1629373397-425d4de58b7aea127ffc7c337418ea8d (846389), страница 70

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

Текст из файла (страница 70)

Finally, the fraction ofelectrons that can ionize isΔneV ioniz 1=2¼ erfc:nKT e12.1 (a) E ¼ mv2⊥ ∴v⊥ ¼ ð2E=mÞ1=2 , r L ¼ mv⊥ =eB:21=2 4 ð2Þ 10 1:6 1019v⊥ ¼¼ 5:93 107 m=s9:11 10319:11 1031 5:93 107 ¼ 6:75 mrL ¼ 1:6 1019 0:5 104(b)(c)(d)v⊥ ¼ ð300Þð1000Þ ¼ 3 105 m=s1:67 1027 3 105 ¼ 6:26 105 m ¼ 626 kmrL ¼ 1:6 1019 5 109" #1=2ð2Þ 103 1:6 1019¼ 2:19 105 m=sv⊥ ¼ð4Þ 1:67 1027ð4Þ 1:67 1027 2:19 105 ¼ 0:183 mrL ¼ 1:6 1019 5:00 1021=2ð2Þð4Þ 1:67 1027 3:5 106 1:6 10192ME¼rL ¼qBð2Þ 1:6 1019 ð8Þ¼ 3:38 102 m2.4 Let initial energy be E 0, and Larmor radii r1 and r2, as shown. Energy at ① isE 1 ¼ E 0 + eEr1; energy at ② is E 2 ¼ E 0eEr2. (It would be acceptable to say:E 1,2 ¼ E 0 eEr L here.) Also v2⊥1, 2 ¼ 2E 1, 2 =M.

We are asked to make theapproximation434Appendix D: Answers to Some ProblemsMv⊥1, 2 M 2E 1, 2 1=2¼eB MeB 1=21 2E 0 1=2eE¼1 þ r 1, 2Ωc ME0r 1, 2 ¼For small E, expand the square root in a Taylor series: 1 2E 0 1=21 eEr 1, 2 ’1r 1, 2Ωc M2 E0 " #11 2E 0 1=21 eE 1 2E 0 1=21r 1, 2 ¼Ωc M2 E 0 Ωc M " #1 2E 0 1=21 eE 1 2E 0 1=21’Ωc M2 E 0 Ωc MThus eE 1 2E 02eEr1 r2 ¼¼E 0 Ω2c MMΩ2cindependent of E 0. The guiding center moves a distance 2(r1r2) in a time2π/Ωc, sovgc ¼ 2ðr 1 r 2 ÞðΩc =2π Þ ¼4eE 12E E¼’MΩc 2π π B BThus the guiding center drift is independent of the ion energy E 0.

The factor 2/πwould be 1 if we did not make the crude approximation.Appendix D: Answers to Some Problems4352.5(a)n ¼ n0 eeϕ=KT e ∴E¼(b) vE ¼ ϕ ¼ ðKT e =eÞlnðn=n0 Þ∂ϕKT e 1 ∂nKT e^r ¼ ^r ¼^r∂re n ∂reλEr ^KT e ^θ¼ θBeBλConsider electrons:2KT e 1=2 KT e m 1 1 v2th 1¼∴ vE ¼vth ¼mm eB λ 2 ωc λNow, rL ¼ mv⊥/eB, so for a distribution of velocities we must find an averagerL. Since v⊥ contains two degrees of freedom, we have122 mv⊥¼ 2 12 KT eThe most convenient average ishv⊥ irms ¼ ð2KT e =mÞ1=2 ¼ vth 1 vth v⊥ 1 vth r LUsing this for v⊥ in rL, we have vE ¼¼2 λ ωc 2 λso that jvEj ¼ vth implies rL ¼ 2λ.(c) If we take ions instead of electrons, we have vthi ¼ (2KTi/M )1/2 ¼ v⊥i, 1 2KT eM1 T e vth i v⊥i1 TerLi ¼ v⊥i/ωci, and vE ¼¼vth i r Li .¼2λeB2λ T i ωci2λ T iMIf jvEj ¼ vthi, it is still true that rLi ¼ 2λ provided that Ti ¼ Te. 2 2e22ϕðr Þ ¼ er =a 12.6 (a) n ¼ n0 exp er =a 1 ¼ n0 eeϕ=KT e ∴KT e∂ϕ KT e 2r r2 =a2∂ϕ¼e^rE r ðr Þ ¼ ∂re a2∂rdEr 2KT e2r 2 r2 =a 2r2 1¼¼01¼edrea2a2a2 2KT e 2 1=2 ð0:2Þ 1:6 1019 pffiffiffi 1=2pffiffiffi e2e¼¼¼ 17 V=mea 21:6 1019 ð:01ÞE ¼ ∇ϕ ¼Emax436Appendix D: Answers to Some ProblemsvE ¼ Er ^θBV Emax ¼Emax17¼ 8500 m=s¼0:2B(b) Compare the force Mg with the force eE for an ion.

