= Solution
To calculate <magnetic-field cancellation of an arc and tangent leads> away from the plane, put $a^2=R^2+h^2$. Each point on the arc is distance $a$ from $(0,0,h)$, and the axial component of $d\mathbf l\times\mathbf s$ still has magnitude $R^2d\phi$. Thus its axial <magnetic field> is
$$
\boxed{B_{z,\rm arc}(h)=-\frac{\mu_0I\theta R^2}{4\pi(R^2+h^2)^{3/2}}.}
$$
For either tangent lead the axial numerator remains $R\,ds$, while the cubed distance is $(R^2+h^2+s^2)^{3/2}$. Hence
$$
\boxed{B_{z,\rm lead}(h)=\frac{\mu_0I}{4\pi}\int_0^\infty\frac{R\,ds}{(a^2+s^2)^{3/2}}=\frac{\mu_0IR}{4\pi(R^2+h^2)}.}
$$
Adding both leads gives
$$
\boxed{B_z(h)=\frac{\mu_0I}{4\pi R}\left[\frac2{1+(h/R)^2}-\frac{\theta}{[1+(h/R)^2]^{3/2}}\right].}
$$
For $\epsilon=h/R\ll1$, a <binomial series> yields
$$
B_z(h)=\frac{\mu_0I}{4\pi R}\left[(2-\theta)+\left(-2+\frac32\theta\right)\epsilon^2+O(\epsilon^4)\right].
$$
The exact condition for cancellation of the axial component is
$$
\theta(h)=2\sqrt{1+\epsilon^2},\qquad\boxed{\theta(h)=2+\frac{h^2}{R^2}+O\!\left(\frac{h^4}{R^4}\right)\quad\text{in radians}.}
$$
For example, keeping $\theta=2$ would leave a small positive axial <magnetic field> proportional to $h^2/R^2$. Both contributions weaken with height, but the arc factor is $(1+\epsilon^2)^{-3/2}$ whereas the lead factor is $(1+\epsilon^2)^{-1}$. The arc therefore loses a greater fraction of its axial <magnetic field>. Increasing its angle compensates for this loss. Physically, the arc is all at the same finite radius, whereas distant elements of the semi-infinite leads undergo a smaller fractional change in distance on moving a little out of the plane.
The cancellation here concerns $B_z$ only; off the plane it does not assert that every component of the <magnetic field> vanishes. The small-height result also stays within the tangent-lead geometry shown in the original paper.
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