



26. Inductance of a wire: As should be apparent from all of the foregoing discussion, every electrical conductor exhibits inductance when it carries a current, and it is useful to have an idea of what that inductance will be. Many radio textbooks address this issue by quoting a formula for the inductance of a straight cylindrical wire, derived by Edward B. Rosa, which appeared in the Bulletin of the American Bureau of Standards in 1908 [Rosa 1908][see also "Inductance of a Straight Wire", by Tim Healy]. We will look at this formula in a somewhat more detailed way than usual; so that we can try to find out why it is that such a widely quoted and used (and misquoted, and misused) equation tends to lead people to the wrong answers (even though it is correct). Rosa's formula can be written:
The quantity on the lefthand side L/l is the inductance per unit length; which means that the number given by calculating the right hand side must be multiplied by the length of the wire l in order to obtain an inductance in Henrys. The right hand side of the equation is separated into two parts: The term "Li" is the socalled internal inductance of the wire, i.e., the inductance per metre due to the magnetic fields inside the wire; and the term after it is the external inductance, which we might also call "Le", i.e., the inductance per metre due to the fields outside the wire. "ln" means "take the natural logarithm (log to the base e, or Loge) of the quantity in the brackets directly after it, and "d" is the diameter of the wire. The quantity inside the logarithm brackets must be dimensionless, and so d must be in the same units as l. μ(e) is the magnetic permeability of the medium external to the wire. It is of course composed of a relative permeability multiplied by the permeability of free space, i.e.: μ(e) = μ0 μr(e) Thus if the wire is surrounded by air, which has effectively the same permeability as free space (i.e., μr=1), then μ(e)=μ0. If, on the other hand, the length of wire for which we intend to calculate the inductance is surrounded by a tightfitting ferrite bead of (say) μr=850, then the external inductance will be 850 times greater than it would have been for the same wire with no magnetic materials in its immediate vicinity. At very low frequencies (DC), the internal inductance per unit length is given by: Li(DC) = μ(i) / 8π Henrys / metre (see, for example., ref [9a], section 5.17) where μ(i) is the permeability of the material from which the wire is made. If the wire is nonferromagnetic (e.g., copper), then we know from previous discussion that we may assume, to an excellent approximation, that μ(i)=μ0. Now, if we assume that the wire is nonmagnetic, and it is surrounded by air, we may rewrite Rosa's equation: L(DC) / l = (μ0 /8π) + (μ0/2π)[ln(4l/d)  1] Henrys / metre and observing that the internal and external inductances have a common factor μ0/2π, we obtain: L(DC) / l = (μ0/2π)[ln(4l/d)  1 + ¼] Henrys / metre. Now, since μ0=4π×10^{7}H/m, we can reduce the equation further to:
