Maxwell's Equations PDF

Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Chapter-06 Maxwell's equation We have the basic equation of electricity and magnetis...

Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Chapter-06 Maxwell's equation We have the basic equation of electricity and magnetism are, 1. