Inductive coupling coefficient of two coaxial coils is derived. By integration of the magnetic flux over the coil section area are obtained the intrinsic inductance and the mutual inductance of the coils.

Fig.1 Inductive coupling of two coaxial coils. The magnetic flux produced by the coil to the left pierces through the section of the coil to the right resulting in the inductive coupling of the coils |

We consider two coils separated by distance $d$. The radii of the coils are $a_1$ and $a_2$. The number of turns of the 1st coil is $n_1$ and of the 2nd coil is $n_2$.

The flow of magnetic field through the section of the coil is given by the integral

$$\Phi = \iint_S{\vec{B}\cdot\vec{\,\mathrm{d}S}} = \int_0^{2\pi}{\int_0^{a-\delta}{B_z(\rho,z)\rho\,\mathrm{d}\rho\,\mathrm{d}\varphi}} = 2\pi\int_0^{a-\delta}{B_z(\rho,z)\rho\,\mathrm{d}\rho}$$

We are using the solution for the magnetic field of a loop current

$$B_z(\rho, z) = \frac{\mu_0 I}{4\pi} \int_0^{2\pi}{\frac{\left( a^2 - \rho a\cos{\varphi}\right)\,\mathrm{d}\varphi}{\left(\rho^2 + a^2 + z^2 -2\rho a\cos{\varphi}\right)^{\frac{3}{2}}}}$$

For mutual inductance we need to find the linkage of the magnetic flow

$$\Psi = \frac{1}{2}n_1 n_2\mu_0 I \int_0^{a_2-\delta}{\int_0^{2\pi}{\frac{\left( a_1^2 - \rho a_1\cos{\varphi}\right)\,\mathrm{d}\varphi}{\left(\rho^2 + a_1^2 + z^2 -2\rho a_1\cos{\varphi}\right)^{\frac{3}{2}}}}\rho\,\mathrm{d}\rho}$$

By definition inductance is a quantity which gives the proportion between the magnetic flow $\Phi$ and the current $I$, i.e. $\Phi = LI$, or in our case $\Psi = MI$.

$$M = \frac{1}{2}n_1 n_2\mu_0 \int_0^{a_2-\delta}{\int_0^{2\pi}{\frac{\left( a_1^2 - \rho a_1\cos{\varphi}\right)\,\mathrm{d}\varphi}{\left(\rho^2 + a_1^2 + d^2 -2\rho a_1\cos{\varphi}\right)^{\frac{3}{2}}}}\rho\,\mathrm{d}\rho}$$

$\delta$ being the radius of the wire cross-section.

Intrinsic inductance parameters are then found by analogy

$$L_1 = \frac{1}{2}n_1^2\mu_0 \int_0^{a_1-\delta}{\int_0^{2\pi}{\frac{\left( a_1^2 - \rho a_1\cos{\varphi}\right)\,\mathrm{d}\varphi}{\left(\rho^2 + a_1^2 -2\rho a_1\cos{\varphi}\right)^{\frac{3}{2}}}}\rho\,\mathrm{d}\rho}$$

$$L_2 = \frac{1}{2}n_2^2\mu_0 \int_0^{a_2-\delta}{\int_0^{2\pi}{\frac{\left( a_2^2 - \rho a_2\cos{\varphi}\right)\,\mathrm{d}\varphi}{\left(\rho^2 + a_2^2 -2\rho a_2\cos{\varphi}\right)^{\frac{3}{2}}}}\rho\,\mathrm{d}\rho}$$

Inductive coupling coefficient is defined as

$$k = \frac{M}{\sqrt{L_1L_2}}$$

Fig.2 Inductive coupling coefficient depending on the distance between two coaxial coils. Both coils have the following parameters: radius of a turn $a=0.1$ m, number of turns $n = 100$, radius of the wire cross-section $\delta=0.1$ mm, intrinsic unductance $L = 8.775$ mH. |

From the graph in Fig.2, it can be seen that coupling coefficient decays fast with distance. For instance, at a separation distance of approximately 25 cm away from each other, two identical coils would have coupling of as small as $k = 0.01$. Due to the complicated nature of the phenomenon, it is difficult to give an analytical expression for $k$, however for any given set of parameters (radius of turns, number of turns, etc.) it can be computed numerically.