1629373397-425d4de58b7aea127ffc7c337418ea8d (846389), страница 73

Просмтор этого файла доступен только зарегистрированным пользователям. Но у нас супер быстрая регистрация: достаточно только электронной почты!

Текст из файла (страница 73)

(2):ð1Þð2Þð3ÞAppendix D: Answers to Some Problems469E1 × B0 + (v1 × B0) × B0 = hj1 × B02−v1⊥B0v1⊥ ¼E1 B0 η j1 B0B20B20Substitute in Eq. (1), which has no parallel component anyway:iωρ0E1 B0 η j1 B0B20B20¼ j1 B0Since, by Eq. (3), E and j1 are in the same direction, take them both to be inthe ^x -direction. Then the y-component isE1¼B0iB0ηþjωρ0 B0 1Equation (3) becomesE1 iB0η 1þB0 ωρ0 B0 21BE1¼ μ0 ω2 0 iηωρ0ω2B2¼ μ0 0 iωη2ρ0kk2 E1 ¼ μ0 iω(b)1=2k ¼ ðμ0 ω Þ iωη!1=2μ0 ρ0 1=2iωηρ0¼ω1B20B201=2ωηρ0 μ0 ρ0ω2 η 1ImðkÞ ¼ ω¼2 v3A2B20B202 1=2B20ρ0But for small η, ω kvA, where k ¼ Re (k) ðη Þ k 2∴ImðkÞ 2vA470Appendix D: Answers to Some Problems6.4 (a)j B ¼ ∇ p ¼ KT∇n ðKT ¼ KT e þ KT i hereÞð j BÞ B ¼ KT∇n B ¼ Bð j BÞ jB2The parallel component is 0 ¼ j║B2 j║B2 ∴ j║ is arbitrary. The perpendicular component isj⊥ ¼(b)KTKT ∂n ^θB ∇n ¼2B ∂rBðð∇ B dS ¼ μ0 j dSþðð1B dL ¼ μ0 j dS ¼ μ0 Ljθ dr0since j and dS are both in the ^θ direction, and L is the width of the loop inthe ẑ direction.

By symmetry, there can be no Br, so only the two z-legs ofthe loop contribute to the line integral. Substituting for jθ, we haveðBax B0 ÞL ¼ μ0 LKTð10∂n=∂rdrBðr Þ(c) ∂n/∂r ¼ n0δ(r a), since ∂n/∂r is a function that is zero everywhereexcept at r ¼ a, is 1 there, and has an integral equal to n0. ThusBax B0 ¼ μ0 KTð10n0δ ð r aÞdrBðr ÞSince all the diamagnetic current is concentrated at r ¼ a, B takes a jumpfrom a constant value Bax inside the plasma to another constant value B0outside. (Remember that the field inside an infinite solenoid is uniform.)Upon integrating across the jump, one obtains the average value of B onthe two sides, i.e., BðaÞ ¼ 12 ðBax þ B0 Þ. Thus11ðBax þ B0 Þ2B2ax B20 ¼ 2μ0 n0 KTBax B0 ¼ μ0 KTn01B2ax 2μ0 n0 KT β ¼ 1∴Bax ¼ 0¼B2B20Appendix D: Answers to Some Problems4716.5 (a) By Faraday’s law, V ¼ dΦ/dtððdΦdt ¼ NΔΦ∴ V dt ¼ NdtSince ΔΦ is the flux change due to the diamagnetic decrease in B,ðNΔΦ ¼ N ðB B0 Þ dSThe sign depends on which side of V is considered positive.

