I wonder how difficult it is to create an artificial planetary magnetic field with generators? What power they would need?

The question is inspired by thinking about possible colonization of Jupiter's moons Io and Europa which are located inside the Jovian radiation belt. Is it possible to create with easy means an artificial magnetic field such that it to shield the surfaces of these moons from radiation? Or it would require astronomical amount of power?

By easy means I mean a device that would not require power greater that normal industrial power plant, best of all, solar-powered or based on once-charged superconductor coils.

  • $\begingroup$ Related question physics.stackexchange.com/q/53184 $\endgroup$ – Jitter Jan 13 '14 at 8:00
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    $\begingroup$ Well another question that opens up now, is what is the benefit of a (weak) planetary field, in comparison to stronger local fields (like just for the settlements)? And at which level would a local field have the same (shielding) effect? $\endgroup$ – WalyKu Aug 4 '15 at 9:52

According to this article the energy stored in the Earth's magnetic field is about $10^{26}$ ergs or $10^{19}$J. According to Wikipedia the annual global generation of electricity is about 20,000TWh, which is between $10^{19}$ and $10^{20}$J, so actually we already produce enough power to generate the Earth's magnetic field.

Actually doing it on a moon of Jupiter would be another problem. I suppose you could use nuclear power to avoid having to build an oil pipeline between the Earth and Europa, but even if the power were available I don't know if current technology is up to generating magnetic fields with that much energy.

  • $\begingroup$ Well, humans produce enough energy, to cover the energy of the magnetic field. However it's not clear, which energy conversion efficiency can we reach, and how much energy should we spend to sustain created magnetic field. If we do use a superconducting wire, do we have enough resources to produce that much of it (and the cooling system)? $\endgroup$ – Fiktor Aug 12 '13 at 17:51
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    $\begingroup$ You seem to be comparing quantities with different units. By itself, the fact that there are $ \ge 10^{19}\,\mathrm{J}$ stored in the Earth's magnetic field tells you nothing about the power needed to generate it, because you also have to know how rapidly that energy is dissipated. If the dissipation time scale is about a year then fair enough, but otherwise comparing our annual energy production (Joules per year) to the total energy in the field (Joules) isn't terribly meaningful. $\endgroup$ – Nathaniel Sep 9 '13 at 8:05
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    $\begingroup$ @Nathaniel If we use energy to create current in a (super)conducting ring which makes a magnetic field, how does it "dissipate"? $\endgroup$ – Keith McClary Oct 1 '15 at 4:50
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    $\begingroup$ @KeithMcClary in the superconducting case it doesn't, as far as I know, though I am not an expert on superconductors. (In the case of a normal conductor it dissipates via Joule heating.) But that's just my point - if you're storing the energy in a system with no dissipation then you can generate the energy at as slow a rate as you like (slower just means it takes longer to spin it up) so the rate at which we generate energy isn't the relevant figure. $\endgroup$ – Nathaniel Oct 1 '15 at 4:59

A superconducting current loop could perhaps do the trick. Let's consider a very simplified analysis for creating a protective magnetic field for Venus.

We're going to be essentially imitating earth's magnetic field.

I don't know much about the solar wind, so I'll assume the strength varies per the inverse square law. Further, I assume the magnetic field strength required for planetary protection is directly proportional to solar wind strength.

Venus orbits at 0.72 AU from the sun. Then, in order to have the same protection on Venus against the solar wind as Earth does now, the magnetic field should be $(1/0.72)^2 = 1.93$ times as strong as Earth's field.

The earth's magnetic field is usually approximated as a dipole. The magnetic field at a given radius R and magnetic latitude theta is given by $\left|B\right| = \frac{B_0}{R^3} \sqrt{1 + 3\sin^2 \theta }$ Tesla

Planetary magnetic fields vary across radius and latitude. For the purpose of this analysis, let us take as reference the value of the magnetic field at the pole. In other words, the target is to get Venus' polar magnetic field strong enough, and hope the rest of the field is also strong enough, since that kinda works for earth.

Plugging $B_0 = 3.12 \times 10^{-5} T $, $R=1$ and $\theta = 90 °$, we get a neat figure of $B_{pole-earth} = 6.24 \times 10^{-5} T$. The required $B_{pole-venus}$ then comes out to be $1.2 \times 10^{-4} T$.

Again, to simplify calculations, I'm going to assume a single equatorial loop of superconductor resulting in the magnetic poles coinciding with the rotational poles. The radius of Venus is 6052 km, and thus the length of the superconducting cable is around $38000 km$.

