Controlled Thermonuclear Fusion — Question 7

Section 43-7 · Physics pages 7284–7285

Question 7 Magnetic-mirror reflection follows from the adiabatic invariant A particle moves from field B0 toward Bmax = 4B0. Assume no electric field, conserved kinetic energy and conserved magnetic moment µ = mv2 ⊥/(2B). Initial pitch angle θ0 is between velocity and magnetic field. s B Bm Schematic; labels, not drawing lengths, specify the data. (a) Derive the condition for reflection before reaching the maximum field. (b) Find the loss-cone half-angle and decide whether initial angles 20◦and 45◦reflect.
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Question 7 - Solution Magnetic-mirror reflection follows from the adiabatic invariant s B Bm (a) Method and calculation v2 ⊥ B = v2 sin2 θ0 B0 , v2 ∥= v2 −v2 ⊥. v∥= 0 ⇒Bm = B0/ sin2 θ0 ≤Bmax. (b) Interpretation and further calculation sin2 θ0 ≥B0/Bmax = 1/4, θlc = 30◦. 20◦: passes the mirror; 45◦: reflects . Physical check Equality gives a turning point at the field maximum. The invariant requires gradual field variation over an orbit and negligible scattering; the opposite end has a corresponding loss cone.

Question 7

Magnetic-mirror reflection follows from the adiabatic invariant

A particle moves from field B0B_0 toward Bmax=4B0B_{\max}=4B_0. Assume no electric field, conserved kinetic energy and conserved magnetic moment μ=mv⟂2/(2B)\mu=mv_\perp^2/(2B). Initial pitch angle θ0\theta_0 is between velocity and magnetic field.

Diagram shown in the original SVG worksheet.

  1. Derive the condition for reflection before reaching the maximum field.

  2. Find the loss-cone half-angle and decide whether initial angles 20∘20^\circ and 45∘45^\circ reflect.

Question 7 - Solution

Magnetic-mirror reflection follows from the adiabatic invariant

Diagram shown in the original SVG worksheet.

(a) Method and calculation

v⟂2B=v2sin⁡2θ0B0,v∥2=v2−v⟂2.\frac{v_\perp^2}{B}=\frac{v^2\sin^2\theta_0}{B_0},\quad v_\parallel^2=v^2-v_\perp^2.v∥=0⇒Bm=B0/sin⁡2θ0≤Bmax.v_\parallel=0\Rightarrow B_m=B_0/\sin^2\theta_0\le B_{\max}.

(b) Interpretation and further calculation

sin⁡2θ0≥B0/Bmax=1/4,θlc=30∘.\boxed{\sin^2\theta_0\ge B_0/B_{\max}=1/4,\quad\theta_{\mathrm{lc}}=30^\circ}.20∘: passes the mirror;45∘: reflects.\boxed{20^\circ:\text{ passes the mirror};\qquad45^\circ:\text{ reflects}}.

Physical check

Equality gives a turning point at the field maximum. The invariant requires gradual field variation over an orbit and negligible scattering; the opposite end has a corresponding loss cone.