How does the divergence of the B field is zero imply that neutrons only "see" spin components perpendicular to the scattering vector? The following statement is made in a text with no further explanation:
"...a direct consequence of Maxwell's Law $\nabla \cdot B = 0$ that implies $B(q)$ has no component parallel to $q.$  This means that neutrons only 'see' spin components perpendicular to the scattering vector Q = q."
Here $B(q)$ is the Fourier transform of $B(r)$ and $q$.
I know one intuitive explanation for this - the effect of the magnetic field is always perpendicular to its direction, so a neutron deflected by a magnetic field was deflected with a scattering vector perpendicular to the field.  But I do not know how to connect the fact that the magnetic field is divergenceless to what is "seen" by a scattered neutron.
Any help is appreciated, thanks!
 A: So the magnetic field is given, in term of its momentum space representation (Fourier transform):
$\vec{B}(x) = \int{d^3q \tilde{B}(q)e^{i\vec{q}\cdot\vec{x}}} $
so that:
$\nabla \vec{B}(x) = \int{d^3q (\tilde{B}(q)\cdot \vec{q}) e^{i\vec{q}\cdot\vec{x}}} $ = 0
where the gradient operates only on the exponential (the only function of $x$, pulling out a vector $\vec{q}$).
Hence:
$\tilde{B}(q)\cdot \vec{q} = 0 $
must be true for the integral to zero generally.
Now identify $\vec{q} =\vec{p}_{\rm final} - \vec{p}_{\rm intital} = \vec{Q} $ as the momentum transfer in the scattering.
The confusion regarding the  statement "the neutron only sees spin components perpendicular to the magnetic field" arises from the fact that it is true in momentum space. When you scatter with momentum transfer $\vec{Q}$, you only "see" components of the field with spatial frequency $1/||Q||$--that's the whole point of form factors and considering scattering in momentum space. The additional constraint provided by Gauss's Law for Magnetism, mean you're only sensitive to fields components (in momentum space, with the right spatial frequency), perpendicular to $ \hat{Q} $. 
Finally, for 'contact' scattering, $\vec{B}$ is parallel to the magnetic moment, which is of course aligned with the spin. After all that, maybe the statement in question becomes clearer.
A: 
Here B(q) is the Fourier transform of B(r) and q.

It looks like they are just linearizing the problem, which means one assumes any quantity can be expressed as $Q \sim e^{i \ \mathbf{q} \cdot \mathbf{r}}$.  Then one can easily see that $\nabla \rightarrow i \ \mathbf{q}$.

But I do not know how to connect the fact that the magnetic field is divergenceless to what is "seen" by a scattered neutron.

The linearization I mentioned above leads to the following approximation:
$$
\nabla \cdot \mathbf{B} \simeq i \ \mathbf{q} \cdot \mathbf{B} \simeq 0
$$
This is another way of saying that $\mathbf{q}$ must be orthogonal to $\mathbf{B}$, thus no neutron can be scattered parallel to $\mathbf{B}$.
