chap12 - CHAPTER 12 + + 12.1. A uniform plane wave in air,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
CHAPTER 12 12.1. A uniform plane wave in air, E + x 1 = E + x 10 cos ( 10 10 t βz) V/m, is normally-incident on a copper surface at z = 0. What percentage of the incident power density is transmitted into the copper? We need to find the reflection coefficient. The intrinsic impedance of copper (a good conductor) is η c = r jωµ σ = ( 1 + j) r ωµ 2 σ = ( 1 + s 10 10 ( 4 π × 10 7 ) 2 ( 5 . 8 × 10 7 ) = ( 1 + j)(. 0104 ) Note that the accuracy here is questionable, since we know the conductivity to only two significant figures. We nevertheless proceed: Using η 0 = 376 . 7288 ohms, we write 0 = η c η 0 η c + η 0 = . 0104 376 . 7288 + j. 0104 . 0104 + 376 . 7288 + 0104 =− . 9999 + 0001 Now | 0 | 2 = . 9999, and so the transmitted power fraction is 1 −| 0 | 2 = . 0001, or about 0 . 01% is transmitted. 12.2. The plane y = 0 defines the boundary between two different dielectrics. For y< 0, ± 0 R 1 = 1, µ 1 = µ 0 , and ± 00 R 1 = 0; and for y> 0, ± 0 R 2 = 5, µ 2 = µ 0 , and ± 00 R 2 = 0. Let E + z 1 = 150 cos (ωt 8 y) V/m, and find a) ω :Have β = 8 = ω/c ω = 8 c = 2 . 4 × 10 9 sec 1 . b) H + 1 : With E in the z direction, and propagation in the forward y direction, H will lie in the positive x direction, and its amplitude will be H x = E y 0 in region 1. Thus H + 1 = ( 150 0 ) cos (ωt 8 a x = 0 . 40 cos ( 2 . 4 × 10 9 t 8 a x A / m . c) H 1 : First, E z 1 = 0E + z 1 = η 0 / 5 η 0 / 1 η 0 / 5 + η 0 / 1 = 1 5 1 + 5 E + z 1 0 . 38 E + z 1 Then H x 1 =+ ( 0 . 38 0 )E + z 1 = 0 . 38 ( 150 ) 377 cos (ωt + 8 So finally, H x 1 = 0 . 15 cos ( 2 . 4 × 10 9 t + 8 a x A / m . 12.3. A uniform plane wave in region 1 is normally-incident on the planar boundary separating regions 1 and 2. If ± 00 1 = ± 00 2 = 0, while ± 0 R 1 = µ 3 R 1 and ± 0 R 2 = µ 3 R 2 , find the ratio ± 0 R 2 0 R 1 if 20% of the energy in the incident wave is reflected at the boundary. There are two possible answers. First, since | 0 | 2 = . 20, and since both permittivities and permeabilities are real, 0 0 . 447. we then set up 0 0 . 447 = η 2 η 1 η 2 + η 1 = η 0 q R 2 0 R 2 ) η 0 q R 1 0 R 1 ) η 0 q R 2 0 R 2 ) + η 0 q R 1 0 R 1 ) = q R 2 3 R 2 ) q R 1 3 R 1 ) q R 2 3 R 2 ) + q R 1 3 R 1 ) = µ R 1 µ R 2 µ R 1 + µ R 2 200
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
12.3. (continued) Therefore µ R 2 µ R 1 = 1 0 . 447 1 ± 0 . 447 = ( 0 . 382 , 2 . 62 ) ± 0 R 2 ± 0 R 1 = ± µ R 2 µ R 1 ² 3 = ( 0 . 056 , 17 . 9 ) 12.4. The magnetic field intensity in a region where ± 00 = 0 is given as H = 5 cos ωt cos βz a y A/m, where ω = 5 Grad/s and β = 30 rad/m. If the amplitude of the associated electric field intensity is 2kV/m, find a) µ and ± 0 for the medium: In phasor form, the magnetic field is H ys = H 0 e jβz + H 0 e + = 5 cos H 0 = 2 . 5. The electric field will be x directed, and is E xs = η( 2 . 5 )e 2 . 5 )e + = ( 2 j)η( 2 . 5 ) sin . Given the electric field amplitude of 2 kV/m, we write 2 × 10 3 = 5 η ,or η = 400  .Now η = 400 = η 0 q µ r 0 R and we also have β = 30 = (ω/c) q µ R ± 0 R .W e solve these two equations simultaneously for µ R and ± 0 R to find µ R = 1 . 91 and ± 0 R = 1 . 70. Therefore µ = 1 . 91 × 4 π × 10 7 = 2 . 40 µ H / m and ± 0 = 1 . 70 × 8 . 854 × 10 12 = 15 .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 13

chap12 - CHAPTER 12 + + 12.1. A uniform plane wave in air,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online