# 磁通量

## 描述

${\displaystyle \displaystyle \Phi _{B}=\mathbf {B} \cdot \mathbf {a} =Ba\cos \theta }$

${\displaystyle \Phi _{B}=\iint \limits _{S}\mathbf {B} \cdot d\mathbf {S} }$

## 通过闭曲面的磁通量

${\displaystyle \Phi _{B}=\int \!\!\!\int \mathbf {B} \cdot d\mathbf {S} =0,}$

## 通过开曲面的磁通量

${\displaystyle {\mathcal {E}}=\oint _{\partial \Sigma (t)}\left(\mathbf {E} (\mathbf {r} ,\ t)+\mathbf {v\times B} (\mathbf {r} ,\ t)\right)\cdot d{\boldsymbol {\ell }}=-{d\Phi _{B} \over dt},}$

${\displaystyle {\mathcal {E}}}$ 電動勢
${\displaystyle \Phi _{B}}$ 为通过开曲面的磁通量，这一开曲面的边界为${\displaystyle \partial \Sigma (t)}$
${\displaystyle \partial \Sigma (t)}$ 为一个随时间变化的闭曲线
${\displaystyle d{\boldsymbol {\ell }}}$ 是边界${\displaystyle \partial \Sigma (t)}$ 无穷小向量元
${\displaystyle \mathbf {v} }$ 是线段${\displaystyle d{\boldsymbol {\ell }}}$ 的速度
${\displaystyle \mathbf {E} }$ 为电场
${\displaystyle \mathbf {B} }$ 磁场

## 与电通量的比较

${\displaystyle \Phi _{E}=\int \!\!\!\int _{S}\mathbf {E} \cdot d\mathbf {S} ={Q \over \epsilon _{0}},}$

${\displaystyle \mathbf {E} }$ 為電場
${\displaystyle S}$ 為任意闭曲面
${\displaystyle Q}$ 为曲面${\displaystyle S}$ 包围的电荷
${\displaystyle \epsilon _{0}}$ 真空電容率

## 参考文献

1. ^ Douglas C Giancoli. Physics for scientists & engineers : with modern physics. 培生集團. 2009: 第760頁. ISBN 0131578499.