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Well let's start off with that I'm not a physicist but I'd like some thoughts on something I came across in my hometown.

This guy:

enter image description here

Is it possible that due to the electrical charge of magnets this guy can make the illusion that he can float ? Or is this probably a cheap trick that fools the eye ? I was standing there for quite some time watching the guy and he keep moving his feet. The resistance that he appeared to have was from a magnet force keeping him afloat. So after I passed this guy I did some physics searches on the web and the first thing that caught my eye was the electrical charge of magnets.

So the question is : Is this related to the electrical charge of a magnet or a cheap trick ?

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    $\begingroup$ I cannot see how we could answer that. I've seen these guys myself, but it is a magician's trick - not a well documented phenomenon of nature. $\endgroup$
    – ACuriousMind ♦
    Commented Aug 20, 2014 at 1:19
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    $\begingroup$ @ACuriousMind, magician's tricks are phenomena of nature (documented or not is another matter). Quoting Poincare on the Michelson-Morley experiment. "It is as if nature conspires for the MM experiment to fail". Poincare: "Then this is a law of nature". See how simple it is? $\endgroup$
    – Nikos M.
    Commented Aug 20, 2014 at 1:31
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    $\begingroup$ I hope you tipped him! If a street busker makes you stop what you're doing and admire, you owe him at least a dollar. $\endgroup$
    – rob ♦
    Commented Aug 20, 2014 at 2:46
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    $\begingroup$ This question appears to be off-topic because it is about guessing how a magician's trick works. $\endgroup$
    – ACuriousMind ♦
    Commented Aug 20, 2014 at 13:08
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    $\begingroup$ Please understand that close votes are not personal. I am not saying you cannot be interested in how this trick works (I'm a bit intrigued by it myself). I am not saying you should stop trying to find answers to your questions (you shouldn't). But this specific question does, in my view, not belong on this specific site, as I don't see a physical explanation arising. And I feel justified in that since the top answer at the moment is probably right but enlightens no one about a bit of actual physics. $\endgroup$
    – ACuriousMind ♦
    Commented Aug 20, 2014 at 13:49

3 Answers 3

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The "trick" is that the cane he is apparently holding is actually firmly attached to the platform. A rigid piece goes up his sleave, then to a harness that holds his whole body up. For more about this type of magic trick device, google "broom suspension" or "aerial suspension harness".

No electric or magnetic fields were abused here.

Image Credit: TwentyTwoWords Image Credit: TwentyTwoWords

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    $\begingroup$ Very clear explanation. If you look closely at the photo in the question you can see that the cane is bending... $\endgroup$
    – Floris
    Commented Aug 20, 2014 at 12:32
  • $\begingroup$ check out this video here, which shows a performer setting the trick up. youtube.com/watch?v=ZwvYkRzpq4g $\endgroup$
    – cantsay
    Commented Aug 21, 2014 at 1:17
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    $\begingroup$ You can run, Olin, but you can't hide. A hit squad from the Magic Circle has been despatched. $\endgroup$
    – peterG
    Commented Aug 21, 2014 at 1:53
  • $\begingroup$ maybe the discussion about the closing of this question will interest you here $\endgroup$
    – Nikos M.
    Commented Aug 21, 2014 at 22:26
  • $\begingroup$ @NikosM. - it certainly did: thanks for pointing it out. I have posted a response. $\endgroup$
    – Floris
    Commented Aug 21, 2014 at 23:15
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Olin Lanthrop clearly gave the most plausible explanation. But just for fun, let's just assume this was an electromagnetic trick. Would that be possible?

First - let's do this using electrical charge: how much charge would you need to allow levitation, and what would the potential have to be?

