I think the matter would be made more intuitive by considering radial and tangential components. You’d have the following (first component: radial, second component: tangential, third component: binormal): $$\mathbf{x} = (r,0,0)$$ $$\mathbf{p} = (0,m c \alpha,0)$$ Combining the two we get: $$\mathbf{N} = (mcr\sqrt{1+\alpha^2},mc^2\alpha t,0)$$ Since $r = a_0$, this can be seen as (at least in cgs-Gaussian units) $$\mathbf{N} = \left(\frac{\hbar}{\alpha} \sqrt{1+\alpha^2},\frac{\hbar c}{a_0} t,0\right) = \left(\hbar \sqrt{1+\frac{1}{\alpha^2}},\frac{\hbar c}{a_0} t,0\right)$$ in SI units you might have some extra $4\pi\epsilon_0$ terms popping up, but the juicy bits shouldn’t change.
Now, what I can gather from here is that the radial component is roughly $137 \hbar$, but indeed the tangential component keeps on growing (although it might be interesting to note that the factor multiplying time would be the energy of a photon with wavelength equal to the Bohr radius?).
One last thing I can infer is that its time derivative is constant, but I don’t know what, if any, physical meaning such a quantity would have. $$\frac{\mathrm{d}\mathbf{N}}{\mathrm{d}t} = \left(0,\frac{\hbar c}{a_0},0\right)$$