= Solution
Write $f=1-2M/r$ in the <Schwarzschild metric>. For the stationary <Killing vector field> $V=\partial_t$, its dual is $V^\flat=-f\,dt$ and the <Papapetrou electromagnetic field> is
$$
F=-f'\,dr\wedge dt=\frac{2M}{r^2}\,dt\wedge dr.
$$
Outside the <Schwarzschild event horizon>, use the static <orthonormal coframe> $e^{\hat0}=\sqrt f\,dt$, $e^{\hat r}=dr/\sqrt f$, $e^{\hat\theta}=r\,d\theta$, $e^{\hat\phi}=r\sin\theta\,d\phi$. Then $F=(2M/r^2)e^{\hat0}\wedge e^{\hat r}$. With the <electric field> convention $E_{\hat i}=F_{\hat i\hat0}$ from the preceding part,
$$
\boxed{E_{\hat r}=-\frac{2M}{r^2},\qquad \mathbf B=0.}
$$
Thus the stationary <Killing vector field> gives a \b[Coulomb electric field], directed inward in this convention. If the charge is normalized by $E_{\hat r}=Q/r^2$, its charge is $Q=-2M$. A convention with $F\mapsto-F$ reverses that charge; the invariant content is a radial monopole <electric field> of magnitude $2M/r^2$.
For the axial <Killing vector field> $V=\partial_\phi$, $V^\flat=r^2\sin^2\theta\,d\phi$, giving
$$
F=2r\sin^2\theta\,dr\wedge d\phi+2r^2\sin\theta\cos\theta\,d\theta\wedge d\phi.
$$
In the same static <orthonormal coframe>, this is a purely <magnetic field> with
$$
\boxed{B_{\hat r}=2\cos\theta,\qquad B_{\hat\theta}=-2\sqrt f\sin\theta,\qquad B_{\hat\phi}=0.}
$$
At large $r$ these are the spherical components of $2\mathbf e_z$. This is therefore the \b[regular magnetic test field asymptotic to a uniform axial field], distorted by the <Schwarzschild black hole>. Its net magnetic monopole flux is zero because the integral of $\cos\theta$ over a sphere vanishes. Both <electromagnetic fields> are regular at the future <event horizon>: in <ingoing Eddington-Finkelstein coordinates>, the electric field is $(2M/r^2)dv\wedge dr$, and the magnetic two-form already contains only regular spatial coordinates. Static observers themselves cease to exist on the <event horizon>.
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