# 亥姆霍茲線圈

## 數學描述

$B = {\left ( \frac{4}{5} \right )}^{3/2} \frac{\mu_0 n I}{R}$

### 推導

$B = \frac{\mu_0 I R^2}{2(R^2+z^2)^{3/2}}$

$B = \frac{\mu_0 n I R^2}{2(R^2+z^2)^{3/2}}$

$B = \frac{2\mu_0 n I R^2}{2(R^2+(R/2)^2)^{3/2}} = {\left ( \frac{4}{5} \right )}^{3/2} \frac{\mu_0 n I}{R}$

### 進階推導

$\mathbf{B} =\frac{\mu_0 I R^2}{2} \left\{\left[R^2+(z-h/2)^2\right]^{-3/2} +\left[R^2+(z+h/2)^2\right]^{-3/2}\right\}\hat{\mathbf{z}}$

$\mathbf{B} =\frac{\mu_0 I R^2}{d^3}\left[1+\frac{3(h^2-R^2)z^2}{2d^4}+\frac{15(h^4-6h^2R^2+2R^4)z^4}{16d^8}+\dots\right]\hat{\mathbf{z}}$

$\mathbf{B} ={\left ( \frac{4}{5} \right )}^{3/2} \frac{\mu_0 I}{R}\left[1-\frac{144}{125}\ \left(\frac{z}{R}\right)^4+\dots\right]\hat{\mathbf{z}}$

$\frac{\Delta B_z}{B_z}\approx -\frac{144}{125}\ \left(\frac{z}{R}\right)^4$

$z=\pm R/2$ ，線圈平面與z-軸相交處，磁場數值的差別為

$\frac{\Delta B_z}{B_z}\approx -\frac{144}{125}\ \left(\frac{1}{2}\right)^4\approx -7\%$

## 參考文獻

1. ^ 亥姆霍茲線圈內部磁場分析
2. ^ 地磁場磁力儀：亥姆霍茲線圈" by Richard Wotiz 2004
3. ^ 喬治亞州州立大學Georgia State University）線上物理網頁：
4. ^ Jackson, John David, Classical Electrodynamic. 3rd., USA: John Wiley & Sons, Inc.. 1999:  pp. 226-227, ISBN 978-0-471-30932-1