An introduction to the air cored coil

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See also ...
[Producing wound components] [A guide to the terminology used in the science of magnetism] [Power loss in wound components] [The force produced by a magnetic field] [Faraday's law] [Flux in a multilayer coil]

What is an air core coil?

We'll use the term 'air core coil' to describe an inductor that does not depend upon a ferromagnetic material to achieve its specified inductance. This covers the cases where there really is just air inside as well as windings upon a different insulator such as bakelite, glass or PTFE etc.

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Do you need an air coil?

What are the advantages of an air core coil?

And the 'downside'?

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The single layer coil

An air core coil A single layer coil has two advantages. Firstly, like all air core coils, it is free from 'iron losses' and the non-linearity mentioned above. Secondly, single layer coils have the additional advantage of low self-capacitance and thus high self-resonant frequency. These coils are mostly used above about 3 Mhz.

In the simple case of a single layer solenoidal coil the inductance may be estimated as follows (Wheeler) -

L = 0.001 N2r2 / (228r + 254l)
Equation ACS

where L is the inductance in henrys, r is the coil radius in metres, l is the coil length in metres (>0.8r) and N is the number of turns.

This formula applies at 'low' frequencies. At frequencies high enough for skin effect to occur a correction of up to about -2% is made.

To construct a self-supporting air cored coil take a length of plain 1 millimetre diameter copper wire and hold one end in a bench vice. Take the other end in a pair of pliers and pull until the wire has stretched slightly - this will straighten it. Using a 5 millimetre diameter drill bit wrap the wire around it until enough turns have been applied. Using 'long nosed' pliers bend the ends of the coil to get them into a radial position.

Small reductions in the inductance obtained can be achieved by pulling the turns apart slightly. This will also reduce self-resonance. Other combinations of wire and coil diameter may be tried but best results are usually obtained when the length of the coil is the same as its diameter.

This property also leads to a disadvantage of the air cored coil: microphony. If you need good frequency stability in the presence of vibration then wind the coil on a support made from a suitable plastic or ceramic former.

Software capable of calculating the inductance of a single layer coil is available at

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The multi layer coil

Figure 1 below shows a multi layer air cored coil wound on a circular coil former or bobbin.

The dimensions of an air coil.

This type of winding is very common because it's simple to construct with a winding machine and a mandrel. We'll consider it in some detail. The first question is this: if you have a fixed length of wire then what dimensions of the winding give the greatest inductance? Put another way, what is the most efficient shape?

The ratio of the winding depth to length, which is c/b, needs to be close to unity; so the winding should have a square cross section. This makes sense because only with the square is the average distance between turns at a minimum (a circular cross section would be even better, but that is hard to construct). Keeping the turns close together maintains a high level of magnetic coupling ('flux linkages') between them, and so the general rule that the inductance of a coil increases with the square of the number of turns is maintained.

OK, now what about the mean radius of the turns: dimension a ? Optimum shape for a multi layer coil

In the adjacent Fig. 2 you see three coils in cross section. Each uses the same length of wire but the diameter of the coil varies. The inductance of any one turn is linearly proportional to its diameter; so you want a large diameter to get the most inductance. Also you need all turns in the winding to be as close as possible to all the others. The coil on the left fulfills these requirements, but it has a problem because, in making the diameter large, you don't have sufficient wire to give it many turns. Since the inductance of the winding as a whole varies as the square of the number of turns the left hand coil won't have high L.

The coil on the right does have a high number of turns, but it isn't optimum for two reasons. The diameter of each turn is small (particularly those near the centre of the coil) which leads to a low inductance per turn. Worse, the distance between turns separated by the diagonals of the winding cross section is large. This leads to weak coupling between them (lower flux linkage) and a failure of the N2 relationship with coil inductance.

Can you anticipate a punch line? Brooks, who wrote a paper in 1931, calculated that the ideal value for the mean radius is very close to 3c/2. We call a coil having these dimensions a Brooks coil. It's worth emphasising that the Brooks ratio is not critical. You can have a coil which deviates from it quite significantly before L falls off very much. Also, you may have other considerations than the inductance alone.

If we let S1 = (c/2a)2 then the inductance, in henrys, of any of our three coils can be approximated by

L = 4×10-7πa N2((0.5+S1/12)ln(8/S1) -0.84834+0.2041S1)
Equation ACA

where a is the mean radius of the winding in metres.

