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In "Back to the Future" they use levitating anti-gravo skateboards without wheels which will be difficult to build, because there is no theory on how anti-gravity could be created.

A possible solution in the far future could be to use Earth's magnetic field to create a kind of induction field that keeps the skateboard above the ground.

What would a theory look like to create such a field?

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    $\begingroup$ I think this post has a lot of issues. First, it's unclear. It suggests that we can use the Earth's 10 $\mu T$ magnetic field to generate what's called an "induction field" (Anti-field or something, Is there any as such?), but doesn't account for the "how-to-do such a thing". Moreover, it explicitly asks for a theory on such a phenomenon. In the absence of a proper definition or assumption, people can play around with the post. So, this is also primarily opinion-based (seems like an anomaly to SE policy :P) $\endgroup$ Commented Sep 1, 2013 at 16:55
  • $\begingroup$ I hope this is clear: it is a hypothetical question. I am expecting answers like the one from user6972. It came up on scifi.stackexchange.com $\endgroup$
    – rubo77
    Commented Sep 1, 2013 at 18:43
  • $\begingroup$ What exactly do you mean by "induction field"? Do you mean new physics, or simple electromagnetism? $\endgroup$ Commented Sep 1, 2013 at 21:05
  • $\begingroup$ Your last sentence seems like you're looking for new physics, not a new application of established theories. In that case, this is off topic (see the close reason below and the linked meta post). However, if you want to know if it is possible in the domain of established theories, feel free to clarify that by editing it and it will probably be reopened. $\endgroup$ Commented Sep 8, 2013 at 8:17
  • $\begingroup$ so this would fit to SciFi then as I tried in the first place? Can you reopen it and migrate it there please? $\endgroup$
    – rubo77
    Commented Sep 8, 2013 at 9:14

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When superconductivity was first discovered this was one concept that was quickly explored. Unfortunately the weak field of earth requires a massive circulating current to generate a counter magnetic field with enough force to lift something off the ground.

There are two limiting factors:

  1. The need for more current means more weight which means more current... In fact the first pass analysis shows you need power on the order of 20kW, which is about the same amount of power generated by a locomotive engine.
  2. The Earth's magnetic field is not perpendicular to the surface, in fact near the equator there's almost zero vertical component to work against. This angle reduces the amount of available field strength to lift against gravity.

Perhaps one day if we can develop massive yet micro-sized power sources something like this might be possible, but away from the equator.

Edit: So how this concept works is first look at the vector of how the magnetic field on the earth's surface (away from the equator). There is a vertical component that is parallel to the force due to gravity.

mag field

A loop of superconducting current (assuming there is a way to get it started) would create a field so that if you set it perpendicular to the earths surface the magnetic moment would oppose the vertical component of the earth's field.

The torus fields looks like this.

loop

I unfortunately need to run, but the calculations are similar to superconductor levitating in earth's magnetic field? but for a loop and not just to levitate itself, but additional mass.

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  • $\begingroup$ "requires a massive circulating current" ... "In fact the first pass analysis shows you need power on the order of 20kW" Current is measured in amps not watts. Why would you need to dissipate 20 kW? Superconducting currents don't dissipate any energy. Also superconductors hover in magnetic fields without any current flowing through them, no? $\endgroup$
    – endolith
    Commented Jan 15, 2016 at 3:51
  • $\begingroup$ @endolith current is a measure of energy and watts is a measurement of energy per time. In this application the magnetic field strength needs to be maximized and strong. To achieve this you need a strong current that has to be introduced to the conductor. Once it is circulating and stable it will continue without dissipation. The hovering effect you mention is not related to this application and is caused by the Meissner effect. $\endgroup$
    – user6972
    Commented Jan 29, 2016 at 18:30
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    $\begingroup$ "watts is a measurement of energy per time" Yes. "current is a measure of energy" No, it's a measure of charge per time. $\endgroup$
    – endolith
    Commented Jan 29, 2016 at 18:46
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I did not quite understand the idea that user6972 discusses and how (s)he could use the vertical component of the Earth's magnetic field, however, one could use the horizontal component of the field. For example, if you have a (closed) superconducting cable with a large current around the equator, it can have some lift in the Earth's magnetic field, and it does not need power (other than to maintain sufficiently low temperature). It's more difficult to provide lift for an arbitrary trajectory (so this is probably not for scateboards:-) ), but it can be done in principle, say, using large closed superconducting cables with current and the difference between Earth's magnetic field at different altitudes. Of course, these are hypothetical possibilities, as the Earth's magnetic field is relatively weak, as user6972 pointed out.

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  • $\begingroup$ The "craft" is a superconducting loop (torus) of current. The flux works against the vertical component of the earth's field (the portion that is perpendicular to the surface of earth) which is much higher at the poles and gets less as you approach the equator. $\endgroup$
    – user6972
    Commented Sep 1, 2013 at 20:04
  • $\begingroup$ @user6972: so it would not create lift in homogeneous field? $\endgroup$
    – akhmeteli
    Commented Sep 1, 2013 at 21:08
  • $\begingroup$ It would produce force in any B field to some extent, but to maximize lift you have to work with the vertical $B_e$ component so their magnetic moments oppose each other. $\endgroup$
    – user6972
    Commented Sep 2, 2013 at 1:25
  • $\begingroup$ @user6972: But you need to use nonzero gradient of the Earth's magnetic field. I see. Thank you for the explanation. $\endgroup$
    – akhmeteli
    Commented Sep 2, 2013 at 3:02

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