(mg for an electronwould be 1836 times smaller.) g ¼ 9.80 m/s2. Mg ¼ (39)(1.67 1027)(9.80) ¼ 6.38 1025 N. eEmax ¼ (1.6 1019)(17) ¼ 2.75 1018 N ¼4 106 Mg. Hence gravitational drift is 4 million times smaller.(c)Mv⊥¼ 102 mrL ¼eB"#1=2ð2Þð0:2Þ 1:6 10191=2v⊥ ¼ ð2KT=MÞ ¼ð39Þ 1:67 1027¼ 9:9 102 m=sð39Þ 1:67 1027 9:9 102B¼¼ 4:00 102 T102 1:6 10192.8c0:3 104T¼3r3 ðr=RÞ B ∇B 1 ∇B1 ¼ vr L ¼ v⊥ r L 2B2 2 B B¼v∇B ∇B 3∂Bc3 ¼^r ¼ 3 4 ^r ¼ Bð^r Þ∇B ¼B r∂rrr1:6 1019 ðKT ÞeV 1 ðKT ÞeV1 2KT=m KT11 v2¼¼¼v⊥ r L ¼ ⊥ ¼2 eB=meBBB1:6 101922 ωc(a)Bðr ¼ 5RÞ ¼0:3 104¼ 2:4 107 T535R ¼ (5)(4000 miles)(1.6 km/mile)(103 m/km) ¼ 3.2 107 mAppendix D: Answers to Some Problems437ðKT ÞeV¼ 0:39ðKT ÞeV m=s2:4 107Ions : KT ¼ 1 eVv∇B ¼ 0:39m=sv∇B ¼ 108Electrons : KT ¼ 3 104 eVv∇B ¼ 1:17 104 m=s(b) Ions: westward; electrons: eastward.(c) 2πr ¼ (6.28)(3.2 107) ¼ 2.0 108 mt¼(d)2:0 1082πr ¼ 1:7 104 s ¼ 4:8 h¼v∇B1:17 104j ¼ nev∇Bneglect ions 7 ¼ 10 1:6 1019 1:17 104 ¼ 1:87 108 A=m22.9 (a) vR ¼ 0, since the electron gains no energy in the parallel ^θ direction.Since the electron starts at rest with no thermal energy, it will come back torest after one cycle.

Hence, the orbit has sharp cusps instead of loops. It isclear that the vE drift must dominate, since the electron starts to the left, andthe Lorentz force makes it move upwards.(b) In cylindrical geometry, ϕ ¼ A ln r + B. Since438Appendix D: Answers to Some Problemsϕ 103 ¼ 460V and ϕð0:1 mÞ ¼ 0,460 ¼ A ln 103 þ B0 ¼ A ln ð0:1Þ þB ¼ A ln ð0:1Þ B460 ¼ A ln 103 A ln ð0:1Þ¼ A ln ð0:01ÞA ¼ 460=ln ð0:01Þ460lnð0:1r Þϕð r Þ ¼½lnr ln ð0:1Þ ¼ 460Vln ð0:01Þln100∂ϕ 460 r 0:1460=r VE¼¼¼∂rln100 0:1r2ln100 m4604V¼ 10 at r ¼ 102 m¼ð4:6Þð1ÞmI ðAÞ104 500 104¼¼ 0:01 T5rð5Þð1Þ104 V=cm¼ 106 m=sjvE j ¼ jE=Bj ¼ 1080:01 TB¼To estimate the ∇B drift, we must find v⊥ in the frame moving with theguiding center.

