S = ∑ q 0 i n \oint \boldsymbol D\cdot \text d \boldsymbol S=\sum q_{0in} ∮D⋅dS=∑q0in q 0 i n q_{0in} q0in 是自由电荷
D = ε E = ε 0 ε r E \boldsymbol D=\varepsilon \boldsymbol E=\varepsilon_0\varepsilon_r \boldsymbol E D=εE=ε0εrE
边界条件 E 1 t = E 2 t E_{1t}=E_{2t} E1t=E2t D 1 n = D 2 n D_{1n}=D_{2n} D1n=D2n
电容器 C = Q U C=\frac{Q}{U} C=UQ
电容器并联相加, 串联倒数相加
电介质填充两种规律
(1) 按等势面填充: D \boldsymbol {D} D 不变, E \boldsymbol {E} E 变
(2) 按电场线填充: D \boldsymbol {D} D 变, E \boldsymbol {E} E 的分布“样子”不变
电容器的能量 W = 1 2 C U 2 = 1 2 Q U = 1 2 Q 2 C W=\frac{1}{2}CU^2=\frac{1}{2}QU=\frac{1}{2}\frac{Q^2}{C} W=21CU2=21QU=21CQ2
电场中的能量体密度 w e = 1 2 D E = 1 2 ε E 2 w_e=\frac{1}{2}DE=\frac{1}{2}\varepsilon E^2 we=21DE=21εE2
电场中的能量 W = ∫ 1 2 ε E 2 d V W=\int \frac{1}{2}\varepsilon E^2 \text d V W=∫21εE2dV
平行板电容器 C = ε S d C=\frac{\varepsilon S}{d} C=dεS
圆柱形电容器 C = 2 π L ε ln ( R 2 / R 1 ) C=\frac{2\pi L\varepsilon}{\ln (R_2/R_1)} C=ln(R2/R1)2πLε
球形电容器 C = 4 π R 1 R 2 ε R 2 − R 1 C=\frac{4\pi R_1 R_2 \varepsilon}{R_2-R_1} C=R2−R14πR1R2ε
球形孤立导体电容器 C = 4 π R ε C=4\pi R\varepsilon C=4πRε
电流 I I I 又叫电流强度
电流密度 d I = J ⋅ d S \text d I=\boldsymbol J\cdot \text d \boldsymbol S dI=J⋅dS
电流 / 电流密度通亮 I = ∫ S j ⋅ d S I=\int_S \boldsymbol j \cdot \text d \boldsymbol S I=∫Sj⋅dS
J = q n v \boldsymbol J=qn\boldsymbol v J=qnv
I = ∫ S J ⋅ d S = − d q i n d t I=\int_S \boldsymbol J\cdot \text d \boldsymbol S=-\frac{\text d q_{in}}{\text d t} I=∫SJ⋅dS=−dtdqin
微分形式 ∂ ρ ∂ t + ∇ ⋅ j = 0 \frac{\partial \rho}{\partial t}+\nabla \cdot \boldsymbol j=0 ∂t∂ρ+∇⋅j=0
若 ∮ S J ⋅ d S = 0 \oint_S \boldsymbol J\cdot \text d \boldsymbol S=0 ∮SJ⋅dS=0, 则 I 为恒定电流
恒定电场与静电场都服从高斯定律和场强环路积分为零的环路定理
R = ρ l S = l σ S R=\rho\frac{l}{S}=\frac{l}{\sigma S} R=ρSl=σSl
J = σ E \boldsymbol J=\sigma \boldsymbol E J=σE
物质导电性能方程 j = σ ⋅ E \boldsymbol j=\sigma\cdot\boldsymbol E j=σ⋅E
j = σ ⋅ E \boldsymbol j=\sigma\cdot\boldsymbol E j=σ⋅E 比 U = I R U=IR U=IR 适用范围更广, 对非均匀导体成立, 对非稳恒电流也成立
稳恒电流和静电场的综合求解的基本方程:
稳恒条件 ∮ S J ⋅ d S = 0 \oint_S \boldsymbol J\cdot \text d \boldsymbol S=0 ∮SJ⋅dS=0
环路定理 ∮ E ⋅ d l = 0 \oint \boldsymbol E\cdot \text d\boldsymbol l=0 ∮E⋅dl=0