Example: If the mean radius is 2.4 cm, the width is 9mm and N is 350 then L = 6.977 mH.

If we have a proper Brooks coil then the formula above boils down to

L = 1.6994×10-6aN2  henrys
Equation ACB

Example: If the mean radius is 2.4 cm and N is 350 then L = 4.996 mH.

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Flux density in an air coil

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When the density is important

Sometimes the purpose of an air coil is not to possess inductance but to create a region of space having a definite magnetic flux density. A well known example of this is the deflection coils placed around a cathode ray tube where the direction of the electron beam is controlled by the current driven through the coils. Outlined below are methods by which you can relate the strength of the field to the coil current. When you know the field at all points around a coil then you can also find its inductance.

The Biot-Savart equation

Fig. FCE: The field due to a current element. You can see in these web pages examples of the duality between the electric world and the magnetic world. There is, though, a fundamental distinction: a point electric charge in a vacuum produces a symmetrical field whereas a 'point magnetic charge' which we will call a current element has a direction of flow which leads to an asymmetry in field distribution along that direction.

An important result of the work carried out by André Ampère is the equation which is refered to as the Biot-Savart. This gives the field at any point surrounding a current element. Such an element should have negligible length, ds, and negligible diameter (see figure FCE).

dB = μ0 I ds sinθ / (4π r2)   teslas
Biot-Savart equation

This shows that the magnitude of the field falls with the square of the distance, r, from the current element and also with the sine of the angle , θ, subtended to the current direction. The direction of the magnetic field is at a right angle to the current direction and also at a right angle to the line r. The 'sense' of the field vector is right handed or clockwise looking along the direction of current flow.

At any point along the circular field line labeled F in figure FCE a tangent to it has a direction fulfilling the requirements just given. Imagine a set of such lines surrounding the current element, spaced in accordance with the expression for dB. Lessen the strain on your imagination by looking at a representation of the function sin(θ)/r2 as an intensity plot.

You might complain that the Biot-Savart equation doesn't seem to be buying us very much: an isolated current element is an imposibility in practice and real coils use wires which may be very thick. What saves the day in the case of air coils is the principle of superposition. Without a ferromagnetic core to spoil the linearity of the system we can simply divide the actual current flow up into any number of smaller regions and then sum, or integrate, the individual fields to obtain the total flux density.

Biot-Savart solves a current loop

Let's apply Biot-Savart to a single turn circular loop and derive a formula for the flux density at any point along the axis. We'll assume that the diameter of the wire is negligible in comparison with the diameter of the loop - in the terminology this is known as a current filament.

The field due to a current loop In figure FLP we see the loop, of radius a, in cross section and we want the flux density at the point P1 on the loop axis at a distance r from the conductor. We start by considering the contribution to the field made by a current element at P2. The figure shows a P1 which happens to be quite far from the loop. ψ is rather small and the field vector (BT, shown in blue) is swinging up towards the radial direction. You can resolve BT into two components: BR, shown in red and BA, shown in green. We can disregard the radial component, BR, because of the symmetry of the loop: it is going to be cancelled by the field generated by the current element on the opposite side of the loop at P3. We are left with just the axial component (that we need to find): BA. Its magnitude will be

BA = BT sinψ
Equation ACC

The Biot-Savart equation gives us dBT so we can combine the two to get

dBA = μ0 I ds sinθ sinψ / (4π r2)
Equation ACD

Clearly, any line (such as r) which joins P1 to the conductor will intersect the direction of current flow at a right angle. This means that sin(θ) in the Biot-Savart equation will be 1.0. At points off axis this won't always be the case and the maths gets harder.

dBA = μ0 I ds sinψ / (4π r2)
Equation ACE

Now, all we have here is the field due to the single current element at P2, but we need the total field due to all the elements around the loop. Fortunately, because of the symmetry all the elements contribute the same as P2, so to do the 'integration' we need only multiply by the path length around the loop: 2πa -

BA = a μ0 I sinψ / (2 r2)
Equation ACF

Substituting r = a / sinψ -

BA = a μ0 I sinψ / (2 (a / sinψ)2)
Equation ACG


BA = μ0 I sin3ψ / (2a)
Equation ACH

And if the coil is multi-turn then superposition again gives

BA = μ0 Fm sin3ψ / (2a)
Equation FMC

where Fm is the magneto motive force: N × I. It may be more convenient to specify the field by an equivalent expression in terms of the axial distance, x, of P1 from the centre of the coil -

