EE330 R Prof Viviana Vladutescu Homework 5 1. A semi-infinite slab exists for z > 0 with = 300 S/m, r = 10.2, and r = 1.0. At the surface (z = 0), E(0,t) = 1.0 cos( x 106t) ax V/m. Find the instantaneous expressions for E and H anywhere in the slab. The general expression for E is: E( z, t ) 1.0e z cos x106 t z a x j j x106 4 x107 j 3.948 V m j j x106 10.2 8.854 x1012 j 284 x10 6 Here, (i.e. it is a good conductor), so 1 f 24.3 m 2 e j 45 0.115e j 45 So now we have V E( z, t ) 1.0e 24 z cos x106 t 24 z a x m To find B we’ll work in phasors. Es 1e z e j z a x , H s 1 a P Es 1 a z 1e z e j z a x 1 e z e j z a y 1 A e 24 z cos x106 t 24 z 45 a y 0.115 m A H( z, t ) 8.7e 24 z cos x106 t 24 z 45 a y m 2. A 0.1 m layer of copper is deposited atop a very thick slab of nickel. For a field incident on the copper surface, (a) calculate Rs at 1.0 GHz. Compare this with Rs at 1.0 GHz for (b) a semi-infinite slab of copper and (c) for a 0.1 m thickness of copper by itself. H( z, t ) Refer to to the figure above In the copper portion the field is Ex Exo e Cu z In the nickel portion, Ex Exo e Cu t e Ni z t The current density in the copper is J xCu Cu Exo e Cu z , and in the nickel is J xNi Ni Exo e Cu t e Ni z t . The current is I Cu Exo e Cu z dydz Ni Exo e Cu t e Ni ( z t ) dydz , or t o t I w Cu Exo e Cu z dz w Ni Exo e Cu t e Ni ( z t ) dz , and upon evaluating I wExo Cu 1 eCut Ni eCut , and with V=ExoL, Ni Cu 1 L we have R Rs , where Rs = Cu 1 e Cu t Ni e Cu t . w Ni Cu Now we’re ready to perform the calculations using the following data: S Np Cu 5.8 x107 , r 1, Cu 479 x103 m m S Np Ni 1.5 x107 , r 600, Cu 596 x106 m m (a) 0.1m Cu over Ni: Rs = 176 m (b) Semi-infinite Cu: Rs = 8.3 m (c) 0.1 m Cu: Rs = 177 m 3. Calculate the DC resistance per meter length of a 4.0 mm diameter copper wire. Now find the resistance at 1.0 GHz. DC: R 1 1 1 1 m 1.37 2 2 7 L a m 5.8x10 0.002 R R 1 s ; 1 f 2.09 x106 m L 2 a 2 a R 1 0.66 7 6 L 5.8 x10 2.09 x10 2 0.002 m 1 GHz: 4. Assume distilled water ( = 10-4 S/m, r = 81, r = 1.0) fills the region z > 0. At the surface, we have E(0,t) = 8.0cos(2x108t) ax V/m. Determine, for z > 0, (a) E(z,t), (b) H(z,t), and (c) Pav at z = 1.0 m. (d) Find the power passing through a 10 square meter surface located at z = 1.0 m. (a) The general expression for E is: E( z, t ) Eo e z cos t z a x and we can see from the given information that V rad Eo 8 , 2 x108 , f 108 Hz, 0 . Also m s 2 x108 81 8.854 x1012 0.45, 10 4 , so 1 (low loss dielectric). 104 1 Np 120 0.0021 2 2 81 m rad r 18.8 c m 1 120 41.9 81 so E( z, t ) 8e 0.0021z cos 2 x108 t 18.8 z a x V m (b) E s 8e 0.0021z e j18.8 z a x 1 V , m 8 0.0021z j18.8 z mA e e a y 191e 0.0021z e j18.8 z a y 41.9 m mA so H( z, t ) 191e 0.0021z cos 2 x108 t 18.8 z a y m 2 1 Exo 2 z W (c) Pavg e a z 0.764e2(0.0021)(1)a z 0.761 2 2 m 2 (d) P Pavg (10m ) 7.6W Hs a P Es V , m 5. A 200 MHz uniform plane wave incident on a thick copper slab imparts a 1.0 mV/m amplitude at the surface. How much power passes through a square meter at the surface? How much power passes through a square meter area 10. m beneath the surface? f 200MHz , Eo 1 Cu: 2 mV 1 Eo 2 , Pavg m 2 j 45 Np e , f 214 x103 , so 5.22e j 45 m m 3 1 10 W Pavg 96 2 ; P Pavg S 96W 3 2 5.22 x10 m Now at 10 m beneath the surface, we have 2 E ( z 10 m) Eo e (10 m ) 103 e (214 x10 )(10 m ) 118 x106 3 1 118 x10 W 1.3 2 ; P 1.3W 3 2 5.22 x10 m 6 2 Pavg V m