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Author Topic: Hollow Inductor Generation Coils  (Read 3448 times)

Offline sm0ky2

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Hollow Inductor Generation Coils
« on: March 17, 2015, 05:04:05 AM »
Hi guys,

I am wondering if anyone out there knows, or can help me figure out the answer to this area of physics.
analyzing the inductance effects of two coils, of same diameter, same voltage and current passing through them,
same # of turns, same diameter of coiling.

One a solid copper wire, and other a hollow copper tube.

Now, our modern conception of this effect, tells us the current passes (primarily) on the surface of the conductor.
So, it stood to reason, that there would be no difference in the magnetic induction between a solid and hollow wire coil.
How wrong I was to think this......     when comparing the magnetism produced by the coils, I find that there is a much greater magnetic field from the hollow tube. This didn't make a lot of sense to me, because our energy equations for the inductor are pretty resound, in that Energy in = Field out.
So why the stronger field?  As I began to investigate this, I came across the equation for the hollow inductor. Now, this doesn't directly answer my question, as the equation is based on what goes on INSIDE the tube....  So, I began testing this, and it seems the math is fairly accurate, within the margin of error of multiple gauss meters,
 

digging a little deeper into this, what I find is that the solid conductor seems to "short out" the magnetic field through its center,
while the hollow-tube induces secondary poles on the inside surface.
By this I mean, if the coil is wound (right hand rule) so that the North pole faces up, south down...
Inside the hollow tube, there exists a south-up, north-down pole orientation.
This causes a flux-intensity over time, in the interior of the tube, pretty much in accordance to the equations.
B-field strength can be measured as well as calculated. OK, great...
But why the much larger external field?  Well,.. I can't seem to find any math on this subject.
But test results indicate that the hollow tube is a much more efficient electro-magnet coil,
and it stands to reason that it would be a more efficient generator as well.

My thoughts are that, the internal B-field is acting like its own inductor, similar to placing an iron core in the center, but without the extra energy consumption of charging up the core.

All I can think at this point, is that we need an equation to handle this. And bring this new information into the main-stream of electromagnetic theory. If anyone knows of an equation, or has the equipment, and know how to help in this, please share your thoughts.

Experiments in this area show a considerable difference in the mass that can be lifted / repelled via current*voltage
But this changes the way we think about our power equation, in terms of the magnetic field.....

I've attached a "slice-view" drawing of a piece of each coil, during a moment in time while current is flowing, to give
a visual of the magnetic field orientation what's going on inside and out.
In a straight tube analysis, there is no anomalous effect, but when the tube is coiled, the inductance (Henrys/meter) goes all to shit.. [ and yes I've checked for shorting between coils]
  Any help would be great



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Offline MarkE

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Re: Hollow Inductor Generation Coils
« Reply #1 on: March 17, 2015, 05:14:55 AM »
I take it from the figure that you are driving current into the page.  From Faraday induction and by applying some calculus you can derive the current distribution in the solid wire cylinder, versus the tube.  For AC signals, Faraday induction causes an EMF to develop that opposes the voltage impressed across either conductor.  Near the center of the wire the BEMF will be nearly identical to the driving EMF and so very little net current flows there and the net magnetic field is weak.  Does that seem weird, or at least a bit ironic:  Where the magnetic effects are strongest, there is the least net magnetic field.


 

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