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This current JD isover a box of width L, so the equivalent current density is j ¼ J D =L ¼ n0 ev y DEquation [3.69] gives jjDj ’ jKT∇n/Bj ¼ jKTn0 /Bj; hence, once vy is chosen sothe two formulas agree for one value of L, they agree for all L, sinceL cancels out.3.6 (a)γKT e ^z ∇nvDe ¼ neBIsothermal means γ ¼ 1.∂nn0 2x¼ 2 ^x∂xa11KT e 2n0 xx2KT e 2xx21 21 2¼ ^y¼ ^yeB a2 n0eB a2aa∇n ¼ ^xvDe(b)(c)vDe ¼ (2)/(0.2)ΛΛ1 0n ð2n0 =a2 Þða=2Þ 1=0:04¼ 33:3 m1¼ ¼¼n ð1 a2 =4a2 Þ3=4n0∴vDe ¼ ð10Þð33:3Þ ¼ 333 m=s3.7 n ¼ n0 er2=r20¼ n0 eeϕ=KT eϕ¼ 2KT e nKT erln ¼ 2n0eer0Appendix D: Answers to Some Problems443(a)∂ϕKT 2r^r ¼ e 2 ^r∂re r0EBEr ^KT e 2rvE ¼¼ θ ¼ θ^BzeB r 20B2E¼ B∇pKT e ∂n=∂r ^KT e ∂ðln nÞ¼θ ¼ ^θ2neBeB ∂renB 2KT e ∂ rKT e 2r¼ ^θ¼ vE QED¼^θ2eB ∂r r 0eB r 20vDe ¼ (b) From (a), the rotation frequency is constant whether we take vE, vDe, vDi,or any combination thereof, since ω ¼ vθ/r and vθ / r.(c) In lab frame,v ¼ vϕ þ vE ¼ 0:5vDe þ ðvDe Þ1¼ vDe23.8 (a)jD ¼ neðvDi vDe Þ ¼ ^θ(b)or:n0 ðKT e þ KT i Þ 2r r2 =r20 2eBr01016 ð0:5Þ 1:6 1019jD ¼¼ 0:147 A=m20:4 r 20 =2r ð2:718ÞjD ¼ neðjvDe j þ jvDi jÞðKT Þe V 2r ð0:25Þ2rr¼¼ 1:25 2 m=sjvDe j ¼ jvDi j ¼22B r00:4r 0r0Using e ¼ 1.6 1019 C, 2 ¼ 2.718,r21AjD ¼ 1016 1:6 1019 ð2Þð1:25Þ 2 ¼ 0:147 2mr0444Appendix D: Answers to Some Problems(c)Since ve ¼ vE + vDe ¼ vE vE ¼ 0 in the lab frame, the current is carried entirelyby ions.3.9ð∇ B ¼ μ0ðjDð∇ BÞ dS ¼ μ0 jD dSþðB dL ¼ μ0 jD dSChoose a loop with one leg along the axis (B ¼ B0) and one leg far away,where B ¼ B1.

Since jD is in the ^θ direction, we can choose the direction ofintegration dL as shown, so that jD · dS is positive. There is no Br ∴þB dL ¼ ðB1 B0 ÞLnðKT i þ KT e Þ 2rjD ¼ ^θBr2ðð L ð01n0 ðKT i þ KT e Þ2 2jD dS ¼er =r0 2r dr dzB1 r 200 0Ln0 ðKT i þ KT e Þ h r2 =r0 i1 2Ln0 KT¼e¼0B1B1where Te ¼ Ti. In this integral, we have approximated B(r) by B1, since B isnot greatly changed by such a small jD. Thus,2n0 KTΔB ¼ B1 B0 ¼ μ0 B12 4π 107 1016 ð0:25Þ 1:6 1019¼0:4¼ 2:5 109 TAppendix D: Answers to Some Problems4454.1 (a) Solve for ϕ1:KT e n1 ω þ ia ω* iae n0 ω* þ ia ω* ia2KT e ωω* þ a þ iaðω* ωÞ n1¼en0ω*2 þ a2ϕ1 ¼If n1 is real,Imðϕ1 Þ aðω* ωÞ¼¼ tan δωω* þ a2Reðϕ1 ÞHence,δ ¼ tan1aðω* ωÞωω* þ a2(b) n1 ¼ n1 eiðkxωtÞ , while ϕ1 ¼ An1ei(kxωt+δ), where A is a positive constant. For ω < ω*, we have δ > 0.

