磁场与电磁学
Part of note taken on ZJU Physics Ⅱ (H), 2021 Fall & Winter
Chap29：磁偶极矩
洛伦兹力导出安培力
\[
\begin{aligned}
f&=q\vec v\times\vec B\\
\mathrm{d}\vec F&=\mathrm{d}Nq\vec v\times \vec B\\
&=n(S\mathrm{d}l)q\vec v\times\vec B\\
&=i\mathrm{d}\vec l\times\vec B;
\end{aligned}
\]
尤其直线、90° 条件下有
\[
F = IBL
\]
闭合线圈在均匀磁场中所受力矩
\[
\tau=iA\hat n\times\vec B\\
\mathrm{d}\tau=i\mathrm{d}A\hat n_A\times\vec B\\
\tau=\sum i\mathrm{d}A\hat n_A\times\vec B\\
\]
磁偶极矩
磁偶极矩 (magnetic dipole moment)
\[
\begin{aligned}
\mu&=NiS\\
\tau&=\mu B\sin\theta\\
\vec\tau&=\vec\mu\times\vec B=NiA\hat n\times \vec B\\
W(\theta)&=-\int_\frac{\pi}{2}^\theta\vec\tau\mathrm d\vec\theta\\
&=-\int_\frac{\pi}{2}^\theta NiAB\sin\theta\mathrm d\theta\\
&=NiAB\cos\theta\\
&=\vec\mu\cdot\vec B\\
U(\theta)&=-W=-\vec\mu\cdot\vec B=-NiA B\cos\theta\\
\end{aligned}
\]
对比电偶极矩 (electric dipole moment)
\[
\begin{aligned}
\vec\tau&=\vec p\times\vec E\\
U(\theta)&=-\vec p\cdot\vec E=-Eqd\cos\theta\\
\end{aligned}
\]
以上的 \(U\) 均为能量含义。
Chap30：毕奥 - 萨伐尔定律
Laplace 介入 Biot and Savart Law
\[
\begin{aligned}
\mathrm d H&=\mathrm d H(r, \alpha)\\
H&=k\frac{I}{r}\tan\frac{\alpha}{2}\\
\mathrm d H&=\frac{\mathrm d H}{\mathrm d l}\mathrm d l\\
&=\left(\frac{\partial H}{\partial \alpha}\cdot\frac{\mathrm d\alpha}{\mathrm dl}+\frac{\partial H}{\partial r}\cdot\frac{\mathrm dr}{\mathrm dl}\right)\mathrm dl\\
\frac{\partial H}{\partial \alpha}&=\frac{kI}{2r\cos^2\frac{\alpha}{2}},\frac{\partial H}{\partial r}=-k\frac I{r^2}\tan\frac{\alpha}{2}\\
\mathrm{d}l\sin\alpha&=r\mathrm{d}\alpha,\mathrm{d}l\cos\alpha=\mathrm{d}r\Rightarrow
\frac{\mathrm{d}\alpha}{\mathrm{d}l}=\frac{\sin\alpha}{r},\frac{\mathrm{d}r}{\mathrm{d}l}=\cos\alpha\\
\mathrm dH&=\left(\frac{kI}{2r\cos^2\frac{\alpha}{2}}\cdot\frac{\sin\alpha}{r}-k\frac I{r^2}\tan\frac{\alpha}{2}\cdot\cos\alpha\right)\mathrm dl\\
&=\left(\frac{\sin\alpha}{2\cos^2\frac{\alpha}{2}}-\tan\frac{\alpha}{2}\cos\alpha\right)k\frac{I\mathrm dl}{r^2}\\
&=\tan\frac{\alpha}{2}\left(1-\cos\alpha\right)k\frac{I\mathrm dl}{r^2}\\
&=\tan\frac{\alpha}{2}\cdot2\cos^2\frac{\alpha}{2}k\frac{I\mathrm dl}{r^2}\\
