For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

(A) Radius of circular coil $= R$
Number of turns on the coil $= N$
Current in the coil $= I$
Magnetic field at a point on its axis at distance $x$ is given by the relation,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
Where,$\mu_{0} =$ Permeability of free space.
$(a)$ If the magnetic field at the centre of the coil is considered,then $x=0$.
$\therefore B=\frac{\mu_{0} I R^{2} N}{2 R^{3}}=\frac{\mu_{0} I N}{2 R}$
This is the familiar result for the magnetic field at the centre of the coil.
$(b)$ Radii of two parallel co-axial circular coils $= R$. Number of turns on each coil $= N$. Current in both coils $= I$. Distance between both coils $= R$.
Let us consider point $Q$ at distance $d$ from the midpoint. One coil is at a distance of $\frac{R}{2}+d$ and the other is at $\frac{R}{2}-d$ from point $Q$.
Magnetic field at point $Q$ is $B = B_{1} + B_{2}$.
$B = \frac{\mu_{0} N I R^{2}}{2} \left[ \left( (\frac{R}{2}+d)^{2} + R^{2} \right)^{-3/2} + \left( (\frac{R}{2}-d)^{2} + R^{2} \right)^{-3/2} \right]$
Using binomial expansion for $d \ll R$ and neglecting higher-order terms of $d/R$:
$B \approx \frac{\mu_{0} N I R^{2}}{2} \left( \frac{5R^{2}}{4} \right)^{-3/2} \left[ (1 - \frac{4d}{5R})^{-3/2} + (1 + \frac{4d}{5R})^{-3/2} \right]$
Using $(1+x)^{n} \approx 1+nx$:
$B \approx \frac{\mu_{0} N I R^{2}}{2} (\frac{4}{5R^{2}})^{3/2} [1 + \frac{6d}{5R} + 1 - \frac{6d}{5R}]$
$B = \frac{\mu_{0} N I R^{2}}{2} \cdot \frac{8}{5\sqrt{5}R^{3}} \cdot 2 = \frac{4}{5\sqrt{5}} \frac{\mu_{0} N I}{R} \approx 0.72 \frac{\mu_{0} N I}{R}$