In this paper we will be describing standing wave self resonance
in an inductor using ideal inductor arguments. We will develop a
generalized equation to describe the formation of any number of
nodes within an inductor. We will show that under self resonance
the frequency of each quarter wave region will be in agreement
with the frequency of the inductor as a whole. We will also show
how the velocity factor (which others have measured) is describing
a relativistic observation.
Part 1 - B field and inductance
We will start with the classic B field of an inductor with the
reminder that this equation assumes that we will be measuring the
B field in the middle of an infinitely long solenoid where the
lines of magnetic force entirely line up with the long axis of the
solenoid. With a short solenoid, lines of magnetic force are not
entirely aligned along the axis of the solenoid and we do not get
B = μ0 i n (where N = nh)
B = (μ0 i N)/ h
We will multiply the numerator and denominator by (2 π r)
B = (μ0 i (2 π r N)) / (2 π r h)
This expression tells us that the B field of a solenoid can be
B = (μ0 i (wire length))/ (surface area of the solenoid).
This formulation for a solenoid clearly shows the geometric
concerns. For any solenoid you will always have a given length of
wire which covers a particular surface area.
Now we will use the previous formulation for the B field of a
solenoid and put it in the inductance formulation.
Again we need to remind ourselves that the ideal equation of
inductance assumes that we have an infinitely long solenoid where
all of the lines of magnetic force align with the long axis of the
solenoid. With a short solenoid towards the ends, your lines of
magnetic force are not aligned with the long axis of the solenoid
and you do not get ideal results. Interestingly, approximation
equations for inductance will typically converge on the ideal for
L = [N][B][A] / [i]
Substituting our previous expression for B, and the
cross-sectional area of the solenoid, we find:
L = ( [N] [μ0 i (2π r N)) / (2π r H)] [π (r)2] ) / [i]
Notice that our current term has canceled out at this point. We
are left with only geometric variables which are not time varying.
Multiplying numerator and denominator by 4π then reducing we
observe that the wire length can be written as a square:
L = μ0 (2π r N)2 / (4π H)
This formulation emphasizes the geometric aspects of inductance.
We also know by definition that:
μ0 = 1 / (C2 ε0)
L = ((2π r N / C)2) / (4π ε0 H)
This formulation for inductance works nicely and is in full
agreement with the classic formulation. This formulation is the
ratio of the square of a period (determined by the traversal of
the wire length at the speed of light) divided by a capacitance
(determined by the height of the solenoid). A circuit traversal
description of inductance shows that we will run into problems if
we attempt to outflank the speed of light with an RF signal.
Properties of Inductance
We think that there is confusion about the properties of
inductance. Some think that frequency or current and voltage
distributions can alter the properties of inductance; but that’s
not quite right. Inductance in an ideal inductor is always a
constant. With standing waves we can partition an inductor into
regions defined by voltage maxima and minima where the inductance
of an individual region is less than the inductance of the entire
inductor. But when we sum these regions in series, the overall
inductance will remain a constant.
Inductance is not ruled by current. We can place a small current
or a large current through an inductor and it will have exactly
the same inductance. Having current bunched up in a particular
place at a given instant does not change the inductance.
Part 2 - Single quarter wave region
We will now examine a single quarter wave region of an ideal
Frequency of Travelling E-M Waves
The equation for describing the frequency of traveling
electromagnetic waves is:
C / wave length = frequency
Frequency of Standing E-M Waves
For standing waves however, we are considering the situation where
wire length = wave length / 4.
With standing waves:
Frequency = C / (4 (2π r N))
Frequency of an L-C System
For an LC system: Frequency = 1 / (2π sqrt(LC))
Substituting L from above into the equation for LC frequency we
Frequency = 1/ (2π sqrt (((2π r N/ C)2 / (4π ε0 H)) capacitance))
(The capacitance in the above equation is the self capacitance of
the inductor at resonance.)
Now we have two expressions for the same frequency; we will set
them to be equal.
[C / (4 (2π r N))] = [C / (2π r N)] [1 / (2π sqrt(capacitance/ (4π ε0 H)))]
At resonance in the above equation, the self capacitance must
satisfy the equation such that:
[1/ (2π sqrt(capacitance/ (4π ε0 H)))] = 1/4.
We can now calculate the self capacitance of our quarter wave
inductor at resonance as:
Capacitance = 16 e H / π
The frequency equation reduces to:
[C / (4 (2π r N))] = [C / (2π r N)] [1/4].