In practice, thisis of no consequence because the oscilloscope trace can easily be invertedby using the polarity switch.(b) In Problem 6.4b, we can draw the loop so that its inner leg lies at anarbitrary radius r rather than on the axis. We then haveBðr Þ B0 ¼ μ0 KTð1r∂n=∂r 0dr μ0 KTBð r 0 Þð1r0∂n=∂r 0drB0where again KT is short for Σ KT∂n2r r2 =r20¼ n0e∂rr 20ðμ0 KT n0 r r0 2 =r2 0 00 2r drBð r Þ B0 ¼eB0 r 20 1μ n0 KT h r0 2 =r2 i1 μ0 n0 KT r2 =r200¼ 0¼eerB0B0This is the diamagnetic change in B at any r. To get the loop signal, wemust integrate over the plasma cross section.ððððV dt ¼ N ðB B0 Þ dS ¼ N½Bðr Þ B0 r dr dθwhere both B and dS are in the ẑ direction. Substituting for B(r) B0 andassuming the coil lies well outside the plasma, we haveÐð1μ0 n0 KT2 2V dt ¼ Ner =r0 rdr2πB00μ0 n0 KT 2 h r2 =r2 i012 2μ0 n0 KT0r0 e¼ Nπr 0B0¼ Nπ1B02B20(c) The quantity in parentheses is β by definition; hence,472Appendix D: Answers to Some Problemsð1V dt ¼ Nπr 20 βB02Both sides of this equation have units of flux,6.6 (a) For each stream, we have∂v1mþ v0 ∇v1 ¼ eE1 ¼ ðiω þ ikv0 Þv1∂tieE1v1 ¼mðω kv0 Þ∂n1þ n0 ð∇ v1 Þ þ ðv0 ∇ Þn1 ¼ 0∂tkv1ðiω þ ikv0 Þn1 þ ikn0 v1 ¼ 0 n1 ¼ n0ω kv0ikE1 e∴n1 j ¼ n0 j 2m ω kv0 jPoisson: ikE1 ¼ (e/E0)(n1a + n1b), where stream a has v0a ¼ v0 ^x , n0a ¼ 12n0 ;stream b has v0b ¼ v0 ^x , n0b ¼ 12n0 : Thus2311nn00eikeE1 6 272þikE1 ¼ 4520mðω kv0 Þ2 ðω þ kv0 Þ2"#n0 e 2 1111¼þ20 m 2 ðω kv0 Þ2ω þ kv20"#1 2111 ¼ ωpþ2ðω kv0 Þ2 ðω þ kv0 Þ2(b)ω2 þ k2 v201 ¼ ω2p 2ω2 k2 v20ω4 ω2p þ 2k2 v20 ω2 þ k2 v20 k2 v20 ω2p ¼ 0 11=21ω2 ¼ ω2p þ 2k2 v20 ω4p þ 8ω2p k2 v2022Letx¼2k2 v20 2 2ω2y ¼ 2ω2pωpAppendix D: Answers to Some Problems473Theny2 ¼ 1 þ x ð1 þ 4xÞ1=2y can be complex only if the () sign is taken.

This y is pure imaginary,and we can let y ¼ iγ:γ2 ¼ ð1 þ 4xÞ1=2 ð1 þ xÞd 2γ ¼ 2ð1 þ 4xÞ1=2 1 ¼ 0dxx¼34Thus7 1γ2 ¼ ð1 þ 3Þ1=2 ¼4 4pffiffiffi2 ImðωÞ1ωpγ¼ ¼ImðωÞ ¼ 3=2ωp226.8 (a)1¼ω2p"1δþ2ωðω kuÞ2#where ω2p n0 e2 =є0 m:(b) This equation is the same as Eq. (6.30) except that m/M is replaced by δ,which is also small, and that the rest frame has changed to one movingwith velocity u. The maximum growth rate does not depend on frame, ascan be seen from Fig. 6.11 by imagining γ to be plotted in the z directionvs. x and y; a shift in the origin of x will not affect the peak.

Analogy withEq. (6.35) then givesγmax δ1=3 ω p(The exact constant that should appear here is 31/224/3 ¼ 0.69. Thederivation of γmax, which is difficult because the dispersion relation iscubic, and the proof that it is independent of frame for real k are left asexercises for the advanced student.)6.9 (a) Since only the y component of vj and E are involved, the given relation iseasily found from Eqs. (4.98b) and (6.23), plus continuity and Poisson’sequation. Note that Ωp is defined with n0, not (1/2)n0.11 (b) Let α Ω2p 1 þ ω2p =ω2c, β k2 v20 : Then the dispersion relation2reduces to474Appendix D: Answers to Some Problemsω4 2ðα þ βÞω2 þ β2 2αβ ¼ 0The dispersion ω(k) is given by1=2ω2 ¼ α þ β α2 þ 4αβInstability occurs if (α2 + 4αβ)1/2 > α + β, or β < 2α, i.e.,1k2 < Ω2p =v20 1 þ ω2p =ω2cWhere this is satisfied, the growth rate is given byγ¼7.3 (a)hα2 þ 4αβ1=2 ðα þ β Þi1=2n p v2 =a2eaπ 1=222nbf b ðvÞ ¼ 1=2 eðvV Þ =bbπf p ðvÞ ¼(b)nb 2ðv V Þ ðvV Þ2 =b2ebπ 1=2b2"#2nb2ðv V Þ2 ðυV Þ2 =b200f b ðvÞ ¼ 3 1=2 1 ¼0eb πb2pffiffiffipffiffiffiv V ¼ b= 2 vϕ ¼ V b= 2 1=2 nb 1=20 f b vϕ ¼ π2eb20f b ðvÞ ¼(c) pffiffi 2 2n p 2b0 V 1=2 eðVb= 2Þ =af b υϕ ¼ 1=2 2aaπ22n p V V 2 =a2 3 1=2 eVbaπAppendix D: Answers to Some Problems475(d) 1=22nb 1=22n p V22e¼ 3 1=2 eV =a2πaπb2nbb2 T b1=2 bV 2 =a2¼ ð2eÞVe¼3npaa2 T pT Vnb 22beV =a∴ ¼ 2e1=2nT pa7.8 From Eq.

