Module 3 : MAGNETIC FIELD
Lecture 15 : Biot- Savarts' Law
  The net field at P, due to both coils add up and is given by $B(z)$
\begin{eqnarray*} &=& \frac{\mu_0 Ia^2}{2}\left[\frac{1}{[a^2+ (b+z)^2]^{3/2}} +... ...+b^2}]^{3/2}}+\frac{1}{[1+\frac{z^2-2zb}{a^2+b^2}]^{3/2}}\right] \end{eqnarray*}
  We can express the above in a power series using a binomial expansion. Up to $z^4$, the terms in the expansion may be written as
 
\begin{displaymath}\left[\frac{1}{[1+ \frac{z^2+2zb}{a^2+b^2}]^{3/2}}\right] = 1... ...)^3 + \frac{315}{128} \left(\frac{z^2+2zb}{(a^2+b^2)}\right)^4 \end{displaymath}
  and a similar expression for the second term with $-2zb$replacing $2zb$. Adding the terms and retaining only up to powers of $z^4$we get
 
\begin{displaymath}B(z) = \frac{\mu_0 Ia^2}{(a^2+b^2)^{3/2}}\left[1-\frac{3}{2}\... ... + \frac{15}{8}\frac{a^4+8b^4-12a^2b^2}{(a^2+b^2)^4}z^4 \right]\end{displaymath}

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