&=k\frac{I\mathrm dl}{r^2}\sin\alpha,\text{ 结合方向得到}\\
\mathrm d\vec H&=k\frac{I\mathrm d\vec l\times\vec r}{r^3}\\
\mathrm d\vec H&=\frac{1}{4\pi}\frac{I\mathrm d\vec l\times\vec r}{r^3}\\
\mathrm d\vec B&=\frac{\mu}{4\pi}\frac{I\mathrm d\vec l\times\vec r}{r^3}
\end{aligned}
\]
匀速运动电荷在空间产生磁场 ( 考虑相对论 )
\[
\begin{aligned}
\vec B&=\frac{\mu_0}{4\pi}\frac{q\vec v\times\vec r}{r^3(1-\frac{v^2}{c^2}\sin^2\theta)^{\frac{3}{2}}}(1-\frac{v^2}{c^2})\\
v&\ll c,\vec B=\frac{\mu_0}{4\pi}\frac{q\vec v\times\vec r}{r^3}
\end{aligned}
\]
载流直导线产生磁场
\[
\begin{aligned}
B=\int\mathrm d B&=\int\frac{\mu_0}{4\pi}\frac{I\mathrm dz}{r^2}\frac{d}{r}\\
&=\int\frac{\mu_0}{4\pi}\frac{Id\mathrm dz}{(z^2+d^2)^{\frac{3}{2}}}\\
&\xlongequal{z=d\tan\theta}\int\frac{\mu_0I}{4\pi d}\cos\theta\mathrm d\theta\\
&=\frac{\mu_0I}{4\pi d}(\sin\theta_1-\sin\theta_2)\\
\theta_1\to\frac{\pi}{2},\theta_2\to-\frac{\pi}{2},B&=\frac{\mu_0I}{2\pi d}
\end{aligned}
\]
环形线圈产生磁场
\[
\begin{aligned}
B=\int\mathrm d B_z&=\int\frac{\mu_0}{4\pi}\frac{I\mathrm dl}{R^2+z^2}\cdot\frac{R}{\sqrt{R^2+z^2}}\\
&=\frac{\mu_0}{4\pi}\frac{I\cdot2\pi R^2}{(R^2+z^2)^{\frac{3}{2}}}\\
&=\frac{\mu_0 I}{2}\frac{R^2}{(R^2+z^2)^{\frac{3}{2}}}\\
z=0,B&=\frac{\mu_0I}{2R}\\
z>>R,B&=\frac{\mu_0IR^2}{2z^3}=\frac{\mu_0\mu(电偶极矩)}{2\pi z^3}
\end{aligned}
\]
Chap31-2：电感
Example1 螺线管电感
\[
L=\frac{\Phi}{i}=\frac{N\phi}{i}=\frac{N(\mu_0ni)S}{i}=\mu_0nNS=\mu_0n^2lS
\]
Example2 环形加速器 Toroid
\[
\phi=\int_a^b(\mu_0\frac{N}{2\pi r}I)h\mathrm dr=\frac{\mu_0NIh}{2\pi}\ln\frac{b}{a}\\
L=\frac{N\phi}{I}=\frac{\mu_0N^2h}{2\pi}\ln\frac{b}{a}
\]
LRCircuits
\[
\begin{aligned}
\varepsilon-iR-L\frac{\mathrm di}{\mathrm dt}&=0\\
\frac{\mathrm di}{\varepsilon-iR}&=\frac{\mathrm d t}{L}\\
-\frac{1}{R}\ln\frac{\varepsilon-iR}{\varepsilon}&=\frac{t}{L}\\
\varepsilon-iR&=\varepsilon e^{-\frac{t}{\tau}},\tau=\frac{L}{R}\\
i&=\frac{\varepsilon}{R}(1-e^{-\frac{t}{\tau}})\\
\end{aligned}
\]
移除电源 ( 放电？)
\[
\begin{aligned}
-iR-L\frac{\mathrm di}{\mathrm dt}&=0\\
\frac{\mathrm di}{i}&=-\frac{R\mathrm dt}{L}\\
i&=\frac{\varepsilon}{R}e^{-\frac{t}{\tau}}\\
\end{aligned}
