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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]
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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What are the advantages of an air core coil?
And the 'downside'?
You may be using an air cored coil not because you require a circuit element with a specific inductance per se but because your coil is used as a proximity sensor, loop antenna, induction heater, Tesla coil, electromagnet, magnetometer head or deflection yoke etc. Then an external field may be what you want.
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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) -
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 http://www.qsl.net/k6mlo/coilpage.html
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Figure 1 below shows a multi layer air cored coil wound on a circular coil former or bobbin.
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 ?
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
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
Example: If the mean radius is 2.4 cm and N is 350 then L = 4.996 mH.
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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.
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).
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.
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.
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
The Biot-Savart equation gives us dBT so we can combine the two to
get
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.
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 -
Substituting r = a / sinψ -
so
And if the coil is multi-turn then superposition again gives
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 -
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 -
This function has a minimum at x = a / 2. At this point a magnetic particle
is attracted most strongly. The gradient here is -
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.
| 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) |
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 -
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
Substituting this into eqn. FMC
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
Substituting this into the equation for dB gives
For the part of the tube between P1 and the left hand end ψ will take values
between π/2 and ψ1
The part of the tube to the right of P1 will superpose an additional flux
at at P1 -
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
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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E-mail:
R.Clarke@surrey.ac.uk
Last revised: 2010 April 4th.