欧姆定律 j = σ ⋅ E \boldsymbol j=\sigma\cdot\boldsymbol E j=σ⋅E
界面关系 j 1 n = j 2 n j_{1n}=j_{2n} j1n=j2n, E 1 t = E 2 t E_{1t}=E_{2t} E1t=E2t
充电
q = C ε ( 1 − e − t R C ) q=C\varepsilon(1-e^{-\frac{t}{RC}}) q=Cε(1−e−RCt)
i = ε R e − t R C i=\frac{\varepsilon}{R}e^{-\frac{t}{RC}} i=Rεe−RCt
u c = ε ( 1 − e − t R C ) u_c=\varepsilon(1-e^{-\frac{t}{RC}}) uc=ε(1−e−RCt)
放电
q = Q e − t R C q=Qe^{-\frac{t}{RC}} q=Qe−RCt
i = Q R C e − t R C i=\frac{Q}{RC}e^{-\frac{t}{RC}} i=RCQe−RCt
u c = Q C e − t R C u_c=\frac{Q}{C}e^{-\frac{t}{RC}} uc=CQe−RCt
电容器时间常量 τ = R C \tau =RC τ=RC 若回路的线度比距离 c τ c\tau cτ 小得多,电场可按恒定电场处理
在平行板电容器内填充两层导电介质, 厚度、介电常数和电导率分别为 ( d 1 , ε 1 , σ 1 ) (d_{1}, \varepsilon_{1}, \sigma_{1}) (d1,ε1,σ1)
和( d 2 , ε 2 , σ 2 d_2, \varepsilon_{2}, \sigma_{2} d2,ε2,σ2 ),设电容器两端电压为 U \boldsymbol{U} U
求:
(1)两介质中的电流密度和电场强度。
(2)介质分界面上的总电荷面密度 σ e \sigma_{e} σe 和自由电荷面密度 σ e 0 \sigma_{e 0} σe0
解:
(1)由对称性和界面关系: j 1 = j 2 = j j_{1}=j_{2}=j j1=j2=j
电场强度: E 1 = j σ 1 , E 2 = j σ 2 \quad E_{1}=\frac{j}{\sigma_{1}}, \quad E_{2}=\frac{j}{\sigma_{2}} E1=σ1j,E2=σ2j
电压关系: U = E 1 d 1 + E 2 d 2 \quad U=E_{1} d_{1}+E_{2} d_{2} U=E1d1+E2d2
解得: j 1 = j 2 = j = σ 1 σ 2 σ 1 d 2 + σ 2 d 1 U \quad j_{1}=j_{2}=j=\frac{\sigma_{1} \sigma_{2}}{\sigma_{1} d_{2}+\sigma_{2} d_{1}} U j1=j2=j=σ1d2+σ2d1σ1σ2U
E 1 = σ 2 σ 1 d 2 + σ 2 d 1 U , E 2 = σ 1 σ 1 d 2 + σ 2 d 1 U E_{1}=\frac{\sigma_{2}}{\sigma_{1} d_{2}+\sigma_{2} d_{1}} U, \quad E_{2}=\frac{\sigma_{1}}{\sigma_{1} d_{2}+\sigma_{2} d_{1}} U E1=σ1d2+σ2d1σ2U,E2=σ1d2+σ2d1σ1U
(2)在界面选扁柱画作少高斯面 S S S
分别用 E \boldsymbol{E} E 和 D \boldsymbol{D} D 的高斯定理:
σ e = ε 0 ( E 2 − E 1 ) = ε 0 ( σ 1 − σ 2 ) σ 1 d 2 + σ 2 d 1 U \sigma_{e}=\varepsilon_{0}\left(E_{2}-E_{1}\right)=\frac{\varepsilon_{0}\left(\sigma_{1}-\sigma_{2}\right)}{\sigma_{1} d_{2}+\sigma_{2} d_{1}} U σe=ε0(E2−E1)=σ1d2+σ2d1ε0(σ1−σ2)U
σ e 0 = D 2 − D 1 = ε 2 E 2 − ε 1 E 1 = ε 2 σ 1 − ε 1 σ 2 σ 1 d 2 + σ 2 d 1 U \sigma_{e 0}=D_{2}-D_{1}=\varepsilon_{2} E_{2}-\varepsilon_{1} E_{1}=\frac{\varepsilon_{2} \sigma_{1}-\varepsilon_{1} \sigma_{2}}{\sigma_{1} d_{2}+\sigma_{2} d_{1}} U σe0=D2−D1=ε2E2−ε1E1=σ1d2+σ2d1ε2σ1−ε1σ2U
在所有情况下, 磁力都是运动电荷之间相互作用的表现.