L = l ( Le + Li ) Internal Impedance To find the internal inductance at a particular frequency, we may start with the internal impedance of a wire; which is given (for example) by Ramo, Whinnery and Van Duzer (ref [9a], section 5.17) for the 'high frequency' case as: Zhf / l = Rs (1 + j ) / (π d) [Ohms / metre] Where Rs is the surface resistivity of the wire, and is given by the expression: Rs = √(π f μ(i) ρ ) [Ohms] μ(i) being the permeability of the wire (=μ0 for nonferromagnetic), and ρ being the ordinary volume resistivity (in Ohm metres). Alternatively: Rs = √(π f μ(i) / σ ) [Ohms] where σ is the conductivity (in Siemens / metre). For those who seek confirmation that Rs really is measured in Ohms, note that the reactance formula XL=2πfL tells us that Henrys (neglecting the dimensionless 2π term) are equivalent to Ohm seconds, and so the units of permeability (H/m) can be given as Ohm seconds / metre. The square root of [ / second]×[Ohm seconds / metre]×[Ohm metres] is Ohms. The expression above for internal impedance is stipulated to apply only at 'high frequencies', because it uses the approximation that the current carrying layer is sufficiently thin that that it can be regarded as a sheet of width equal to the circumference of the wire (this being the πd term on the bottom of the fraction). We can therefore qualify the statement a little, by saying that the approximation is good for 'high frequencies' or 'large diameters'. It is incidentally, exactly the same approximation as was used earlier when calculating the RF resistance of wires, and we may gain an idea of when it will fail us by comparing the wire diameter with the table of conduction layer thicknesses for various metals at various frequencies given earlier. We may note, for example, that copper has a skin depth of 50 μm at 1.75 MHz, and so for a copper wire of 1 mm diameter the approximation will be very good, but for a wire of 0.2 mm (200μm) diameter, it will introduce a significant error. Now we, of course, want the internal inductance of the wire. We may therefore take the internal reactance, which is the imaginary part of the internal impedance, and obtain the inductance from the reactance formula XL=2πfL, i.e.: Xi(hf) / l = Rs / (π d) = [ √(π f μ(i) ρ ) ] / (π d) [Ohms / metre] and Li(hf) / l = Xi(hf) / 2πf = [ √(π f μ(i) ρ ) ] / (2π² f d) [Henrys / metre] Now, recalling the trick for rearranging equations involving square roots, i.e., any number is the square of its own square root; and factoring out μ/2π (which, as you may recall, is the common factor for the two parts of Rosa's formula when the internal end external permeabilities are the same), we obtain: Li(hf) / l = (μ(i)/2π) / [d √(π f μ(i) σ ) ] Which takes us all the way back to equation (2.2), for the skin depth, i.e., δi = 1 / √( π f μ σ ) Thus:
Li(DC) / l = (μ(i)/2π) ¼ Since Li(hf) can never be greater than Li(DC), then our expression for Li(hf) can only be valid when d is considerably greater than 4δi. It should also be apparent that the internal inductance starts to diminish at frequencies a lot lower than "VHF" in the modern sense. Using the result just obtained, we can rewrite Rosa's formula properly for the 'high frequency' (or large diameter) case. We will do so first on the assumption that the wire is not ferromagnetic, and the external medium is air; thus:





