= Solution
Put the two <coil> centres at $z=\pm a/2$. By <superposition>, their axial <magnetic field> is
$$
B(z)=\frac{\mu_0NIa^2}{2}\left\{\frac1{[a^2+(z-a/2)^2]^{3/2}}+\frac1{[a^2+(z+a/2)^2]^{3/2}}\right\}.
$$
Both <magnetic field> contributions point the same way. Evaluation gives
$$
\boxed{B(0)=\left(\frac45\right)^{3/2}\frac{\mu_0NI}{a},\qquad
B(\pm a/2)=\frac{\mu_0NI}{2a}\left(1+\frac1{2\sqrt2}\right).}
$$
The 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 $B'(0)=0$ by reflection <symmetry> and $B''(0)=0$ by its chosen separation; all odd <derivatives> vanish there. In fact,
$$
\frac{B(z)}{B(0)}=1-\frac{144}{125}\left(\frac za\right)^4+O\!\left((z/a)^6\right).
$$
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.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2019-helmholtz-field-and-electron-turn.png]
{title=Helmholtz axial magnetic field and a semicircular electron trajectory}
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