(7.127), we obtain ∑αjZ0 (ζ j) ¼ 2Ti/Te, where αj ¼ n0j/n0e, ζ j ¼ ω/kυthj.Assume at first that αH is small, so that αA 1, αH ¼ α; furthermore, smallα means that υϕ will be nearly unchanged from vs of argon. Then doubling the0Landau damping rate means Im Z0 (ζ H) ¼ Im Z0 (ζ A), where Im Z ζ j ¼pffiffiffi22i π ζ j eζ j . ¼ Thus22ζ A eζA ¼ αζ H eζHζA¼ζHMAMH1=2α¼ζ A ðζ2A ζ2H ÞeζH2α ¼ ð40Þ1=2 eζA ð11=40ÞKT e þ 3KT i MA13¼2MA2KT ipffiffiffiffiffi 6:5ð0:975Þα ¼ 40e¼ 1:12 102 1%ζ 2A ¼Thus α is so small that our initial assumptions are justified.7.9 (a)2k21α 0α 00Z ðζ e Þ þ Z ðζ h Þ¼ Z ðζ i Þ þ2θθkDieh(b)Since ζ h ζ e 1,pffiffiffi20Z ðζ i Þ 2 2i π ζeζImZ 0 ðζ h Þ ImZ0 ðζ e Þ(c) Since Z0 (ζ h) Z0 (ζ e) 2), the ζ h term in (a) is negligible compared withthe ζ e term if θh θe and α < 1/2.

Now the dispersion relation is 2k2 2ð1 αÞ 2T iT e k2Z ðζ i Þ ¼ 2 þ1⨪α þ¼θeTekDiT i k2Di0!476Appendix D: Answers to Some ProblemsThe last term is k2 λ2D and is negligible when quasineutrality holds. Thus theion wave dispersion relation is the same as usual, except that Ti/Te has beenreplaced by (1 α) Ti/Te. Since small Ti/Te means less Landau damping, thehot electrons have decreased ion Landau damping.8.3 Refer to Fig. 8.4. Take a number of ions with v ¼ u0 and split them into twogroups, one with v ¼ u0 + Δ and one with v ¼ u0 Δ.

After acceleration in apotential ϕ, the faster half will have less fractional energy gain (because itstarted with more energy) and, hence, will have less fractional densitydecrease. The opposite is true for the slower half, and to first order the totaldensity decrease is the same as if all ions had v ¼ u0. However, there is asecond-order effect which makes the slower group dominate. This can be seenby making Δ so large that v 0 for the slower half, which clearly must thensuffer a huge density decrease.

