Motion of Charges in Unifor E and Fields Assue an ionized gas is acted upon by a unifor (but possibly tie-dependent) electric field E, and a unifor, steady agnetic field. These fields are assued to be externally supplied, and to be unaffected by the presence or the otion of the charges theselves. In reality, the charges ay generate significant space charge, and odify the potential according to Poisson s equation ( 2 φ = ρ/ɛ 0 ), or generate significant net current, and hence odify through Apere s equation ( /μ 0 = j + ɛ0 E ). If so, the analysis t in this section would have to be viewed at ost as part of a ore coplete forulation that would incorporate these odifications in a self-consistent fashion. For a particle of ass and charge q, oving at velocity w, the equation of otion is, dw dt = q(e + w ) () The projection on (w,e ) is siply, dw dt = qe (2) and this part of the otion is unaffected by. We will concentrate here on the projection perpendicular to (w, E ). We can take advantage of the constancy of and the fact that E is unifor, and use coplex algebra to strealine the analysis. Take axis (Oxyz), where lies along O z, and define, w = w x + iw y (3) E = E x + ie y (4) So that w =(w x + iw y ) î z =(w y iw x ) = iw. The perpendicular part of () is then, dw = q(e iw) dt or, dw dt + iω cw = q E (5) where, q ω c = (6) is the cyclotron frequency, or gyro frequency, of the particle. (a) Case with no electric field The general solution of (5) when E = 0 is, w = Ae iωct (7) and since w = dz with z = x + iy, then dt A z = i e iωct (8) ωc
y r L y w x q > 0 q < 0 w r L x The constant A in (7) can be taken to be real (this defines t = 0 as the tie of crossing through the y =0 axis), and can be identified as the constant agnitude A = w of the perpendicular velocity. Fro (8), the position vector of the particle is 90 rotated with respect to the velocity, and has a agnitude r L = A ω c = w, q/ r L = w q (9) This is called the Laror radius of the particle, for a velocity w. This is also called the gyro radius. oth vectors w and z are seen fro (7) and (8) to rotate at angular velocity ω c. The otion is in the sense shown in the figure above, depending on polarity. The gyro frequency for electrons can be fairly high, usually second only to the plasa frequency, aong the iportant scales. For exaple, in a Hall thruster, with T e =20eV = 20 e =20, 600 = 232, 000K, the ean theral speed is, k c e = 8 kt e = π e 8.38 0 23 2.32 0 5 π 9. 3 =2.99 0 6 s With a agnetic field of the order of 0.0 Tesla, we find, e ω ce = =.6 0 9 0.0 =.8 0 9 rad/s ; r e 9. 0 3 L = c e ω ce =.7 The ions are uch heavier, especially in propulsion devices using Xenon. Fro (6) and (9), ω ci /ω ce = e / i and r Li /r Le = i / e. In the Hall thruster exaple, with Xenon, i / e = 3 850 = 2.42 0 5,soω ci =7.4 0 3 rad/s =.2kHz and r Li =0.84. Since the device is only a few c in typical diension, the ion trajectories are nearly straight in it (weak ion agnetization), but the electron trajectories (unless disrupted by collisions) are doinated by very tight gyro rotations (strong electron agnetization). These conclusions would be strongly odified if the ions were hydrogen and were higher. For exaple, in fusion, 5T, and we find for H + ions at kev (c i =5 0 5 /s), r Li =, eaning that even the ions are now strongly agnetized. Fro the sense of the gyro otion of both the electrons and ions, we see that this otion will induce in both cases soe agnetic field in a direction opposite the original. This is called diaagnetis. For a general distribution of current densities j, the induced is, ind = (μ0 μ) (0) particles where μ 0 =4π 0 7 Hy/ is the pereability of vacuu and μ is the agnetic oent, defined as, μ = r jdv () 2 2
For an isolated gyrating charge, the current is concentrated along the circular path, and has a ean value q/(period) =qω c /2π. Dividing by the infinitesial area δa swept by q, we get j, the current density, and so, qω c r j = r L 2π δa The volue to be integrated over is 2πr L δa, therefore, μ = r 2 qω c 2 L 2π δa 2πδA = ( ) 2 ( ) w q q 2 q or in agnitude, 2 w2 μ = where the nuerator is the kinetic energy of the perpendicular otion of the particle. (2) Assue n particles per unit volue are agnetized (usually, in Electric Propulsion, this is n e, the electron density, but in a fusion plasa, n = n e + n i, as we saw). Then, fro (0), = nμ 0 μ ind and, fro the definition of teperature, 2 w2 accounting for the two diensions involved in w, or, ind ind kt = μ0n = particles 2 kt, with the factor of 2 2 = nkt (kinetic pressure) = (3) 2 /μ 0 agnetic pressure For the Hall