洛伦兹力 磁感应强度 F = q v × B \boldsymbol F=q\boldsymbol v\times\boldsymbol B F=qv×B
1 T = 1 0 4 G 1\text T=10^4\rm G 1T=104G
磁通量 Φ = ∫ S B ⋅ d S \Phi=\int_S \boldsymbol B\cdot\text d\boldsymbol S Φ=∫SB⋅dS
毕奥 - 萨伐尔定律 d B = μ 0 4 π I d l × e r r 2 \text d \boldsymbol B=\frac{\mu_0}{4\pi}\frac{I\text d\boldsymbol l\times\boldsymbol e_r}{r^2} dB=4πμ0r2Idl×er
真空磁导率 μ 0 = 1 ε 0 c 2 = 4 π × 1 0 − 7 N / A 2 \mu_0=\frac{1}{\varepsilon_0 c^2}=4\pi\times 10^{-7}\rm N/A^2 μ0=ε0c21=4π×10−7N/A2
c = 1 μ 0 ε 0 c=\frac{1}{\sqrt{\mu_0\varepsilon_0}} c=μ0ε01
磁通连续性定理 ∮ B ⋅ d S = 0 \oint \boldsymbol B\cdot\text d\boldsymbol S=0 ∮B⋅dS=0
d B = μ 0 4 π q v × e r r 2 \text d \boldsymbol B=\frac{\mu_0}{4\pi}\frac{q\boldsymbol v\times\boldsymbol e_r}{r^2} dB=4πμ0r2qv×er
安培环路定理 ∮ B ⋅ d r = μ 0 ∑ I i n \oint \boldsymbol B\cdot \text d \boldsymbol r =\mu_0\sum I_{in} ∮B⋅dr=μ0∑Iin
∮ B ⋅ d r = μ 0 ∫ S ( J c + ε 0 ∂ E ∂ t ) ⋅ d S \oint\boldsymbol B\cdot\text d\boldsymbol r=\mu_0\int_S\left(\boldsymbol J_c+\varepsilon_0\frac{\partial \boldsymbol E}{\partial t}\right)\cdot \text d \boldsymbol S ∮B⋅dr=μ0∫S(Jc+ε0∂t∂E)⋅dS
传导电流 I c I_c Ic
位移电流 I d = ε 0 d Φ d t = ε 0 d d t ∫ S E ⋅ d S I_d=\varepsilon_0 \frac{\text d\Phi}{\text d t}=\varepsilon_0\frac{\text d}{\text d t}\int_S\boldsymbol E\cdot \text d\boldsymbol S Id=ε0dtdΦ=ε0dtd∫SE⋅dS
位移电流密度 J d = ε 0 ∂ E ∂ t \boldsymbol J_d=\varepsilon_0\frac{\partial \boldsymbol E}{\partial t} Jd=ε0∂t∂E
全电流 I = I c + I d I=I_c+I_d I=Ic+Id
无限长直电流 B = μ 0 I 2 π r B=\frac{\mu_0 I}{2\pi r} B=2πrμ0I
一段直导线 B = μ 0 I 4 π r ( cos θ 1 − cos θ 2 ) B=\frac{\mu_0 I}{4\pi r}(\cos \theta_1-\cos \theta_2) B=4πrμ0I(cosθ1−cosθ2)
无限长均匀载流薄圆筒 B 内 = 0 , B 外 = μ 0 I 2 π r B_ 内 =0, B_ 外 =\frac{\mu_0 I}{2\pi r} B内=0,B外=2πrμ0I
无限长直载流密绕螺绕管 / 螺绕环 B 内 = μ 0 n I , B 外 = 0 B_ 内 =\mu_0 n I, B_ 外 =0 B内=μ0nI,B外=0 对于螺绕环 n = N 2 π r n=\frac{N}{2\pi r} n=2πrN
无限大平面电流 B ⋅ 2 l = μ 0 j l B\cdot 2l=\mu_0 j l B⋅2l=μ0jl
圆电流圈中心点和轴线上的磁场 B 中 心 = μ 0 I 2 R , B 轴 线 = μ 0 I S 2 π ( R 2 + x 2 ) 3 / 2 B_{中心}=\frac{\mu_0 I}{2R}, B_{轴线}=\frac{\mu_0 IS}{2\pi(R^2+x^2)^{3/2}} B中心=2Rμ0I,B轴线=2π(R2+x2)3/2μ0IS
