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I was wondering how realistic it is to build an artificial magnetic field around Mars using electromagnetism.

So my question is; what kind of current would be needed in a loop around Mars' equator to generate a magnetic field similar to that of Earth's?

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    $\begingroup$ Hi @朱軒德 and Welcome to Space! It'a good question and it's similar to several others asked here before. space.stackexchange.com/a/27539/12102 $\endgroup$
    – uhoh
    Feb 16, 2019 at 9:18
  • $\begingroup$ @NathanTuggy I think the question might be closed, but I'm not sure that's the right choice. The OP asked for a numerical value for a current (even before grammar adjustment) and I'm not sure that question provides anything along those lines. I left a link to a list of questions so that some time and care could be taken to ensure that a proper answer could be linked to if it existed. Really, that seems not to be a good choice at all. $\endgroup$
    – uhoh
    Feb 16, 2019 at 9:54
  • $\begingroup$ fixed title as well... $\endgroup$
    – uhoh
    Feb 16, 2019 at 10:32
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    $\begingroup$ Similar question suggesting a better approach than a loop around the equator: space.stackexchange.com/questions/13022/… $\endgroup$
    – Hobbes
    Feb 16, 2019 at 10:36

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This page gives a formula $$B = \frac{\mu_0 I}{2R}$$ for the field at the centre of a current carrying loop in vacuum. Earth's magnetic field is about $10^{-4}T$, the radius of Mars is $3.4\times 10^6 m$ and $\mu_0$ is $4\pi\times 10^{-7}$. Solving for $I$ we get about $5\times 10^8$ -- 500 million amps.

However, Mars is not a vacuum, and its core is iron. Depending on its purity and temperature (which are not, I think, very well known) that could increase the field by up to 1000-fold or more, so a relatively modest current of a few hundred thousand amps might suffice, or even no current once the core was magnetised.

All of which said, I am not convinced that a magnetic field is a major short-term consideration in terraforming Mars. In the very long term it might reduce loss of light gasses from the top of the atmosphere and steer away some kinds of radiation. It doesn't help with UV though, and a thicker atmosphere would deal with charged particles anyway.

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  • $\begingroup$ Side note: a thin layer of glass, what we have in any window on the Earth, filters practically all the UV. We don't need any more, than big glass houses. $\endgroup$
    – peterh
    Feb 16, 2019 at 11:56
  • $\begingroup$ Another side note: the current highest temperature superconductor is now $\rm -70^\circ \rm C$, although it requires $\approx$ 1000 bar for that. It is not unreachable in an environment where the highest temperature is $\approx 10 ^\circ C$. I suspect, having such a superconducting wire over the Martian Equator wouldn't cost much more than the Large Hadron Collider or a large continental oil pipe. $\endgroup$
    – peterh
    Feb 16, 2019 at 12:03
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    $\begingroup$ Wait, I don't think a solid iron core will boost the field in the atmosphere by "1000-fold or more". Can you support how that's going to work with an authoritative source, or if none is to be found, a thorough physical explanation? Imagine a 2 cm diameter iron ball and a 4 cm loop of wire, where exactly to you think the field will jump by a factor of 1000? $\endgroup$
    – uhoh
    Feb 16, 2019 at 12:59
  • $\begingroup$ @peterh Mars' circumference is 21,000 km. Projects similar in scope on Earth are the longest undersea fiberoptic cables (e.g. en.wikipedia.org/wiki/Fiber-Optic_Link_Around_the_Globe). $\endgroup$
    – Hobbes
    Feb 16, 2019 at 13:14
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    $\begingroup$ The iron core of Mars does not help much. Only a short part of the magnetic field lines will be within the iron core, but the longer part are in vacuum, just like the Earth's magnetic field. Even a very small air gap dominates the magnetic resistance. $\endgroup$
    – Uwe
    Feb 16, 2019 at 14:11

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