= Solution
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.
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