B = μ o 4 π r 3 [ 3 ( r ⋅ m ) r r 2 − m ] , ( r > > 磁 矩 线 度 ) \boldsymbol B=\frac{\mu_o}{4\pi r^3}[\frac{3(\boldsymbol r\cdot \boldsymbol m)\boldsymbol r}{r^2}-\boldsymbol m], (r>> 磁矩线度) B=4πr3μo[r23(r⋅m)r−m],(r>>磁矩线度)
磁矩、电流圈在外磁场中的势能 W = − m B 外 = − I S ⋅ B 外 W=-\boldsymbol m\boldsymbol B_ 外 =-IS\cdot \boldsymbol B_ 外 W=−mB外=−IS⋅B外
半径 R R R 的圆形平行板电容器内充满介电 常数 ε \varepsilon ε 、磁导率 μ \mu μ 的均匀介质,如图已知电容器充电时的 d E d t \frac{\mathrm{d} E}{\mathrm{d} t} dtdE 及其方向,忽略边缘效应
求: i d i_d id和 B p B_p Bp ( r < R ) (r
对圆面 S S S 有:
I d = ∬ S ∂ D ∂ t ⋅ d S = ∬ s ε d E d t d S = ε d E d t π R 2 I_{d} =\iint_{S} \frac{\partial \boldsymbol{D}}{\partial t} \cdot \mathbf{d} \boldsymbol{S}=\iint_\boldsymbol{s}\varepsilon\frac{\mathbf{d} \boldsymbol{E}}{\mathbf{d} t}\mathbf{d} \boldsymbol{S}=\varepsilon \frac{\mathbf{d}\boldsymbol{E}}{\mathbf{d} t}\pi R^2 Id=∬S∂t∂D⋅dS=∬sεdtdEdS=εdtdEπR2
过 P点垂直轴线作环形回路 L L L, 方向和圆面 S ′ S^{\prime} S′ 成右手关系:
∮ L H ⋅ d l = H ⋅ 2 π r = ∑ I d 内 \oint_{L} \boldsymbol{H} \cdot \mathbf{d} \boldsymbol{l}=\boldsymbol{H} \cdot \boldsymbol{2} \pi \boldsymbol{r}=\sum \boldsymbol{I}_{d内} ∮LH⋅dl=H⋅2πr=∑Id内
∑ I d 内 = ∬ S ′ ∂ D ∂ t ⋅ d S = π r 2 ε d E d t \sum I_{d内}=\iint_{S^{\prime}} \frac{\partial \boldsymbol{D}}{\partial t} \cdot \mathbf{d} \boldsymbol{S}=\pi r^{2} \boldsymbol{\varepsilon} \frac{\mathbf{d} \boldsymbol{E}}{\mathbf{d} t} ∑Id内=∬S′∂t∂D⋅dS=πr2εdtdE
H P = ε r 2 d E d t H_{P}=\frac{\varepsilon r}{2} \frac{d E}{d t} HP=2εrdtdE
B P = μ H P = μ ε r 2 ⋅ d E d t B_{P}=\mu H_{P}=\frac{\mu \varepsilon r}{2} \cdot \frac{d E}{d t} BP=μHP=2μεr⋅dtdE
r = m v B q r=\frac{mv}{Bq} r=Bqmv
T = 2 π m B q T=\frac{2\pi m}{Bq} T=Bq2πm
螺旋运动的螺距 h = 2 π m B q v / / h=\frac{2\pi m}{Bq}v_{//} h=Bq2πmv//
霍尔效应 U H = I B n q b U_H=\frac{IB}{nqb} UH=nqbIB
F = ∫ L I d l × B \boldsymbol F=\int_L I\text d \boldsymbol l\times \boldsymbol B F=∫LIdl×B
磁矩 m = S I e n \boldsymbol m=SI\boldsymbol e_n m=SIen
力矩 M = m × B \boldsymbol M=\boldsymbol m\times \boldsymbol B M=m×B
B = μ r B 0 B=\mu_r B_0 B=μrB0, μ 0 \mu_0 μ0 为相对磁导率
磁化强度 M = ∑ m i Δ V \boldsymbol M=\frac{\sum \boldsymbol m_i}{\Delta V} M=ΔV∑mi