Let the phase of n1 be 0 at (x0, t0):kx0 ωt0 ¼ 0. If ω and k are positive and x0 is fixed, then the phase of ϕ1is 0 at kx0 ωt + δ ¼ 0 or t > t0. Hence ϕ1 lags n1 in time. If t0 is fixed,kx ωt0 + δ ¼ 0 at x < x0, so ϕ1 lags n1 in space also (since ω/k > 0 andthe wave moves to the right, the leading wave is at larger x). If k < 0 andω > 0, the phase of ϕ1 would be 0 at x > x0; but since the wave nowmoves to the left, ϕ1 still lags n1.4.31eðni1 ne1 Þ20iωmve1 ¼ eE1 ðelectronsÞikE1 ¼ iωMvi1 ¼ eE1 ðionsÞiωne1 ¼ ikn0 ve1 ðelectronsÞ iωni1 ¼ ikn0 vi1 ðionsÞkiekieE1 ni1 ¼ n0E1ne1 ¼ n0ωmωωMω1 k ie 11ikE1 n0þikE1 ¼E1 ¼ 2 Ω2p þ ω2p20 ω ω M mωω2 ¼ ω2p þ Ω2p446Appendix D: Answers to Some Problems4.4 Find ϕ1, E1, and v1 in terms of n1:ω n1k n0ieEq:½4-23 : E1 ¼n120 kEq:½4-22 : v1 ¼But E1 ¼ ikϕ1,e∴ϕ1 ¼ n120 k 2Hence, E1 is 90 out of phase with n1; ϕ1 is 180 out of phase; and v1 is either inphase or 180 out of phase, depending on the sign of ω/k.

In (a), E1 is foundfrom the slope of the ϕ1 curve, since E1 ¼ ∂ϕ1/∂x. In (b), E1/n1 / i sgn (k)∴ δ ¼ π/2. If ω/k > 0,E1 / exp iðkx jωjt π=2Þthe standing for the sign of k. Hence, E1 leads n1 by 90 . Opposite if ω/k < 0.4.511k1k ieikE1 ¼ en1 ¼ en0 v1 ¼ en0E12020 ω20 ω! mωω2pn0 e2¼0or∇1ik 1 E1 ¼ 0E120 mω2ω2ω2p∴2 ¼ 1 2ωAppendix D: Answers to Some Problems4474.7 (a)mn0 ðiωÞυ1 ¼ en0 E1 mn0 vυ1ivieE1υ1 1 þ¼ωmω1kikE1 ¼ en1 n1 ¼ n0 υ1 ðcontinuityÞ20ω1 k ieE1iv 1ikE1 ¼ e n01þ20 ω mωωiv ω2p ω2 þ ivω ¼ ω2pω2 1 þω(b) Let ω ¼ x + iy. Then the dispersion relation is x2 y2 + 2ixy + ivx vy ¼ ω2p .We need the imaginary part: 2xy + vx ¼ 0, y ¼ (1/2)v ∴ Im (ω) ¼ v/2.Since x ¼ Re (ω), v > 0, andE1 / eiωt ¼ eiωt e yt ¼ eixt eð1=2Þvtthe oscillation is damped in time.4.8 mn0(iω)v1 ¼ en0E1 en0(v1 B0).