Remember that in deriving v∇B, v⊥ was taken as thevelocity in the undisturbed circular orbit. Here, the latter is moving withvelocity vE, so that it does not look circular in the lab frame. Nonetheless,it can still be decomposed into a circular motion with velocity v⊥ plus anE B drift of the guiding center. Consider the z component of velocity(along the wire). At point ① on the orbit, vz ¼ vE + v cos ωct ¼ 0, wherecos ωct ¼ 1, its maximum negative value; hence, vE ¼ v⊥. The sameresult can be obtained by considering 2 that at point ② vz ¼ vE + v⊥1(cos ωct ¼ 1).

The energy there, 2 mvz , must equal the energy gained infalling a distance 2rL in an electric field. Thus1mv⊥EmðvE þ v⊥ Þ2 2r L eE ¼ 2eE¼ 2mv⊥ ¼ 2mv⊥ vE2BeBv2E þ 2v⊥ vE þ v2⊥ ¼ 4v⊥ vEðvE v⊥ Þ2 ¼ 0vE ¼ v⊥Now we can calculate v∇B:v∇B 1:6 1019 102eB1 v2⊥ ∇Bωc ¼¼ ¼ 1:76 109 s1¼m9:11 10312 ωc B ∇BdB I ð1Þ104B ¼ 102 m1¼¼B2drrr21v1¼ 2:8 104 m=sv∇B ¼ E ¼2 ωc 21016 1:8 109This amounts to a slowing down of the vE drift due to a distortion ofthe orbit into a hairpin shape because of the change in Larmor radius.Appendix D: Answers to Some Problems439The undisturbed orbit is the path taken by the valve on a bicycle wheel asit rolls along:Finally, we note that the finite Larmor radius correction to vE isnegligible:1 2 2 E 1 r 2L Er ∇ ’4 L B 4 r2 B 9:11 1031 106rL ¼ ¼ 5:7 104 m1:6 1019 ð0:01Þ1 r2r ’ 102 m ∴ L2 ¼ 0:08%4r2.12 Let all velocities refer to the midplane, and let subscripts i and f refer to initialand final states (before and after acceleration).(a) Given: Rm ¼ 5, v⊥i ¼ v║i since μ is conserved, v⊥f ¼ v⊥i, and only v║ willincrease.

It will increase until the pitch angle θ reaches the loss cone:sin 2 θm ¼v2⊥ fv2⊥ fþv2k f¼111¼¼221 þ vk f =v⊥i Rm 5Hence v2k f =v2⊥i ¼ 4, vk f ¼ 2v⊥i : Energy is 11 5E f ¼ M v2k f þ v2⊥ f ¼ Mð4 þ 1Þv2⊥i ¼ mv2⊥i222 11 222Ei ¼ M vki þ v⊥i ¼ Mð1 þ 1Þv⊥i ¼ Mv2⊥i22∴E f ¼ 2:5Ei ¼ ð2:5Þð1Þ ¼ 2:5keV(b) (1) Let particle have v0 > 0 and hit piston moving at velocity vm < 0. In theframe of the piston, the particle bounces elastically and comes off withits initial velocity, but in the opposite direction. Let 0 refer to the frameof the piston.