M = μ r − 1 μ 0 μ r B \boldsymbol M=\frac{\mu_r-1}{\mu_0\mu_r}\boldsymbol B M=μ0μrμr−1B
面束缚电流密度 j ′ = M × e n \boldsymbol j'=\boldsymbol M\times \boldsymbol e_n j′=M×en
总束缚电流 I ′ = ∮ d I ′ = ∮ L M ⋅ d r I'=\oint \text d I'=\oint_L\boldsymbol M\cdot \text d \boldsymbol r I′=∮dI′=∮LM⋅dr
磁感应强度 B = B 0 + B ′ \boldsymbol B=\boldsymbol B_0+\boldsymbol B' B=B0+B′
磁场强度 H = B μ = B μ 0 − M \boldsymbol H=\frac{\boldsymbol B}{\mu}=\frac{\boldsymbol B}{\mu_0}-\boldsymbol M H=μB=μ0B−M
∮ L H ⋅ d r = ∑ I 0 i n \oint_L \boldsymbol H\cdot \text d \boldsymbol r=\sum I_{0in} ∮LH⋅dr=∑I0in
磁场的边界条件 H 1 t = H 2 t H_{1t}=H_{2t} H1t=H2t, B 1 n = B 2 n B_{1n}=B_{2n} B1n=B2n
磁感线穿过两介质分界面 tan θ 1 tan θ 2 = μ r 1 μ r 2 \frac{\tan \theta_1}{\tan \theta_2}=\frac{\mu_{r1}}{\mu_{r2}} tanθ2tanθ1=μr2μr1
用封闭铁盒可以实现磁屏蔽
感应电动势 E = d Ψ d t = − N d Φ d t \mathscr{E}=\frac{\text d \Psi}{\text dt}=-N\frac{\text d \Phi}{\text d t} E=dtdΨ=−NdtdΦ
当穿过各匝线圈的磁通量相等时,N 匝线圈的全磁通为 Ψ = N Φ \Psi=N\Phi Ψ=NΦ
动生电动势 E = ∮ L ( v × B ) d l \mathscr E=\oint_L(\boldsymbol v\times\boldsymbol B)\text d\boldsymbol l E=∮L(v×B)dl
∣ E ∣ = B l v |\mathscr E|=Blv ∣E∣=Blv
感生电动势 ∮ L E i ⋅ d l = − d Φ d t = − ∫ S ∂ B ∂ t ⋅ d S \oint_L\boldsymbol E_i\cdot \text d\boldsymbol l=-\frac{\text d\Phi}{\text dt}=-\int_S \frac{\partial \boldsymbol B}{\partial t}\cdot \text d\boldsymbol S ∮LEi⋅dl=−dtdΦ=−∫S∂t∂B⋅dS
其中 E i E_i Ei 表示感生电场, 由于静电场的环路积分为零, 所以
∮ L E ⋅ d r = − ∫ S ∂ B ∂ t ⋅ d S \oint_L\boldsymbol E\cdot \text d\boldsymbol r=-\int_S \frac{\partial \boldsymbol B}{\partial t}\cdot \text d\boldsymbol S ∮LE⋅dr=−∫S∂t∂B⋅dS
Ψ 21 = M 21 i 1 \Psi_{21}=M_{21}i_1 Ψ21=M21i1
E 12 = − d Ψ 21 d t = − M 21 d i d t \mathscr E_{12}=-\frac{\text d\Psi_{21}}{\text dt}=-M_{21}\frac{\text di}{\text dt} E12=−dtdΨ21=−M21dtdi
M 21 M_{21} M21 是回路 L 1 L_1 L1 对回路 L 2 L_2 L2 的互感系数, 固定回路的互感系数是一个常数, M 21 = M 12 = M M_{21}=M_{12}=M M21=M12=M, M M M 称作这两个导体回路的互感系数, 简称他们的互感
E L = − d Ψ d t = − L d i d t \mathscr E_{L}=-\frac{\text d\Psi}{\text dt}=-L\frac{\text di}{\text dt} EL=−dtdΨ=−Ldtdi, L = Ψ i L=\frac{\Psi}{i} L=iΨ 为自感系数, 简称自感
