= Magnetic-field cancellation of an arc and tangent leads
{title2=$\theta(h)=2\sqrt{1+(h/R)^2}$}
For a circular arc of radius $R$ carrying clockwise <electric current> $I$ and subtending angle $\theta$, with two tangent semi-infinite leads extending on the crossing side of the arc, the axial <magnetic field> at height $h$ above the centre is
$$
B_z(h)=\frac{\mu_0I}{4\pi R}\left[\frac2{1+(h/R)^2}-\frac{\theta}{[1+(h/R)^2]^{3/2}}\right].
$$
The crossing is insulated. The positive term is from the two leads and the negative term from the arc, with $+z$ out of the clockwise-current plane. The <Biot-Savart law> yields the lead integral $\int_0^\infty R(R^2+h^2+s^2)^{-3/2}\,ds=R/(R^2+h^2)$. Cancellation of the axial component gives the title formula, expanding to $\theta=2+h^2/R^2+O(h^4/R^4)$ in <radians>. This is cancellation of one component away from the plane, not necessarily the entire <magnetic field>.
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