Self Resonance Of Solenoids And Of Straight Wires, Including Radio Antenna (fwd)

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---------- Forwarded message ----------
Date: Sat, 10 Nov 2007 21:57:23 -0500
From: Jared Dwarshuis <jdwarshuis@xxxxxxxxx>
To: Pupman <tesla@xxxxxxxxxx>
Subject: Self Resonance Of Solenoids And Of Straight Wires,
     Including Radio Antenna

Self Resonance Of Solenoids And Of  Straight Wires, Including Radio Antenna



Written By:

Jared Dwarshuis and Lawrence Morris

November 2007


Abstract:

We will demonstrate that the self resonant frequency of each quarter
wavelength region in an inductor is exactly equal to the self resonant
frequency of the entire inductor.

……....................................................................................................................................…………………………………………………………………………………….

Part (1)

We start with Lenz ideal equation for inductance of an air cored solenoid:
L = u Nsqrd Area / H

We multiply the numerator and denominator by 4pi and regroup:

L = u (2pi R N)sqrd / 4pi H

L = u (wire length)sqrd / 4pi H

Since: C = 1/ sqrt( u e )

L =  (wire length/ C)sqrd  /    4pi e H

Notice that the inductance of a straight wire occurs when you set: H =
wire length.

Now we have an expression for inductance in terms of conducting path
length/ C,  that also shows us the inverse capacitance component of
inductance.
…………………………………………………………………………………….......................................................................................................................................……..

Part (2)

Under conditions of standing waves it is always true that any uniform
medium will be partitioned into even intervals. So we will need an
expression that reflects this division of the solenoid or wire.

L = ( wire/ 2n C)sqrd    2n/ 4pi e H           [Where: n = 1/2, 2/2, 3/2,……]
                                                            [Where
Wire length/ (n/2) = wavelength]
.......................................................................................................................................
..…………………………………………………………………………………………….

Part (3)

We need a driven frequency to get our inductor to manifest its self
resonant frequencies.

 Frequency =  n/2   C/ wire length          [Where: n = 1/2, 2/2, 3/2,……]
                                                          [Where
Wire length/ (n/2) = wavelength]
…………………………………………….......................................................................................................................................………………………………………………

Part (4)

We must also satisfy:

Frequency = 1 / 2pi  sqrt(LC)
……………………………………………………………………………………………...................................................................................................................................

Part(5)

 Set the expressions equal:

n/2 C/wire length = 1 / 2pi sqrt(   (wire length/ 2n C)sqrd     2n/
4pi e H  ) ( self capacitance/2n) )

We will pull  (2n C/ wire length) from the radical and will place the
2pi inside the radical on the right side and do a bit of canceling.

n/2 C/ wire = 2n C/ wire  sqrt(     eH / pi    1/capacitance  )

Or:

n/2 C/wire =  C/ wire sqrt ( 4eH/ pi   (n)sqrd    1/cap )

So we can now express the self capacitance of a region  in a solenoid
inductor or straight wire as:

Self capacitance = 4 e H (n)sqrd / pi
……………………………………………………………………………………………..........................................................................................................................................

 Part (6)

 Departures from ideality:

It matters little that "real inductors" have less inductance then the
theoretical inductance of Lenz inductance equation.  The "real" self
capacitance exists only to be annihilated by the inverse self
capacitance of the exact same region. Subsequently for self resonance,
we need only to consider the wire lengths of quarter wavelength
sections.

………………………………………………………………………………………………........................................................................................................................................

Part (7)

Lets look at an example:

Let the coil have a thousand meters of wire and a solenoid height (H )
of one meter
Then Ltotal  = 100 mh

Lets put a full wavelength on a coil. Since two complete nodes exist
with a full wave our (n) = 2

Each quarter wavelength region has a height (H) of ¼ meter with  an
inductance of 25 mh   and the predicted self capacitance of each
region = 16 e H /pi  with H= ¼ meter

Then the frequency of a region:

300, 000 Hz = 1 / 2pi sqrt ( 25mh  4e/pi)
……………………………………………………………………………………………...........................................................................................................................................

Part (8)

Our inductor partitions are inductors in series so the sum of L is
simply the addition of the partitions.

In our example:

 Ltotal = .25mh + .25mh + .25mh + .25mh

Our self capacitance  is also in series:

Ctotal = 1/     1/(4e/pi)  +  1/(4e/pi)  +   1/(4e/pi)  +1/(4e/pi)

So:

Ctotal = e/pi

Then frequency = 1/ 2pi sqrt ( Ltotal Ctotal)

300, 000hz = 1/ 2pi sqrt (100mh  e/pi )

We have just demonstrated the claim of our abstract!

End.