1.63  142 

0.376  201 

0.193  235.5 
So how do these results compare to our modified version of Rosa's
formula (equation 6.2)? To
find out, we start by calculating the skin depth for copper (ρ=17.241
nΩm, μr=1) at 5 MHz, i.e.: δi = 1 / √( π f μ0 / ρ ) = 1 / √[ π × 5 × 10^{6} × 4 × π × 10^{7} / ( 17.241 × 10^{9} ) ] = 29.55 μm This quantity (before truncation to two decimal places) is used to calculate the δi/d internal inductance correction term for equation (6.2), and the results are shown in the table below: 

d / mm 

L / nH 
L / nH 

1.63 

142 


0.376 

201 


0.193 

235.5 

The δi/d figures in the table show
that the internal inductance contribution to the calculated inductance
is very small except for the case of the ultrathin 36SWG wire;
i.e., "VHF" has already happened at 5 MHz. The calculated
inductances however are much too high, and so we need to find
out what went wrong. The problem, as pointed out by Ian Hickman,
and of which Rosa was certainly aware, is that it is impossible
to make an unambiguous definition of inductance without considering
the electrical circuit that gives rise to it. A formula for the
inductance of an isolated length of wire begs the question: "how
can the current (which gives rise to the inductance) get into
the wire without creating fields that interact with the fields
used to define the inductance?" The answer is that it can't.
Ian Hickman made his measurements by forming the wire into a
loop, so that he could connect it to the terminals of a network
analyser; and although the curvature of such a loop is sufficiently
low that we might consider a small segment of it to be reasonably
straight; what he actually measured is the inductance of a circuit.
So, can we take the magnetic fields due to the whole circuit
into account? Loop Inductance The clue to how we might improve the accuracy of the calculation is given by the well known formula for the external inductance of a circular loop (see for example, ref [9], section 5.24, or ref [18] p146). The expression is: Le = R μ(e) [ln(8R/r)  2] [Henrys]. R>>r Where Le is the external inductance of the loop, R is the radius of the loop, and r is the radius of the wire. Now, we would like to relate this expression to Rosa's formula, and so straight away we will equate: l=2πR, i.e., the length of the wire is the circumference of the loop. Hence: Le = l (μ(e)/2π) [ln(8l / 2πr)  2] [Henrys] and we can replace the radius of the wire with its diameter using d=2r, i.e.: Le = l (μ(e)/2π) [ln(8l /πd)  2] [Henrys]. This looks very much like the external inductance part of equation(6.2), except that the term inside the logarithm bracket has changed from 4l/d to 8l/πd, and the 1 has become a 2. Now recall that two numbers can be multiplied by adding their logarithms and taking the antilogarithm, i.e.: Log(ab) = Log(a) + Log(b). This works for natural logarithms, not just base 10, and so we can remove an additive term from the logarithm in the expression for external inductance. In this case the substitution we want is: ln(8l /πd) = ln(4l/d) + ln(2/π) i.e., ln(8l /πd) = ln(4l/d) 0.451582705 and so the formula for the external inductance of a wire loop can be written: Le = l (μ(e)/2π) [ln(4l/d)  2.451583 ] [Henrys]. l>>d What we see here is that the 1 from Rosa's formula has changed to 2.45 by the act of forming the wire into a loop. All we have to do now is add the internal inductance, which is the same as before, to obtain (assuming a nonferromagnetic wire in air):


d / mm 

L / nH 
L / nH 
L /nH 

1.63 

142 



0.376 

201 



0.193 

235.5 


What we find is that the results from equation (6.3) have predicted the inductances within a few percent (and there would have been something badly wrong if they hadn't). So Rosa's formula is definitely wrong for this application. The two formulae are shown below placed next to each other: 

L = 200 l [ln(4l/d)  1 + (δi/d)] nH, l >> d >> 4δi 


L = 200 l [ln(4l/d)  2.4516 + (δi/d)] nH, l >> d >> 4δi 

Now notice that if the length of the wire l becomes
very long, the term ln(4l/d) will become considerably
larger than either 1 or 2.45, in which case it will make little
difference which formula is used. Hence it appears that we can
define an inductance per unit length for infinitely large loops
(except that the concept of lumped inductance presumes an infinite
speed of light). In realistic situations, the loop formula always
gives a lower inductance, and Rosa himself would not have expected
anything else. The truth of the matter is that Rosa's formula
is not a formula for the inductance of a wire, and he would no
doubt have been horrified to think that it would come to be used
in such a way. It is instead the starting point for the calculation
of inductance, and it is perfectly correct when so interpreted.
It sets up the magnetic field around the wire; and can be used
in conjunction with other formulae (for mutual inductance), which
establish the extent to which that field is cancelled or augmented
by the rest of the circuit (All of the relevant techniques were
developed by E B Rosa and F W Grover during the early part of
the 20th Century, and were collected by Grover in his book "Inductance
Calculations: Working Formulas and Tables" [19]).
If the wire is formed into a loop, the magnetic field due to
the outgoing conductor is partially cancelled by the field due
to the return conductor (and vice versa) and the inductance is
reduced. For a singleturn loop, the largest inductance is obtained
when the area enclosed by the loop is at its maximum, i.e., when
the loop is circular, because this is the configuration that
gives the least cancellation of the magnetic field. This maximum
possible value of inductance is given by the singleturn loop
formula (6.3). The maximum
inductance can only be exceeded by stacking conductors in such
a way that their fields add together, i.e., by coiling the loop
into several turns; in which case, if the turns are stacked closely,
the inductance compared to that of a single turn is increased
by a factor of N², where N is the number of turns. If a
single turn loop is not circular, its inductance will be less
than that predicted by equation (6.3),
the extreme (but impossible) limit being reached when the outgoing
and returning conductors are superimposed, the enclosed area
is zero, and the inductance is zero. 



