 Solution

For a steady thin-wire electric current, the Biot-Savart law iswhere $\mathbf{R}$ points from the source element to the field point and $d\mathbf{\ell }$ follows positive electric current. On a circular loop's axis the transverse contributions cancel by symmetry. The axial numerator is $a^{2}d\phi$ and the distance is $\sqrt{a^2+z^2}$, givingIts graph is positive and even, with maximum ${\mu }_{0}I/(2a)$ at the centre and a $|z{|}^{-3}$ far-field decay. Reversing the electric current reverses its sign.

For two equally oriented $N$-turn coils centred at $\pm D$, superposition givesConsequentlyThis is an even differentiable function, so every odd derivative at zero vanishes, including the first and third. To check the second derivative explicitly, let $f(s)=(a^2+s^2{)}^{-3/2}$. ThenThus $2D=a$ makes the quadratic variation vanish, which is the Helmholtz coil condition. The central value and leading axial variation are

Near the centre, the coils' magnetic field lines are nearly straight and parallel to the axis. On the axis rotational symmetry makes $B_x=B_y=0$ exactly; in the central plane the radial components from the two coils cancel. More quantitatively, in the current-free central volume, axisymmetry and $\nabla \cdot \mathbf{B}=0$ implyThe suppressed axial derivatives therefore also suppress transverse magnetic fields near the centre. The smallness applies locally compared with the coil radius; it is not a claim of an exactly uniform field throughout the apparatus.