# How much power would be needed to make a substitute for natural Earth magnetic field?

Imagine the Earth's magnetic field will take a "maintainance pause" for 1000 years and there is gigant 10 years "doomsday countdown timer" ticking in the sky. The magnetic field will be just normal for 9 years and begin to fade down linearly during the last year.

How much power whould be needed to provide artificial substitute for natural Earth field?

• What's the best rescue plan for this? Can we just make a giant [superconductive] coil spanning over equator?
• For how long the Earth will survive without the magnetic field?

/* I'm not sure whether this question suits better here or, for example, on Worldbuilding SE */

• Although (some of) your separate questions are fine, it's not a good idea to ask all of them at the same time. This renders the question too braod; this may change if you choose to either reword or rewrite (parts of) your question. – Danu Oct 10 '14 at 22:18
• @Danu, Chosen the main question, demoting others to "bonus questions". – Vi. Oct 10 '14 at 22:33
• Related question on Space Exploration. – HDE 226868 Oct 10 '14 at 23:33

There are two parts to this question (even when you cut out the bonus bits).

1. How much energy is stored in the earth's magnetic field (ramp up the magnet)
2. How much power to keep that field going (drive current through big loop)

The former is given by the $\frac12 L I^2$ - so we need to estimate the inductance of the coil needed and its current.

A single loop around the equator would have an inductance approximately

$$L = \mu_0 b \log(\frac{b}{a})$$

where b = radius of coil, and a = radius of wire. For a 1 cm thick wire we get

$$L = 4 \pi \cdot 10^{-7} \cdot 6.3 \cdot 10^6 \log(6.3 \cdot 10^8)\\ \approx 160 H$$

Now for a field of 0.5 Gauss, we would need

$$I = \frac{2\pi r B}{\mu_0} = \frac{4\cdot 10^7\cdot 0.5\cdot 10^{-4}}{4\pi\cdot 10^{-7}}1.6GA$$

Wow - Giga amperes. I may have to rethink that 1 cm copper wire... if it has about 0.2 ohms resistance per km (from resistivity of 17 nOhm meter), resistance of the loop is 4 kOhm. So power to keep the current flowing would be about $I^2 R = 1.5 E 22 W$. That's a bit steep. Let's increase the copper wire by 100,000 (making it a 10 square meter section) and drop that to a more reasonable (?) 1.5 E 17 W. Because that thin wire would need 100x more power per second than is used by the USA in a year... With a tip of the hat to @CuriousOne who noticed I had a few zeros missing.

But that's not yet estimating the power to ramp the magnetic field up... because that's given by $\frac12 L I^2$, so from the above requires an energy of

$$W = 0.5 \cdot 160 \cdot (1.6 \cdot 10^9)^2 = 2 \cdot 10^{20} J$$

Interestingly, according to Wolfram alpha that's almost exactly twice the total energy use of the USA per year. Better turn off that air conditioning unit and start saving for Armageddon.

That's a whole lot of power... somewhat more than David Hammen estimated in the similar question. And David knows a thing or two about these things, so I'm hoping he will find this and fix my mistake. You will need a very thick wire (or more turns) to keep the power dissipation to something that can be handled by, say, boiling the ocean. Actually, using the ocean as your conductor might just work - as long as you can prevent the current from short circuiting. Conductivity of sea water is about 20 million times worse than for copper, but suddenly it's not so hard to have a conductor with a cross section of $10 km^2$

Superconductors

The question was asked "how about using superconductors"? Here are a couple of thoughts.

First - superconductors have a critical current density above which they stop working. A typical value is $20 kA/cm^2$. At that value, you need 8 square meters of cross section to carry 1.6 GA (whether you do this as a single turn or multi turn), so the volume of conductor is $4\cdot 10^7 \cdot 8 = 3.2\cdot 10^8 m^3$. And you need to cool that volume of conductor to supercooling temperature (and then keep it there...). Thermodynamics is not your friend, and although heat capacity drops with temperature, the energy needed to get 1 J of heat from 4 K to 300 K is about 100 J (75 for "perfect" heat engine, but who has one of those). So getting that much superconductor might be a problem, and cooling it to liquid helium temperatures would be a big problem too... Oh - and keeping it cool: that would be a problem too.

As for the cost; according to http://large.stanford.edu/courses/2011/ph240/kumar1/docs/62-03.pdf cable carrying 200A costs \$20/m. So that's 10 cents per amp-meter. We need 1.6 GA x 40,000 km, which is 6 E 16 amp meter, or a cost of 6 E 15 US dollars. The fact that the US GDP is about 1.6E13 dollars means that this would be 400 years of total economic output - not counting the fact that there is just not that much material available.

I think we need another plan...

PS After thinking about this some more, I have come to the conclusion that I should not have ignored the presence of significant amounts of iron in the core of the earth - this changes the magnetic properties, as we no longer have an "air core" magnet. This probably reduces the steady state current requirements significantly but my EM is rusty when it comes to stored energy for that situation.

• Is there any chance with superconductors? – Vi. Oct 11 '14 at 1:10
• Will solar wind "stress" our magnet and require even more energy? – Vi. Oct 11 '14 at 1:11
• Cool answer! You may definitely want to go with MUCH thicker wire, because at 1.6Ga and 0.2 Ohm/km the electric voltage to drive this thing would be 320MV/km, at least two orders of magnitude more than what's "healthy". The force on the wire would be 80kN/m, right? Better bring a few strong ground anchors, too. PS: wouldn't the loop resistance be 8kOhm and the power required be 1e22W? Did you miss an I^2? – CuriousOne Oct 11 '14 at 1:34
• @Vi.: You got it. The only way to build this realistically would be with superconductors and the interaction with solar flares would be absolutely brutal. The system would have to be able to absorb enormous amounts of energy by compensating for induction voltages in the MV range. That said, by breaking up the system in many parallel windings, the requirements could potentially be achieved, but at what cost? – CuriousOne Oct 11 '14 at 1:39
• "at what cost" -> Two cost options: "cheap" = can be done with 9 years of united mankind flat-out best-efforts work; "expensive" = can't be done by today's humans; (One more cost option: more expensive than evacuating the mankind to other planet vs less expensive) – Vi. Oct 11 '14 at 1:58