PlanetPhysics/Off Axis Example of Biot Savart Law

$$r = x \hat{x} + z \hat{z}$$ \\ $$r' = r \cos \phi' \hat{x} + r \sin \phi' \hat{y}$$ \\ $$dl = -r d\phi' \sin \phi' \hat{x} + r d\phi' \cos \phi' \hat{y} $$\\ $$ r - r' = (x - r' \cos \phi') \hat{x} + r \sin \phi' \hat{y} + z \hat{z}$$ \\ $$ dl \times (r - r') = zr d\phi \cos \phi' \hat{x} + zrd\phi \sin \phi' \hat{y} + [- x r d\phi cos\phi' + r^2 d\phi] \hat{z} $$ \\

$$ \left | r - r' \right | = \sqrt{ x^2 - 2rx \cos \phi'+ r^2 + z^2}$$ \\

with

$$\alpha = x^2 + r^2 + z^2$$ \\ $$\beta = 2rx$$ \\

expand

$$( \alpha - \beta \cos \phi')^{-\frac{3}{2}} $$\\

rewrite as

$$\alpha^{-\frac{3}{2}}( 1 - \frac{\alpha}{\beta}x)^{-\frac{3}{2}} $$\\

use expansion formula $$(1 - x)^{-n} = 1 + \frac{nx}{1!} + \frac{n(n+1)x^2}{2!} $$\\

$$1 + \frac{3}{2}\frac{\beta}{\alpha} \cos \phi' + \frac{15}{8}\frac{\beta^2}{\alpha^2} \cos^2 \phi' $$

$$B_r = \frac{2 \pi I a^2 \cos \theta}{c(a^2 + r^2)^{3/2}} \left [ 1 + \frac{15 a^2 r^2 sin^2\theta}{4(a^2 + r^2)^2} + ... \right ] $$\\

$$B_{\theta} = -\frac{\pi I a^2 \sin \theta}{c(a^2 + r^2)^5/2} \left [ 2a^2 - r^2 + \frac{15 a^2 r^2 \sin^2 \theta(4a^2 - 3r^2)}{8(a^2 + r^2)^2} + ... \right] $$\\