labview1958

Anyone still interested in discussing maglev?

Anyone still interested in discussing electromagnetic field theory?

A strong magnetic field destroys the magnetic field of bulk superconductors, I am using very weak forces.

"* They consider magnets with L~R, while I had L>>R. This means that only my case (3), with F~1/x^4 for large x, is appropriate for their magnets. For one case, they get a pull of 7.2# at 10" and 0.6# at 20". This is approaching a 1/x^4 fall off. " Actuall y I was thinking of the cases where L<<<R and L<z<R and also the cases where L<<<R and z>>R>>L. Again z = separation distance between magnets.

Can the difference register on a weighing machine which has an accuracy of 0.0001g ?

"The magnetic field of the first magnet falls off as the inverse third power of distance, " If we have two similiar magnets with the magnetic field falling off as the inverse third power then the force beteen magnets would be ....1/x^3 * 1/x^3 = ....1/x^6 !

Here is an interesting site. http://www.exo.net/~pauld/activities/magnetism/forcebetweenmagnets.html According to this site the forces between two magnets can be F = ...1/x^7

The basic queistion is: Can the superconductor shield the weak lower magnet from the weak upper magnet! Both magnets are much smaller then the superconductor.

My "thought" experiment is to explore or study the eddy current generated in the copper disc and the meisnner effect in superconductors. I want to show that both are two sides of a coin.

Q. Is there a metal ball inside the copper ring which is allowed to move upon rotation? The copper ring surrounds a plastic disc.The center of the plastic disc is attached to a spindle of a high speed rotating drill. Q. Is the rotatig hollow copper ring's radius parallel (preferred) or perpendicular to the external magnet? From the diagram, it appears to be perpendicular. It is perpendicular. Q. Where are your two termination contact points located? There are no termination contact points. Q. What is your reasoning behind using a 'hollow' copper ring? How do you believe this will improve the eddy current effect, as ooposed to using Faradys flat copper disk? The "hollow" copper ring is easier to spin then a disc. The eddy current penetration levels off about 1cm thickness. Thus a copper ring with thickness of more then 1cm is not necessarry. Q. Are you familiar with electric-magnetic breaks? Yes. Q. In what way do you believe the hollow copper ring will show drag? As the copper ring moves downwards, the magnet will relatively move upwards with respective to the copper. Thus if the magnet moves "upwards" then the electromagnetic drag is "downwards". Thus registering a reading on the balance.

How do you know that is 1/x^4? Any proof?

I have a new link. Try this! http://zone.ni.com/devzone/conceptd.nsf/webmain/DA3932D02D45350786256854005646F3

If the limiting factor is not 1/x^5 then what it is?

When a magnet approaches a spinning copper disc it experiences both electromagnetic lift and drag forces. It is mentioned here. http://www.superlife.info/en/book/victor/levitation.htm

Is there a limit of F = ..1/x^4 for larger distance? Should an even larger distance give F = ..1/x^5 where x is the separation. At what point does F = 0? Is it possible to calculate that?

The North pole is facing the rotatating hollow copper ring. The rotating hollow copper ring is 6 cm in diameter and has a thickness of 1mm.

Here is a site that is interesting! http://www.magnetsales.com/Design/Calc_filles/PullAndPushBetween2RectMagnets.asp There is some empirical data here, however I find that the FORCE rapidly becomes zero.

I am trying to recreate faraday's disc experiment but I am using a rotating hollow copper ring about 6 cm, diameter. I am trying to measure the electromagnetic drag. Is the setup feasible? The copper sphere should be a hollow copper ring.

As the upper magnet is moved down, the SC prevents the magnetic field from penetrating it as the small magnet has a very weak field. Thus the upper part of the SC repels ALL the upper field lines. Thus the lower magnet is not affected by the movement of the upper magnet. Both magnets are small and weak.

What I undestand is that if r=radius of the pole and z is the separation distance then: If z << r, then F = ..1/z^1 If z = r , then F =..1/z^2 If z > r, then F = ....1/z^3 If z>> r, then F = ..1/z^4 Definetly, F = A EXP(-zx)

Can the following setup measure drag force? I hope the image comes?

"To calculate the force when they are still close together requires using a Legendre polynomial expansion of the field from the end of one magnet." What about my formula. Is it reasonable? "If z is the separation distance & x is a constant, does it follow F=AEXP(-zx)" With your reasoning the force Vs distance is NOT a power series, then mine is more logical?

I am a school teacher and a science hobbyist. Right now I am dabbling in superconductivity. However I am quite busy and I have not yet gone around to do my thought experiment. My assumption is that as the SC becomes a shield, thus moving the top magnet downwards have no effect upon the force reading. That is the superconductor shields the attraction between the upper and lower magnet. If the SC is fixed and the upper magnet is moved there is no change in the force reading. Is it a valid assumption. The magnets are much smaller than the SC.

Remember Faradays experiment with a spinning copper disc and a permanent magnet? I want to recreate faraday's experiment but instead of using a spinning copper disc, I want to use a spinning superconducting disc. What would happen?

Let's assume that my superconductor is in a plastic tube held in place and cannot move. I record the repulsive force at room temperature. I pour in the plastic tube 100ml of liquid nitrogen. As the SC reaches 77K would the force reading increase or decrease or reamains the same?