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Unformatted text preview: Phys 2214 Homework #7 Solutions November 1, 2011 1. EM Waves in Conductors. The current density is ~ J = ~ E/ , where is the electrical resistivity. Amperes law in differential form is ~ B = ~ E + ~ E t . If is small (a good conductor), then the first term on the right hand side will dominate and we can neglect the second term. (a) Faradays law is ~ E = ~ B/t . Taking the curl of this gives ( ~ E ) = ~ B t = t ~ B , where the spatial and temporal derivatives have been interchanged. Replace the left hand side using the vector identity ( ~ E ) = ( ~ E ) 2 ~ E . Then ( ~ E ) 2 ~ E = t ( ~ B ) . We know that ~ E = 0, from Gauss law with no charge density, and ~ B = ( / ) ~ E from Amperes law in a conductor. Then 2 ~ E = ~ E t . 1 (b) Let ~ E ( z,t ) = E x e i ( z t ) i . Sugstituting into the above wave equation gives ( i ) 2 ~ E = ( / )( i ) ~ E . Then 2 = i/ , or = q i/ . Using i = (1+ i ) / 2, we get = q / 2 (1+ i ) (1 + i ). (c) Substituting this equation for into the plane wave solution gives ~ E = E x e i [ (1+ i ) z t ] i = E x e z e i ( z t ) i , which is a plane traveling wave whose amplitude decays with dis tance z into the conductor. (d) The electric field decays to 1 /e of its initial value when = 1, or = q 2 / (the skin depth). (e) Assume = = 4 10 7 Tm/A . i. For Cu, = 1 . 7 10 8 m . For f = 60 Hz , = 8 . 47 10 3 m = 8 . 47 mm . For f = 840 MHz , = 2 . 3 10 6 m = 2 . 3 m ....
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This note was uploaded on 01/02/2012 for the course PHYSICS 2214 at Cornell University (Engineering School).
 '11
 DAVIS,J.C.
 Current, Work, Waves And Optics

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