How strong a magnetic field would be needed to deflect cosmic rays?

For example, lets imagine we wanted to protect the occupants of the International Space Station from cosmic rays. I assume we would need to create some kind of electromagnetic shield (or electrostatic shield). How powerful would these need to be?

  • 2
    $\begingroup$ Depends on the energy (more precisely the spectrum and the cutoff energy that you consider "safe") of the cosmic rays and the size of the field. For practical purposes in low Earth orbit astronauts are still sufficiently shielded by Earth's magnetic field. It's when you go significantly beyond the first few hundred miles altitude where things get problematic. $\endgroup$
    – CuriousOne
    Oct 27, 2015 at 0:43
  • $\begingroup$ The question is not about shielding. It is about DEFLECTION. $\endgroup$ Oct 27, 2015 at 13:12

1 Answer 1


Background Info
The first thing to do is to consider the relativistic gyrofrequency, given by: $$ \Omega_{cs} = \frac{ Z_{s} \ e \ B_{o} }{ \gamma \ m_{s} } $$ where $Z_{s}$ is the charge state of species $s$, $e$ is the fundamental charge, $B_{o}$ is the quasi-static magnetic field magnitude, $\gamma$ is the relativistic Lorentz factor, and $m_{s}$ is the mass of species $s$.

Next, we consider the gyroradius or Larmor radius is given by: $$ \rho_{cs} = \frac{ v_{\perp} }{ \Omega_{cs} } = \frac{ \gamma \ m_{s} \ v_{\perp} }{ Z_{s} \ e \ B_{o} } = \frac{ m_{s} \ c \ \sqrt{\gamma^{2} - 1} }{ Z_{s} \ e \ B_{o} } $$ where $v_{\perp}$ is the velocity of the charged particle orthogonal to $\mathbf{B}_{o}$ and $c$ is the speed of light in vacuum.

If we want to deflect a particle from its nearly ballistic trajectory, we need to make the gyroradius (hopefully much) smaller than the size of the region we want to protect.

For a 10 MeV electron, the gyroradius is ~88 km in a 400 nT field (i.e., typical field strengths at ~4-5 $R_{E}$ altitudes). The Earth's field is roughly 30,000 nT (it varies from equator to pole, but just use this to make things easy). Thus, a 10 MeV electron has a gyroradius of ~1.2 km, still much larger than the ISS.

If we kick up the fields to 1 T, then that same 10 MeV electron's gyroradius changes to ~35 mm (or ~$3.5 \times 10^{-5}$ km), which seems more reasonable. However, generating 1 T fields typically requires very massive magnets and protons will have a larger gyroradius by a factor of nearly 1800 at the same speed. The cost of launching things into space is very expensive and the price depends upon mass. So perhaps magnets are not the best option?

Generally, people on the ISS have small shelters in which they can take refuge during geomagnetic storms, but they are protected by mostly particles below about 10 MeV only. This is okay because below this is energy are the highest fluxes, thus the largest exposure.

Above 10 MeV, there is little to do and truthfully, there are reasons why you do not want to try to stop such particles. For instance, look up articles on linear energy transfer. At extremely high energies, the particle will actually impart very little energy to you as it passes through (though it will destroy everything in its path). Though I think that destroying a cell is better than damaging because the body will just kill and absorb the destroyed cell. Whereas repairs may lead to mistakes that can extrapolate to things like cancer (Note: I am not an oncologist or cell biologist, so I would recommend reading more on the subject yourself and verify my statements.).

Another method is to use materials with high amounts of hydrogen (e.g., polyethylene) as form of protection. Such materials tend to absorb charged particle radiation (and neutrons as well) quite well but do not necessarily have to be dense, therefore not massive. High $Z$ materials like lead, tungsten, and tantalum are massive and, thus, expensive to launch into space.

Therefore, the most cost effective and practical approach is the one that is currently used, which is basically using styrofoam.

Smaller Applications
On small instruments for spacecraft, such as on the Van Allen Probes, they use multiple layers of various materials like tungsten, aluminium, niobium, and tantalum. There are occasional uses for heavier metals like gold due to its stability.

Below you will find a list of energies, Lorentz factors, and corresponding speeds for a proton given the energies in the first column.

Energy [MeV] | Lorentz Factor | Speed [km/s]
  1.0000000  |   1.0010658    |   13830.070
  10.000000  |   1.0106579    |   43423.141
  100.00000  |   1.1065789    |   128369.78
  1000.0000  |   2.0657890    |   262326.09
  10000.000  |   11.657890    |   298687.48
  100000.00  |   107.57890    |   299779.51
  1000000.0  |   1066.7890    |   299792.33

The corresponding relativistic gyroradii of a proton in a 1 T, 100 T, and 1000 T magnetic field (assuming all kinetic energy is in a speed orthogonal to $\mathbf{B}_{o}$) are:

Energy [MeV] | p gyroradius | p gyroradius | p gyroradius
             |   [m, 1 T]   |  [m, 100 T]  | [m, 1000 T]
  1.0000000  |   0.1445355  |   0.0014454  |   0.0001445
  10.000000  |   0.4581554  |   0.0045816  |   0.0004582
  100.00000  |   1.4829707  |   0.0148297  |   0.0014830
  1000.0000  |   5.6573732  |   0.0565737  |   0.0056574
  10000.000  |   36.351669  |   0.3635167  |   0.0363517
  100000.00  |   336.67930  |   3.3667930  |   0.3366793
  1000000.0  |   3338.7693  |  33.3876930  |   3.3387693
  • $\begingroup$ @KyleKanos - These do not block the CRs, they just detect them so the observer has a "background" or "noise" level to subtract. They are usually located behind the main detector stack and are typically made of silicon (or some other semi-conductor) like the main solid state wafers. $\endgroup$ Oct 27, 2015 at 12:24
  • 1
    $\begingroup$ Hmm, (briefly) reading more on this shows I am indeed wrong on my understanding of how the ACD's "block" CRs. I'll delete my comment now :| $\endgroup$
    – Kyle Kanos
    Oct 27, 2015 at 12:41
  • $\begingroup$ No worries, it's a subtle issue and is even misunderstood by some experimentalists, unfortunately... $\endgroup$ Oct 27, 2015 at 12:44
  • $\begingroup$ The question is not about shielding. $\endgroup$ Oct 27, 2015 at 13:11
  • $\begingroup$ @TylerDurden - I know, but to create a magnetic field strong enough to deflect cosmic rays (generally limited to 10's to 100's of MeV) is completely impractical and unrealistic. My answer also ignored other effects such as the magnetic torquing that would occur from having such a large magnetic field source moving through the Earth's background field and effect on instrumentation/electronics. $\endgroup$ Oct 27, 2015 at 19:12

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