Take B0 in the ẑ direction and E1 and k inthe ^x direction. Then the y-component isiωmv y ¼ evx B0vxω¼ iωcvySince ω ¼ ωh > ωc, jvx/vyj > 1; and the orbit is elongated in the ^x direction,which is the direction of k.4.9 (a)1∇ E1 ¼ en1 k ¼ kx ^x þkz^z E y ¼ k y ¼ 0201iðkx Ex þ kz Ez Þ ¼ en120We need n1:∂n1þ n0 ∇ v1 ¼ 0∂t iωn1 þ n0 iðkx vx þ kz vz Þ ¼ 0We need vx, vz:Mn0 ðiωÞv1 ¼ en0 E1 en0 ðv1 B0 Þ448Appendix D: Answers to Some ProblemsieiωcEx vymωωiωcvxy- component : v y ¼ 0 þω1ieω2ieω2Ex þ c2 vx ¼Ex 1 c2vx ¼ mωmωωωx -component :vx ¼ ieEzvz ¼ mω"#1n0 ieω2cContinuity : n1 ¼þ k z Ezk x Ex 1 2ω mωω#1"en0 ieω2cþ k z Ezk x Ex þ k z E z ¼ ik x Ex 1 2ωe0 ω mωz-component :kx ¼ k sin θkz ¼ k cos θ"#12ωpω2c2222∴E1 sin θ þ kE1 cos θ ¼ 2 kE1 sin θ 1 2þ kE1 cos θωω"#1ω2pω2c221 ¼ 2 sin θ 1 2þ cos θωωω2c ω2pω2c221 2 ¼ 2 1 cos θ þ 1 2 cos θωωω(b)ω2p ω2cω2 ω2c ω2p ¼ 2 cos 2 θω 2222 2ω ω ωh þ ω p ωc cos 2 θ ¼ 0 QEDω4 ω2h ω2 þ ω2p ω2c cos 2 θ ¼ 01=22ω2 ¼ ω2h ω4h 4ω2p ω2c cos 2 θFor θ ! 0, cos 2 θ ! 1;1=22ω2p þ ω2c 4ω2p ω2c¼ ω2p þ ω2c ω2p ω2c2ω2 ¼ ω2h ω2 ¼ ω2p , ω2cThe ω ¼ ωp root is the usual Langmuir oscillation.

The ω ¼ ωc root is spuriousbecause at θ ! 0, B0 does not enter the problem. For θ ! π/2, cos2 θ ! 0, 2ω2Appendix D: Answers to Some Problems449¼ ω2h ω2h , ω ¼ 0, ωh : The ω ¼ ωh root is the usual upper hybrid oscillation.The ω ¼ 0 root has no physical meaning, since on oscillating perturbation wasassumed.(c)11ω4 ω2h ω2 þ ω4h ¼ ω4h ω2p ω2c cos 2 θ44 2 1 21ω2 ω2h þ ω p ωc cos θ ¼ ωh22ð y 1Þ 2 þ(d)ω2 ¼12x¼1a2QEDa ¼ 1/2(ωc/ωp + ωp/ωc)15/41ωp/ωc121(e)2 2 1=2ω2p þ ω2c ω2p þ ω2c 4ω2p ω2c cos 2 θLower root: Take () sign; ω is maximum when cos2 θ is maximum (¼1).Thusi h1ω2 <ω2p þ ω2c ω2p ω2c 2¼ ω2c¼ ω2pif ω p > ωcif ωc > ω pUpper root: Take (+) sign; ω is maximum when cos2 θ ¼ 0, ω2 ¼ ω2h : Thusω2þ < ω2h : This root is minimum when cos2 θ ¼ 1; thusω2þ >i hω2p þ ω2c þ ω2p ω2c 12¼ ω2p¼ ω2cif ω p > ωcif ωc > ω p450Appendix D: Answers to Some Problems4.10 Use V+, N+ for proton velocity and densityV, Nfor antiprotonsv, nfor electronsv+, n+ for positrons(a)∇ E ¼ B_€E_ ∇ ∇ E ¼ μ _jþ E∇ B¼μ0 j þ 20c2cw2..−(k × k × E) = − m0n0e (v+ − v−) − 2 Eco= k 2E − k(k • E)1ω2 c 2 k 2 E ¼n0 eðv_ þ v_ Þ20emn0 v ¼ en0 E v_ ¼ Em1eω2 c 2 k 2 ¼n0 e ð1 þ 1Þ ¼ 2ω2p20mω2p ¼n0 e220 mω2 ¼ 2ω2p þ c2 k2(Or the 2 can be incorporated into the definition of ωp.)(b) ∇ · E1 ¼ (1/E0)(N+N + n+n)1, where n+ ¼ n0eeϕ/KT+, n¼ n0eeϕ/KT.Let T+ ¼ T ¼ Te n1 ¼ n0eϕ/KTe.

