Your assumption that neutron-nucleus scattering is primarily via magnetic interaction is not correct. Neutron-nucleus scattering is primarily from the strong interaction, which is (as its imaginative name suggests) stronger than electromagnetism. While neutrons can interact by scattering magnetically from electrons, as currently suggested by another answer, the neutron-electron interaction is much smaller than the neutron-nuclear interaction.
Most nuclei have a “neutron separation energy” of order $\sim 8\rm\,MeV$: if a free neutron were to settle down and be trapped inside of a nucleus, that’s the amount of energy it would have to release. Nuclei are smaller than atoms, and the nuclear interaction is short-range. So to a free neutron, a crystal lattice looks more or less like empty space with a network of delta-function-like attractive potentials at the locations of the nuclei.
When you solve the Schrödinger equation for a square well,
$$
\left(\frac{\hat p^2}{2m} + V(\vec r)\right)\psi = E\psi
$$
you find that a free particle has more momentum (and therefore shorter wavelength) inside of the well than outside. For thermal-or-cold neutrons the difference is enormous: the free particle has milli-eV kinetic energy and angstrom-scale wavelength, while the piece of the wavefunction within the nucleus is mega-eV above the bottom of the well and has femtometer-scale wavelength.
If you go through the freshman-quantum exercise of matching the wavefunction at the boundary of the square well, so that it is continuous and smooth, with spherical symmetry, you find something interesting. For nearly all widths and depths of the square well, the free part of the wavefunction will have a zero-crossing somewhere near the nuclear radius. (I remember a good illustration of this in Golub’s textbook.) You can produce the same external wavefunction if, instead of a finely-tuned attractive well, you imagine an infinite, repulsive barrier whose radius is the location of this zero in the wavefunction. The radius of this made-up repulsive potential is called the “scattering length,” and the whole model is called the Fermi pseudopotential. The scattering length is directly related to the scattering cross section.
If you look at a table of scattering lengths and cross sections (like this one, which used to be available as a little booklet that fit in your pocket), you’ll see that all of the even-even nuclei (like $\rm{}^{12}C_6$) have zero incoherent scattering cross section, because even-even nuclei have no spin to exchange with the neutron. But all of the coherent scattering cross sections are a a few barns, because all of the coherent scattering lengths are a few femtometers. (There are a couple of low-mass nuclei where the scattering length is negative, because the zero in the wavefunction shows up on the wrong side of the origin, but even then the length scale is femtometers.)
Contrast this with the neutron-electron scattering length of $10^{-3}$ femtometers: the electrons are basically invisible, compared to the nuclei.
Even for spin-incoherent scattering, it’s not right to think of the neutron-nucleus interaction as magnetic. The energy scales are wrong. The nuclear magneton $e\hbar/2m_N$ is much smaller than the Bohr magneton $e\hbar/2m_e$, but incoherent neutron-nucleus scattering lengths are still mostly femtometer-ish. Incoherent neutron scattering comes from spin exchange mediated by the strong force.
So, that’s the answer to your title question. Neutron scattering from carbon-12 isn’t about magnetism at all. Neutron scattering from carbon-12 is about the neutron looking down into the abyss, wondering what its life would be like as part of a carbon-13, deciding that there is no place for it there, but having its long-distance behavior changed by the experience.
