cha1_4 - which if is larger and therefore the i term is...

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PHYSICS 880.06 (Fall 2005) Proble Set 1 Solution (1.1) A&M Problem 1.4 d p d t = - e E + p mc × H · - p τ , H = H z ˆ z , E ( t ) = Re( E ( ω ) e - iωt ) . (a) Seek steady-state solutoin of this form p ( t ) = Re( p ( ω ) e - iωt ) , - p ( ω ) = - e E ( ω ) + p ( ω ) mc × H - p ( ω ) τ . - + 1 τ p x ( ω ) = - e E x ( ω ) + 1 mc p y ( ω ) H z , - + 1 τ p y ( ω ) = - e E y ( ω ) - 1 mc p x ( ω ) H z , - + 1 τ p z ( ω ) = - eE z ( ω ) . E ( ω ) = E x ( ω x + E y ( ω y , E y = ± iE x , E z = 0 . The solution is p x = - 1 - i ( ω ω c ) τ E x , p y = ± ip x , p z = 0 , where ω c = eH z mc . The current density is j = - ne p m , j x = σ 0 1 - i ( ω ω c ) τ E x , j y = ± ij x , j z = 0 , where σ 0 = ne 2 τ m . 1
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(b) From Maxwell equations, -∇ 2 E = ω 2 c 2 1 + 4 πiσ ω E , σ xx = σ yy = σ 0 1 - i ( ω ω c ) τ . Look for a solution of this form E x ( k, t ) = E 0 e - i ( kz - ωt ) . Plugging in, k 2 c 2 = ω 2 ˆ 1 - ω 2 p ω 1 ω ω c + i/τ ! = ω 2 ² ( ω ) , where ² ( ω ) = 1 - ω 2 p ω 1 ω ω c + i/τ , ω 2 p = 4 πne 2 m . (c) For polarization E y = iE x , ² ( ω ) = 1 - ω 2 p ω 1 ω - ω c + i/τ . (SKETCH/PLOT?...) Assuming ω p c 1 and ω c τ 1, for large ω , one can rewrite the above eq. as ² ( ω ) = 1 - ω p ω 1 ω ω p - ω c ω p + i τω p 1 - ω p ω 1 ω ω p = 1 - ω 2 p ω 2 , which is positive for ω > ω p , and real solutions for k exist. For small but positive ω , one has ² ( ω ) = 1 + ω 2 p ω 1 ω c - ω - i/τ , which, if
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Unformatted text preview: which, if is larger and therefore the i/ term is ignored, is positive for < c , and consequently real solutions for k exist. (d) For c (but still > 0), ( ) 1- 2 p 1- c 2 p c , k 2 c 2 = 2 2 p c 2 = 2 p c , = c k 2 c 2 2 p . = 1 cm, T = 10 kilogauss. c = 3 10 10 cm/s, e = 4 . 8 10-10 esu. Taking a typical metalic electron density of 10 23 / cm 3 , the helicon frequency is f = 2 = eH mc k 2 c 2 4 ne 2 m 1 2 = Hc 8 2 ne k 2 = Hc 8 2 ne 2 2 = Hc 2 ne 1 2 = (10 4 )(3 10 10 ) (2)(10 23 )(4 . 8 10-10 ) 1 (1) 2 = 3 . 1 Hz . 2...
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