thruster exaple, with n = ne 0 8 3, T =20eV, =0.0T, ind/ 4 0 3, which can be ignored. For fusion, n 0 9 3, T 0 3 ev, 5T, so ind / 8 0 5, also weak diaagnetis. Diaagnetic effects could be large in an electric propulsion plue propagating relatively far fro the source, under the effect of the geoagnetic -field ( 3 0 5 T ). Assuing n 0 5 3, T ev, we find ind / 0.2, so that the plue will tend to exclude the abient field. A non-unifor distribution of agnetic oents, such as fro a density gradient, gives rise collectively to a acroscopic current, called agnetization current or diaagnetic current. An illustration is shown in the figure below: the eleentary gyro currents in the vertical direction cancel pairwise inside the box, but there is a net upwards current on the left edge, and a greater downwards current on the right edge. Overall, a region with a positive dn/dx, as indicated, sees a negative j y due to this effect. A ore detailed analysis follows. 3
We begin by recasting the agnetic oent in the for, μ = 2 r2 LI2π = IA =+IA (4) y where A = πr 2 L is the area of the sall gyro loop, and, as before, I = q/(gyro period). x Consider a acroscopic surface Σ containing a contour C, and let us calculate the net current which crosses Σ due to the gyro otion of the particles lying near Σ. Only particles actually near the contour C itself contribute, since otherwise the current loops coe into and out of the enclosed portion of Σ. When A is parallel to the line eleent dl, the particle contributes a net passing current I (case (a) in the figure). When A is perpendicular to dl (case (b)) there is no contribution. ds Σ C A A dl A (a) (b) Overall, any particle in the volue eleent A dl = A C dl contributes I to I crossing, I = dl crossing nia dl = M (5) where M = nμ = nia is the so-called Magnetization vector. Converting to a surface integral, ( ) I crossing = M ds (6) where ds is a surface eleent on Σ, and the vector ds points perpendicular to Σ. Clearly, the agnetization current density is then, Σ j M = M (7) This is in addition to other currents which ay arise when the guiding centers, i.e., the centers of the Laror circuits, ove about. These guiding-center otions are called drifts, and will be exained later. A physical interpretation of M is provided by rewriting the induced agnetic field as, ind = +μ 0 M (8) 4
so that +μ 0 M is the contribution to due to eleentary gyro otions. 0 t Consider now a quasi-static situation, where frequencies are low enough that ɛ E << j. This turns out to be applicable to all our probles of interest, except EM wave propagation. We then have, = j (9) μ0 where j is the total current density. If we separate this into the drift current j d and the agnetization current, μ0 = j d + M or M = j d μ 0 This propts the introduction of a vector H (the agnetic field strength) such that, H = and then, H = j d Since in our situation M is itself proportional to, ind M = + = nμ μ0 = n 2 w2 2 M (20) μ 0 we can define a odified pereability of the ediu, μ, such that (20) can be written, H = μ (2) where, = + n w 2 2 = + P (22) μ μ 0 2 μ 0 2 so, in this (diaagnetic) case, μ << μ 0. The opposite is true of ferroagnetic edia, where ind is in the sae direction as. More directly, we can still use μ 0, but reeber to count both drift and agnetization currents. (b) Effect of a unifor electric field (steady) ( ) Returning to Eq. () dw/dt = q E + w, we notice that for that velocity w = v D such that E + v D = 0, the electrostatic and the agnetic force cancel each other, and dv D /dt = 0, which is consistent with that cancellation. To solve for v D, we write, ( E + vd ) = 0 5
E + ( v D ) ( ) v D = 0 The part of v D parallel to is non-zero only if E = 0 (otherwise the particles accelerate along so v D is no longer constant). So, assue v D is to, and v D = 0, with the result that, E v = D (23) 2 which is a constant drift velocity perpendicular to both E and, and independent of the charge, the ass or the charge s sign. In agnitude, E v D = (24) This result can also be obtained fro the coplex forulation of Eq. (5). With E = constant, the equation adits a particular solution, iω c w = q E q w = v D = i E ω c and since ω = q/ then, E v D = i (25) which is equivalent to (23). Adding to this an arbitrary aount of the hoogeneous solution w H = e iωct, the general solution is, w = Ae iωct which is a superposition of the E drift and Laror rotations. E i (26) The nature of this total otion is best understood for a siple case when E = ie y, giving w = Ae iωct + E y /, with a real, positive drift along x (real E y /). Suppose we ipose w =0att = 0, then A = E y / and w = Ey ( e iωct ). Real part: Iaginary part: w x = E y [ cos(ω ct)] x = Ey [ω c t sin(ω c t)] ω c E y w y = sin(ω c t) y = Ey [ ω c cos(ω ct)] 2E y y ax = ωc = 2E y q 2 So, w x is always positive, except it becoes zero at ω c t =0, 2π, 4π,... and w y is zero at ω c t =0,π,2π, 3π, 4π,... and oscillates (+) and ( ). For q>0 (heavy ion) and q<0 (light electron), the otion is sketched below: 6
