# Can a magnet rip protons from a nucleus?

Pretty self explanatory. I’m wondering if the strong nuclear force could be overcome by a strong enough magnet?

• Magnetic fields don't create forces on stationary charges. Commented Sep 9, 2019 at 2:20
• @BowlOfRed they can create forces on stationary magnetic dipoles though – which includes charged particles with nonzero spin. The Stern-Gerlach experiment essentially uses this effect. Commented Sep 9, 2019 at 13:08
• Did you try to do a back-of-an-envelope-calculation? Commented Sep 9, 2019 at 18:01
• Three protons and your magnet is a Philosopher's Stone: Pb to Au!
– MTA
Commented Sep 9, 2019 at 22:47
• If I simply use the most basic estimate for magnetic energy on a dipole of $\mu B$, then for a proton with $\mu\approx 10^{-3} \mu_B$, it would take a field of $10^{13}$ Tesla to start becoming comparable to nuclear binding energies of a few MeV. For comparison, the field from magnetars are hundreds of times smaller than that. Commented Sep 10, 2019 at 7:41

Protons and neutrons are in orbitals within the nucleus which have angular momentum , so the statement of "stationary charges" is true only to first order.

The magnetic fields in laboratory experiments are not strong enough to induce a proton or a neutron to exit the nucleus.

In astronomical observations, neutron stars and magnetars are studied and there the magnetic fields are strong enough to change the shape of an atom and affect the nucleus of atoms.

For nuclei in the iron region of the nuclear chart it is found that fields in the order of magnitude of $$10^{17}G$$ significantly affect bulk properties like masses and radii.

It is possible that if stronger astrophysical fields exist, the nucleus may break apart due to the magnetic field. This is studied in astrophysics as the "Coulomb breakup" of the nucleus, for example here.

• I guess the question boils down to know whether it's possible to create a strong field enough before the B field density is so strong that it creates a black hole. The B field upper limit should be easily calculable I guess. Commented Sep 10, 2019 at 8:06

Maybe. The action of a magnetic field would change the nuclear orbitals if it were strong enough, and would determine a new region of stability for the nucleus. A 'destabilized' nucleus could then decay in a variety of ways, including spontaneous fission.

Well, theoretically, it could. The magnetic field requirements for any such effect are EXTREMELY high, and any apparatus capable of generating such a field would also have its atomic (electron) orbitals disturbed so as to destabilize any and all of its material structure. It's not something I'd know how to build.

• I guess the question boils down to know whether it's possible to create a strong field enough before the B field density is so strong that it creates a black hole. The B field upper limit should be easily calculable I guess. Commented Sep 10, 2019 at 8:06
• I know where to find though. We call these magnatars. Commented Sep 10, 2019 at 15:41
• A magnetar, as I understand it, implements a kind of fusion, not fission (emitting a proton is not to be expected). It's a strong magnet, with no evidence of fission products in the vicinity. Commented Sep 10, 2019 at 19:03

One way this could happen is if you have a highly relativistic (very fast) nucleus travelling in a large magnetic field, such that in the nucleus' rest frame the magnetic field is partially transformed into an electric field. For a very rough calculation, I'll say that the force required to remove a proton from a nucleus is $$F_{strip} = 25 \textrm{ kN}$$ (based on the plot from the wikipedia page for the nuclear force, because my nuclear physics knowledge is nearly non-existent). Assuming the nucleus is travelling perpendicular to the magnetic field, the field magnitude is $$E = \gamma v B$$ in the nuclear rest frame, and we can just multiply by the proton charge to get the force, then substitute a value of $$B$$ representing the highest known magnetic fields in the universe, magnetars at around $$10^{11} \textrm{ T}$$:

$$$$F_{strip} = \frac{qvB}{\sqrt{1-v^2/c^2}}, \quad v^2 = \frac{\left(\frac{F_{strip}}{qB}\right)^2}{1+\left(\frac{F_{strip}}{qBc}\right)^2}$$$$

Doing these substitutions gives $$v/c = 0.999999981568$$, or a Lorentz factor of $$\gamma \approx 5200$$ which is surprisingly feasible.