一、磁的基本现象和规律

磁场强度与磁的库仑定律

$$ \begin{gather} \begin{aligned} \pmb{ H } &= \frac{ F }{ q_{m_0}} \\ &= - \nabla U_{m0} \end{aligned} \end{gather} $$

安培定律

1. 对于两段载流导线上的电流元来说, $dF_{12}\propto \dfrac{I_1dl_1\cdot I_2dl_2}{r_{12}^2}$

• $d\pmb l_1=dl_1\cdot sin\theta_1，d\pmb l_2=dl_2\cdot sin\theta_2$

2. $\therefore dF_{12}=k \dfrac{I_1I_2dl_1dl_2sin\theta_1sin\theta_2}{r_{12}^2}$

$$ \begin{gather} \begin{aligned} d\pmb F_{12}=k\frac{I_1I_2d\pmb l_2\times(d\pmb l_1\times\pmb r_{12})}{r_{12}^2} \end{aligned} \end{gather} $$

• 经实验得出, $k=\frac{\mu_0}{4\pi}$

磁感应强度 Biot-Savart Law

磁感应强度B

$$ \begin{gather} \begin{aligned} \pmb B=\mu_0\pmb H \end{aligned} \end{gather} $$

Biot-Savart Law

推导

$$ \begin{aligned} d\pmb F_{12} &= \frac{\mu_0}{4\pi}\frac{I_2d\pmb l_2\times(I_1d\pmb l_1\times \pmb r_{12})}{r_{12}^2} \\ &= I_2 d\pmb{l_2} d \pmb B \\ \therefore d\pmb B &=\frac{\mu_0}{4\pi}\frac{I_1d\pmb l_1\times \pmb r_{12}}{r_{12}^2} \end{aligned} $$

$$ \begin{gather} \begin{aligned} \pmb B=\frac{\mu_0}{4\pi}\oint\frac{I_1d\pmb l_1\times\pmb { \overrightarrow r}}{r_{12}^2} \end{aligned} \end{gather} $$

• B的单位为$N/A\cdot m$, 称为$T$, $1T=10^4Gs$

• 其中，$\pmb{ \overrightarrow r}$是$r$的向量(不是单位)

应用

1. 载流直导线

$$ B = \int_{A_{1}}^{A_2}\frac{\mu_0}{4\pi}\frac{Idlsin\theta}{r^2}\\\ \\ 根据几何关系:tan(\pi-\theta)=\frac{r_0}{l}\\\ \\ \therefore dl=\frac{r_0d\theta}{sin^2\theta}\\\ \\ \therefore B=\frac{\mu_0}{4\pi}\int_{\theta_1}^{\theta_2}\frac{Isin\theta d\theta}{r_0}\\\ \\ \pmb{B=\frac{\mu_0I}{4\pi r}(cos\theta_1-cos\theta_2)}\\\ \\ 当直线无限长时:\pmb{B=\frac{\mu_0 I}{2\pi r}}\\\ $$

2. 载流圆线圈

$$ \\ dB=\frac{\mu_0}{4\pi}\frac{Idl}{r^2}=\frac{\mu_0}{4\pi}\frac{Idl}{r_0^2}sin^2\alpha\\\ \\ B=\oint dBcos\alpha=\oint\frac{\mu_0}{4\pi}\frac{Idl}{r_0^2}sin^2\alpha cos\alpha\\\ \\ \therefore \pmb {B=\frac{\mu_0R^2I}{2(R^2+r_0^2)^{3/2}}}\\\ \\ 当P点在球心时:\pmb{B=\frac{\mu_0I}{2R}}\\\ \\ 当P点在无穷远处:\pmb{B=\frac{\mu_0R^2I}{2r_0^3}} $$

3. 载有环向电流的圆筒

$$ \\ 设半径为R，总长度为L,电流的线密度为\iota\\\ \\ 取轴线为x轴，中点为原点\\\ \\ 对于x处的P点\\\ \\ dB=\frac{\mu_0R\iota \cdot dl}{2(R^2+(x-l)^2)^{3/2}}\\\ \\ B = \int_{-L/2}^{L/2}dB\\\ \\ 其中: 距离r=\sqrt{R^2+(x-l)^2}\\\ \\ \frac{x-l}{R}=cot\beta\longrightarrow dl=\frac{Rd\beta}{sin^2\beta}\\\ \\ \therefore \pmb{B=\frac{\mu_0\iota}{2}(cos\beta_1-cos\beta_2)}\\\ \\ cos\beta_1=\frac{x+L/2}{\sqrt{R^2+(x+L/2)^2}}，cos\beta_2=\frac{x-L/2}{\sqrt{R^2+(x-L/2)^2}}\\\ \\ 当螺线管无限长:\pmb{B=\mu_0\iota}\\\ \\ 在无限长螺线管的 一端:\pmb {B=\frac{\mu_0\iota}{2}}\\\ \\ 其中，电流线密度\iota=nI\\\ \\ \therefore 对于无限长螺线管有:B=\mu_0nI\\\ $$

