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,

$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

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 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 the coils $= R$

Let us consider point $Q$ at distance $d$ from the centre.

Then, one coil is at a distance of $\frac{R}{2}+d$ from point $Q$.

Magnetic field at point $Q$ is given as

$B_{1}=\frac{\mu_{0} N I R^{2}}{2\left[\left(\frac{R}{2}+d\right)^{2}+R^{2}\right]^{3 / 2}}$

Also, the other coil is at a distance of $\frac{R}{2}-d$ from point $Q$.

Magnetic field due to this coil is given as

$B_{2}=\frac{\mu_{0} N I R^{2}}{2\left[\left(\frac{R}{2}-d\right)^{2}+R^{2}\right]^{3 / 2}}$

Total magnetic field, $B=B_{1}+B_{2}$

$=\frac{\mu_{0} I R^{2}}{2}\left[\left\{\left(\frac{R}{2}-d\right)^{2}+R^{2}\right\}^{-3 / 2}+\right]\left\{\left(\frac{R}{2}+d\right)^{2}+R^{2}\right\}^{-3 / 2} \times N$

$=\frac{\mu_{0} I R^{2}}{2}\left[\left(\frac{5 R^{2}}{4}+d^{2}-R d\right)^{-3 / 2}+\left(\frac{5 R^{2}}{4}+d^{2}+R d\right)^{-3 / 2}\right] \times N$

$=\frac{\mu_{0} I R^{2}}{2} \times\left(\frac{5 R^{2}}{4}\right)^{-3 / 2}\left[\left(1+\frac{4}{5} \frac{d^{2}}{R^{2}}-\frac{4}{5} \frac{d}{R}\right)^{-3 / 2}+\left(1+\frac{4}{5} \frac{d^{2}}{R^{2}}+\frac{4}{5} \frac{d}{R}\right)^{-3 / 2}\right] \times N$

For $d < < R ,$ neglecting the factor $\frac{d^{2}}{R^{2}},$ we get:

$\approx \frac{\mu_{0} I R^{2}}{2} \times\left(\frac{5 R^{2}}{4}\right)^{-3 / 2} \times\left[\left(1-\frac{4 d}{5 R}\right)^{-3 / 2}+\left(1+\frac{4 d}{5 R}\right)^{-3 / 2}\right] \times N$

$\approx \frac{\mu_{0} I R^{2} N}{2 R^{3}} \times\left(\frac{4}{5}\right)^{3 / 2}\left[1-\frac{6 d}{5 R}+1+\frac{6 d}{5 R}\right]$

$B=\left(\frac{4}{5}\right)^{3 / 2} \frac{\mu_{0} I N}{R}=0.72\left(\frac{\mu_{0} I N}{R}\right)$

Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.