# Could a magnet pull oxygen out of the air?

I read that the $\ce{O2}$ molecule is paramagnetic, so I'm wondering: could a strong magnet pull the $\ce{O2}$ to one part of a room – enough to cause breathing problems for the organisms in the room?

(I'm not a professional chemist, though I took some college chemistry.)

• I'd expect there to be a slight increase in concentration of oxygen gas in the surroundings of a strong magnet, though I'd be really curious to see if anyone can provide some quantification. – Nicolau Saker Neto Jul 15 '15 at 0:20
• Even MRIs, weak as they are, are powerful enough to cause vertigo. I don't think an organism in a magnetic field that powerful would be very happy. – Brian Gordon Jul 15 '15 at 1:11
• Here's a video in case anyone wants clear visual proof that oxygen molecules are attracted to magnets. It also shows the lovely pale blue of liquid oxygen. – Nicolau Saker Neto Jul 15 '15 at 21:04
• I'm pretty sure osmotic pressure will step in and fix the problem long before it gets unbreathable. – Joshua Jul 16 '15 at 4:26
• Its not quite what you asked, but it is a common lecture demonstration that liquid oxygen in a suspended test-tube is easily pulled to one side by a small magnet. Liquid nitrogen, as expected is unaffected. – porphyrin Jul 1 '16 at 7:09

I'm a physicist, so apologies if the answer below is in a foreign language; but this was too interesting of a problem to pass up. I'm going to focus on a particular question: If we have oxygen and nothing else in a box, how strong does the magnetic field need to be to concentrate the gas in a region? The TL;DR is that thermal effects are going to make this idea basically impossible.

The force on a magnetic dipole $\vec{m}$ is $\vec{F} = \vec{\nabla}(\vec{m} \cdot \vec{B})$, where $\vec{B}$ is the magnetic field. Let us assume that the dipole moment of the oxygen molecule is proportional to the magnetic field at that point: $\vec{m} = \alpha \vec{B}$, where $\alpha$ is what we might call the "molecular magnetic susceptibility." Then we have $\vec{F} = \vec{\nabla}(\alpha \vec{B} \cdot \vec{B})$. But potential energy is given by $\vec{F} = - \vec{\nabla} U$; which implies that an oxygen molecule moving in a magnetic field acts as though it has a potential energy $U(\vec{r}) = - \alpha B^2$.

Now, if we're talking about a sample of gas at a temperature $T$, then the density of the oxygen molecules in equilibrium will be proportional to the Boltzmann factor: $$\rho(\vec{r}) \propto \mathrm e^{-U(\vec{r})/kT} = \mathrm e^{-\alpha B^2/kT}$$ In the limit where $kT \gg \alpha B^2$, this exponent will be close to zero, and the density will not vary significantly from point to point in the sample. To get a significant difference in the density of oxygen from point to point, we have to have $\alpha B^2 \gtrsim kT$; in other words, the magnetic potential energy must comparable to (or greater than) the thermal energy of the molecules, or otherwise random thermal motions will cause the oxygen to diffuse out of the region of higher magnetic field.

So how high does this have to be? The $\alpha$ we have defined above is approximately related to the molar magnetic susceptibility by $\chi_\text{mol} \approx \mu_0 N_\mathrm A \alpha$; and so we have1 $$\chi_\text{mol} B^2 \gtrsim \mu_0 RT$$ and so we must have $$B \gtrsim \sqrt{\frac{\mu_0 R T}{\chi_\text{mol}}}.$$ If you believe Wikipedia, the molar susceptibility of oxygen gas is $4.3 \times 10^{-8}\ \text{m}^3/\text{mol}$; and plugging in the numbers, we get a requirement for a magnetic field of $$B \gtrsim 258\ \text{T}.$$ This is over five times stronger than the strongest continuous magnetic fields ever produced, and 25–100 times stronger than most MRI machines. Even at 91 kelvin (just above the boiling point of oxygen), you would need a magnetic field of almost 150 T; still well out of range.