Note: N0 ¼ n0 n0.kk∂N þ N 0 ∇ V ¼ 0 N 1 ¼ N 0 V ¼ n0 V ωω∂tMðiωÞV ¼ eE1 ¼ ikeϕ ðMþ ¼ M ¼ MÞk2 n0 eϕk eϕN 1 ¼ 2V ¼ ω MωM 22e kk n0 eϕeeϕþ ðn0 n0 Þ∇ E1 ¼ k2 ϕ ¼þ20 ω2 ω2M20KT e2n0 e2 2k2n0 e 212 kϕ¼2ϕ¼2ϕΩp 2ω2 0 M ω220 kT eλ2Dk2 λ2D þ 2 ¼2k2 2 22k2Ω p λD ¼ 2 v2s2ωωv2s kT eMAppendix D: Answers to Some Problems4512v2sv2sω2¼¼λD 222 þ k λD 1 þ ð1=2Þk2 λ2Dk2ck4.11 en¼ωkT e 20n0 e21=2ω2pc2 k 2¼1 2 ¼22ωωpffiffiffiffi∴ en¼ 2ω2 ¼ ω2p þ c2 k24.12 In ∇ B ¼ μ0j1, j1 is the current carried by electrons only, since Cl ions aretoo heavy to move appreciably in response to a signal at microwave frequencies.

Hence,j1 ¼ n0e eve ¼ ð1 κÞn0 eve1If ωp is defined with n0 (i.e., ω2p ¼ n0 e2 =20 m), the dispersion relation becomesω2pc2 k 2¼1ð1κÞω2ω2Cutoff occurs for f ¼ (1κ)1/2 fp ¼ (0.4)1/2(9)(n0)1/2, wheref ¼c 3 1010¼¼ 1010λ3Thus1010n0 ¼ð0:63Þð9Þ2¼ 3:1 1018 m34.13 (a) Method 1: Let N ¼ No. of wavelengths in length L ¼ 0.08 m, N0 ¼ No. ofwavelengths in absence of plasma.LN¼λLN0 ¼λ02πλ¼kck¼ωω2p1 2ω!1=2!1=2ω2pL LkLL ω¼ 1 2ΔN ¼ N 0 N ¼ λ0 2π λ0 2π cω2!1=2 3ω2pω1L5 ¼ 0:1¼ ∴ΔN ¼ 41 1 22πc λ0λ0ω452Appendix D: Answers to Some ProblemsL0:08¼ 10¼λ0 0:008ω2 1=2∴ 1 ω2p¼ 1 102f 2p ¼ f 2 2 102n¼1f 2p2¼ 1 2 102f 2c¼2 102 ¼ 2:8 1019λ02:8 1019ð 9Þ 2¼ 3:5 1017 m3Method 2: Let k0 ¼ free-space k.