Some assumptions:

70 kg guy
40 cm levitation (based on apparent height)
equal charges
cane is perfect insulator

The force needed would be about 700N; solving

$$F=\frac{Q^2}{4\pi\epsilon_0 r^2}$$ gives $$Q = r\sqrt{4\pi\epsilon_0 F}$$ and $$V = \frac{Q}{4\pi\epsilon_0 r} \approx 3.5 MV$$

If I did not make a mistake, this makes the voltage on both the performer and the platform above which he is hovering about 3.5 MV. Note that this value is independent of the height he is hovering... just the force required. The breakdown voltage of atmospheric air depends on many factors - but you're not going to hold 3.5 MV on an irregularly shaped object without some serious corona discharge - the electrical breakdown strength of air is around $3.6\cdot10^6 V/m$, which will easily be exceeded in this configuration. I conclude it cannot be static electricity holding him up.

So, could it be magnetism?

With magnetic levitation, there are two problems: the strength of the magnet, and stability (two dipoles cannot provide a stable levitation platform). Let's tackle the stability by making the platform is a superconductor; then the "magnetic pressure" is

$$P_{mag}=\frac{B^2}{2\mu_0}$$

Because I don't want to do the integral, I am going to assume that the field is uniform over a diameter equal to the distance; then we can compute the force:

$$F = P\cdot A = \frac{B^2 \pi d^2 / 4}{2\mu_0}\\ = \frac{B^2 \pi d^2 }{8\cdot 4\pi 10^{-7}}\\ = \frac{B^2 0.16 }{32\cdot 10^{-7}}$$

Setting to 700N and solving for $B$:

$$B = \sqrt{700 \cdot 200 \cdot 10^{-7}}\\ = 0.12 T$$

In principle it is possible to make permanent magnets that strong... but they would weigh quite a bit more than 70 kg (meaning you would have to update this calculation), and you wouldn't want to walk around with them on the market square of Leuven. Oh - and there's the minor problem of the cryogenics needed for your superconducting platform...

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  • $\begingroup$ So if I understand this correctly the magnets would have a mass more then 70kg to achieve levitation through magnets ? $\endgroup$
    – Nicolas Pierre
    Commented Aug 20, 2014 at 13:42
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    $\begingroup$ @NicolasPierre - I did not compute that, but to maintain a field of 0.1 T at a distance of 40 cm (as opposed to "at the surface") you need quite a large magnet, as a dipole field falls off with $r^3$; when you have a large area, you operate in the "near field" region of the magnet so it will drop off more slowly. So while it's possible to have a smaller magnet (e.g. Neodymium) with a quoted strength of 0.1T, it would not work in this situation. $\endgroup$
    – Floris
    Commented Aug 20, 2014 at 13:48
  • $\begingroup$ If you added an electromagnet (which would only need to provide a small portion of the levitation) and a PID controller then minor adjustments could constantly be made to the force stabilising the whole thing: this would remove the requirements for superconducting $\endgroup$
    – Richard Tingle
    Commented Aug 20, 2014 at 15:34
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    $\begingroup$ I think you're being over-hasty in rejecting the magnetism hypothesis. His cryogenic kit is in the sewers: the purpose of his little "stage" is to conceal the open manhole cover. The "cane" is used to pipe coolant to the magnet on his body. Easy when you know how. ;-) $\endgroup$
    – David Richerby
    Commented Aug 20, 2014 at 17:31
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    $\begingroup$ @DavidRicherby I see... And this clever fellow invented a silent cryo cooler which can run off a portable power supply . Must be galling to be that talented and still end up begging for pennies in the market square... There is no justice in this world. $\endgroup$
    – Floris
    Commented Aug 20, 2014 at 17:35
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Spiders can fly by charging a piece of web and using the 120 V/m field to go up.

But i suspect that for a human-size weight, one would need too big of a charge to use electrostatic forces, and such would discharge through the air.

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    $\begingroup$ is this from spiderman? i am kidding of course, but i think the mechanics of how spiders jump was on an early spiderman volume $\endgroup$
    – Nikos M.
    Commented Aug 20, 2014 at 6:10
  • $\begingroup$ And how does this answer the question? $\endgroup$
    – yuritsuki
    Commented Aug 20, 2014 at 19:25
  • $\begingroup$ maybe the discussion about the closing of this question will interest you here $\endgroup$
    – Nikos M.
    Commented Aug 21, 2014 at 22:27

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