Now the magnetic field of a single circular loop of current, at a point on the axis of the circle is given by $$B = \frac{\mu_0}{4 \pi} \frac{2 \pi R^2 I}{(z^2 + R^2)^{3/2}} T$$

Since the pole is at ground level, the expression simplifies quite a bit to yield $$B_{pole-venus} = \frac{\mu_0}{4 \pi} \frac{pi I}{R} T$$

Solving for current, we get $$I = B_{pole-venus} \frac{R}{\pi} \frac{4 \pi}{\mu_0} A$$ yielding a value of 2.3 GA . Huge, but surely not universe wrecking.

The inductance of a coil of wire does not have an accurate closed-form expression even in the ideal scenario, so we will use Kirchhoff's approximation given by $$L = 4 \pi R (\ln{\frac{8 R}{p}} - 1.75)$$

Let us assume a generous conductor of $p = 3 m$ radius. We get an inductance of $L = 415 MH$ - an enormous value, but entirely expected given the planetary scale.

Finally, the energy in a loop of current is given by $E = \frac{1}{2} L I^2$ , giving a value of $1.12 \times 10^{27} J$. This is a pretty big number - obviously we would want to generate this from Venus' abundant solar energy.

Venus' insolation, another figure that definitely has an inverse square law variation, should be around ${1/0.72}^2 = 1.93$ times the earth's insolation which is the solar constant, equal to $1362 W/m^2$. This gives us a very encouraging $P = 2627 W/m^2$ as Venus' insolation, and $P \pi R^2 = 3.023 \times 10^{17} W$ as Venus' total solar budget. Assuming 100% efficiency shows us that it would require around 37 years to fully charge up the magnetic field. Not trivial at all, but certainly practically doable on a timescale of millenia.

Addendum : Since both solar wind (presumably) and insolation vary as the inverse square of distance to the sun, the number of years of capturing solar energy should be independent of distance from the sun. The inductance of the loop however is proportional to $R \, \text{ln}R$, and current required is proportional to $R$, so dependence of energy required on the radius ($0.5 L I^2$)is more than cubic. As suggested by RiskyScientist below, running the numbers for Callisto shows it would require $2.4 \times 10^{22} J$ or $0.26 y = 95$ days for solar charging of the magnetosphere generator.

The major assumptions here are :

  1. Dipole approximation
  2. Polar magnetic field strength
  3. Inverse square law variation in solar wind
  4. Necessity/sufficiency of earth's magnetic field for protection against atmospheric stripping
  5. Single current loop

Most of these serve to decrease the required energy, so there should be some hope here.

The magentic field at the surface of the conductor is $154 T$ which is far greater than any human-created field. The superconducting cable would occupy a volume of at least $38000 \pi p^2 = 1.075 km^3$. weighing around $8.6 \times 10^{12} kg$ (assuming density of $8000 kg/m^3$).

However, such a loop offers a critical advantage should industrial civilisation ever arise : electric storage AND transmission across the planet, with basically unlimited discharge and charge rates.

  • $\begingroup$ Whilst Venus is nicely placed for solar power, it is inconveniently warm for superconductivity - the Jovian moons are in the opposite situation, and also suffer far less solar wind. It would be interesting to run the numbers for that case. $\endgroup$ – RiskyScientist Jul 2 '16 at 17:01
  • $\begingroup$ @RiskyScientist making that excel sheet is paying off :) however, all but Callisto suffer from the Jovian radiation which is itself of extremely high intensity. $\endgroup$ – Milind R Jul 2 '16 at 18:56

This is assuming the earths magnetosphere is the minimum required to shield an object from cosmic rays which is, in fact, incorrect. Here is the what you need to figure out to calculate a satellite system to do what your proposing .. 1-what is the required minimum strength of a magnetic field so that it deflects virtually all cosmic rays and the solar wind. 2-how much power is needed to generate a field. 3-the amount of power solar collectors can generate at that distance or using radioisotope generators. 4- how many satellites will be needed to create an effective shield.

It should be noted that if Mars where to be colonized to make it habitable would require us to generate such a field as it has no internal Dynamo :3 the large hydron collider researchers should be able to answer this, check out their forums

On that note, I'm sure it's actually feasible if you can figure out a cheaper way to get into orbit then rockets.


The most feasible method by which the artificial magnetic field can be generated is through placement of hundreds (if not thousands) of magnet satellites into the orbit of the target moon or planet all of which revolve at certain distance and velocity.These kind of satellites may use solar energy as their main source of energy which in fact lowers the energy costs for their global constant function. However, manufacturing and launching of this huge number of satellites will definitely remain a matter of enormous costs and resources that for sure ought to be mulled over.

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    $\begingroup$ -1, there's no quantitative reasoning here. One could at least estimate a lower bound for solar panel surface area. $\endgroup$ – user12029 Jan 13 '14 at 3:10

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