\]
考虑能量
\[
\begin{aligned}
\varepsilon-iR-L\frac{\mathrm di}{\mathrm dt}&=0\\
\varepsilon i&=i^2R+Li\frac{\mathrm di}{\mathrm dt}\\
\varepsilon i:\text{电源功率}\ \ &i^2R:\text{热功率}\ \ Li\frac{\mathrm di}{\mathrm dt}:\text{电感储能功率}\\
\varepsilon i=\frac{\varepsilon \mathrm dq}{\mathrm dt}&:\text{因此有电源功率的意义}\\
U_L&=\int_0^tLi\frac{\mathrm di}{\mathrm dt}\mathrm dt=\int_0^iLi\mathrm di=\frac{1}{2}Li^2\\
\end{aligned}
\]
磁场能量密度 ( 从螺线管推导 )
\[
\begin{aligned}
u_B=\frac{U}{V}&=\frac{\frac12LI^2}{lA}\\
&=\frac{\frac12\mu_0n^2lAI^2}{lA}\\
&=\frac{(\mu_0nI)^2}{2\mu_0}=\frac{B^2}{2\mu_0}\\
\text{对比 }u_E&=\frac{1}{2}\epsilon_0E^2
\end{aligned}
\]
Example3
\[
\begin{aligned}
\phi&=\int_a^{d-a}(\frac{\mu_0i}{2\pi r}+\frac{\mu_0i}{2\pi(d-r)})l\mathrm dr\\
&=\frac{\mu_0il}{2\pi}\ln\frac{r}{d-r}\bigg|_a^{d-a}=\frac{\mu_0il}{\pi}\ln\frac{d-a}{a}\\
L&=\frac{\phi}{i}=\frac{\mu_0l}{2\pi}\ln\frac{d-a}{a}
\end{aligned}
\]
互感
互感 (Mutual Induction)
\[
\begin{gathered}
i,B\propto i_1\Rightarrow \Phi_{12}=M_{12}i_1\\
\text{同理 }\Phi_{21}=M_{21}i_2\\
\text{变 }i_1\text{ 以动 }i_2,\; M_{12}\frac{\mathrm di_1}{\mathrm dt}=\frac{\mathrm d\Phi_{12}}{\mathrm dt}=-\varepsilon_2\\
\text{同理，变 }i_2\text{ 以动 }i_1,\varepsilon_1=-M_{21}\frac{\mathrm di_2}{\mathrm dt}
\end{gathered}
\]
证明互感系数相等
\[
\begin{gathered}
\phi_{12}=\oint_{L_2}\vec{A_1}\cdot\mathrm d\vec l_2,\vec A_1=\frac{\mu_0I_1}{4\pi}\oint_{L_1}\frac{\mathrm d\vec l_1}{r_{12}}\\
\therefore\phi_{12}=\frac{\mu_0I_1}{4\pi}\oint\limits_{L_1}\oint\limits_{L_2}\frac{\mathrm d\vec l_1\cdot\mathrm d\vec l_2}{r_{12}}\\
M_{12}=M_{21}=\frac{\mu_0}{4\pi}\oint\limits_{L_1}\oint\limits_{L_2}\frac{\mathrm d\vec l_1\cdot\mathrm d\vec l_2}{r_{12}}
\end{gathered}
\]
Example 4
\[
\begin{gathered}
B=\mu_0\mu_r\frac{N_1}{l}i_1=\mu\frac{N_1}{l}i_1\\
\Phi_{12}=N_2 BS=\mu\frac{N_1N_2S}{l}i_1=Mi_1\\
M=\mu\frac{N_1N_2S}{l}\\
L_1=\mu\frac{N_1^2S}{l},L_1=\mu\frac{N_2^2S}{l}\\
M=\sqrt{L_1L_2}\\
M=K\sqrt{L_1L_2},K\in[0,1],\text{ 同轴 }K=1,\text{ 垂直 }K=0
\end{gathered}
\]
L-C circuit
\[
\begin{gathered}
\frac{q}{C}=-L\frac{\mathrm di}{\mathrm dt}\Rightarrow \frac{\mathrm d^2q}{\mathrm dt^2}+\frac{q}{LC}=0\\