自感磁能 W m = 1 2 L I 2 W_m=\frac{1}{2}LI^2 Wm=21LI2
磁场的能量 W m = B 2 2 μ V = ∫ B H 2 d V W_m=\frac{B^2}{2\mu}V=\int \frac{BH}{2}\text dV Wm=2μB2V=∫2BHdV
磁能量密度 w m = 1 2 B H w_m=\frac{1}{2}BH wm=21BH
真空中的电磁场规律
{ ∮ S E ⋅ d S = q ε 0 = 1 ε 0 ∫ V ρ d V ∮ S B ⋅ d S = 0 ∮ L E ⋅ d r = − d Φ d t = − ∫ S ∂ B ∂ t ⋅ d S ∮ L B ⋅ d r = μ 0 I + 1 c 2 d Φ e d t = μ 0 ∫ s ( J + ε 0 ∂ E ∂ t ) ⋅ d S \left\{\begin{array}{l} \oint_{S} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{S}=\frac{q}{\varepsilon_{0}}=\frac{1}{\varepsilon_{0}} \int_{V} \rho \mathrm{d} V \\ \oint_{S} \boldsymbol{B} \cdot \mathrm{d} \boldsymbol{S}=0 \\ \oint_{L} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{r}=-\frac{\mathrm{d} \Phi}{\mathrm{d} t}=-\int_{S} \frac{\partial \boldsymbol{B}}{\partial t} \cdot \mathrm{d} \boldsymbol{S} \\ \oint_{L} \boldsymbol{B} \cdot \mathrm{d} \boldsymbol{r}=\mu_{0} I+\frac{1}{c^{2}} \frac{\mathrm{d} \Phi_{\mathrm{e}}}{\mathrm{d} t}=\mu_{0} \int_{s}\left(\boldsymbol{J}+\varepsilon_{0} \frac{\partial \boldsymbol{E}}{\partial t}\right)\cdot \mathrm{d} \boldsymbol{S} \end{array} \right. ⎩⎪⎪⎨⎪⎪⎧∮SE⋅dS=ε0q=ε01∫VρdV∮SB⋅dS=0∮LE⋅dr=−dtdΦ=−∫S∂t∂B⋅dS∮LB⋅dr=μ0I+c21dtdΦe=μ0∫s(J+ε0∂t∂E)⋅dS
有介质的情况下
{ ∮ S D ⋅ d S = ∫ V ρ 0 d V ∮ S B ⋅ d S = 0 ∮ L E ⋅ d r = − ∫ S ∂ B ∂ t ⋅ d S ∮ L H ⋅ d r = ∫ S ( j 0 + ∂ D ∂ t ) ⋅ d S \left\{\begin{array}{l} \oint_{S} \boldsymbol{D} \cdot \mathrm{d} \boldsymbol{S}=\int_{V} \rho_0 \mathrm{d} V \\ \oint_{S} \boldsymbol{B} \cdot \mathrm{d} \boldsymbol{S}=0 \\ \oint_{L} \boldsymbol{E} \cdot \mathrm{d} \boldsymbol{r}=-\int_{S} \frac{\partial \boldsymbol{B}}{\partial t} \cdot \mathrm{d} \boldsymbol{S} \\ \oint_{L} \boldsymbol{H} \cdot \mathrm{d} \boldsymbol{r}=\int_{S}\left(\boldsymbol{j}_0+\frac{\partial \boldsymbol{D}}{\partial t}\right)\cdot \mathrm{d} \boldsymbol{S} \end{array} \right. ⎩⎪⎪⎨⎪⎪⎧∮SD⋅dS=∫Vρ0dV∮SB⋅dS=0∮LE⋅dr=−∫S∂t∂B⋅dS∮LH⋅dr=∫S(j0+∂t∂D)⋅dS
表述为微分形式
