1629373397-425d4de58b7aea127ffc7c337418ea8d (846389), страница 69
Текст из файла (страница 69)
The resistivity of A relative to that of B is(A)(B)(C)(D)(E)100 times smaller10 times smallerapproximately the same10 times larger100 times larger9. The average electron velocity jvj in a 10-keV Maxwellian plasma is(A)(B)(C)(D)(E)7 102 m/s7 104 m/s7 105 m/s7 106 m/s7 107 m/s10. Which of the following waves cannot propagate when B0 = 0?(A) electron plasma wave(B) the ordinary waveAppendix C: Sample Three-Hour Final Exam425(C) Alfvén wave(D) ion acoustic wave(E) Bohm–Gross wave11. A “backward wave” is one which has(A)(B)(C)(D)(E)k opposite to B0ω/k < 0dω/dk < 0vi = vevϕ opposite to vg12.
“Cutoff” and “resonance,” respectively, refer to conditions when the dielectricconstant is(A)(B)(C)(D)(E)0 and 11 and 00 and 11 and 0not calculable from the plasma approximation13. The lower and upper hybrid frequencies are, respectively,(A) (ΩpΩc)1/2 and (ωpωc)1/21=21=2and ω2p þ ω2c(B) Ω2p þ Ω2c1=2(C) (ωcΩc)1/2 and ω2p þ ω2c1=21=2(D) ω2p ω2cand ω2p þ ω2c(E) (ωRωL)1/2 and (ωpωc)1/214. In a fully ionized plasma, diffusion across B is mainly due to(A)(B)(C)(D)(E)ion–ion collisionselectron–electron collisionselectron–ion collisionsthree-body collisionsplasma diamagnetism15. An exponential density decay with time is characteristic of(A)(B)(C)(D)(E)fully ionized plasmas under classical diffusionfully ionized plasmas under recombinationweakly ionized plasmas under recombinationweakly ionized plasmas under classical diffusionfully ionized plasmas with both diffusion and recombination426Appendix C: Sample Three-Hour Final Exam16.
The whistler mode has a circular polarization which is(A)(B)(C)(D)(E)clockwise looking in the +B0 directionclockwise looking in the B0 directioncounterclockwise looking in the +k directioncounterclockwise looking in the k directionboth, since the wave is plane polarized17. The phase velocity of electromagnetic waves in a plasma(A)(B)(C)(D)(E)is always >cis never >cis sometimes >cis always <cis never <c18. The following is not a possible way to heat a plasma:(A)(B)(C)(D)(E)Cyclotron resonance heatingAdiabatic compressionOhmic heatingTransit time magnetic pumpingNeoclassical transport19. The following is not a plasma confinement device:(A)(B)(C)(D)(E)Baseball coilDiamagnetic loopFigure-8 stellaratorLevitated octopoleTheta pinch20.
Landau damping(A)(B)(C)(D)(E)is caused by “resonant” particlesalways occurs in a collisionless plasmanever occurs in a collisionless plasmais a mathematical result which does not occur in experimentis the residue of imaginary singularities lying on a semicirclePart B (Two Hours, Open Book; Do 4 Out of 5)1.
Consider a cold plasma composed of n0 hydrogen ions, 12 n0 doubly ionized Heions, and 2n0 electrons. Show that there are two lower-hybrid frequencies andgive an approximate expression for each. [Hint: You may use the plasmaapproximation, the assumption m/M 1, and the formulas for v1 given in theAppendix C: Sample Three-Hour Final Exam427text. (You need not solve the equations of motion again; just use the knownsolution.)]2.
Intelligent beings on a distant planet try to communicate with the earth bysending powerful radio waves swept in frequency from 10 to 50 MHz everyminute. The linearly polarized emissions must pass through a radiation beltplasma in such a way that E and k are perpendicular to B0.
It is found thatduring solar flares (on their sun), frequencies between 24.25 and 28 MHz do notget through their radiation belt. From this deduce the plasma density andmagnetic field there. (Hint: Do not round off numbers too early.)3. When β is larger than m/M, there is a possibility of coupling between a drift waveand an Alfvén wave to produce an instability. A necessary condition for this tohappen is that there be synchronism between the parallel wave velocities of thetwo waves (along B0).(a) Show that the condition β > m/M is equivalent to vA < vth.(b) If KTe = 10 eV, B = 0.2 T, ky = 1 cm1, and n = 1021 m3 find the requiredvalue of kz for this interaction in a hydrogen plasma.