\omega=\sqrt{\frac1{LC}},q=q_0\cos(\omega t+\phi)\\
i=-q_0\omega\sin(\omega t+\phi)\\
U_E=\frac{1}{2}\frac{q_0^2}{C}\cos^2(\omega t+\phi),U_B=\frac12Lq_0^2\omega^2\sin^2(\omega t+\phi)
\end{gathered}
\]
RLC circuit
\[
\varepsilon=iR+\frac qC+L\frac{\mathrm di}{\mathrm dt}\\
\]
Example
\[
\begin{aligned}
E\cdot2\pi rl&=\frac{q}{\epsilon_0}\Rightarrow E=\frac{q}{2\pi r\epsilon_0 l}\\
U_E&=\int_a^b\frac{1}{2}\epsilon_0(\frac{q}{2\pi r\epsilon_0 l})^2(2\pi r\mathrm dr l)\\
&=\frac{q^2}{4\pi \epsilon_0 l}\int_a^b\frac1r\mathrm dr\\
&=\frac{q^2}{4\pi \epsilon_0 l}\ln\frac{b}{a}\\
或C&=\frac{2\pi\epsilon_0l}{\ln\frac ba},U_E=\frac{q^2}{2C}=\frac{q^2}{4\pi\epsilon_0l}\ln\frac ba
\end{aligned}
\]
\[
\begin{aligned}
B\cdot2\pi r&=\mu_0I\Rightarrow B=\frac{\mu_0I}{2\pi r}\\
U_B&=\int_a^b\frac{1}{2\mu_0}(\frac{\mu_0I}{2\pi r})^2(2\pi r\mathrm dr)l\\
&=\frac{\mu_0I^2l}{4\pi}\int_a^b\frac1r\mathrm dr\\
&=\frac{\mu_0I^2l}{4\pi}\ln\frac ba\\
\text{或 }L&=\frac\phi I=\frac{\displaystyle \int_a^b\frac{\mu_0I}{2\pi r}l\mathrm dr}{I}\\
&=\frac{\displaystyle \frac{\mu_0Il}{2\pi }\ln\frac ba}{I}=\frac{\mu_0l}{2\pi }\ln\frac ba\\
U_B&=\frac12LI^2=\frac{\mu_0I^2l}{2\pi }\ln\frac ba
\end{aligned}
\]
Chap31-3：磁化
从分子电流产生磁矩
\[
\begin{gathered}
i=\frac{e}{T}=\frac{e}{\frac{2\pi r}{v}}=\frac{ev}{2\pi r}\\
|\vec \mu|=i(\pi r^2)=\frac12evr\\
\end{gathered}
\]
根据量子模型，电子轨道并不稳定，不能指望 \(v\)、\(r\)，唯角动量量子化守恒。
轨道磁矩 (Orbital magnetic dipole moment)
\[
|\vec{\mu_L}|=\frac{e}{2m}(mvr)=\frac{e}{2m}L
\]
\(L\) 为角动量，而 \(\displaystyle L=n\hbar=n\frac{h}{2\pi}\)，则
\[
|\vec{\mu_L}|=\frac{e}{2m}n\frac{h}{2\pi}
\]
\(n=1\) 时，为玻尔磁子 Bohr magneton。自旋磁矩 (Spin magnetic dipole moment)
\[
\vec{\mu_s}=-\frac{e}{2m}\vec{s},\; s=\pm\frac12\hbar
\]
有 \(\vec\mu=\vec\mu_L+\vec\mu_S\approx\vec\mu_L\)。外加磁场 \(\vec B\)，会受到力矩 \(\vec\tau=\vec\mu\times\vec B\) 的作用。定义 \(\vec{\mu}=\sum_i\vec\mu_i\)。
考察不外加磁场的电子轨道
\[
\begin{gathered}
\frac{Ze^2}{4\pi\epsilon_0 r^2} = m\omega_0^2r \\
\omega_0 = \sqrt{\frac{Ze^2}{4\pi\epsilon_0mr^3}}
\end{gathered}
\]
考察外加磁场后的角速度变化
当 \(\vec{\omega_0}\parallel\vec B\)，
\[
