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磁场高斯定理的推导

本节中,我们基于毕奥-萨法尔定律推导磁场的高斯定理

如下图所示,激发磁场的电流元和要测量磁场的目标点相对于坐标原点 \(O\) 的位矢分别为 \(\vec{q}\)\(\vec{r}\),目标点相对于电流元的位矢 \(\vec{k} = \vec{r} - \vec{q}\)

Image title
磁场高斯定理推导

根据毕奥-萨法尔定律

\[ \vec{B} = \frac{\mu_0}{4\pi} \int_V \frac{\vec{j}\big(\vec{q}\tm\big) \times \vec{k}}{k^3} \,\d{v} = \frac{\mu_0}{4\pi} \int_V \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} \,\d{v} \]
\[ \begin{align} \vec{B} & = \frac{\mu_0}{4\pi} \int_V \frac{\vec{j}\big(\vec{q}\tm\big) \times \vec{k}}{k^3} \,\d{v} \\ & = \frac{\mu_0}{4\pi} \int_V \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} \,\d{v} \end{align} \]

这里需要明确的是,\(\vec{r}\)\(\vec{q}\) 是两套不同的坐标,上式中对体积求积分是对坐标 \(\vec{q}\) 求积分,得到的磁感应强度为依赖 \(\vec{r}\) 的场函数 \(\vec{B}\big(\vec{r}\tm\big)\),其散度

\[ \nabla \cdot \vec{B} = \frac{\mu_0}{4\pi} \nabla \cdot \int_V \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} \,\d{v} = \frac{\mu_0}{4\pi} \int_V \nabla \cdot \sb{ \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} } \,\d{v} \]
\[ \begin{align} \nabla \cdot \vec{B} & = \frac{\mu_0}{4\pi} \nabla \cdot \int_V \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} \,\d{v} \\ & = \frac{\mu_0}{4\pi} \int_V \nabla \cdot \sb{ \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} } \,\d{v} \end{align} \]

由公式

\[ \nabla \cdot \rb{\vec{a} \times \vec{b}\tm} = \vec{b} \cdot \rb{\nabla \times \vec{a}\tm} - \vec{a} \cdot \rb{\nabla \times \vec{b}\tm} \]

可得

\[ \nabla \cdot \rb{ \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} } = \underbrace{\frac{\vec{r}-\vec{q}}{\abs{\vec{r}-\vec{q}\tm}^3} \cdot \sb{\nabla \times \vec{j}\big(\vec{q}\tm\big)}}_{\displaystyle \rb{1}} - \underbrace{\vec{j}\big(\vec{q}\tm\big) \cdot \rb{\nabla \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3}}}_{\displaystyle \rb{2}} \]
\[ \begin{align} & \nabla \cdot \rb{ \vec{j}\big(\vec{q}\tm\big) \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3} } = \underbrace{\frac{\vec{r}-\vec{q}}{\abs{\vec{r}-\vec{q}\tm}^3} \cdot \sb{\nabla \times \vec{j}\big(\vec{q}\tm\big)}}_{\displaystyle \rb{1}} \\ & - \underbrace{\vec{j}\big(\vec{q}\tm\big) \cdot \rb{\nabla \times \frac{\vec{r} - \vec{q}}{\abs{\vec{r} - \vec{q}\tm}^3}}}_{\displaystyle \rb{2}} \end{align} \]

\(\vec{B}\) 求散度和旋度,\(\nabla\) 只对坐标 \(\vec{r}\) 求导,与坐标 \(\vec{q}\) 及其场函数无关。因此,上式中

  • \(\nabla \times \vec{j}\rb{\vec{q}\tm} = \vec{0}\),因此,上式中 \(\rb{1} = 0\)
  • \(\rb{2}\) 中,\(\displaystyle \frac{\vec{r}-\vec{q}}{\abs{\vec{r}-\vec{q}\tm}^3} = - \nabla \frac{1}{\abs{\vec{r}-\vec{q}\tm}}\),因此

    \[ \rb{2} = \vec{j}\rb{\vec{q}\tm} \cdot \nabla \times \rb{\nabla \frac{1}{\abs{\vec{r}-\vec{q}\tm}}} \]

    由于标量场的梯度是无旋场,上式中 \(\rb{2} = 0\)

综上所述,基于毕奥-萨法尔定律,磁感应强度的散度为零,即

\[ \nabla \cdot \vec{B} = 0 \]

另外,根据高斯定理,磁通量的积分形式

\[ \oint_S \vec{B} \cdot \d{\vec{s}} = \int_{V} \nabla \cdot \vec{B} = 0 \]

其中,\(S\) 为包围空间区域 \(V\) 的边缘曲面