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Case 2: $ B^{(0)}(\mathbf{x}) = (z,0,x)$

 

Two terms need to be calculated. Firstly,

$\displaystyle F_1(\mathbf{x}';\mathbf{x})$ $\displaystyle =$ $\displaystyle \iiint (z - z') \frac{\partial^2 G}{\partial x' \partial z'} \, dx' \, dy' \, dz'$  
  $\displaystyle =$ $\displaystyle \int z G \, dy' - \iint z' \frac{\partial G}{\partial z'} \, dy' \, dz'$  
  $\displaystyle =$ $\displaystyle \int z G \, dy'
- \int z' G \, dy' + \iint G \, dy' \, dz'$  
  $\displaystyle =$ $\displaystyle \frac{z'-z}{4\pi} \, \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}^2 + {z'}^2}}\right)
- \frac{1}{4\pi} \iint \frac{1}{r'} \, dy' \, dz'$  
  $\displaystyle =$ $\displaystyle \frac{z'-z}{4\pi} \, \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}^2 + {z'}^2}}\right)$  
    $\displaystyle \quad - \frac{1}{4\pi} \left( y' \, \mathrm{sinh}^{-1}\left(\frac...
...2 + {z'}^2}}\right)
- x' \, \mathrm{atan}\left(\frac{z'y'}{x'r'}\right) \right)$  
  $\displaystyle =$ $\displaystyle \frac{x'}{4\pi} \, \mathrm{atan}\left(\frac{z'y'}{x'r'}\right)
- ...
...frac{y'}{4\pi} \mathrm{sinh}^{-1}\left(\frac{z'}{\sqrt{{y'}^2 + {x'}^2}}\right)$ (35)

and secondly
$\displaystyle F_2(\mathbf{x}';\mathbf{x})$ $\displaystyle =$ $\displaystyle \iiint (x - x') \frac{\partial^2 G}{\partial {z'}^2} \, dx' \, dy' \, dz'$  
  $\displaystyle =$ $\displaystyle \iint x \frac{\partial G}{\partial z'} \, dx' \, dy' - \iint x' \frac{\partial G}{\partial z'} \, dx' \, dy'$  
  $\displaystyle =$ $\displaystyle \frac{x}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) - \iint x' \frac{z'}{4\pi {r'}^3} \, dx' \, dy'$  
  $\displaystyle =$ $\displaystyle \frac{x}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) + \int \frac{z'}{4\pi r'} \, dy'$  
  $\displaystyle =$ $\displaystyle \frac{x}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right) + \frac{z'}{4\pi} \mathrm{sinh}^{-1}\left(\frac{y'}{\sqrt{{x'}^2 + {z'}^2}}\right).$ (36)

Due to the linearity of equations 14 and 21 these two terms can be combined at this stage to give

$\displaystyle F(\mathbf{x}';\mathbf{x})$ $\displaystyle =$ $\displaystyle F_1(\mathbf{x}';\mathbf{x}) + F_2(\mathbf{x}';\mathbf{x})$  
  $\displaystyle =$ $\displaystyle \frac{x}{4\pi} \mathrm{atan}\left(\frac{x'y'}{z'r'}\right)
+ \fra...
...frac{y'}{4\pi} \mathrm{sinh}^{-1}\left(\frac{z'}{\sqrt{{y'}^2 + {x'}^2}}\right)$  

which can be substituted directly to give $ B^{(1)}_z$.

Once again, note that for $ F_1$ and $ F_2$ here both primed and unprimed coordinates appear, whereas for the constant fields the unprimed coordinates did not appear in the expressions for $ F$.


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