 By P.S. Dhogal

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Example text

2 The conductivity σ does not enter this equation directly, although it may be written as σ = ωp2τ/4π. 5 × 1016 s −1 . The homogeneous equation has a solution of the form x ( t > 0 ) = Ae −λt sin ( ωt + φ ) , 12 2 2 where ω = ⎡ωp + ( ρ 2 ) ⎤ and λ = ρ/2. To this we add the particular solution x = ⎣ ⎦ –e/mω and find A and φ to satisfy the initial conditions x(0) = 0 and x ( 0 ) = 0. 11. The Laplacian ∇ 2 ϕ = 0, whence d 2f − K 2f = 0 . 2 dz This has solutions f = AeKz for z < 0 f = Ae − K ( z −d ) for z > d f = B cosh K ( z − d 2 ) for 0 < z < d .

2 s −1 . It is true that ωτ will be <<1 for any reasonable relaxation time, but ωc τ > 1 can be shown to be the applicable criterion for helicon resonance. 5. md 2r/dt 2 = − mω2r = −eE = 4πeP/3 = −4πne 2r/3 . Thus ωο2 = 4πne 2 3m. 2 6. md 2r dt 2 = −mω2r = −(e c) ( v × Bzˆ ) − mωο r, where ωo2 = 4πne2/3m, from the solution to A. Thus, with ωc ≡ eB/mc, 2 −ω2 x = iωωc y − ωo x ; 2 −ω2 y = −iωωc x − ωo y . 14-1 2 2 Form ξ ≡ x + iy; then −ω2ξ − ωωc ξ + ωο ξ = 0, or ω2 + ωωc − ωο = 0, a quadratic equation for ω.

2 3. (a) The equation of motion of the electrons is −ω2 x e = −(e/me )E x + iωωe ye ; − ω2 ye = −(e/me )E y − iωωe x e . For the holes, −ω2 x h = (e/m h )E x + iωωh y h ; − ω2 y h = (e/m h )E y − iωωh x h . The result follows on forming ξe = xe + iye and ξh = xh + iyh. (b) Expand −1 (ωe + ω) −1  ωe (1 − ω / ωe ) and (ωh − ω)  ωh −1 (1 + ω / ωh ) . In this approximation −1 −1 (ξh − ξe ) / E +  (c/B)(ωh + ωe ) = (c2 / eB2 ) (m h + me ) . 4. From the solution to Problem 3 we have P + = pe 2 E + / m h ωh ω , where we have dropped a term in ω2 in comparison with ωhω.