\frac{Ze^2}{4\pi\epsilon_0r^2}+e\omega rB = m\omega^2r
\]
根据 \(\omega=\omega_0+\Delta\omega\)，代入展开，舍去二阶小量有
\[
\frac{Ze^2}{4\pi\epsilon_0r^2}+e\omega_0rB+e\Delta\omega rB = m\omega_0^2r+2m\omega_0\Delta\omega r
\]
则
\[
\begin{gathered}
e\omega_0rB\approx e\omega_0rB+e\Delta\omega rB=2m\omega_0\Delta\omega r\\
\Delta\omega=\frac{eB}{2m}
\end{gathered}
\]
当 \(\vec{\omega_0}\parallel-\vec B\)，类似地有
\[
\begin{aligned}
&\frac{Ze^2}{4\pi\epsilon_0r^2}-e\omega rB=m\omega^2r\\
&\frac{Ze^2}{4\pi\epsilon_0r^2}-e\omega_0rB-e\Delta\omega rB=m\omega_0^2r-2m\omega_0\Delta\omega r\\
&e\omega_0rB\approx e\omega_0rB+e\Delta\omega rB=2m\omega_0\Delta\omega r\\
&\Delta\omega=\frac{eB}{2m}
\end{aligned}
\]
由之前结论，\(\displaystyle\vec\mu_0=-\frac12evr=-\frac12er^2\vec\omega_0\)，\(\displaystyle\Delta\vec\mu=-\frac12er^2\Delta\vec\omega=-\frac{e^2r^2}{4m}\vec B\)
所以磁矩的变化量总是反抗外加磁场的方向，且 \(|\Delta\vec\mu|\ll|\vec\mu|\)
定义磁化强度 (Magnetization) \(\displaystyle\vec M=\frac{\sum_i\vec\mu_i}{\Delta V}\) ( 对比极化强度 \(\displaystyle\vec P=\frac{\sum_i\vec p_i}{\Delta V}\))
\[
\vec B=\vec B_0+\vec B_M\text{ (对比 }\vec E=\vec E_0+\vec E')
\]
Ampere's Law
令 \(j_M\) 为磁化电流线密度，则
\[
\bigg|\sum_i\vec \mu_i\bigg|=i_MS=j_MlS
\]
随后则有
\[
\Delta V=lS,\bigg|\vec M\bigg|=\frac{\displaystyle\bigg|\sum_i\vec \mu_i\bigg|}{\Delta V}=j_M
\]
即 \(\vec j_M=\vec M\times\vec n\)。对比电场，\(\sigma'=|\vec P|\cos\theta=\vec P\cdot\vec n\)。随后
\[
\begin{gathered}
\oint_L\vec M\cdot\mathrm d\vec l=\sum\int_{a_i}^{b_i}\vec M\cdot\mathrm d\vec l=\sum M\overline{a_ib_i}=\sum j_M\overline{a_ib_i}=\sum i_M\\
\oint_L\vec B\cdot\mathrm d\vec l=\oint_L\vec B_0\cdot\mathrm d\vec l+\oint_L\vec B_M\cdot\mathrm d\vec l=\mu_0(\sum i_0+\sum i_M)\; (\text{原 Ampere's Law})\\
\end{gathered}
\]
即得
\[
\begin{gathered}
\sum i_0=\oint_L\frac{\vec B}{\mu_0}\cdot\mathrm d\vec l-\sum i_M=\oint_L\left(\frac{\vec B}{\mu_0}-\vec M\right)\cdot\mathrm d\vec l=\oint_L\vec H\cdot\mathrm d\vec l\\
\vec M=\chi_m\vec H\left(=\chi_m\frac{\vec B}{\mu}\right)\;(对比P=\chi_e \vec D=\chi_e\epsilon E)\\
\vec B=\mu_0\vec H+\mu_0\vec M=\mu_0(1+\chi_m)\vec H\xlongequal{\mu_r=1+\chi_m}\mu_0\mu_r\vec H(=\mu\vec H)=\mu_r\vec B_0=\kappa_m\vec B_0
