UCSB Science Line Hi! I'm having some trouble with the formula to find the force between the Bismuth and the magnet: F = mu_0/2*M*A*H and figuring out the size and mass of the bismuth. Could you show me how it's done, so I can use that as a model for further calculations? The unit conversions and numbers are a little confusing and I want to make sure I'm doing it right. Thanks. Question Date: 2008-08-14 Answer 1:I was actually incorrect in sending you that equation before. There is a better equation that would be used for diamagnetic levitation, although it's a bit trickier. I'm going to make a few simplifications to the equation, and we can take a look at what the different things mean:B2 / z = mu_0 * p * g / xLooking at the left hand side of the equation, B is the magnetic field we're using, and has units of Teslas; z is how thick the sample of Bismuth is in your experiment, which will have units of meters. When measuring thickness, you want to look at how thick the Bismuth is in the direction perpendicular to the magnet. This is the part of the equation that I've simplified a bit - I'll talk a bit more about that later.Now let's look at the right hand side of the equation. As I mentioned before, mu_0 is a constant that is always the same, no matter what. The value is 1.2566 * 10-6 T m / A (Tesla meter per Amp). p is the density of whatever material we're dealing with. For Bismuth, the density is 9800 kg/m3 (kilograms per meter3). g is a constant you might have seen before - it's the acceleration on Earth due to gravity (for example, weight = m * g). Its value is 9.8 kg m/s2 (kilograms meters per second squared). Lastly, x is the susceptibility of the material we're looking at. As you mentioned before, for Bismuth, it's -1.66 * 10-4 (with no units). These numbers are just constants - the tricky stuff was on the left hand side of the equation.So if you want to figure out how much Bismuth you need for a given magnetic field, you would plug in the value of B for whatever magnet you have, and then you can calculate z. As I mentioned, I simplified the equation a bit. Because of this, the value you get probably won't be exactly right. However, this is OK! I would suggest getting some rare-earth magnets - they're very strong, and should work well enough for what you are doing. Then, you could try calculating how thick the Bismuth should be given that, and then try an experiment to see if it works. If it doesn't, that's OK, too. Just try some different thicknesses to see what works the best.Also, note that this equation doesn't tell us how much Bismuth we need. That's because it's actually not important how much Bismuth we have. If we have more Bismuth, it will be heavier, but it will also have a stronger magnetic field coming from it to react with the magnet. The important thing is how thick the Bismuth is.Also, when getting a magnet, I would suggest trying to get one that has a big area. This is mostly just so that it would make it a lot easier to try to balance the Bismuth over it, although smaller ones would work well, too. Answer 2:I would assume that the mass is taken into account in that equation somewhere, but frankly, I cannot remember the units on all that myself (I can't remember exactly what a Tesla is, for example). A decent college-level physics textbook would have the units for you, and explain what they are so you can figure out if the units don't match or something. Unfortunately, I just haven't been able to find that information online, and my text when I took undergraduate physics is locked away in storage. My suggestion is to see if you can get someone in the UCSB physics department to lend you the textbook they use for undergraduate classes in electricity and magnetism. All of the pieces of that equation should be in there.Click Here to return to the search form.    Copyright © 2020 The Regents of the University of California, All Rights Reserved. UCSB Terms of Use