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Magnetic Levitation 
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#1
Aug2512, 10:49 AM

P: 6

OK, this'll be my first post here.
I was writing on a topic about magnetism when I realized that I simply don't know enough to answer the question effectively. The question is, How much energy does it take to magnetically levitate a 180.0 kg mass using the Earth's gravitational field? (That is, using electricity to power an electromagnet to generate another magnetic field) Some searching gave me the equation for Magnetic Pressure and Electrical force, but I don't understand how to solve this problem. A prompt reply would be appreciated. Thanks! 


#2
Aug2512, 01:16 PM

Mentor
P: 11,835

Hmmm. The problem is that the Earth's magnetic field is very very weak. I saw a thread or two on this in the past and while I don't remember the details I do remember that it required a huge amount of power. So much that I don't think it was possible to levitate something that carried it's own power source.
Unfortunately I don't know how to figure the force required either lol. 


#3
Aug2512, 04:01 PM

P: 57

i guess use a solenoid to create a strong / almost uniform magnetic field, hang the mass by a wire which is partly in the magnetic field created by the solenoid and run a current through the wire?



#4
Aug2512, 05:32 PM

Sci Advisor
Thanks
PF Gold
P: 12,130

Magnetic Levitation
I remember setting a similar problem which involved calculating how much current a DC electricity cable would have to carry in order to support its own weight in the Earth's field. The BIL = Force formula was all you needed. You could do the same for your solenoid.
BTW, the answer was a ridiculous value  as you'd expect and which I can't remember  and tells you that the arrangement is selfdefeating as you need thick (=heavy) cable to carry the current without melting. 


#5
Aug2512, 10:15 PM

P: 6

Thanks guys, a ridiculous number is what I'm looking for.
Could you use electrostatic forces to levitate instead and would it be more efficient? 


#7
Aug2612, 05:52 AM

P: 6

Riiiight, silly me.
Given that I'm calculating for a charged piece of flat metal, not a solenoid, do I need to modify the equation F=IBL Or can I just get rid of L? 


#9
Aug2612, 06:27 AM

P: 6

*shrugs*
Uh, does it affect my calculations? 


#10
Aug2612, 07:26 AM

Mentor
P: 11,828

High current density, and problems with electric resistance? Use superconductors.
However, the magnetic field of earth is too weak and too homogeneous to levitate an object just based on this field. It is possible to levitate objects with other magnets. With superconductors, this is quite easy and does not require any power to keep it levitating (well ok, you have to cool the superconductor). 


#11
Aug2612, 07:33 AM

P: 449

Hi.
You can just use pair of permanent magnet for magnetic levitation. See Science Olympiad page http://scioly.org/wiki/Magnetic_Levitation. Regards. 


#12
Aug2612, 07:50 AM

Mentor
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#13
Aug2612, 08:29 AM

P: 449

Hi.
Regards. 


#14
Aug2612, 09:38 AM

P: 6

OK, let's clarify some things here.
MFB, thanks for the recommendation; For the time being I'm just trying to prove how ludicrous it is to do it without the superconductors. I'm getting a result of about 5.7*10^7 Amperes required to counteract a gravitational force of 1765 Newtons, but I'm not sure my equation is correct. I'm just calculating for F=IB because I'm not using a solenoid, so I don't have a figure for L; does this give me a wrong answer? sweet springs, I know what the force should be, but I'm trying to figure out if I can simply use F=IBL to calculate Amperage required to levitate magnetically. 


#15
Aug2612, 10:56 AM

Mentor
P: 11,828

I*B is not a force. You need some length scale of your current. However, there is a problem: You cannot create or destroy charges. You always need some way for the charges to return  in the opposite direction, which gives a force downwards.
Therefore, you cannot simply use the homogeneous part of the magnetic field to levitate something. You have to use its variation in space, and this is extremely small (of the order of (50µT)/(5000km)). To get a force of 1N, you already need something like 10^9 Ampere in a loop with 100m^2 internal area (WolframAlpha query) or 10^5 A with an area of 1km^2. Superconductors can get current densities of about 10^6A/cm^2, for 10^5 A you need ~0.1cm^2 cable crosssection or ~10g/m cable length (probably more). Ignoring stability issues, this 1kmloop will have a mass of ~50kg and a corresponding force of 500N. As the area scales quadratic with the cable length, scaling the whole system by a factor of 500 would fit  but how do you handle a loop of a really thin wire with a diameter of ~500km? Maybe you can improve the setup a bit  use better superconductors, use regions with a larger variation of the magnetic field and so on. But the numbers are so far away from any reasonable setup that I would be extremely surprised if that would help. For regular conductors, something like 10^3 A/cm^2 is more realistic. Don't even try ;). 


#16
Aug2612, 08:05 PM

P: 6

Ah, thanks.
So it has to be a solenoid, and one with a huge diameter, too. Let me see if I understand your point. 10^9 Amperes * 100 meters^2 * magnetic field of Earth (50 microTeslas/5000km) = 1 Newton of force Because I want to use a 25meter^2 area loop, I can just multiply the Amperage by 4 to get the value for 1 Newton, right? That would give me something ridiculous on the order of 4*10^10 Amperes for a single Newton. (o.O) Am I right? Right now I'm just calculating for how ridiculous the preposition is. How would using a superconductor affect how practical the premise is? 


#17
Aug2712, 11:35 AM

Mentor
P: 11,828

4 times 10^9 A is just 4*10^9 A, not 4*10^10. Apart from that, you are right, the product of current and area determines the force, if you keep the position of the setup the same and neglect effects of higher order.
With magnets in the lab as field source, things are much better. You can easily get gradients of 1T/m (and more), which is better by a factor of 10^11. Instead of superconducing rings, you can use a simple superconducting object of any shape, and use the Meissner effect for levitation. 


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