The phase shift isΔϕ ¼ðL0Δk dx ¼ ðk0 kÞL ¼ ð0:1Þ2πThis leads to the same answer.(b) From above, ΔN is small if ω2p /ω2 is small; hence expand square root:"L1ΔN λ01 ω2p12 ω2!#¼L 1 ω2p/nλ 0 2 ω2QED4.14 From Eq. (4.101a), we have for the X-waveω2p ωcEy ¼ 0ω2 ω2h Ex þ iωAt resonance, ω ¼ ωh) Ey ¼ 0, E ¼ Ex ^x : Since k ¼ kx ^x , Ek; and the wave islongitudinal and electrostatic.4.15 Since ω2h ¼ ω2c þ ω2p ; clearly ωp < ωh. Further,1=2 1ωc þ ω2c þ 4ω2p21=2 122< ωc þ ωc þ 4ωc ω p þ 4ω p21¼ ωc þ ωc þ 2ω p ¼ ω p ∴ ωL < ω p2ωL ¼Also,ωR ¼and121=2 22ωc þ ωc þ 4ω p> ωcAppendix D: Answers to Some Problems453ω2R ωR ωc ω2p ¼ 0∴ðEq: ½4-107Þω2R ¼ ωR ωc þ ω2p > ω2c þ ω2p ¼ ω2h4.17 (a) Multiply Eq. (4.112b) by i and add to Eq.

(4.112a):ωc Ex þ iE y ¼ 0ω2 c2 k2 α Ex þ iE y þ αωNow subtract from Eq. (4.112a):ωc Ex iE y ¼ 0ω2 c2 k2 α Ex iE y αωThusFðωÞ ¼ ω2 c2 k2 αð1 þ ωc =ωÞGðωÞ ¼ ω2 c2 k2 αð1 ωc =ωÞSinceα FðωÞ ¼ ω2G ð ωÞ ¼ ω2ω2p1 ω2c =ω2ω2p =ω2c2 k 2 211 ωc =ωω!ω2p =ω2c2 k 2 211 þ ωc =ωω!From Eqs. (4.116) and (4.117),FðωÞ ¼ 0 for the R wave andGðωÞ ¼ 0 for the L wave(b) Ex ¼ iEy) Ey ¼ iEx. Let Ex ¼ f(z) eiωt. ThenEy ¼ f(z)i eiωt ¼ f(z) eiωt+i(π/2) ¼ f(z) ei[ωt(π/2)]454Appendix D: Answers to Some ProblemsEy lags Ex by 90 .

Hence E rotates counterclockwise on this diagram.This is the same way electrons gyrate in order to create a clockwisecurrent and generate a B-field opposite to B0. For the L wave, Ey ¼ iExso thatEy ¼ f(z) ei(ωt+π/2) and Ey leads Ex by 90 .(c) For an R-wave, Ey ¼ iEx. The space dependence is Ex ¼ f(t) eikz, Ey ¼ f(t)ieikz ¼ f(t) ei(kz+π/2) For k > 0, Ey leads Ex (has the same phase at smaller z).For k < 0, Ey lags Ex (has the same phase at larger z).4.19ω2p =ω2c2 k 2¼1ω21 ωc =ωc2 ð2Þv3ϕ¼1c2 v2ϕω2p =ω21 ωc =ωdvϕ1¼ ω2pð2ω ωc Þ ¼ 02dωðω ωωc Þ2Appendix D: Answers to Some Problems45512∴2ω ωc ¼ 0 ω ¼ ωcAt ω ¼ 1/2ωc,ω2p4ω2pc2>1¼1¼1þ1 2 1 2ω2cv2ϕωc ω c24∴ vϕ < c:ω2p =ω2c2 k 2¼1ω21 ωc =ω4.20c 2 k 2 ¼ ω2 ωω2pω ωcð ω ωc Þ ω 2c2 2k dk ¼ 2ω dω ω p dωðω ω#c Þ2"ωc ω2p¼ 2ω þdωð ω ωc Þ 2dωkc2kc2¼2dk ω þ ωc ω2p =2ðω ωc Þω þ ω2p =2ωcif ω ωcBut!1=2!22 1=2ωωωppck ¼ ω2 ω2 þif ω ωc1 ωc =ωωc1=21=2ω2 þ ωω2p =ωc1 þ ω2p =ωωcdω¼c∴¼cdkω þ ω2p =2ωc1 þ ω2p =2ωωcTo prove the required result, one must also assume v2ϕ c2 ; as is true forwhistlers, so that ω2p =ωωc 1 (from line 1).

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