The left side of the equation describes frequency as 1 / (4 times
the traversal time of the wire length). The right side of the
equation describes frequency as [1 / (the traversal time of the
wire length)] multiplied by a unit-less correction factor. The
parts of the equation that have units of frequency are related by
a factor of 4.
A Relativistic Interpretation
We consider this factor of 4 to be a relativistic gamma factor
because we are making an observation of the relativistic
interference pattern of a standing electromagnetic wave.
By applying the Lorentz transformation (with Gamma = 4) we find that:
4 = 1/ sqrt(1- ((v/C)2)).
This gives us an observed wave velocity of sqrt(15/16)C. This
velocity is the velocity that others have been measuring.
Part 3 - Multiple quarter wave regions
Now we will describe multiple node formation within an inductor at
resonance. Regardless of the number of voltage nodes we put in our
inductor we must adhere to the same velocity factor of:
velocity = sqrt(15/16)C
First we will demonstrate that our accounting procedures must be
based on quarter wave regions. We know that for an inductor;
voltage from point A to point B = L di/dt. Consider points A to B
as being between two current nodes. Now we have a problem since
there is no difference in voltage between current nodes, but
clearly there must be inductance within these regions. We must
consider quarter wave regions since our voltage A to B must be
between a minimum and a maximum for our regional inductance to
We will need to add features to the frequency equation for a
quarter wave region to account for node formation.
Frequency = (n/2) C/(2π r N) = 1/ (2π sqrt( ((2π r N) / (2 n C))2 (2n/(4π ε0 H)) (self capacitance) ))
Where n = 1/2, 2/2, 3/2, 4/2 ... (the number of 1/2 waves)
The features added allow us to divide the wire length (2π r N)
and solenoid height (H) when we go to the higher harmonics. The
right side of the equation will only tally the inductance of a
single quarter wave region found within the inductor.
A Concrete Example
Let's pick an inductor with easy dimensions to work with. Let the
wire length = 1000 meters and let the height (H) of the solenoid =
A Quarter Wave
We will start with the quarter wave region where n = 1/2 (the
fundamental quarter wave frequency).
Frequency = C/ 4 (2πr N) = 1/ (2π sqrt (((1000/C)2 / (4π ε0 )) self capacitance))
From this we can calculate the self capacitance as:
capacitance = 16 ε0 / π
and the resonant frequency as:
frequency = 75,000 hz
A Half Wave
Now let's see what happens when we take the very same inductor and
double the frequency to get a half wave, where: n = 1
Frequency = (n/2) C/(2π r N) = 1/ [ 2π sqrt( ((2π r N) / (2nC))2 2n/(4π ε0 H) (self capacitance)) ]
Substituting in: n = 1 (Because it is a half wave)
F = C / [ 2 (2π r N) ] = 1/ [ 2π sqrt( (2π r N /(2C))2 2/4π ε0 H (self capacitance)) ]
Then: Frequency = C/ 2 (2πr N) = 1/ [ 2π sqrt( ((500/C)2 2 / (4π ε0) ) (self capacitance)) ]
From this we can calculate the self capacitance of the region as:
Cregion = 8 ε0 / π
Our resonant frequency is now:
Frequency = 150,000 hz
When we examine the LC frequency of the region we have half the
inductance of the previous quarter wave example (which we will
call L*/2) and half the capacitance (which we will call C*/2). We
can simplify this by showing the frequency of the region as:
Fregion = 1/ [ 2π sqrt( (L*/2) (C*/2) ) ]
or twice the frequency of the quarter wave. Now we can examine the
two regions in series. The rules say that inductors in series
Ltotal = L*/2 + L*/2
As we can see our total inductance has not changed it is exactly
the same as our first example with just a quarter wave. Now let's
examine the capacitance in series which add by the reciprocal of
the sum of the reciprocals.
Ctotal = 1/ [ 1/(C*/2) + 1/(C*/2) ]
This gives us a series capacitance that now totals C*/4. When we
examine the LC frequency of the entire inductor we get:
Ftotal = 1/ [ 2π sqrt( (L*)( C*/4) ) ]
or twice the frequency of our original quarter wave.
So now we see that for self resonance it is a requirement that
both the quarter wave region and the entire structure, resonate at
the same frequency. This pattern is true of all of the
harmonics. We will show a last example