Controlled Thermonuclear Fusion — Question 6

Section 43-7 · Physics pages 7282–7283

Question 6 Plasma beta compares particle pressure with magnetic energy density A quasineutral D-T plasma has ion and electron densities each 1020 m−3, with kT=10.0 keV for both. Let B=5.00 T, µ0 = 4π × 10−7 SI and 1 keV = 1.602 × 10−16 J. Define β = p/[B2/(2µ0)]. pressure p B B Schematic; labels, not drawing lengths, specify the data. (a) Calculate total particle pressure and beta. (b) At fixed density and temperature, find the change in beta if B increases by 20 percent.
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Question 6 - Solution Plasma beta compares particle pressure with magnetic energy density pressure p B B (a) Method and calculation p = (ni + ne)kT = 2(1020)(10)(1.602 × 10−16) = 3.204 × 105 Pa . β = 2µ0p B2 = .0322101 = 3.22101% . (b) Interpretation and further calculation β′/β = (B/B′)2 = 1/(1.20)2 = .694444 . β′ = .0223681 = 2.23681% . Physical check Beta is a pressure ratio, not by itself a stability criterion. Geometry and pressure profiles matter in determining allowable equilibria.

Question 6

Plasma beta compares particle pressure with magnetic energy density

A quasineutral D-T plasma has ion and electron densities each 1020m−310^{20}\,\mathrm{m^{-3}}, with kT=10.0 keV for both. Let B=5.00 T, μ0=4π×10−7\mu_0=4\pi\times10^{-7} SI and 1keV=1.602×10−161\,\mathrm{keV}=1.602\times10^{-16} J. Define β=p/[B2/(2μ0)]\beta=p/[B^2/(2\mu_0)].

Diagram shown in the original SVG worksheet.

  1. Calculate total particle pressure and beta.

  2. At fixed density and temperature, find the change in beta if B increases by 20 percent.

Question 6 - Solution

Plasma beta compares particle pressure with magnetic energy density

Diagram shown in the original SVG worksheet.

(a) Method and calculation

p=(ni+ne)kT=2(1020)(10)(1.602×10−16)=3.204×105Pa.\boxed{p=(n_i+n_e)kT=2(10^{20})(10)(1.602\times10^{-16})=3.204\times10^5\,\mathrm{Pa}}.β=2μ0pB2=.0322101=3.22101%.\boxed{\beta=\frac{2\mu_0p}{B^2}=.0322101=3.22101\%}.

(b) Interpretation and further calculation

β′/β=(B/B′)2=1/(1.20)2=.694444.\boxed{\beta\prime/\beta=(B/B\prime)^2=1/(1.20)^2=.694444}.β′=.0223681=2.23681%.\boxed{\beta\prime=.0223681=2.23681\%}.

Physical check

Beta is a pressure ratio, not by itself a stability criterion. Geometry and pressure profiles matter in determining allowable equilibria.