You may assume00n0 =n0 ¼ 1 cm1 , where n0 ¼ dn0 =dr.4. When anomalous diffusion is caused by unstable oscillations, Fick’s law ofdiffusion does not necessarily hold. For instance, the growth rate of driftwaves depends on ∇n/n, so that the diffusion coefficient D⊥ can itself dependon ∇n. Taking a general form for D⊥ in cylindrical geometry, namely,D⊥ ¼ Ar s n p q∂n:∂rShow that the time behavior of a plasma decaying under diffusion follows theequation∂n¼ f ðr Þn pþqþ1∂tShow also that the behavior of weakly and fully ionized plasmas is recoveredin the proper limits.5.
In some semiconductors such as gallium arsenide, the current–voltage relationlooks like this:428Appendix C: Sample Three-Hour Final ExamThere is a region of negative resistance or mobility. Suppose you had asubstance with negative mobility for all values of current. Using the equationof motion for weakly ionized plasmas with KT = B = 0, plus the electron continuity equation and Poisson’s equation, perform the usual linearized wave analysis to show that there is instability for μe < 0.Appendix D: Answers to Some Problems1.1 (a) At standard temperature and pressure, a mole of an ideal gas contains6.022 1023 molecules (Avogadro’s number) and occupies 22.4 L.
Hence,the number per m3 is 6.022 1023/2.24 102 ¼ 2.69 1025 m3.(b) Since PV ¼ NRT, n ¼ N/V ¼ P/RT. Hence n1/n0 ¼ P1T0/P0T1. Taking n0 tobe the density in part (a) and n1 to be that in part (b), we have 103273¼ 3:30 1019 m3n1 ¼ 2:69 1025760 ð273 þ 20Þ1.2Note that a diatomic gas such as H2 will have twice as many atoms per torras, say, He.ð1ð122^f ðuÞdu ¼ Ah1¼eu =h d½u=h11ð1pffiffiffi2¼ Ahex dx ¼ Ah π1A ¼ ð2πKT=mÞ1=21.2aðð^f ðu; vÞdudv ¼ 1 ¼ A eðu2 þv2 Þ=h2 dudvð1ð122eðv=hÞ dv ¼ Ah2 πeðu=hÞ du1¼Aðð11A ¼ ð2πKT=mÞ1© Springer International Publishing Switzerland 2016F.F. Chen, Introduction to Plasma Physics and Controlled Fusion,DOI 10.1007/978-3-319-22309-4429430Appendix D: Answers to Some ProblemsIn cylindrical coordinates,1¼Að ð11222eðu þv Þ=h du dv ¼ A1 ¼ 2πAh2ð10ex2ð 1 ð 2πxdx ¼ πAh02ð10er=h20rdrdϕ ¼ 2πAð12eðr=hÞ rdr01e dy ¼ πAh e ¼ πAh22 y yA ¼ ð2πKT=mÞ1.4210p ¼ nðKT e þ KT i Þ ¼ 1021 4 104 1:6 1019¼ 6:4 106 N=m21 atm ’ 105 N=m2∴ p ¼ 64 atm1 atm ’ 14:7 lb=in2 : ¼ ð14:7Þð144Þ=ð2000Þ¼ 1:06 tons=ft22p ’ 68 tons=ft1.5d2 ϕe ð ni ne Þ1eϕ=KT ieϕ=KT e¼¼neee1dx2ε0ε0n1 e eϕeϕ’þε0 KT i KT e1n1 e 211jxj=λD, where 2 ¼þϕ ¼ ϕ0 eKT e KT iε0λD1=2If T i T eλD ’ ðε0 KT i =n1 e2 ÞIf T e T iλD ’ ðε0 KT e =n1 e2 Þ1=2However, this result is deceptive because in most experiments the ionsmove too slowly to shield charges.