\end{gathered}
\]
国外常用 \(\kappa_m\) 而非 \(\mu_r\)，用 \(\kappa_e\) 而非 \(\epsilon_r\)，对比 \(\vec E=\frac{\vec E_0}{\epsilon_r}=\frac{\vec E_0}{\kappa_e}\)。
最终磁场大于外加磁场，电场小于外加电场。
\[
\chi_m\vec B_0=\mu_0(\chi_m\vec H)=\mu_0\vec M=\vec B-\mu_0\vec H=\kappa_m\vec B_0-\vec B_0=(\kappa_m-1)\vec B_0
\]
因此 \(\chi_m=\kappa_m-1\Rightarrow\kappa_m(\mu_r)=\chi_m+1\)，对比 \(\kappa_e(\epsilon_r)=\chi_e+1\)
考虑如下四种情况：
- 顺磁性 (Paramagnetism): \(\kappa_m>1\), \(\vec B>\vec B_0\)
- 逆磁性 (Diamagnetism): \(\kappa_m<1\), \(\vec B<\vec B_0\)
- 铁磁性 (Ferromagnetism): \(\kappa_m\gg 1\), \(\vec B\gg\vec B_0\)
- 超导体 (Superconductor): \(\kappa_m=0\), \(\vec B=0\)
Example1 旋转环形电荷
\[
i=\frac{\Delta q}{\Delta t}=\frac{q}{\frac{2\pi}{\omega}}=\frac{q\omega}{2\pi}\\
\mu=i(\pi R^2)=\frac{q\omega R^2}{2}
\]
Example1-plus 旋转圆盘电荷
\[
\begin{gathered}
d\mu=\pi r^2(\frac{\sigma\cdot2\pi r\mathrm d r}{\frac{2\pi}{\omega}})=\sigma\pi\omega r^3\mathrm d r\\
\mu=\int_0^R\sigma\pi\omega r^3\mathrm d r==\frac{\sigma\pi\omega R^4}{4}=\frac{q\omega R^2}{4}\\
\end{gathered}
\]
Exmple2 旋转球电荷
\[
\begin{gathered}
\rho=\frac{q}{\frac{4}{3}\pi R^3}=\frac{3q}{4\pi R^3}\\
\mathrm d\mu=\frac{[\rho\pi (R\sin\theta)^2\mathrm dz]\omega (R\sin\theta)^2}{4}\\
\mathrm dz=\mathrm d(R\cos\theta)=-R\sin\theta\mathrm d\theta\\
\end{gathered}
\]
\[
\begin{aligned}
\mu&=\int_R^{-R}\frac{[\rho\pi (R\sin\theta)^2\mathrm dz]\omega (R\sin\theta)^2}{4}\\
&=\int_0^\pi\frac{[\rho\pi (R\sin\theta)^2R\sin\theta\mathrm d\theta]\omega (R\sin\theta)^2}{4}\\
&=\frac14\rho\pi\omega R^5\int_0^\pi\sin^5\theta\mathrm d\theta\\
&=\frac{3q\omega R^2}{16}\cdot2\frac{4\cdot2}{5\cdot3\cdot1}=\frac{q\omega R^2}{5}\\
\end{aligned}
\]
\[
\begin{gathered}
\mu=\left\{\begin{matrix}
\displaystyle\frac{q\omega R^2}{2}\text{，圆环}\\
\displaystyle\frac{q\omega R^2}{4}\text{，圆盘}\\
\displaystyle\frac{q\omega R^2}{5}\text{，球}
\end{matrix}\right.\quad\quad\quad
J\text{ (转动惯量)}=\left\{\begin{matrix}
MR^2\text{，圆环}\\
\displaystyle\frac{1}{2}MR^2\text{，圆盘}\\
\displaystyle\frac25MR^2\text{，球}
\end{matrix}\right.
