01电磁学——电磁感应

Lingfeng2024-06-10

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)}