# Electromagnetic levitation

1. Nov 4, 2008

### dbrown7

Looking at A2 courcework im using soleniods and a iron core to magnitally levitate a aluminium ring.
However what relevent readinging into relevent theories and what formulas do i need to look at?
Is there a forumla that links can predict the height of the levitating magnet? with say the field strength?
I know that the current, no. coils and weight of the ring effect the height but are there any equations?

Thanks

2. Jan 13, 2009

### bodkin77

I'm pretty sure you need to look at faradays laws of induction and diamagnetic levitation. I think though that you will need to use tremendous power to diamagnetically levitate stuff, even aluminium. Note that an aluminium ring is not a magnet, until you induce magnetism in it with the current. If you want to levitate with less power, you want to look at turning you apparatus upside down and using a copper or aluminium sheet base and levitating the solenoid itself. Though I not sure if it would be feasable with the iron core in the centre of it. This way it can move around without falling off also. I think that giving it velocity will also give it height. For this you want to look at Lorentz forces I think which are discussed at http://en.wikipedia.org/wiki/Lorentz_force. But you need to be quite adept at Calculus to get anywhere at this site. You need to balance the force of the gravity (the weight, or mg) with the derivate of the magnetic flux with respect to time. ie. mg=-(d/dt)*fluxB, where 'flux' is Sigma. Then you substitute the double integral equation for Sigma B in which has Area (A) and mag field strength (B) in it. It think that you can just take out the integrals and dA and replace them with just A. Then sub in V=It and also B=(u0*I)/(2*pi*r) in (Bio-Savart law?), where u0 is 'mew naught' or the vacuum permeability (4*pi*E-7). Then you can sub in the Area of the ring (pi R squared). Simplify, derive and then rearrange for small r and you get the height (r). Then sub V=It back in and should get r= (R2 * u0 * I) / (2 * t * W). The R2 is the radius of the ring squared. You will know everything except for the t and I'm not sure how to work that out, but I've heard a couple of times that it takes about 15 seconds. At least that will get you in the ballpark, if it's correct!!?? convert everything to standard units (m,s,kg, Amps). I just worked did an example and it's smaller than expected. But at least I know that you have to balance the grav force and mag force to solve for height/radius.

3. Jan 15, 2009

### turin

I'm assuming that you are driving the solenoid with AC. Otherwise, then you will have the problem that bodkin mentioned about diamagnetism. In fact, if I'm not mistaken, aluminum is paramagnetic, so you will have a huge problem trying to levitate the aluminum ring diamagnetically. If you use AC, then the magnetic response of the material is negligible compared to the Faraday response. Of course, if the magnetic force is equal to the weight, you've got levitation. You can use Faraday's law to determine the current in the aluminum ring. If you know the magnetic field as a function of height outside of the iron core, then you can calculate the magnetic force on the ring. In this case, you need to integrate the force from a bunch of segments around the ring. And, since I and B are oscillating, you should average the force over one period of the oscillation.

Being able to calculate this hinges on your knowledge of the magnetic field in and out of the iron core. Unfortunately, I can't help you with that.

Last edited: Jan 15, 2009
4. Jan 16, 2009

### bodkin77

for what you wanna do, maybe it's just mg=BIL=(u0*I/2*pi*r)IL. rearrange for r. This actually gives believable figures. assuming L=0.2m and I=100amps you'll get 1.55cm levitation. note u0 is magnetic permeability. I don't know much about generating amperage. how do you get 100Amps?

5. Jan 16, 2009

### bodkin77

oh, i assumed m=0.1kg. and I think the right part of the equation applies for a flat disk or a square. you could probably use less amperage for a ring? maybe I used the wrong value for permeability. i can't find the value for aluminium. if it's not right then maybe find an equation with susceptibility in it. susceptibility of aluminium is 2.3x10^-5

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