\end{gathered}
\]
Example3
\[
\begin{gathered}
2\pi rB_{10}=\mu_0\frac{\pi r^2}{\pi R^2}I\Rightarrow B_{10}=\frac{\mu_0 Ir}{2\pi R^2},B_{1}=\mu_{r}B_{10}=\frac{\mu_{r1}\mu_0Ir}{2\pi R^2}\\
2\pi rB_{20}=\mu_0I\Rightarrow B_{20}=\frac{\mu_0I}{2\pi r},B_2=\mu_{r2}B_{20}=\frac{\mu_{r2}\mu_0I}{2\pi r}\\
M_1=\frac{B_1-B_{10}}{\mu_0}=\frac{\mu_{r1}-1}{\mu_0}B_{10}=\frac{(\mu_{r1}-1)Ir}{2\pi R^2}\\
M_2=\frac{B_2-B_{20}}{\mu_0}=\frac{\mu_{r2}-1}{\mu_0}B_{20}=\frac{(\mu_{r2}-1)I}{2\pi r}
\end{gathered}
\]
Example4
已知 \(L,N,I\)。
\[
\begin{gathered}
B_0=\mu_0nI=\frac{\mu_0NI}{L}\\
\mu_r=\frac{B}{B_0}\\
I=\frac{B_0L}{\mu_0N}\\
M=\frac{B-B_0}{\mu_0}=\frac{(\mu_r-1)NI}{L}=\frac{i}{l}\\
\therefore i=(\mu_r-1)NI=\frac{(\mu_r-1)B_0L}{\mu_0}
\end{gathered}
\]
Example5
已知 \(n,I,B\)。
\[
\begin{gathered}
B=\mu_rB_0\Rightarrow\mu_r=\frac{B}{B_0}=\frac{B}{\mu_0nI}\\
H\cdot 2\pi r=NI\Rightarrow H=\frac{NI}{2\pi r}=nI\\
j_m=M=\frac{B}{\mu_0}-H\\
\end{gathered}
\]
Chap32：综合应用
Example1 平行板电容器
\[
\Delta V=4+3t(电势差),\phi_E=?,i_d=?,B=?
\]
解：
\[
\begin{gathered}
E=\frac{\Delta V}{D}=\frac{4+3t}{D},\phi_E=E\pi R^2=\frac{\pi R^2}{D}(4+3t)\\
i_d=\varepsilon_0\iint\frac{\mathrm dE}{\mathrm dt}\mathrm dS=\frac{3\varepsilon_0\pi R^2}{D}\\
B\cdot2\pi r=\mu_0\frac{\pi r^2}{\pi R^2}i_d=\frac{3\mu_0\varepsilon_0\pi r^2}{D}\Rightarrow B=\frac{3\mu_0\varepsilon_0 r}{2D}\\
\end{gathered}
\]
新设：
\[
q=q_m\sin\omega t,\phi_E=?,i_d=?,B=?
\]
解：
\[
\begin{gathered}
E=\frac{q_m\sin\omega t}{CD}=\frac{q_m\sin\omega t}{\epsilon_0 \pi R^2}\\
i_d=\varepsilon_0\cdot\frac{q_m\omega\cos\omega t}{\epsilon_0 \pi R^2}\cdot\pi R^2=q_m\omega\cos\omega t
\end{gathered}
\]
Example3
\[
\begin{gathered}
j=\sigma E\Rightarrow E=\rho \frac i{\pi R^2}\\
B\cdot2\pi R=\mu_0 i
\end{gathered}
\]