# Could I really use bar magnets in space to protect myself from solar wind?

The magnitude at the magnetic equator is given by the approximation: $$\lvert \mathbf{B} \rvert \left( r \right) \approx B_{o} \left( \frac{R_{E}}{r} \right)^{3} \tag{1}$$ where $$R_{E}$$ is the Earth's radius, $$B_{o}$$ is roughly 31,200 nT (i.e., the average field magnitude at the Earth's surface near the magnetic equator), and $$r$$ is the distance from the center of Earth. As you can see, by the time you reach ~4 $$R_{E}$$ the magnetic field has dropped to ~490 nT.

How strong is Earth's magnetic field in space?

And this one claims the Earth's magnetic field already at the surface to be weaker then a bar magnet:

How can a refrigerator magnet be stronger than the Earth's magnetic field?

Now the Earth's magnetic field is supposed to protect us from solar wind.

There is an intrinsic magnetic field generated somehow in Earth's core (dynamo discussion could fill volumes) and that field interacts with the magnetic field and charged particles of the solar wind.

How does the Earth's magnetic field protect it from the solar wind?

Based on these, if I have strong enough magnets (based on the fact that they are stronger then Earth's magnetic field in space) with me in space, can I really protect myself from solar wind?

Question:

1. Could I really use bar magnets in space to protect myself from solar wind?
• The earth's magnetic field doesn't drop off as sharply with distance as your bar magnet. Go up 100 kilometers, and the Earth's magnetic field is still about as strong as it was at the surface. At a thousand, it's still over half as strong. Go one kilometer from any reasonable bar magnet, and you won't be able to detect its field. Sep 14, 2022 at 1:19

Yes, the magnetic field magnitude of Earth, which forms the magnetosphere, is small compared to say, a good bar magnet. However, the magnetic moment of Earth is on the order of $$\sim 8 \times 10^{22}$$ $$A \ m^{2}$$. For comparison, the magnetic moment of a bar magnet is on the order of 1 $$A \ m^{2}$$ (from Why is Earth's magnetic field not strong?).
The functional form of a magnetic dipole is given by: $$\mathbf{B}\left( \mathbf{x} \right) = \frac{ \mu_{o} }{ 4 \ \pi } \left[ \frac{ 3 \ \hat{\mathbf{n}} \left( \hat{\mathbf{n}} \cdot \mathbf{m} \right) - \mathbf{m} }{ \lvert \mathbf{x} \rvert^{3} } \right] \tag{0}$$ where $$\mathbf{B}$$ is the magnetic field [$$T$$], $$\mathbf{x}$$ is some 3-vector position [$$m$$] pointing from an origin to some observation point, $$\mu_{o}$$ is the permeability of free space [$$T \ m \ A^{-1}$$], $$\hat{\mathbf{n}}$$ is the unit vector parallel to $$\mathbf{x}$$, and $$\mathbf{m}$$ is the dipole magnetic moment [$$A \ m^{2}$$].
The $$B_{o}$$ from your Equation 1 is an approximation from $$\mu_{o} \ \lvert \mathbf{m} \rvert$$, with a few constants and normalization factors. The corresponding $$B_{o}$$ for a bar magnetic with distance normalized to one Earth radii, $$R_{E}$$, would be something like 20+ orders of magnitude smaller.