Cyclotrons and Synchrotrons — Question 10

Section 28-6 · Physics pages 4590–4591

Question 10 The synchronous RF phase must supply both planned acceleration and per-turn losses An ion of positive charge q receives energy qV0 sin ϕ per turn from an RF system and loses energy Wloss per turn. The desired net energy increase is G per turn. Treat both G and Wloss as prescribed constants. ϕ RF gain G + Wloss π Schematic; labels, not drawing lengths, specify the data. (a) Derive the phase condition and the feasibility bound on G. (b) For qV0=10.0 keV, Wloss=2.00 keV and G=3.00 keV, find all phases between zero and pi that satisfy energy balance. State what the energy equation alone says about phase stability.
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Question 10 - Solution The synchronous RF phase must supply both planned acceleration and per-turn losses ϕ RF gain G + Wloss π (a) Method and calculation ∆K = qV0 sin ϕ −Wloss = G ⇒sin ϕ = G + Wloss qV0 . 0 ≤G + Wloss ≤qV0 for 0 ≤ϕ ≤π. (b) Interpretation and further calculation sin ϕ = (3 + 2)/10 = 1/2 ⇒ϕ = π/6 or 5π/6. d(∆K) dϕ = qV0 cos ϕ has opposite signs at the two phases. Stability additionally requires the relation between an energy error and the next arrival-phase error. Physical check An equilibrium energy balance identifies candidate synchronous phases but does not by itself establish stable longitudinal motion.

Question 10

The synchronous RF phase must supply both planned acceleration and per-turn losses

An ion of positive charge q receives energy qV0sin⁡ϕqV_0\sin\phi per turn from an RF system and loses energy Wloss per turn. The desired net energy increase is G per turn. Treat both G and Wloss as prescribed constants.

Diagram shown in the original SVG worksheet.

  1. Derive the phase condition and the feasibility bound on G.

  2. For qV0=10.0 keV, Wloss=2.00 keV and G=3.00 keV, find all phases between zero and pi that satisfy energy balance. State what the energy equation alone says about phase stability.

Question 10 - Solution

The synchronous RF phase must supply both planned acceleration and per-turn losses

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔK=qV0sin⁡ϕ−Wloss=G⇒sin⁡ϕ=G+WlossqV0.\Delta K=qV_0\sin\phi-W_{\mathrm{loss}}=G\Rightarrow\sin\phi=\frac{G+W_{\mathrm{loss}}}{qV_0}.0≤G+Wloss≤qV0for 0≤ϕ≤π.0\le G+W_{\mathrm{loss}}\le qV_0\quad\text{for }0\le\phi\le\pi.

(b) Interpretation and further calculation

sin⁡ϕ=(3+2)/10=1/2⇒ϕ=π/6or5π/6.\sin\phi=(3+2)/10=1/2\Rightarrow\phi=\pi/6\quad\text{or}\quad5\pi/6.d(ΔK)dϕ=qV0cos⁡ϕhas opposite signs at the two phases.\frac{d(\Delta K)}{d\phi}=qV_0\cos\phi\quad\text{has opposite signs at the two phases}.Stability additionally requires the relation between an energy error and the next arrival-phase error.

Physical check

An equilibrium energy balance identifies candidate synchronous phases but does not by itself establish stable longitudinal motion.