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As the field lines move through the plasma, the induced currents cause ohmicheating of the plasma. This energy comes from the energy of the field. The energylost per m3 in a time τ is ηj2τ. Sinceμ0 j ¼ ∇ B BLð6:17Þfrom Maxwell’s equation with displacement current neglected, the energy dissipation is1946 Equilibrium and Stabilityη j2 τ ¼ ηBμ0 L2 2μ0 L2 B2B¼¼2ημ02μ0ð6:18ÞThus τ is essentially the time it takes for the field energy to be dissipated intoJoule heat.Problems6.1. Suppose that an electromagnetic instability limits β to (m/M)1/2 in a D–Dreactor.
Let the magnetic field be limited to 20 T by the strength of materials.If KTe ¼ KTi ¼ 20 keV, find the maximum plasma density that can becontained.6.2. In laser-fusion experiments, absorption of laser light on the surface of apellet creates a plasma of density n ¼ 1027 m3 and temperature Te ’ Ti ’104 eV. Thermoelectric currents can cause spontaneous magnetic fields ashigh as 103 T.(a) Show that ωcτei 1 in this plasma, and hence electron motion is severelyaffected by the magnetic field.(b) Show that β 1, so that magnetic fields cannot effectively confine theplasma.(c) How do the plasma and field move so that the seemingly contradictoryconditions (a) and (b) can both be satisfied?6.3. A cylindrical plasma column of radius a contains a coaxial magnetic fieldB ¼ B0 z^ and has a pressure profilep ¼ p0 cos 2 ðπr=2aÞ(a) Calculate the maximum value of p0.(b) Using this value of p0, calculate the diamagnetic current j(r) and the totalfield B(r).(c) Show j(r), B(r), and p(r) on a graph.(d) If the cylinder is bent into a torus with the lines of force closing uponthemselves after a single turn, this equilibrium, in which the macroscopicforces are everywhere balanced, is obviously disturbed.
Is it possible toredistribute the pressure p(r, θ) in such a way that the equilibrium isrestored?6.4 Diffusion of Magnetic Field into a Plasma1956.4 Consider an infinite, straight cylinder of plasma with a square density profilecreated in a uniform field B0 (Fig. P6.4). Show that B vanishes on the axis ifβ ¼ 1, by proceeding as follows.(a) Using the MHD equations, find j⊥ in steady state for KT ¼ constant.Fig. P6.4(b) Using ∇ B ¼ μ0j and Stokes’ theorem, integrate over the area of the loopshown to obtainX ð 1 ∂n=∂rdr, Bax Br ¼0KTBax B0 ¼ μ0Bð r Þ0(c) Do the integral by noting that ∂n/∂r is a δ-function, so that B(r) at r ¼ a isthe average between Bax and B0.6.5 A diamagnetic loop is a device used to measure plasma pressure by detectingthe diamagnetic effect (Fig.
P6.5). As the plasma is created, the diamagneticcurrent increases, B decreases inside the plasma, and the flux Φ enclosed by theloop decreases, inducing a voltage, which is then time-integrated by an RCcircuit (Fig. P6.5).(a) Show thatððV dt ¼ N ΔΦ ¼ N Bd dS,loopBd B B 01966 Equilibrium and StabilityFig. P6.5(b) Use the technique of the previous problem to find Bd(r), but now assume n(r) ¼ n0 exp [(r/r0)2]. To do the integral, assume β 1, so that B can beapproximatedð by B0 in the integral.12(c) Show that Vdt ¼ Nπr 20 β B0 , with β defined as in Eq. (6.8).6.5Classification of InstabilitiesIn the treatment of plasma waves, we assumed an unperturbed state which was oneof perfect thermodynamic equilibrium: The particles had Maxwellian velocitydistributions, and the density and magnetic field were uniform.
In such a state ofhighest entropy, there is no free energy available to excite waves, and we had toconsider waves that were excited by external means. We now consider states thatare not in perfect thermodynamic equilibrium, although they are in equilibrium inthe sense that all forces are in balance and a time-independent solution is possible.The free energy which is available can cause waves to be self-excited; the equilibrium is then an unstable one. An instability is always a motion which decreases thefree energy and brings the plasma closer to true thermodynamic equilibrium.Instabilities may be classified according to the type of free energy available todrive them.
There are four main categories.6.5 Classification of Instabilities197Fig. 6.7 HydrodynamicRayleigh–Taylor instabilityof a heavy fluid supportedby a light one6.5.1Streaming instabilitiesIn this case, either a beam of energetic particles travels through the plasma, or acurrent is driven through the plasma so that the different species have drifts relativeto one another. The drift energy is used to excite waves, and oscillation energy isgained at the expense of the drift energy in the unperturbed state.6.5.2Rayleigh–Taylor instabilitiesIn this case, the plasma has a density gradient or a sharp boundary, so that it is notuniform.
In addition, an external, non-electromagnetic force is applied to theplasma. It is this force which drives the instability. An analogy is available inthe example of an inverted glass of water (Fig. 6.7). Although the plane interfacebetween the water and air is in a state of equilibrium in that the weight of the wateris supported by the air pressure, it is an unstable equilibrium. Any ripple in thesurface will tend to grow at the expense of potential energy in the gravitation alfield.
This happens whenever a heavy fluid is supported by a light fluid, as is wellknown in hydrodynamics.6.5.3Universal instabilitiesEven when there are no obvious driving forces such as an electric or a gravitationalfield, a plasma is not in perfect thermodynamic equilibrium as long as it is confined.The plasma pressure tends to make the plasma expand, and the expansion energycan drive an instability. This type of free energy is always present in any finiteplasma, and the resulting waves are called universal instabilities.1986.5.46 Equilibrium and StabilityKinetic instabilitiesIn fluid theory the velocity distributions are assumed to be Maxwellian. If thedistributions are in fact not Maxwellian, there is a deviation from thermodynamicequilibrium; and instabilities can be driven by the anisotropy of the velocitydistribution.
