At the inner edge, the imposed torque is
$$
\Theta=\mathcal G(r_{\rm in})
=\dot M\sqrt{GM}\left(\sqrt{r_{\rm out}}-\sqrt{r_{\rm in}}\right).
$$
Therefore
$$
\boxed{r_{\rm out}=
\left(\sqrt{r_{\rm in}}+
\frac{\Theta}{\dot M\sqrt{GM}}\right)^2}.
$$
The angular-momentum conservation law says that $\mathcal Fh+\mathcal G$ is constant. The magnetic process supplies angular momentum $\Theta$ at the inner boundary; viscous stress passes it outward, and the mass leaving at $r_{\rm out}$ carries the injected angular momentum together with the angular momentum $\dot Mh_{\rm in}$ that entered with the mass. A larger torque must therefore move the removal radius outward so that each unit mass can carry more <specific angular momentum>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 321 / 1 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-321/1/b/ii/solution