The velocity representation gives
$$
u_\phi-r\omega=\frac{kB_\phi}{\rho},
\qquad
u_p=\frac{kB_p}{\rho}.
$$
Therefore
$$
\boxed{R=\frac{|u_\phi-r\omega|}{u_p}
=\frac{|B_\phi|}{B_p}}.
$$
In the launching region, $r\simeq r_0\ll r_a$ and $A\ll1$, so part (d) gives
$$
|B_\phi|\simeq\frac{\mu_0k\Omega_0r_a^2}{r_0}.
$$
Consequently
$$
\boxed{R\simeq\widetilde R
=\frac{r_a^2\Omega_0\mu_0k}{r_0B_p}}.
$$

The modified Bernoulli kinetic energy is
$$
\frac12u_p^2+\frac12(u_\phi-r\omega)^2
=\frac12(1+R^2)u_p^2.
$$
Thus a decrease in $\Phi_{cg}$ is shared between poloidal acceleration and relative toroidal motion. If $R$ is approximately constant, the same potential drop produces a poloidal kinetic increase smaller by the factor $1/(1+R^2)$ than the estimate that neglects the toroidal term. If $R$ varies strongly, that term can absorb or return energy, so monotonic decrease of $\Phi_{cg}$ alone no longer proves an equally direct increase of $u_p^2/2$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 314 / 4 / f / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-314/4/f/solution