# Wireless power transmission through magnetically coupled inductive coils

## Introduction

I think that many of the readers have seen at least one video on popular video services, where electricity is transmitted through empty space using inductive coils.

In this article, we want to address the fundamentals of the process of wireless energy transmission using a magnetic field. Starting with consideration of the simplest inductive coil, and calculating its inductance, we will gradually move on to the theory of electrical circuits, in the framework of which the method of transmitting maximum power will be shown and justified under other equal conditions. So, let's begin.

## Magnetic field of a single coil with current

Consider the magnetic field of a single coil with a current. Find the magnetic field of the coil at any point in space. Why is such a review necessary? Because in almost all books, at least in those that the author of the article was able to find, the solution to this problem is limited to finding only one component of the magnetic field and only along the axis of the coil - ${B}_{z}\left(z\right)$$B_z (z)$, while we will find the law for the magnetic field in the whole space.

Illustration to the law of Bio-Savard-Laplace

To find the magnetic field, we use the law of Bio-Savart-Laplace (see Wikipedia - the Law of Bio-Savard-Laplace ). The figure shows that the center of the coordinate system $O$$O$coincides with the center of the coil. The contour of the circle of the coil is indicated as $C$$C$, and the radius of the circle - as $a$$a$. During the coil current flows $I$$I$. $\ vec {r}$- is a variable-radius-vector from the origin to an arbitrary point of the coil. $\ vec {r} _0$Is the radius vector to the observation point. We also need a polar angle. $\ varphi$- the angle between the radius vector $\ vec {r}$and axis $OX$$OX$. The distance from the axis of the turn to the observation point is denoted by $\ rho$. And finally $\ mathrm {d} \ vec {r}$- elementary increment of the radius-vector $\ vec {r}$.

According to the law of Bio-Savard-Laplace, the circuit element with current $\ mathrm {d} \ vec {r}$creates an elementary contribution to the magnetic field, which is given by the formula



Now we’ll dwell on the variables and expressions included in the formula. Given the axial symmetry of the problem, we can write













In order to find the resultant magnetic field, it is necessary to integrate over the whole loop contour, i.e.



After the substitution of all expressions and some identical transformations, we obtain expressions for the axial and radial components of the magnetic field, respectively





To find the absolute value of the magnetic field, it is necessary to sum up the components according to the Pythagorean theorem $B = \ sqrt {B_r ^ 2 + B_z ^ 2}$.

Let us demonstrate the solution obtained using the example of a radius coil $a=0.1$$a = 0.1$(m) and $I=1$$I = 1$(BUT).

The amplitude of the axial component of the magnetic field

The amplitude of the radial component of the magnetic field

Absolute magnetic field amplitude

Note that for a coil of arbitrary shape, at large distances $z \ gg a$i.e. much larger than the characteristic size of the coil, the behavior of the magnetic field will tend to the solution found.

Hint...
For such calculations and graphing is convenient to use MathCad 15

## Inductor. Magnetically coupled coils

Now that we know the solution for the magnetic field of a single turn, we can find the inductance of the coil, consisting of $n$$n$turns. By definition, inductance is the coefficient of proportionality between the current in a coil and the magnetic flux through the cross section area. Here we use the ideal coil model, which is dimensionless in the direction of its axis of symmetry. Of course, in practice this does not happen. However, as approximations, the formulas obtained will be good enough. Although coils are considered dimensionless along $OZ$$OZ$, you must specify a non-zero radius of the wire cross section. Denote it $\ delta$and example equal $\ delta = 0.$(mm) Otherwise, when integrating the magnetic flux, the integrand will turn to infinity.

Inductively connected coils

The figure shows two magnetically coupled coils. Let the first coil have a radius ${a}_{1}$$a_1$and contains ${n}_{1}$$n_1$turns, and the second - ${a}_{2}$$a_2$and ${n}_{2}$$n_2$respectively. Then, in order to find the own inductances, it is necessary to calculate the magnetic flux of each coil through its own section.



Since there are many turns in the coil, we find the quantity called the flux linkage, by multiplying twice the number of turns



By definition, inductance is a coefficient of proportionality $L$$L$in the formula $\ Psi = LI$. Thus, we obtain the own inductances of the coils





Let the centers of the coils be separated by distance $d$$d$, lie on the same axis, and their plane of turns is oriented parallel. To find the mutual inductance, it is necessary to calculate the flux linkage formed by one coil through the cross section of another, i.e.