Electrons do the shielding, so λDdepends on Te, even when Te Ti, which is the usual case.1.6 (a)Appendix D: Answers to Some Problems431d2 ϕnq¼2dxε0Let ϕ ¼ Ax2 + Bx + C; ϕ0 ¼ 2Ax + B; ϕ00 ¼ 2A. At x¼0, ϕ0 ¼ 0 by symmetry∴ B ¼ 0. At x ¼ d, ϕ ¼ 0; therefore, 0 ¼ Ad2 + C and C ¼ Ad2.nq1∴ A¼nq;ε02ε01ϕ ¼ Ax2 Ad 2 ¼nq d2 x22ε0ϕ00 ¼ 2A ¼ (b) Energy E to move a charge q from x1 to x2 is the change in potential energyΔ(qϕ) ¼ q(ϕ2ϕ1). Let ϕ1 ¼ 0 at x ¼ d and ϕ2 ¼ (12 ε0)nqd2 at x ¼ 0. ThenE¼1nq2 d2 :2ε0Let d ¼ λD; thenE¼1KTε0¼ ½KT ¼ Eavnq22ε0nq2for a one-dimensional Maxwellian distribution. Hence, if d > λD, E > Eav.
Ifthe velocities are distributed in three dimensions, we have Eav ¼ 32 KT andE > 13 Eav . The factor 3 is not important here. The point is that a thermalparticle would not have enough energy to go very far in a plasma (d > > λD)if the charge of one species is not neutralized by another species.1.7 (a) λD ¼ 7400(2/1016)1/2 ¼ 104 m, ND ¼ 4.8 104.(b) λD ¼ 7400(0.1/1012)1/2 ¼ 2.3 103 m, ND ¼ 5.4 104.(c) λD ¼ 7400(800/1023)1/2 ¼ 6.6 107 m, ND ¼ 1.2 105.1.8N D ¼ 1:38 106 T 3=2 =n1=23=2 33 1=2¼ 1:38 106 5 107= 10¼ 15:41.9 From Eq. (1.18), λD ¼ 69ðT=nÞ1=2 m,T in K6 1 n1011þ¼ 4:20From Problem 1.5, λD 2 ¼ 2 ¼69 T 4760 100 100Hence, λD ¼ 0:49 m.
The particle masses do not matter.432Appendix D: Answers to Some Problems e eϕ=KT2n0eeϕ ¼ ϕ=λ2D ¼ κ2 ϕ1.10 ∇2 ϕ ¼ r12 drd r 2 dϕ 1 εe0 KTdr ¼ ε0 ðne ni Þ ¼ ε0 n0 eewhere κ 1=λD .krLet ϕ ¼ Φ e r , where Φ is a constant in units of V-m.dϕ1kΦdϕ¼ Φ 2 ekr þ ekr ¼ ekr 2 ð1 þ kr Þ r 2¼ Φð1 þ kr Þekrdrrrrdrddϕr2¼ Φ kekr kð1 þ kr Þekr ¼ kΦekr ½1 ð1 þ kr Þ ¼ k2 r 2 ϕdrdr1 d2 dϕr¼ k2 ϕ ∴ k2 ¼ κ 2r 2 drdrϕð r Þ ¼ Φeκrer=λD,¼Φrrϕ ¼ aϕ0 eκaϕ0 ¼ Φeκa,aΦ¼aϕ0eκaeκraa¼ ϕ0 eκðraÞ ¼ ϕ0 eðraÞ=λDrrr1=2¼ 1:20 108 ¼1.11 Let Te ¼ 300 K. Then λD ¼ 69ðT=nÞ1=2 ¼ 69 300= 102212 nm1.12 From Eq.
(1.13), f ðuÞ ¼ A exp ðmu2 =2KT e Þ. From Eq. (1.6) and Problem 1.2,pffiffiffivth ð2KT e =mÞ1=2 and A ¼ nð2πKT=mÞ1=2 ¼ n=vth πpffiffiffiSo f ðuÞ ¼ ðn=vth π Þexp u2 =v2th . We wish to integrate f(u) from ucritð1ð1n2 2f ðuÞdu ¼ pffiffiffito 1 and from ucrit to 1.eu =vth du ¼vπthucrituc ritðn 1 y2pffiffiffie dy, where y u=vth .π ycritLet ½mu2crit ¼ eV ioniz , ucrit ¼ ð2eV ioniz =mÞ1=2 , where V ioniz ¼ 15:8 eV.ð2 x 2The error function erf(x) is defined aserf ðxÞ ¼ pffiffiffi et dt,π 0ð2 1 t2erfcðxÞ ¼ pffiffiffie dtπ xðn 1 y2nThe integral from ucrit to 1 is then pffiffiffie dy ¼ erfcð ycrit Þ, whereπ ycrit21=21=2 1=2 2eV ionizmeV ioniz. This density has¼ycrit ¼ ucrit =vth ¼2KT emKT eAppendix D: Answers to Some Problems433to be doubled to account for negative velocities u.
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
