电磁感应定律

法拉第电磁感应定律

$$ \begin{gather} \begin{aligned} \xi &= - N \frac{ d \phi }{ dt} \end{aligned} \end{gather} $$

楞次定律

闭合回路中感应电流的方向总是使得它所激发的电场阻碍引起感应电流的磁通量的变化

动生电动势和感生电动势

动生电动势

定义

恒磁场内导体运动所产生的电动势

推导

自由电子受到的力$\pmb F=-e(\pmb{v\times B})$
单位正电荷的非静电“力” $\pmb K=\frac{\pmb F}{-e}=\pmb{v\times B}$

$$ \begin{gather} \begin{aligned} \xi = \int_C^D\pmb K\cdot d\pmb l=\int_C^D (\pmb{v\times B})d\pmb l \end{aligned} \end{gather} $$

交流发电机原理

$$ \\ 设AB长为l,CD长为s\\\ \\ \xi_{AB} = \xi_{CD} = \int_A^B(\pmb{v\times B})d\pmb l=vBlcos\theta\\\ \\ \xi = 2\xi_{AB} = 2vBlcos\theta\\\ \\ \theta = \omega t, v = \frac{s}{2}\omega\\\ \\ \pmb{\xi =\omega BS cos\omega t}\ $$

感生电动势和涡旋电场

定义

导体不动，磁场变化产生的电动势

$$ \\ \xi = \int\pmb K\cdot d\pmb l\\\ \\ \xi = -\frac{d\phi}{dt}=-\frac{d}{dt}\int \pmb B\cdot d\pmb S = -\frac{d}{dt}\int \pmb A\cdot d\pmb l = -\oint \frac{\partial A}{\partial t}d\pmb l\\\ \\ \therefore \pmb K=-\frac{\partial \pmb A}{\partial t}\\\ \\ 麦克斯韦发现,\pmb K=\pmb E_旋\\\ \\ \therefore \pmb E_旋=-\frac{\partial \pmb A}{\partial t}\\\ \\ 又\because \pmb E_势 = -\nabla U\\\ \\ \therefore \pmb E = \pmb E_势 + \pmb E_旋 = -\nabla U - \frac{\partial \pmb A}{\partial t}\ $$

电子加速器

$$ 设轨道处的磁场为B(R),运动轨道内的平均磁场为\overline B\\\ \\ 有evB(R) = \frac{mv^2}{R}\longrightarrow B(R)=\frac{mv}{eB}\\\ \\ \xi = -\frac{d\phi}{dt}=\int \pmb E_旋\cdot d\pmb l \\\ \\ \therefore \pmb E_旋 = -\frac{1}{2\pi R}\frac{d\phi}{dt}\\\ \\ 根据牛顿第二定律 \frac{d(mv)}{dt}=F=-e E_旋 = \frac{e}{2\pi R}\frac{d\phi}{dt}\\\ \\ d(mv) = \frac{e}{2\pi R}d\phi\\\ \\ 设v_0=0,\phi_0=0,有\\\ \\ mv = \frac{e}{2\pi R}\phi = \frac{e}{2\pi R}\pi R^2 \overline B\longrightarrow \overline B = \frac{2mv}{eB}\\\ \\ \therefore B(R) = \frac{1}{2}\overline B\ $$

互感和自感

互感

推导

$$ 易知: \Phi_{12}=M_{12}I_1,\Phi_{21}=M_{21}I_2 \\\ \\ \Phi_{12}=\oint \pmb A\cdot d\pmb l=\oint\frac{\mu_0I_1}{4\pi}\oint\frac{d\pmb l_1}{r_{12}}d\pmb l_2\\\ \\ \Phi_{12}=\frac{\mu_0I_1}{4\pi}\oint\oint\frac{d\pmb l_1\cdot d\pmb l_2}{r_{12}}\\\ \\ \therefore M_{12}=\frac{\Phi_{12}}{I_1}=\frac{\mu_0}{4\pi}\oint\oint\frac{d\pmb l_1\cdot d\pmb l_2}{r_{12}}\\\ \\ 同理：M_{12}=M_{21} $$

单位

$$ 1H=\frac{1Wb}{1A}=\frac{1V\cdot 1s}{1A}\\\ \\ 1mH=10^{-3}H,1\mu H=10^{-6}H $$

自感

定义

电生磁，磁再生电，影响原电动势

推导

$$ \begin{gather} \begin{aligned} \xi &= - \frac{ d \phi }{ dt } \\ &= -L \frac{ dI }{ dt } \end{aligned} \end{gather} $$

互感与自感的关系(无漏磁)

$$ 根据(1)式：M=\frac{N_1\Phi_{21}}{I_2}=\frac{N_2\Phi_{12}}{I_1}\\\ \\ 又\because L_1=\frac{N_1\Phi_1}{I_1},L_2=\frac{N_2\Phi_2}{I_2}\\\ \\ \therefore M^2=L_1L_2\longrightarrow \pmb{M=\sqrt{L_1L_2}}\\\ \\ 在有漏磁的情况下，M\lt \sqrt{L_1L_2} $$

两个线圈串联的总自感

顺接与反接

产生的磁场方向相同为顺接，相反为反接

推导

$$ 顺接时\\\ \\ \xi_1+\xi_{21}=-L_1\frac{dI}{dt}-M\frac{dI}{dt}\\\ \\ \xi_2+\xi_{12}=-L_2\frac{dI}{dt}-M\frac{dI}{dt}\\\ \\ 总电动势\xi=-(L_1+L_2+2M)\frac{dI}{dt}\\\ \\ \therefore 总自感\pmb{L = L_1+L_2+2M}\\\ \\ 反接时 \\\ \\ \pmb{L = L_1+L_2-2M} $$

自感磁能与互感磁能

自感磁能

$$ 对于一个线圈来说,在电压增大过程中抵抗其自感电动势所做的功为\\\ \\ dW = -\xi_i i dt\\\ \\ 其中 \xi_i=-L\frac{dI}{dt}\\\ \\ \therefore dW = Lidi\\\ \\ \pmb{W=\int dW=\int_0^I Lidi=\frac{1}{2}LI^2},称为自感磁能 $$

互感磁能

$$ 同理，互感磁能应为\pmb{W=MI_1I_2}\\\ \\ 因此互感线圈的总磁能为\\\ \\ \pmb{W=\frac{1}{2}L_1I^2+\frac{1}{2}L_2I^2+MI_1I_2} $$