Jared Dwarshuis & Lawrence Morris

Then

B = (μ_{0} i N)/ h

We will multiply the numerator and denominator by (2 π r)

Then

B = (μ_{0} i (2 π r N)) / (2 π r h)

This expression tells us that the B field of a solenoid can be written as:

B = (μ_{0} 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.

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 long solenoids.

L = [N][B][A] / [i]

Substituting our previous expression for B, and the cross-sectional area of the solenoid, we find:

L = ( [N] [μ_{0} 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 = μ_{0} (2π r N)^{2} / (4π H)

This formulation emphasizes the geometric aspects of inductance.

We also know by definition that:

μ_{0} = 1 / (C^{2} ε_{0})

So:

L = ((2π r N / C)^{2}) / (4π ε_{0} 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.

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.

C / wave length = frequency

With standing waves:

Frequency = C / (4 (2π r N))

Substituting L from above into the equation for LC frequency we get:

Frequency = 1/ (2π sqrt (((2π r N/ C)^{2} / (4π ε_{0} H)) capacitance))

(The capacitance in the above equation is the self capacitance of the inductor at resonance.)

[C / (4 (2π r N))] = [C / (2π r N)] [1 / (2π sqrt(capacitance/ (4π ε_{0} H)))]

[1/ (2π sqrt(capacitance/ (4π ε_{0} H)))] = 1/4.

We can now calculate the self capacitance of our quarter wave inductor at resonance as:

Capacitance = 16 e H / π

[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.

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.

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 have meaning.

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 ε_{0} 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.

Then:

Frequency = C/ 4 (2πr N) = 1/ (2π sqrt (((1000/C)^{2} / (4π ε_{0} )) self capacitance))

From this we can calculate the self capacitance as:

capacitance = 16 ε_{0} / π

and the resonant frequency as:

frequency = 75,000 hz

Frequency = (n/2) C/(2π r N) = 1/ [ 2π sqrt( ((2π r N) / (2nC))^{2} 2n/(4π ε_{0} 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 ε_{0} H (self capacitance)) ]

Then: Frequency = C/ 2 (2πr N) = 1/ [ 2π sqrt( ((500/C)^{2} 2 / (4π ε_{0}) ) (self capacitance)) ]

From this we can calculate the self capacitance of the region as:

C_{region} = 8 ε_{0} / 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:

F_{region} = 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 add. So

L_{total} = 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.

C_{total} = 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:

F_{total} = 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

F_{region} = 1/ [ 2π sqrt( (L*/4) (C*/4) ) ]

or four times the frequency of our original quarter wave inductor. Examination of the regions in series shows that our total inductance is

L_{total} = L*/4 + L*/4 + L*/4 + L*/4.

Our capacitance in series now sums to C*/16 and the LC frequency for the entire structure is:

F_{total} = 1/ [ 2π sqrt( (L*) (C*/16) ) ]

or four times the frequency of the original quarter wave.

Copyright © 2012 Jared Dwarshuis & Lawrence Morris