# Can it be realistic to build a artificial magnetic field around Mars? [duplicate]

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?

• 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 – uhoh Feb 16 at 9:18
• @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. – uhoh Feb 16 at 9:54
• fixed title as well... – uhoh Feb 16 at 10:32
• Similar question suggesting a better approach than a loop around the equator: space.stackexchange.com/questions/13022/… – Hobbes Feb 16 at 10:36

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.
• 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. – peterh says reinstate Monica Feb 16 at 12:03