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According to this video:

http://www.youtube.com/watch?v=BqKeiiezqzc

it should be possible to levitate humans because the water in our bodies is diamagnetic - but it is stated (at 1:14) "although the magnetic fields required would be enormous".

So I wonder: how strong of a magnetic field would you need to levitate a human (using only the diamagnetism of the water in our bodies)? And (maybe easier to do) - how strong was the field used to levitate the frog in the video?

There must be a way to compute the amount of magnetization that occurs, and how that results in a repulsive force; but I don't know where to begin. Any suggestions?

Qmechanic
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Nakul
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1 Answers1

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A bit of clicking gets you to http://www.ru.nl/hfml/research/levitation/diamagnetic/ which tells us that the frog was levitating in a field of 16 Tesla. They give the math as well:

Therefore, the vertical field gradient ∇B2 required for levitation has to be larger than $2µ_0ρg/χ$. Molecular susceptibilities χ are typically 10$^{-5}$ for diamagnetics and 10$^{-3}$ for paramagnetic materials and, since ρ is most often a few g/cm3, their magnetic levitation requires field gradients ~1000 and 10 T$^2$/m, respectively. Taking $\ell$ = 10cm as a typical size of high-field magnets and ∇B$^2$ ~ B$^2/\ell$ as an estimate, we find that fields of the order of 1 and 10T are sufficient to cause levitation of para- and diamagnetics. This result should not come as a surprise because, as we know, magnetic fields of less than 0.1T can levitate a superconductor (χ= -1) and, from the formulas above, the magnetic force increases as B2.

The key to note is that to calculate the force it's not the magnetic field itself, but the square of the gradient of the magnetic field, that matters. As an object becomes larger (like a human) your factor $\ell$ in the above becomes larger, and this means that $B$ needs to be larger too. So you would need an enormously large magnet, with fields on the order of several 10's of Tesla, to levitate a human. Even a small one.

Floris
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