Then the mutual inductance of the coils is given by



As far as the author knows, such integrals can be taken only numerically.
Note that as a rule $\ Psi_ {12} = \ Psi_ {21}$and ${M}_{12}={M}_{21}=M$$M_ {12} = M_ {21} = M$. Coil coupling coefficient is called magnitude



We investigate the dependence of the coupling coefficient of the coils on the distance. To do this, consider two identical coils with a radius of turns ${a}_{1}={a}_{2}=0.1$$a_1 = a_2 = 0.1$(m) and the number of turns ${n}_{1}={n}_{2}=100$$n_1 = n_2 = 100$. In this case, the inductance of each of the coils will be ${L}_{1}={L}_{2}=8.775$$L_1 = L_2 = 8.775$(mH).

Coil coupling ratio of the distance between them

The schedule will not change if the number of turns in both coils is equally changed, or the radius of both coils is equally changed. The coupling coefficient is conveniently expressed as a percentage. From the graph it is seen that even with the distance between the coils of 1 (mm), the coupling coefficient is less than 100%. The coefficient drops to 10% at a distance of about 60 (mm), and to 1% at 250 (mm).

## Wireless power transmission

So, we know inductance and coupling coefficient. Now let us use the theory of electrical circuits of alternating current to find the optimal parameters at which the transmitted power would be maximum. To understand this paragraph, the reader must be familiar with the concept of electrical impedance, as well as Kirchhoff's laws and Ohm's law. As is known from the theory of circuits, two inductively coupled coils form an air transformer. For the analysis of transformers convenient T-shaped equivalent circuit.

Air transformer and its equivalent circuit

The transmitting coil on the left will be called the "trasmitter", and the receiving coil on the right will be called the "receiver". Between coils coupling coefficient $k$$k$. On the side of the receiver is the consumer represented by the load ${z}_{L}$$z_L$. The load can generally be complex. Input voltage on the transmitter side ${u}_{1}$$u_1$and input current - ${i}_{1}$$i_1$. The voltage transmitted to the receiver - ${u}_{2}$$u_2$and current transmitted ${i}_{2}$$i_2$. The full impedance on the transmitter side is denoted as ${z}_{1}$$z_1$, and the full impedance on the side of the receiver ${z}_{2}$$z_2$.

It is assumed that a sinusoidal voltage is applied to the input of the circuit. $u_1 = u_ {1m} \ sin {\ omega t}$.

Denote ${R}_{coil\phantom{\rule{thinmathspace}{0ex}}1},{R}_{coil\phantom{\rule{thinmathspace}{0ex}}2},{L}_{coil\phantom{\rule{thinmathspace}{0ex}}1},{L}_{coil\phantom{\rule{thinmathspace}{0ex}}2},M$$R_ {coil \, 1}, R_ {coil \, 2}, L_ {coil \, 1}, L_ {coil \, 2}, M$- resistance and inductance of the coils (two own and one mutual), respectively. Then, according to transformer theory





On the other hand, according to our designations



${z}_{1}={r}_{1}+j{x}_{1}$



${z}_{2}={r}_{2}+j{x}_{2}$

Where ${r}_{1},{r}_{2}$$r_1, r_2$- total resistances on the side of the transmitter and receiver, respectively, and ${x}_{1},{x}_{2}$$x_1, x_2$- full reactive resistances.

Communication impedance is $z_3 = j \ omega M = j x_3$.

Find the input current of the circuit



where is the sign $||$$||$denotes a parallel connection of resistances. Then the voltage transmitted to the receiver



And induced current



We can find the integrated power transmitted to the receiver



${s}_{2}={u}_{2}{i}_{2}^{\ast }={p}_{2}+j{q}_{2}$

Thus, we have an expression for the complex power



Expression for active power component



Expression for reactive power components



In most practical tasks, the maximum active power is required, therefore



Either that same







For convenience, we introduce the function



$f\left({x}_{1},{x}_{2}\right)=\left({r}_{1}{x}_{3}+{r}_{2}{x}_{3}+{r}_{1}{x}_{2}+{r}_{2}{x}_{1}\right)+j\left({x}_{1}{x}_{3}+{x}_{2}{x}_{3}+{x}_{1}{x}_{2}-{r}_{1}{r}_{2}\right)$

and examine it for extremes



Where do we get a system of two equations





This system has five solutions, two of which are nonphysical, since they lead to imaginary values ​​of quantities that are supposed to be valid. Three other physical solutions are given below, together with the corresponding power formulas.
Solution 1



Power



Solution 2 and 3