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