For instance, if T|| and T⊥ are different, an instability called themodified Harris instability can arise. In mirror devices, there is a deficit of particleswith large v||/v⊥ because of the loss cone; this anisotropy gives rise to a “loss coneinstability.”In the succeeding sections, we shall give a simple example of each of these typesof instabilities. The instabilities driven by anisotropy cannot be described by fluidtheory and a detailed treatment of them is beyond the scope of this book.Not all instabilities are equally dangerous for plasma confinement. A highfrequency instability near ωp, for instance, cannot affect the motion of heavyions. Low-frequency instabilities with ω Ωc, however, can cause anomalousambipolar losses via E B drifts.
Instabilities with ω Ωc not efficiently transportparticles across B but are dangerous in mirror machines, where particles are lost bydiffusion in velocity space into the loss cone.6.6Two-Stream InstabilityAs a simple example of a streaming instability, consider a uniform plasma in whichthe ions are stationary and the electrons have a velocity v0 relative to the ions.That is, the observer is in a frame moving with the “stream” of ions. Let the plasmabe cold (KTe ¼ KTi ¼ 0), and let there be no magnetic field (B0 ¼ 0).
The linearizedequations of motion are then∂vi1¼ en0 E1∂t∂ve1þ ðv0 ∇ Þve1 ¼ en0 E1mn0∂tMn0ð6:19Þð6:20ÞThe term (ve1 · ∇)v0 in Eq. (6.20) has been dropped because we assume v0 to beuniform. The (v0 · ∇)v1 term does not appear in Eq. (6.19) because we have takenvi0 ¼ 0. We look for electrostatic waves of the formE1 ¼ EeiðkxωtÞ x^ð6:21Þwhere x^ is the direction of v0 and k. Equations (6.19) and (6.20) becomeiωMn0 vi1 ¼ en0 E1 ,vi1 ¼ieE^xMωð6:22Þ6.6 Two-Stream Instability199mn0 ðiω þ ikv0 Þve1 ¼ en0 E1ve1 ¼ ie E^xm ω kv0ð6:23ÞThe velocities vj1 are in the x direction, and we may omit the subscript x.
The ionequation of continuity yields∂ni1þ n0 ∇ vi1 ¼ 0∂tkien0 kni1 ¼ n0 vi1 ¼EωMω2ð6:24ÞNote that the other terms in ∇ · (nvi) vanish because ∇n0 ¼ v0i ¼ 0. The electronequation of continuity is∂ne1þ n0 ∇ ve1 þ ðv0 ∇ Þne1 ¼ 0∂tð6:25Þðiω þ ikv0 Þne1 þ ikn0 ve1 ¼ 0ð6:26Þne1 ¼kn0iekn0Eve1 ¼ ω kv0mðω kv0 Þ2ð6:27ÞSince the unstable waves are high-frequency plasma oscillations, we may not usethe plasma approximation but must use Poisson’s equation:ε0 ∇ E1 ¼ eðni1 ne1 Þð6:28Þ"11ikε0 E ¼ eðien0 kEÞþMω2 mðω kv0 Þ2#ð6:29ÞThe dispersion relation is found upon dividing by ikε0E:1 ¼ ω2p"m=M1þω2ðω kv0 Þ2#ð6:30ÞLet us see if oscillations with real k are stable or unstable. Upon multiplyingthrough by the common denominator, one would obtain a fourth-order equation forω.
If all the roots ωj are real, each root would indicate a possible oscillationE1 ¼ Eeiðkxω j tÞ x^If some of the roots are complex, they will occur in complex conjugate pairs. Letthese complex roots be writtenω j ¼ α j þ iγ jð6:31Þ2006 Equilibrium and StabilityFig. 6.8 The function F(x,y) in the two-streaminstability, when the plasmais stablewhere α and γ are Re(ω) and Im(ω), respectively. The time dependence is nowgiven byE1 ¼ Eeiðkxα j tÞ eγ j t x^ð6:32ÞPositive Im(ω) indicates an exponentially growing wave; negative Im(ω) indicatesa damped wave. Since the roots ωj occur in conjugate pairs, one of these will alwaysbe unstable unless all the roots are real.
The damped roots are not self-excited andare not of interest.The dispersion relation (Eq. (6.30)) can be analyzed without actually solving thefourth-order equation. Let us definex ω=ω py kv0 =ω pð6:33ÞThen Eq. (6.30) becomes1¼m=M1þ Fðx; yÞ:2xðx y Þ2ð6:34ÞFor any given value of y, we can plot F(x, y) as a function of x. This function willhave singularities at x ¼ 0 and x ¼ y (Fig. 6.8). The intersections of this curve withthe line F(x, y) ¼ 1 give the values of x satisfying the dispersion relation. In theexample of Fig. 6.8, there are four intersections, so there are four real roots ωj.However, if we choose a smaller value of y, the graph would look as shown inFig.
6.9. Now there are only two intersections and, therefore, only two real roots.The other two roots must be complex, and one of them must correspond to anunstable wave. Thus, for sufficiently small kv0, the plasma is unstable. Forany given v0, the plasma is always unstable to long-wavelength oscillations(small k and y). The maximum growth rate predicted by Eq. (6.30) is, form/M 1 (cf. Problem 6.6), ωm 1=3ð6:35ÞImωpMSince a small value of kv0 is required for instability, one can say that for a given k,v0 has to be sufficiently small for instability. This does not make much physicalsense, since v0 is the source of energy driving the instability.
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