01电磁学——电磁感应

Lingfeng 2024-06-10

01电磁学——电磁感应

==不要忘记线圈匝数!!!==

楞次定律 变化磁通量产生电动势 $\varepsilon= - \frac{d\phi}{dt}=-\frac{d}{dt} \iint_{S} \vec{B}\cdot d\vec{S}$ 感生电动势 $\varepsilon_{ind}= -\iint_{S} \frac{ \partial \vec{B} }{ \partial t } \cdot d\vec{S}$ 动生电动势 $\varepsilon_{mot}= \int_{a(-)}^{b(+)} \vec{v}\times \vec{B} \cdot \, d\vec{l}$

例1 长直导线载有导线 $I$ , 旁边有一与它共面的矩形线圈. 求感应电动势.

法一注意到 $B= \frac{\mu_{0}I}{2\pi(x+l)}$ 故 $d\phi = BdS = \frac{\mu_{0}bI}{2\pi} \frac{1}{x+l} dl$ 此时 $\begin{align*} \phi & = \frac{\mu_{0}bI}{2\pi} \int _{0}^{a} \frac{1}{x+l} \, dl \\ & = \frac{\mu_{0}bl}{2\pi}( \ln(a+x)-\ln x) \end{align*}$ 故 $\begin{align*} \varepsilon= -N\frac{d\phi}{dt} & = N\frac{\mu_{0}bI}{2\pi}\left( \frac{1}{x}- \frac{1}{a+x}\right) \cdot v \\ & = \frac{\mu_{0}NIab}{2\pi x(x+a)}v \end{align*}$
法二用动生电动势 $\begin{align*} \varepsilon_{AB} & = \int_{A}^{B} \vec{v}\times \vec{B} \cdot d\vec{l} \\ & = \frac{\mu_{0}I}{2\pi x} bv \end{align*}$ 同理 $\begin{align*} \varepsilon _{CD} & = \int _{C}^{D} \vec{v}\times \vec{B} \cdot d\vec{l} \\ & = -\frac{\mu_{0}I}{2\pi(x+a)}bv \end{align*}$ 故 $\begin{align*} \varepsilon & =N(\varepsilon_{AB}+\varepsilon_{CD}) \\ & = \frac{\mu_{0}NI}{2\pi x}\left( \frac{1}{x} -\frac{1}{x+a} \right)bv \\ & = \frac{\mu_{0}NIab}{2\pi x(x+a)}v \end{align*}$

例2 长度为 $L$ 在磁场中旋转, 求电动势.

解
由动生电动势 $\begin{align*} \varepsilon_{OA} & =\int_{0}^{L} \vec{v}\times \vec{B} \cdot \, d\vec{l} \\ & = - Bw\int_{0}^{L} l \, dl \\ & = -\frac{1}{2}BwL^{2} \end{align*}$

例3 无限长直载流螺线管 $n$ , 半径为 $R$ , 电流为 $I=\alpha t$ . 求感应电动势.

解
由先前结论 $B= \mu_{0}nI$ 故 $\begin{align*} \varepsilon & = -\iint \frac{ \partial \vec{B} }{ \partial t } \cdot d\vec{S} \\ & = \mu_{0}n\alpha \cdot \pi r^{2} \end{align*}$ 注意到 $\varepsilon= \oint_{L}\vec{E} \cdot d\vec{l}=E \cdot 2\pi r$ 故 $E= \frac{\mu_{0}n\alpha r}{2}$ 当在螺线管外面时 $\varepsilon=\mu_{0}n\alpha \cdot \pi R^{2} = E\cdot 2\pi r$ 故此时 $E= \frac{\mu_{0}n\alpha R^{2}}{2r}$

例4 例1中改成 $I=I_{0}\cos \omega t$ .

解
此时 $\begin{align*} \varepsilon_{ind} & = - \iint \frac{ \partial \vec{B} }{ \partial t } \cdot d\vec{S} \\ & = - \int _{0}^{a} \frac{ \partial }{ \partial t } \left( \frac{\mu_{0}I}{2\pi(x+l)} \right)\, bdl \\ & = \frac{\mu_{0}b}{2\pi}\omega I_{0}\sin\omega t \int _{0}^{a} \frac{1}{l+x} \, dl \\ & = \frac{\mu_{0}bI_{0}\omega \sin\omega t}{2\pi} \ln\left( \frac{x+a}{x} \right) \end{align*}$ 动生 $\varepsilon_{mot}= \frac{\mu_{0}Ibav}{2\pi x(x+a)}$