Let's see if we can estimate this mutual inductance from first principles, the Neumann formula:
M = (μ
0/4π) ∫ d
l1.d
l2 / r
For simplicity take two coplanar square wire loops (one turn only) of side 'a' and distance 'd' between centers. By hand calculation I get
k = M/L ≈ a
2 / (π.d.(d+a))
For d = a (worst case) I get k = 1/2π = 0.16. If d = 2a then k = 0.05.
N turns should not matter as the N
2 factor would enter both M and L. Now that we know the spatial dependence we can sense that our estimate will upper bound the k for a spiral. It's a pessimistic upper bound, though. I'm guessing for d = 2a, a typical spiral would have k = 0.01.
Bottom line: if d = a (and it would be, right?) then you can't ignore mutual.
M.G.Rajan
www.eecalc.com