Take $v>0$ to mean the magnitude of the inward radial velocity. In the <Gaussian units> used for the <magnetic pressure>, approximate the <surface density of a disk> by $\Sigma\simeq2H\rho$. The <mass accretion rate> and <free-fall speed> then give
$$
\rho\simeq\frac{\dot M}{4\pi rHv}=\frac{\dot M}{4\pi\epsilon r^2v},\qquad
v=\left(\frac{2GM}{r}\right)^{1/2}.
$$
At the <magnetospheric truncation radius>, balance the <magnetic pressure> of the <magnetic dipole field> with the specified <ram pressure>:
$$
\frac{\mu^2}{8\pi r_m^6}\simeq\frac12\rho v^2\simeq\frac{\dot M v}{8\pi\epsilon r_m^2}.
$$
Thus
$$
r_m^{7/2}\simeq\frac{\epsilon\mu^2}{\dot M\sqrt{2GM}},\qquad
r_m\simeq\left(\frac{\epsilon^2\mu^4}{2GM\dot M^2}\right)^{1/7}.
$$
Discarding the convention-dependent numerical factor gives
$$
\boxed{r_m\sim\left(\frac{\epsilon^2\mu^4}{GM\dot M^2}\right)^{1/7}.}
$$
The <magnetic pressure> grows inward as $r^{-6}$, faster than the <ram pressure> $\propto r^{-5/2}$, so the field disrupts the flow inside this radius. The numerical coefficient depends on the vertical averaging and on treating the inflow as radial <free fall>, as stipulated here; a detailed <thin disk> boundary model need not have the same coefficient.

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 321 / 1 / b / Solution

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 Incoming links: Gaussian units