E v D The figure shows why the drift is in the sae direction regardless of the charge s sign. Ions and electrons start out under E alone (zero velocity gives zero agnetic force) in opposite directions. Then, as the vertical velocity develops (with opposite signs), the agnetic force qw turns the trajectories to the right in both cases, since qv y > 0 for both. We can also see that the penetration in the y direction is uch greater for the ions, and that the E y field does separate the charges, as one would expect, but only to a finite extent (with no field, they would siply accelerate away fro each other forever). In a Hall thruster, where r Li >> r Le and r Li >> L, only the electrons drift. In a fusion plasa, both ions and electrons do, and as a consequence the E drift current is zero. In the Hall thruster case, the ion current in the E direction is nearly zero, but electrons drift at the full E/, and there is a nonzero E drift current, which is called the Hall current: j Hall = en e E 2 (27) (c) Effect of a tie-varying (but unifor) E field Returning to Eq. (5), if E = E(t), the general solution can be obtained by the ethod of variation of the constant in the hoogenous part. The result is, [ t ] q w = A + Eeiωct dt e iωct (28) The integral can be evaluated explicitly for sinusoidal E(t) of any frequency. ut before looking at that case, consider the case where E(t) has any shape, but has a slow variation, when copared to the cyclotron otion. Since ω c is very high for electrons in general, these slow changes ay not be very slow after all. A foral solution can then be obtained using integration by parts in (28). Using, w = and repeating the process, [ e iωct dt = A + E i eiωct i iq d ( e iωct) then, w = Ae iωct i E + i 7 t e iω de ] ct dt dt e iωct de iq dt +...
w = Ae iωct i E + de q 2 +... (29) dt Higher ters would contain higher derivatives of E, and are neglected for these slow variations. The first two ters are failiar (gyro otion, E drift). The third ter is a new drift, called Polarization drift, v p = de (30) q 2 dt The reason of this nae is the following: As seen fro (30), ions and electrons will drift in opposite directions, and so there will always be a current associated with this drift. Assuing n e = n i, q e = e, and q i = e, this current is, ( i j p = ev pi n e ev pe n e = en e e + ) e de 2 e 2 dt The e part is negligible, so the ion polarization current doinates. The overall plasa ass density is ρ p = n e ( i + e ) n e i, so, ρp de j p = 2 dt This is the sae for as Maxwell s polarization (or displaceent) current j Maxw = ɛ 0 E/ t, hence the nae. It is natural to ask whether j p is or is not uch larger than j Maxw, since 2 8 3 j Maxw is usually negligible. Their ratio is ρ p /ɛ 0 ; for a Hall thruster plasa (n e 0, 25 i = Xe =2.2 0 kg, 0.0T ) we find, j p = ρ p ɛ 0 0 8 2.2 0 25 = 2.5 0 8 2 8.854 0 2 (0 2 ) 2 j Maxw This is a clear warning that the displaceent current (3) ay not be negligible, even when Maxwell s is. Carrying the thought forward, since we know that, in vacuu, Maxwell s currents are responsible for EM waves, we now realize that in a plasa, where j p replaces j Maxw, there should be forally identical waves due to j p. To see this, assue j p is the only current present, and ignore j Maxw : (3) ρ p E μ0 2 t (32) For sall perturbations, = 0 + and ρ p = ρ p0 + ρ p. Since << 0 and assuing there is no electric field in the background E = E 0 + E, Eq. (32) reads, = ρ p0 E μ0 0 2 t Take the curl of this equation, and use Faraday s law E = / t, μ0 = ρ p0 2 0 8 2 t 2
and since ( )= ( ) 2, we obtain the wave equation, with a propagation velocity, 2 t 2 = 2 0 2 μ 0 ρ p0 (33) v A = 2 0 (34) μ 0 ρ p0 which is called the Alfven velocity. These waves are, like ordinary EM waves, transverse, in the sense that E and are perpendicular to the propagation vector, and to each other. Since we found that ρ p0 /ɛ 0 0 2 >>, we can put, μ 0 ρ p0 / 2 0 ɛ 0 μ 0 >> or c 2 >> v 2 A i.e., the Alfven velocity is generally well below the speed of light. Note: If we include the charge otion along 0, we can obtain a different type of copressive Alfven waves. 9
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