{ ∇ ⋅ D = ρ 0 ∇ ⋅ B = 0 ∇ × E = − ∂ B ∂ t ∇ × H = j 0 + ∂ D ∂ t \left\{\begin{array}{l} \nabla \cdot \boldsymbol{D}=\rho_0 \\ \nabla \cdot \boldsymbol{B}=0 \\ \nabla \times \boldsymbol{E}=-\frac{\partial \boldsymbol{B}}{\partial t} \\ \nabla \times \boldsymbol{H}=\boldsymbol{j}_0+\frac{\partial \boldsymbol{D}}{\partial t} \end{array} \right. ⎩⎪⎪⎨⎪⎪⎧∇⋅D=ρ0∇⋅B=0∇×E=−∂t∂B∇×H=j0+∂t∂D
对于各向同性的线形介质, 有
D = ε 0 ε r E , B = μ 0 μ r H , j 0 = σ E \boldsymbol{D}=\varepsilon_{0} \varepsilon_{\mathrm{r}} \boldsymbol{E}, \quad \boldsymbol{B}=\mu_{0} \mu_{\mathrm{r}} \boldsymbol{H}, \quad \boldsymbol{j}_0=\sigma \boldsymbol{E} D=ε0εrE,B=μ0μrH,j0=σE
洛伦兹力公式
F = q E + q v × B \boldsymbol{F}=q\boldsymbol{E}+q\boldsymbol{v}\times\boldsymbol{B} F=qE+qv×B
界面关系
{ E 1 t = E 2 t D 1 n − D 2 n = σ 0 H 1 t − H 2 t = ( i 0 × e n ) ⋅ e t B 1 n = B 2 n \left\{\begin{array}{l} E_{1t}=E_{2t}\\ D_{1n}-D_{2n}=\sigma_0\\ H_{1t}-H_{2t}=(\boldsymbol{i_0}\times\boldsymbol{e_n})\cdot\boldsymbol{e_t}\\ B_{1n}=B_{2n} \end{array} \right. ⎩⎪⎪⎨⎪⎪⎧E1t=E2tD1n−D2n=σ0H1t−H2t=(i0×en)⋅etB1n=B2n
右手关系 E E × H H = u u \frac{\boldsymbol{E}}{E}\times\frac{\boldsymbol H}{H}=\frac{\boldsymbol u}{u} EE×HH=uu
振幅关系 μ H = ε E \sqrt{\mu}H=\sqrt{\varepsilon}E μH=εE
E B = 1 ε μ = u \frac{E}{B}=\frac{1}{\sqrt{\varepsilon\mu}}=u BE=εμ1=u
波速、折射率 u = 1 ε μ u=\frac{1}{\sqrt{\varepsilon\mu}} u=εμ1 c = 1 ε 0 μ 0 c=\frac{1}{\sqrt{\varepsilon_0\mu_0}} c=ε0μ01 n = ε r μ r n=\sqrt{\varepsilon_r\mu_r} n=εrμr u = c n u=\frac{c}{n} u=nc (对于非铁磁质 n = ε r μ r ≈ ε r n=\sqrt{\varepsilon_r\mu_r}\approx\sqrt{\varepsilon_r} n=εrμr≈εr)平面电磁波的能量密度 w = ε E 2 = E H u w=\varepsilon E^2=\frac{EH}{u} w=εE2=uEH
单位时间内通过与传播方向垂直的单位面积的能量, 叫电磁波的能流密度, 其时间平均值就是电磁波的强度. 能流密度矢量 S \boldsymbol S S 又被称作波印亭矢量
S = E × H \boldsymbol S=\boldsymbol E\times\boldsymbol H S=E×H
S / / \boldsymbol S_{//} S// 沿导线由电源传向负载
S ⊥ \boldsymbol S_{\perp} S⊥ 沿径向由外向内传播,补偿导线的焦耳热损耗
电磁波质量密度 m = w c 2 = E H c 2 u m=\frac{w}{c^2}=\frac{EH}{c^2u} m=c2w=c2uEH
电磁波动量密度 g = m u = 1 c 2 E × H = S c 2 \boldsymbol g=m\boldsymbol u=\frac{1}{c^2}\boldsymbol E\times\boldsymbol H=\frac{\boldsymbol S}{c^2} g=mu=c21E×H=c2S
辐射压强 全反射 p r = 2 g ⋅ c = 2 E H c p_r=2g\cdot c=2\frac{EH}{c} pr=2g⋅c=2cEH 全吸收 p r ′ = g ⋅ c = E H c p_r'=g\cdot c=\frac{EH}{c} pr′=g⋅c=cEH