Initial and final velocities in this frame are440Appendix D: Answers to Some Problems0vi ¼ v0 vm0v f ¼ ðv0 vm Þ(Note: vm is negative.) Transforming back to lab frame,0v f ¼ v f þ vm ¼ v0 þ 2vmSince vm is negative, the change in velocity is 2jvmj. QED(2) At each bounce, the change in momentum is Δpk ¼ 2 mjvmj. If N is thenumber of bounces, pkf ¼ pki + NΔp. Thuspk f pki vk f vki 2v⊥i v⊥i 1v⊥i¼¼¼2 vm2vmΔp2vm 3 192¼ 1:6 1016 JEi ¼ Mv⊥i ¼ 1 keV ¼ 10 1:6 101=21:6 1016∴v⊥i ¼¼ 3:1 105 m=s1:67 1027N¼vm ¼ 104 m=s1 3 105¼ 15 bounces∴N ¼2 104(3) Average vk is 11vki þ vk f ¼ ðv⊥i þ 2v⊥i Þ223¼ v⊥i ¼ 4:6 1052L ¼ 1013 mNL ð15Þ 1013¼∴t ¼¼ 3:2 108 sv4:6 105ð¼ 10yÞv¼However, L changes during this time by a distance ΔL ¼ 2vm t ¼ ð2Þ 104 3:2 108 ¼ 6:4 1012 mso that actual time is more like 2.5 108 s.

Since only factor-of-twoaccuracy is required, it is not necessary to sum the series—the aboveanswer of 3.2 108 s will do.2.13 (a)ðvk ds ’ vk L ¼ constant ∴v_ k L þ vk L_ ¼ 0Appendix D: Answers to Some Problems(b)v_ kL_¼vkL441v_ k ’Δvk vk ¼L_TLΔvk L2v⊥i v⊥i L2 1013¼¼2v3 2 104vk L_ 1ð2v⊥i þ v⊥i Þ m2¼ 3:3 108 sT’2.14 As B increases, Maxwell’s equation ∇ E¼B_ predicts an E-field. Thisinduced E-field has a component along v and accelerates the particle. IfB increases slowly and adiabatically, E will be small; but the integratedeffect over many Larmor periods will be finite. The invariance of μ allows usto calculate the energy increases without doing this integration. _ Hence, σ_ ¼ ∇ ρ=B2 E_ :3.1 ∂σ=∂t þ ∇ j ¼ 0, where j ¼ jP ¼ ρ=B2 E:The time derivative of Poisson’s equation is ∇ E_ ¼ σ=2_ 0 1ρ __∴∇ E ¼ ∇E20B23.2∇ 1þρE_ ¼ 02 0 B2Assuming the dielectric constant 2 to be constant in time, we have ∇ D_ ¼ ∇ 2E_ ¼ 0: By comparison, 2 ¼ 1 + ρ/E0B2.2’1þΩ2pnMne2 M2nM¼¼’22222M20 B20 B2e BΩc0True if 2 1.3.3 Take divergence of Eqs.

(3.56) and (3.58):∂∇ ð∇ EÞ ¼ ∇ B_ ¼ 0 ∴ ð∇ BÞ ¼ 0∂t∴ ∇ · B ¼ 0 if it is initially zero. This is Eq. (3.57),∇ ð∇ BÞ ¼ 0 ¼ μ0 ½qi ∇ ðni vi Þ þ qe ∇ ðne ve Þ þ∇ E_c2from Eq. (3.60), ∇ ðni vi Þ ¼ n_ i , ∇ ðne ve Þ ¼ n_ e∇ E_∴μ0 ðqi n_ i qe n_ e Þ þ 2 ¼ 0c ∂1∇ E ð ni qi þ n e qe Þ ¼ 0∂t20If [] ¼ 0 initially, ∇ · E ¼ (l/20)(niqi + neqe). This is Eq. (3.55).442Appendix D: Answers to Some Problems3.4jD ¼ ðKT i þ KT e ÞB ∇n KT ne/e BLB2Since KT / eϕ and E / ϕ/L, KT/eL / E ∴ jD / neE/B / nev, sinceE/B ¼ vE.3.5 Let jD be constant in the box of width L. Δn ¼ n0 L, jJDj ¼ jΔnevyj ¼ jn0 Levyj:from the difference between the currents on the two walls.

Характеристики

Список файлов книги