= Solution
For a steady thin-wire <electric current>, the <Biot-Savart law> is
$$
d\mathbf B=\frac{\mu_0I}{4\pi}\frac{d\boldsymbol\ell\times\mathbf R}{R^3},
$$
where $\mathbf R$ points from the source element to the field point and $d\boldsymbol\ell$ follows positive <electric current>. On a circular loop's axis the transverse contributions cancel by <symmetry>. The axial numerator is $a^2d\varphi$ and the distance is $\sqrt{a^2+z^2}$, giving
$$
B_z(z)=\frac{\mu_0I}{4\pi}\int_0^{2\pi}\frac{a^2\,d\varphi}{(a^2+z^2)^{3/2}},\qquad \boxed{B_z(z)=\frac{\mu_0I a^2}{2(a^2+z^2)^{3/2}}.}
$$
Its graph is positive and even, with maximum $\mu_0I/(2a)$ 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 gives
$$
B_{\rm ax}(z)=\frac{\mu_0NIa^2}{2}\left[\frac1{(a^2+(z-D)^2)^{3/2}}+\frac1{(a^2+(z+D)^2)^{3/2}}\right].
$$
Consequently
$$
\boxed{B_{\rm ax}(0)=\frac{\mu_0NIa^2}{(a^2+D^2)^{3/2}}.}
$$
This is an even differentiable function, so every odd derivative at zero vanishes, including the first and third. To check the second derivative explicitly, let $f(s)=(a^2+s^2)^{-3/2}$. Then
$$
f''(s)=\frac{3(4s^2-a^2)}{(a^2+s^2)^{7/2}},\qquad B_{\rm ax}''(0)=\frac{3\mu_0NIa^2(4D^2-a^2)}{(a^2+D^2)^{7/2}}.
$$
Thus \b[$2D=a$ makes the quadratic variation vanish], which is the <Helmholtz coil> condition. The central value and leading axial variation are
$$
\boxed{B_0=\left(\frac45\right)^{3/2}\frac{\mu_0NI}{a},\qquad \frac{B_{\rm ax}(z)}{B_0}=1-\frac{144}{125}\left(\frac za\right)^4+O((z/a)^6).}
$$
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-helmholtz-profile.png]
{title=An original comparison of axial field uniformity for several coil separations}
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 $\nabla\cdot\mathbf B=0$ imply
$$
B_r(r,z)=-\frac r2B_{\rm ax}'(z)+O(r^3),\qquad B_z(r,z)=B_{\rm ax}(z)-\frac{r^2}{4}B_{\rm ax}''(z)+O(r^4).
$$
The 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.
Back to article page
