P214soln7 - Problem set 7 solutions by Ali Vanderveld(1 The...

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Unformatted text preview: Problem set 7 solutions by Ali Vanderveld (1) The electric field is ~ E ( ~ r,t ) = < "- d ~ A dt # = < h ˆ ziωAe i ( ~ k · ~ r- ωt ) i . (1) Since the electric field is pointing solely in the z direction, Gauss’s law will be satisfied if ∂E z ∂z = 0 , (2) which means that E z must not be a function of z ; this will happen if ~ k does not have a z component, i.e. ~ k = ( k 1 ,k 2 , 0). In order to find the dispersion relation, plug ~ E into the wave equation ∇ 2 ~ E = 1 c 2 ∂ 2 ~ E ∂t 2 (3) to see that q k 2 1 + k 2 2 = ~ k = ω c . (4) Also, this wave is propagating in the ~ k direction, and it is polarized in the z direction. (2) We are given the vector potential, which has y and z components and thus is only a function of x position ~ A ( x,t ) = ( ˆ y + e iδ ˆ z ) Ae i ( kx- ωt ) , (5) where the physical electric and magnetic fields are ~ E ( x,t ) = < "- d ~ A dt # (6) and ~ B ( x,t ) = < h ~ ∇ × ~ A i , (7) respectively. To this end, we can calculate d ~ A dt =- iω ~ A =- ( ˆ y + e iδ ˆ z ) iωAe i ( kx- ωt ) (8) 1 and ~ ∇ × ~ A = ∂A y ∂x ˆ z- ∂A z ∂x ˆ y = ( ˆ z- e iδ ˆ y ) ikAe i ( kx- ωt ) (9) such that ~ E ( x,t ) = < ( ˆ y + e iδ ˆ z ) iωAe i ( kx- ωt ) =- ωA { ˆ y sin( kx- ωt ) + ˆ z [sin δ cos( kx- ωt ) + cos δ sin( kx- ωt )] } =- ωA [ˆ y sin( kx- ωt ) + ˆ z sin( kx- ωt + δ )] (10) and ~ B ( x,t ) = < ( ˆ z- e iδ ˆ y ) ikAe i ( kx- ωt ) = kA { ˆ y [sin δ cos( kx- ωt ) + cos δ sin( kx- ωt )]- ˆ z sin( kx- ωt ) } = kA [ˆ y sin( kx- ωt + δ )- ˆ z sin( kx- ωt )] . (11) Notice that B y =- k ω E z (12) and B z = k ω E y ; (13) using this information, we can then calculate the dot product ~ E · ~ B = E y B y + E z B z = E y- k ω E z + E z k ω E y = 0...
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This note was uploaded on 09/25/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Spring '07 term at Cornell.

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P214soln7 - Problem set 7 solutions by Ali Vanderveld(1 The...

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