# Repulsion of the Earth's Magnetic Field

• B
I have a question why a magnet's magnetic field or a coil's magnetic field will not oppose the Earth's magnetic field? Is it because the earth's field is not strong enough or the magnet/coil doesn't cover enough square area? What formula's would I use to figure out what strength or area I would need to have a platform rise up/oppose the earth's magnetic field

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Drakkith
Staff Emeritus
All magnet's will 'oppose' the Earth's magnetic field, and practically all commercially available magnets are strong enough to swamp Earth's field while in close proximity to the magnet. Earth's magnetic field is much larger than anything you can buy or make, but that's because it's being created over a huge region inside the mantle/core of the Earth.

What formula's would I use to figure out what strength or area I would need to have a platform rise up/oppose the earth's magnetic field
Any textbook that covers electromagnetism will give you the basic principles needed to do these calculations. Unfortunately I can tell you right now that levitating a platform by using magnets to oppose Earth's magnetic field is simply not feasible. Earth's field is just too weak.

anorlunda
Staff Emeritus
I use to figure out what strength or area I would need to have a platform rise up/oppose the earth's magnetic field
It is not a problem of direction of the field. The Earth's magnetic field is simply not strong enough to levitate heavy objects.

There is no force on a magnet in a uniform magnetic field (there is however a torque that rotates the magnet in the direction of the field, eg. compass). On a small scale (metres) the Earth's field is nearly uniform. To get repulsion, your magnet would need to be on the scale of the Earth (like the repulsion between opposed bar magnets).

.Scott
Homework Helper
There is no force on a magnet in a uniform magnetic field (there is however a torque that rotates the magnet in the direction of the field, eg. compass). On a small scale (metres) the Earth's field is nearly uniform. To get repulsion, your magnet would need to be on the scale of the Earth (like the repulsion between opposed bar magnets).
Depending on the orientation of the magnet within the field, there can certainly be a force applied to the magnet.

I haven't found any demonstration of this on the web. But if you used a float-type compass and allowed it to float in a wind-free very quiet lake, it should slowly move across the water towards north (or the closest Earth pole).

The effect has bee proposed for maneuvering spacecraft within Earth orbit - for example, to collect orbiting space debris. Here is a description of that:
http://www.star-tech-inc.com/papers/EDDE_Update_for_2016_Tether_Conference_2016_June11.pdf

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But if you used a float-type compass and allowed it to float in a wind-free very quiet lake, it should slowly move across the water towards north (or the closest Earth pole).

The effect has bee proposed for maneuvering spacecraft within Earth orbit - for example, to collect orbiting space debris. Here is a description of that:
http://www.star-tech-inc.com/papers/EDDE_Update_for_2016_Tether_Conference_2016_June11.pdf
The net force (not talking about torque) on a dipole is proportional to the rate of change of the field, not the field strength. My point is that the problem is not primarily the weakness of the field as the other comments suggest, but that it is changing very slowly on the scale of the proposed platform.

The spacecraft concept depends on having the return current path through the plasma and, (as I understand it) moving at high speed through the field.

anorlunda
Staff Emeritus
Spacecraft? I thought the question was about a levitating skateboard.

Very rough calculation (mks units):
Earth's magnetic moment ##M## = ##8 \times 10^{22}## Ampere-metre##^2##
Magnetic moment of platform ##m##
Permeability of space ##\mu = 1.3 \times 10^{-6}##
Earth's radius ##r = 6.4 \times 10^6##

The force between coaxial dipoles is
$$F \approx \frac{\mu mM}{r^4}$$ so $$\frac{m}{F} \approx \frac{r^4}{\mu M} \approx 1.6 \times 10^{10}$$
This means that for a force of ##1## Newton (##0.1## kg weight) you would need a ##\approx 1##metre loop with a current of ##10^{10}##A (or ##1##km with ##10,000##A).