1 I'm making an assumption here that the gas is sufficiently diffuse that we can ignore the magnetic interactions between the molecules. A better approximation could be found by using a magnetic analog of the Clausius-Mossotti relation; and if the gas gets sufficiently dense, then all bets are off.

• Interesting calculations but take a look at the patent in ToddMinehart's answer – bon Jul 15 '15 at 20:57
• The patent claims to yield a difference in density of 50% (30% vs 20% oxygen), which would correspond to a Boltzmann factor of 1.5, or $\chi_\text{mol} B^2/(\mu_0 RT) = \ln (1.5) \approx 0.4$. That'll bring the necessary $B$ down to $\sqrt{0.4} \approx$ 63% of the threshold value above, which is still pretty high. And as was pointed out in the comments of @ToddMinehart's answer, the patent office doesn't certify that the ideas will actually work with current technology... – Michael Seifert Jul 15 '15 at 21:10

Physics says yes. And at the US Patent Office, it certainly looks like the answer is yes, according to (at least) this US patent application. The abstract reads:

A process for separating O$_2$ from air, that includes the steps effecting an increase in pressure of an air stream, magnetically concentrating O$_2$ in one portion of the pressurized air stream, the one portion then being an oxygen rich stream, and there being another portion of the air stream being an oxygen lean stream, compressing the oxygen rich stream and removing water and carbon dioxide therefrom, to provide a resultant stream, and cryogenically separating said resultant stream into a concentrated oxygen stream and a waste stream.

I would say the yes is conditional in that you'd probably need a room that is sealed from communication with the atmosphere (otherwise equilibrium will be re-established with respect to oxygen pretty quickly), is vented to the outside in order to dispose of the oxygen, houses one heck of a strong magnet, and has a strongly-locked door to prevent the unfortunate organisms from leaving.

• Once you have a room sealed from communication with the atmosphere and with a strongly-locked door, the unfortunate organisms will cause their own breathing problems eventually ;-) – Steve Jessop Jul 15 '15 at 2:11
• The patent office don't check whether inventions actually work. – Stop Harming Monica Jul 15 '15 at 16:54
• What if you just walk to where the magnet is? Unless it's a REALLY strong magnet that can pull the $O_2$ to a height that the organisms can't reach (in which case organisms like frogs might experience other side effects) chances are you could just stand next to the magnet. Of course your life span depends upon how quickly the $O_2$ is removed from the room and whether or not you can get help within that time limit. – Arc676 Jul 16 '15 at 13:39
• Even if this works it is likely that the only way to achieve notable changes in concentration would involve a multi-step cascade involving thousands of steps (a little like the gas centrifuge process for concentrating specific isotopes of uranium using gas diffusion of UF6). Not, therefore, practical or cheap. – matt_black Nov 26 '16 at 11:30

No, to separate oxygen from air you need extremely high gradients of field strength. in this paper http://link.springer.com/article/10.1007%2Fs11630-007-0079-1#page-1 they used about 0.4T per mm, but to cause breathing problems it is enough to get the O2 concentration below 17%, so lets say 0.1T/mm is needed. To sustain such gradients over a whole room (lets say 4 meters), the field strength on one side of the room needs to bee around 400T, which is very likely to kill humans (dissolved ions begin to circle due to lorentz force instead of going where they should, moving around induces huge currents…), and has never been achieved in a continous field. (from nationalmaglab.com: …45-tesla hybrid magnet, which offers scientists the strongest continuous magnetic field in the world…)

• Doesn't mean its not possible. Just means that other much bigger effects occur. physics.org/facts/frog-really.asp – Aron Jul 15 '15 at 10:24
• I totally forgot about that effect, so in addition to all the things mentioned above everyone in the room woulb be crushed against the wall… – Christian Vögl Jul 15 '15 at 10:31