# Movement of a magnet due to repulsion (help)

1. Feb 18, 2012

### Kieranlavelle

Basically I was wondering if I was to place a magnet inside a tube sealed at both ends via two magnets (as the diagram shows) would the magnet keep on moving up and down or would it find a resting place. I think it would find a resting place but then I also thought there will also be gravity acting on it when it was falling down and I wondered if that would make a difference. Also If it would find a resting place is there anyway I could stop it from doing so without an external power source.http://img535.imageshack.us/img535/7071/diagramv.png [Broken]

Last edited by a moderator: May 5, 2017
2. Feb 18, 2012

### sambristol

It will find a 'resting place' where the effect of gravity is balanced by the magnetic repulsion but it will oscillate for a while until the oscillation is damped by a) air resistance and b) magnetic fields set up by opposing induced currents in the magnets – look up Lenz's law in a search engine

3. Feb 18, 2012

### Kieranlavelle

Ok thanks for your help I appreciate it.

4. Feb 18, 2012

### DarioC

If this is a real project you might want to use a pendulum instead.

The tube has at least one problem. The magnet will try to turn over and doing so will cause friction against the sides of the tube or whatever guide you use. A pendulum will try to twist in a similar way, but should be easier to control with less of a friction problem.

DC

5. Feb 18, 2012

### Naty1

I thought Earnshaw's theorem argues against a stable magnetic levitation...but I don't know the application scope of the theorem nor your assumptions about 'stabilizing' frictional forces.

http://en.wikipedia.org/wiki/Magnetic_levitation

You can get some insights at the same link, above. Usually, absent friction, the issue is not stability but instability.....

In your diagram, my guess is that sufficient friction will likely cause a multiplicity of 'stable' resting places... but the less friction, the less stability.