= Solution
Take $z$ positive upward, let sphere A's centre be $Z(t)\mathbf e_z$, and write its downward <velocity> as $-U\mathbf e_z$, with $\dot Z=-U$. At B's centre the displacement from A is $\mathbf r=\ell\mathbf e_x-Z\mathbf e_z$. The <vorticity of translating-sphere Stokes flow> therefore gives
$$
\boldsymbol\omega_\infty=-\frac{3aU\ell}{2(\ell^2+Z^2)^{3/2}}\mathbf e_y.
$$
B is held in translation but has zero applied <torque>. By <Faxén's rotational law>, it rotates with half the incident <vorticity>:
$$
\boldsymbol\Omega_B=-\frac{3aU\ell}{4(\ell^2+Z^2)^{3/2}}\mathbf e_y.
$$
Integrating the <rotation induced by a passing Stokes sphere> along the complete path cancels $U$:
$$
\Delta\boldsymbol\theta=\frac{3a\ell}{4}\mathbf e_y\int_{+\infty}^{-\infty}\frac{dZ}{(\ell^2+Z^2)^{3/2}}
=\boxed{-\frac{3a}{2\ell}\mathbf e_y.}
$$
Its magnitude is $3a/(2\ell)$ radians, with direction about the negative $y$ axis by the right-hand rule. All instantaneous rotations have this same axis, so the integrated <angular velocity> gives the net angle without a noncommuting-rotation correction. The result is independent of B's radius in the prescribed incident-flow calculation.
\Image[/past-exam-of-the-advanced-fluid-dynamics-course-of-the-university-of-oxford/2023-passing-sphere-rotation.png]
{title=Angular velocity and cumulative rotation from an isolated sphere passing a torque-free target}
The inequality $\ell>a+b$ ensures no overlap. It does not by itself make mutual <hydrodynamic interactions> negligible: the stated flow is the isolated-A solution, and a held finite B also disturbs A's flow. Thus the displayed angle is the result for that supplied incident flow, or the leading one-way approximation to separated bodies. A fully coupled two-sphere calculation uses the <method of reflections for Stokes flow> or an exact pair resistance solution.
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