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

Solution

The velocity representation gives

u_{ϕ} - r ω = \frac{k B _{ϕ}}{ρ}, u_{p} = \frac{k B _{p}}{ρ} .

Therefore

R = \frac{∣ u _{ϕ} - r ω ∣}{u _{p}} = \frac{∣ B _{ϕ} ∣}{B _{p}} .

In the launching region,

r ≃ r_{0} ≪ r_{a}

and

A ≪ 1

, so part (d) gives

∣ B_{ϕ} ∣ ≃ \frac{μ _{0} k Ω _{0} r _{a}^{2}}{r _{0}} .

Consequently

R ≃ R = \frac{r _{a}^{2} Ω _{0} μ _{0} k}{r _{0} B _{p}} .

The modified Bernoulli kinetic energy is

\frac{1}{2} u_{p}^{2} + \frac{1}{2} (u_{ϕ} - r ω)^{2} = \frac{1}{2} (1 + R^{2}) u_{p}^{2} .

Thus a decrease in

Φ_{c g}

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

Φ_{c g}

alone no longer proves an equally direct increase of

u_{p}^{2} /2

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