BA = μ0 Fm a2 / (2(a2 + x2)3/2)
Equation ACU

Observe that, if x is large in comparison with a, the field is in inverse proportion to x3. This is a well known result for the 'far field of a magnetic dipole'. As the diameter of the loop grows the field at some specific large x will increase whilst the field near the centre of the loop decreases. Equation ACU is easily differentiated to obtain the gradient -

dBA/dx = μ0 Fm (-3 / 2) a2 x (a2 + x2)-5/2
Equation ACV

This function has a minimum at x = a / 2. At this point a magnetic particle is attracted most strongly. The gradient here is -

dBA/dx = -24 √5 μ0 Fm / (125 a2)
Equation ACW

The field in a solenoid

It's not too difficult to extend the above result for a current loop to solve the field at any point on the axis of a solenoidal coil (shown in cross section in figure FSD). As usual, I'm going to use a combination of 'hand waving' arguments and integral calculus. If you just want the bottom line then see equation FDS.

Solenoid parameters
Symbol Purpose
N The total number of turns in the solenoid
b The length of the solenoid (metres)
a The radius of the solenoid (metres)
P1 The point on the axis at which we need to
find the flux density: B
z An arbitrary distance along the axis from P1
ψ The angle subtended at P1 by the axis and a
point at the top of the winding at z
ψ1 The angle subtended at P1 by the axis and
the top of the winding at the left hand end
ψ2 Ditto for the RH end
I The solenoid current (amps)

Field within a sloenoid First we have to make a simplification because the flux lines will not run perfectly true along the axis but will spiral slightly around it (reflecting the helical nature of the coil itself). This effect becomes less if the turns are spaced more closely. In fact this leads on to the concept of a current sheet. If we kept the dimensions of the solenoid the same but used, say, three times the number of turns and at the same time reduced the current to one third then the overall magneto motive force is the same and so the field produced will be much as before.

In the limit we no longer worry about the total number of individual turns and the coil current. Mentally replace them with a thin walled copper tube carrying a total current N × I amps flowing around its circumference and distributed evenly along its length. Don't ask how the current gets there; this is pure hypothesis :-) . Now we can talk of an mmf per unit length of the tube, Fl -

Fl = I × N / b   amps per metre
Equation MLT

Having gone to the trouble of smoothing out the turns into a continuous current sheet or tube we now proceed to chop it up again! If we consider an infinitesimally short length of tube, dz, then we have precisely the sort of current loop whose axial field we calculated (equation FMC) via the Biot-Savart equation. The mmf for the loop will be

Fm = Fl dz
Equation ACH

Substituting this into eqn. FMC

dB = μ0 Fl dz sin3ψ / (2a)
Equation ACI

This represents the contribution to the flux density at P1 due to a current loop whose perimeter subtends an angle ψ with the axis at P1. From the geometry

z = a / tanψ
Equation ACJ

dz/dψ = -a cosec2ψ
Equation ACK

dz = -a dψ cosec2ψ
Equation ACL

Substituting this into the equation for dB gives

dB = μ0 Fl (-a dψ cosec2ψ) sin3ψ / (2a)
Equation ACM

dB = -(μ0 Fl / 2) sinψ dψ
Equation ACN

B = -(μ0 Fl / 2) sinψ dψ
Equation ACO

For the part of the tube between P1 and the left hand end ψ will take values between π/2 and ψ1

B = -(μ0 Fl / 2) π/2ψ1 sinψ dψ
Equation ACP

B = (μ0 Fl / 2) [cosψ]π/2ψ1
Equation ACQ

B = (μ0 Fl / 2) cosψ1
Equation ACR

The part of the tube to the right of P1 will superpose an additional flux at at P1 -

B = (μ0 Fl / 2) (cosψ1 + cosψ2)
Equation FDS

If b >> a (the solenoid is long and thin) then the angles ψ1 and ψ2 will both tend to zero for a point in the middle

B = μ0 Fl   teslas
Equation ACT

At the end of a long coil the flux will be exactly half this value. Only towards the ends does the flux 'leak' through the sides.

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Last revised: 2010 April 4th.