Magnetic-field cancellation of an arc and tangent leads ($\theta (h)=2\sqrt{1+(h/R{)}^2}$)

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 isThe 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.