To compensate for this, u0 must be increased tohigher than the Bohm value.8.4 The maximum current occurs when the space charge of decelerated ions neargrid 3 decreases the electric field to zero. Thus we can apply the ChildLangmuir law to the region between grids 2 and 3." #1=2 8:85 1012 ð100Þ3=24 ð2Þ 1:6 1019AJ¼¼ 27:2 2 3 2279 ð4Þ 1:67 10m10π2A ¼ 4 103 ¼ 1:26 105 m24I ¼ JA ¼ 0:34mA8.6 (a) At ωp ¼ ω, 2 0 E2pFNL ¼ ¼ ∇ peff ¼ effL2L∴ peff ¼ 12 20 E2 : But I 0 ¼ c20 E2 ¼ P=A, where P ¼ 1012 and A ¼(π/4) (50 106)2 ¼ 1.96 109 m2P1012N ¼ 8:50 1011 2¼892cA ð2Þ 3 10 1:96 10m118:50 10 ð0:2248Þlb¼¼ 1:23 108 22in:ð39:37Þpeff ¼(b)F ¼ pA P=2c ¼ 1012 =ð2Þ 3 108 ¼ 1667 NF ¼ Mg M ¼ F=g ¼ 1667=9:8 ¼ 170 kg ¼ 0:17 tonnesAppendix D: Answers to Some Problems(c)2nKT ¼ peff∴n ¼8.78:5 1011 ¼ 2:66 1027 m3ð2Þ 103 1:6 1019 ∂n ∂ 2 0 E2FNL ¼ ∇ p∴ ðnKTÞ ¼ ∂rnc ∂r2 220 E1 ∂n20 ∂ 2 þ ln n0E lnn ¼ 2nc KTn ∂r 2nc KT ∂r2n ¼ n eє0 hE i=2nc KT0At r ¼ 0,2nmin ¼ n0 eє0 hE imax=2nc KT ¼ n0 eα 220 E max∴α ¼2nc KT8.9k0 ¼ 2π=λ0 ¼ 2π=1:06 106 ¼ 5:93 106 m1ki 2k0 ¼ 1:19 107 m1" #1=2 103 1:6 1019KT e þ 3KT i 1=23 1=2¼1þvs ¼θMð2Þ 1:67 10271=23ωi ¼ Δω ¼ ki vs ¼ 1:19 107 2:19 105 1 þθ1=23¼ 2:61 1012 1 þθΔωΔλω02πc¼∴Δω ¼ : Δλ ¼ 2 Δλω0λ0λ0λ08 ð2π Þ 3 1010¼ 21:9 106 21:06 10¼ 3:67 1012233:67 1012Te1¼2 θ¼¼ 3 ∴T i ¼ keV1þ ¼12θ3Ti2:61 10477478Appendix D: Answers to Some Problems8.10 (a) 2 1 2 8ω1 ω2 Γ1 Γ2E0 ¼ E ¼2c1 c2c1 c2 ¼20 k21 ω4pn0 ω20 MΓ2 ¼ω2p νω22 2 4ω1 Γ1 ω20 ν n0 M4ω1 Γ1 ω20 νMm¼E20 ¼ω2 k21 20 ω2pω2 k21 e2 2 e2 E204ω1 Γ1 νMv0 ¼ 2 2 ¼m ω0ω2 k21 m 2vω21 ω21 M ω21 v2e M4Γ1 ν2∴ 02 ¼k1 ¼ 2 ¼¼veMω1 ω 2KT evs 2v04Γ1 νei¼v2eω1 ω0(b)since ω2 ω0 when n nc.2πc ð2π Þ 3 108ω0 ¼¼ 1:78 1014 s1¼λ0 106 10:6102 1:6 1019KT em22 ¼ 1:76 1013 2¼ ve ¼30ms0:91 10Γ 1 π 1=2Te¼¼ 10θð3 þ θÞ1=2 eð3þθÞ=2 θ ¼8ω1Ti2¼ 3:40 105:2 105 ð10Þ5 ln Λη ¼ 5:2 10¼¼ 5:2 107 Ω-m3=23=2ð100ÞT eV 23 19 221:610105:2 107ne η¼vei ¼¼ 1:46 109 s130m0:9110 2 ð4Þ 3:4 102 1:46 109 m2v0 ¼1:76 1013 ¼ 1:96 107 214s1:78 10From Problem 8.6(a): m2 ω2 I 0 ¼ c20 E2 ¼ c20 2 0 v20eAppendix D: Answers to Some Problems4792 2 0:91 1030 1:78 1014 1:96 107I 0 ¼ 3 10 8:854 1021:6 1019WW¼ 5:34 1010 2 ¼ 5:34 106 2mcmi 2h28.11ωs þ 2iγωs ω21 ðωs þ iγ ω0 Þ2 ω22 ¼ 14c1 c2 E0 :812If ω2s ¼ ω21 , ðωs ω0 Þ2 ¼ ω22 , and γ=ωs 1, then12ð2iγωs Þ½2iγðωs ω0 Þ ¼ c1 c2 E0 ¼ 4γ2 ωs ω24From Problem 8.10,c1 c2 ¼20 k21 ω4pn0 ω20 M¼k21 ω2p e2ω20 mM2γ2 ¼8.13 (a)k21 ω2p v20 m ð2k0 Þ2 Ω2p v2016ω0 ωs16ωs ω2 ω20 mM 16ωs ω2 M 1=2ω20 Ω2p v20v0 ω0¼ 2Ωp∴γ ¼2 ωs4c ω0 ωsk21 ω2p e2 E0¼∂v¼ en0 E γi KT i ∇ n Mn0 νv þ FNL∂tMn0 ðiω þ νÞv ¼ en0 ðikϕÞ γi KT i ikn1 þ FNLMn0with eϕ/KTe ¼ n1/n0, this becomesðω þ iνÞv ¼ kv2sn1 iFNLþn0 Mn0Continuity:n1 iFNLiωn1 þ ikn0 v ¼ iωn1 þ ikn0 ðω þ iνÞ1 kv2s þ¼0n0 Mn0 2ω þ iνω k2 v2s n1 ¼ ikFNL =MWhen FNL ¼ 0,480Appendix D: Answers to Some Problemsν1νi2 2¼ k vs ∴ω kvs 1 i¼ kvs νω 1þiω2ω2Hence Im ω Γ ¼ ν=2: So ω2 þ 2iΓω k2 v2s n1 ¼ ikFNL =M2(b)FNL ¼ ω2pω2p∇20 hE0 E2 i ¼ ik20 hE0 E2 iω0 ω 2ω0 ω2Thus,ikFNL 1ik ω2pik 20c1 ¼¼M h E 0 E 2 i M ω0 ω2!¼ω2p k2 20ω0 ω2 M8.14 The upper sideband has hω2 ¼ hω0 + hω1, so that the outgoing photon hasmore energy than the original photon hω0.

Характеристики

Список файлов книги