安培环路定律(暂定)

$$ \oint \pmb B\cdot d\pmb l=\mu_0I\\\ \\ 或\pmb B=\frac{\mu_0I}{4\pi}\nabla\Omega $$

磁场对载流导线的作用

安培力与无限长直导线之间的作用

安培力

$$ \begin{gather} \begin{aligned} d\pmb F &= Id\pmb l\times\pmb B \end{aligned} \end{gather} $$

无限长直导线

$$ \begin{aligned} dF_{12} &= I_2dl_2B_1=\frac{\mu_0I_1I_2}{2\pi a}dl_2 \\ dF_{21} &=\frac{\mu_0I_1I_2}{2\pi a}dl_1 \end{aligned} $$

• 单位作用力

$$ \begin{gather} \begin{aligned} f=\frac{dF_{12}}{dl_2}=\frac{\mu_0I_1I_2}{2\pi a} \end{aligned} \end{gather} $$

矩形载流线圈所受力矩

$$ \\ F_{AB}=F_{CD}相互抵消\\\ \\ L = F_{BC}\frac{a}{2}sin\theta+F_{AD}\frac{a}{2}sin\theta\\\ \\ L = BIbasin\theta=IBSsin\theta\\\ \\ \therefore\pmb L=IS(\pmb{n\times B})\\\ $$

载流线圈的磁矩

$$ \\ dl_1sin\theta_1=dl_2sin\theta_2=dh\\\ \\ dF_{12}=Idl_1Bsin\theta_1;dF_{21}=Idl_2Bsin\theta_2\\\ \\ \therefore dF_{12}=dF_{21}=IBdh\\\ \\ \therefore dL=dF_{12}\cdot x_1+dF_{21}\cdot x_2=IBdh(x_1+x_2)\\\ \\ L = IBS\\\ \\ 其中，m=IS\longrightarrow\pmb m=IS\pmb n\\\ \\ \therefore \pmb{L = m\times B}\\\ $$

磁偶极子和载流线圈的等价性(待定)

带电粒子在磁场中的运动

洛伦兹力

$$ \begin{gather} \begin{aligned} \pmb F &= q\ \pmb{v\times B} \\ |F| &= |q|\ vB\ sin\theta \end{aligned} \end{gather} $$

洛伦兹力和安培力的关系

$$ \\ 设自由电子移动速度为u,单位体积的自由电子数为n\\\ \\ 在时间\Delta t内通过面S的电子数为n\Delta V = nSu\Delta t\\\ \\ 电量\Delta q = enSu\Delta t\\\ \\ I = \frac{\Delta q}{\Delta t}= enSu\\\ \\ 每个电子的洛伦兹力为\\\ \\ F=evB\\\ \\ F_{总}=nS\Delta l F=(ensv)\Delta l B=BI\Delta l $$

带电粒子在磁场中的运动

1. 当 $\pmb{v\perp B}$

$$ qvB=m\frac{v^2}{R}\\\ \\ \pmb{R=\frac{mv}{qB}}，\pmb{T=\frac{2\pi m}{qB}} $$

2. 普遍情形

$$ h=v_{//}T=\frac{2\pi mv}{qB} $$

3. 磁聚焦

• 由于$v_{//}$相同，一段时间后这些粒子又会聚集在一起

荷质比

汤姆孙法

$$ \\ eE=evB\longrightarrow v=\frac{E}{B}\\\ \\ R=\frac{mv}{eB}\\\ \\ \frac{e}{m}=\frac{v}{RB}=\frac{E}{RB^2}\\\ \\ 实验结果:\frac{e}{m}=1.759\times 10^{11}C/kg $$

磁聚焦法

$$ \\ \frac{1}{2}mv^2=e\Delta U\\\ \\ v = \sqrt{\frac{2e\Delta U}{m}}\\\ \\ h = \frac{2\pi m v}{eB}\\\ \\ \therefore \frac{e}{m}=\frac{8\pi^2\Delta U}{l^2B^2} $$