﻿ 磁场中的高斯定理另一证明

# 磁场中的高斯定理另一证明Another Proof Theorem of Gauss in the Magnetic Field

Abstract: In this paper, we prove theorem of magnetic field of Gauss with the first principle of the motion charge creating the magnetic field equivalence the law of Biot-Savart. This proves to be a further understanding of the nature of the magnetic field. And it can be used for reference in college physics teaching.

1. 引言

1.1. 毕奥-萨伐尔定律发现简介

1.2. 毕奥-萨伐尔定律

$\text{d}B=\frac{{\mu }_{0}}{4\text{π}}\frac{I\text{d}l×\stackrel{^}{r}}{{r}^{2}}$ (1)

$I=nqvs$ (2)

$B=\frac{\text{d}B}{ns\text{d}l}=\frac{{\mu }_{0}}{4\text{π}}\frac{qv×\stackrel{^}{r}}{{r}^{2}}$ (3)

1.3. 磁场中的高斯定理

$\underset{s}{∯}B\cdot \text{d}S=0$ (4)

$\nabla \cdot B=0$ (5)

$\underset{s}{∯}B\cdot \text{d}S=\underset{V}{\iiint }\left(\nabla \cdot B\right)\text{d}V=0$

2. 磁场中的高斯定理的证明

2.1. 运动电荷产生的磁场的高斯定理的证明

$\nabla \cdot \left(f×g\right)=\left(\nabla ×f\right)\cdot g-f\cdot \left(\nabla ×g\right)$ (6)

$\nabla \cdot B=\frac{{\mu }_{0}q}{4\text{π}}\nabla \cdot \left(v×\frac{\stackrel{^}{r}}{{r}^{2}}\right)=\frac{{\mu }_{0}q}{4\text{π}}\left[\left(\nabla ×v\right)\cdot \frac{\stackrel{^}{r}}{{r}^{2}}-v\cdot \left(\nabla ×\frac{\stackrel{^}{r}}{{r}^{2}}\right)\right]$ (7)

${\left(\nabla ×v\right)}_{x}=\frac{\partial {v}_{z}}{\partial y}-\frac{\partial {v}_{y}}{\partial z}=\frac{\partial v}{\partial y}-\frac{\partial 0}{\partial z}=0$

$\begin{array}{c}{\left(\nabla ×\frac{\stackrel{^}{r}}{{r}^{2}}\right)}_{x}=\frac{\partial {\left(\frac{\stackrel{^}{r}}{{r}^{2}}\right)}_{z}}{\partial y}-\frac{\partial {\left(\frac{\stackrel{^}{r}}{{r}^{2}}\right)}_{y}}{\partial z}\\ =\frac{\partial \left(\frac{z-{z}^{\prime }}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3/2}}\right)}{\partial y}-\frac{\partial \left(\frac{y}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{3/2}}\right)}{\partial z}\\ =\frac{3zy}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5/2}}-\frac{3yz}{{\left({x}^{2}+{y}^{2}+{z}^{2}\right)}^{5/2}}=0\end{array}$

2.2. 电流产生的磁场的高斯定理的证明

$B=\int \text{d}B={\int }_{a}^{b}\frac{{\mu }_{0}}{4\text{π}}\frac{I\text{d}l×\stackrel{^}{r}}{{r}^{2}}$ (8)

$\begin{array}{c}\nabla \cdot B=\nabla \cdot \left({\int }_{a}^{b}\frac{{\mu }_{0}}{4\text{π}}\frac{I\text{d}l×\stackrel{^}{r}}{{r}^{2}}\right)={\int }_{a}^{b}\left(\nabla \cdot \left(\frac{{\mu }_{0}}{4\text{π}}\frac{I\text{d}l×\stackrel{^}{r}}{{r}^{2}}\right)\right)\\ ={\int }_{a}^{b}\left(\nabla \cdot \left(\frac{{\mu }_{0}}{4\text{π}}\frac{nsqv\text{d}l×\stackrel{^}{r}}{{r}^{2}}\right)\right)={\int }_{a}^{b}ns\text{d}l\left(\nabla \cdot \frac{{\mu }_{0}}{4\text{π}}\frac{qv\text{\hspace{0.17em}}×\stackrel{^}{r}}{{r}^{2}}\right)\\ ={\int }_{a}^{b}ns\text{d}l\left(\nabla \cdot B\right)=0\end{array}$

3. 结论

[1] 陈毓芳, 邹延肃. 物理学史简明教程[M]. 北京: 北京师范大学出版社, 1986: 183-190.

[2] 申先甲, 张锡鑫, 束炳如, 陈毓芳. 物理学史教程[M]. 湖南教育出版社, 1987: 288-295.

[3] 谢国亚, 林朝金, 廖其力. 大学物理教程(下) [M]. 吉林: 吉林大学出版社, 2012: 31-36.

[4] Purcell, E.M. (1965) Electricity and Magnetism, Berkeley Physics Course. Vol. 2, McGraw Hill, 183-184.

[5] 郭硕鸿. 电动力学[M]. 北京: 高等教育出版社, 1979: 13-14.

[6] 赵凯华, 陈熙谋. 电磁学[M]. 北京: 高等教育出版社, 1985: 352-353.

[7] 梁灿彬, 秦光戎, 梁竹健. 电磁学[M]. 北京: 高等教育出版社, 1980: 294-302.

[8] 廖其力, 余艳, 邓娅, 邓敏艺, 白克钊. 用Mathematica研究环形电流平面内磁场[J]. 广西物理, 2016, 37(1): 54-56.

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