 Solution

Put the two coil centres at $z=\pm a/2$. By superposition, their axial magnetic field isBoth magnetic field contributions point the same way. Evaluation givesThe corresponding coefficients are $0.715542$ and $0.676777$. Each individual magnetic field falls as one moves away from its own coil; their sum is much flatter near the midpoint.

For a single coil, differentiating $(a^2+x^2{)}^{-3/2}$ twice produces $3(4x^2-a^2)(a^2+x^2{)}^{-7/2}$. At $x=a/2$ this vanishes. Thus the pair has $B^{\prime }(0)=0$ by reflection symmetry and $B^{\prime \prime }(0)=0$ by its chosen separation; all odd derivatives vanish there. In fact,This Helmholtz coil arrangement produces a nearly uniform central magnetic field, useful for controlled charged particle in a uniform magnetic field and magnetic field calibration. The small orbit used below lies within that central region.