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Equation For Levitating Magnets?

 
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Dec23-10, 05:24 PM   #1
 

Equation For Levitating Magnets?


Hello there,

I was wondering how much force can two opposing negative magnets hold. If you had two magnets levitating for example, how much weight would they hold up. So, I was wondering if there is some kind of equation for this with the strength of the magnets as the variable.
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Dec23-10, 05:55 PM   #2
 
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If you are using electro-magnets it might be slightly easier because you can simply work to current values.

More current = stronger magnetic field. So you can increase the lifting force by increasing the current through the coil and simply use current values to show the lifting force.

I'd recommend starting here: http://en.wikipedia.org/wiki/Magnetic_levitation

Perhaps under the lift section.
Dec23-10, 09:31 PM   #3
 
Quote by jarednjames View Post
If you are using electro-magnets it might be slightly easier because you can simply work to current values.

More current = stronger magnetic field. So you can increase the lifting force by increasing the current through the coil and simply use current values to show the lifting force.

I'd recommend starting here: http://en.wikipedia.org/wiki/Magnetic_levitation

Perhaps under the lift section.
Whether it is electromagnet or not doesn't concern me because I am mainly concerned with the Gauss units and how many pounds of force 138000 Gauss units can hold for example. Unless an electromagnet with the same amount of Gauss units would have a different levitating force.
Dec24-10, 05:38 AM   #4
 
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Equation For Levitating Magnets?


No, the force would be the same for stationary or eletro-magnet.

The wiki link gives some equations, but I was never good with magnetism so I can't be of much more help unfortunately.
Jan7-11, 11:11 PM   #5
 
You should get an acrylic 'tube' and two cylindrical magnets and drop them in there and see how much weight it takes to make them touch. If you know the strength of the magnets it would seem pretty easy to calculate. To me that seems easier and more accurate that an equation, though more expensive. I know thats not the answer you were looking for but whats wrong with some real life experimentation.
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