- Assume a long tube with wire / windings on the outside. Now I drop a magnet in the tube (assume the magnet falls straight at center of tube without touching the sides). Assuming the nominal g of 10m/s^2, will the current generated due to this arrangement have a frequency of 10Hz? Are there formulas to calculate the current, voltage etc of such a set up? Responses welcome.
- You'll have to calculate the drop in lenz. The wire size will make a difference. The larger the wire size the higher the amperage/current. The type of materal in the wire makes a difference. You can measure that resistance in the wire for the calculation. The gauss of the magnet will make a big difference too.
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**Sat, 1/1/11, mzalendo2006**wrote:*<mzalendo2006@...>*

From: mzalendo2006 <mzalendo2006@...>

Subject: [free_energy] Frequency Question

To: free_energy@yahoogroups.com

Date: Saturday, January 1, 2011, 8:50 AMAssume a long tube with wire / windings on the outside. Now I drop a magnet in the tube (assume the magnet falls straight at center of tube without touching the sides). Assuming the nominal g of 10m/s^2, will the current generated due to this arrangement have a frequency of 10Hz? Are there formulas to calculate the current, voltage etc of such a set up? Responses welcome. - No, the frequency out depends on the length of the coil and how fast the magnet falls. The speed of the magnet depends not only on gravity but also how much current your coil is delivering (zero current = approximate free fall, short the coil's leads together and the magnet will fall more slowly).

The EXACT answers depend on the gauge of the wire, the coil size, gap size between coil and magnet, type of magnet (field strength to weight ratio). An expert in electromagnet, motor or generator design would be helpful for that level of analysis.

--- In free_energy@yahoogroups.com, "mzalendo2006" <mzalendo2006@...> wrote:

>

> Assume a long tube with wire / windings on the outside. Now I drop a magnet in the tube (assume the magnet falls straight at center of tube without touching the sides). Assuming the nominal g of 10m/s^2, will the current generated due to this arrangement have a frequency of 10Hz? Are there formulas to calculate the current, voltage etc of such a set up? Responses welcome.

> - Adored and Respected MemberJoin Date: Feb 2008Location: Newcastle/ CambridgePosts: 407
**Re: Magnet falling through coil of wire**Peak as it enters, zero emf whilst fully within coil (no change of flux linkage), then negative larger amplitude peak as it leaves. The difference is due to its acceleration whilst within the coil... as the amplitude of the peak is proportional to velocity.http://www.thestudentroom.co.uk/showthread.php?t=1473137*"And of what kind are the men that will strive for this profitable preeminence, through all the bustle of cabal, the heat of contention, the infinite mutual abuse of parties, tearing to pieces the best of characters? It will not be the wise and moderate, the lovers of peace and good order, the men fittest for the trust. It will be the bold and the violent, the men of strong passions and indefatigable activity in their selfish pursuits. These will thrust themselves into your government and be your rulers."**- Excerpt from "Dangers of a Salaried Bureaucracy" addressed to the Constitutional Convention members by Benjamin Franklin in 1787***From:**mzalendo2006 <mzalendo2006@...>**To:**free_energy@yahoogroups.com**Sent:**Sat, January 1, 2011 8:50:10 AM**Subject:**[free_energy] Frequency QuestionAssume a long tube with wire / windings on the outside. Now I drop a magnet in the tube (assume the magnet falls straight at center of tube without touching the sides). Assuming the nominal g of 10m/s^2, will the current generated due to this arrangement have a frequency of 10Hz? Are there formulas to calculate the current, voltage etc of such a set up? Responses welcome.

- Kirk gave a good answer. I will add a little.Assume a coordinate system with zero at the bottom of the coil /tube, and the top of the tube is at height L.Let the magnet drop from height H. As the magnet falls,V = g t is the velocity of the magnet at time tx = 1/2 g t^2 where x is the distance the magnet has fallen.I would like to know the time the magnet enters the tube (x = H-L) and the time that it exits (x = H).This is easy for the open-circuit case because the tube does not affect the rate of fall. Solving for open-circuit,t (in) = SQR [2 (H-L) /g]t (out) = SQR (2 H /g)So, the transit time through the tube is t(out) - t(in). Strictly speaking, there is no "frequency" because there is only one cycle of voltage. But, if you wanted to excite a resonant circuit, the best frequency to tune to would be the inverse of the transit time.Now, if you have a closed circuit from the coil, then energy will be extracted from the system, and the kinetic energy of the magnet will be diminished and the transit time will increase.In no case will the energy extracted from the coil exceed the available energy, which is E = m g H.Ernie Rogers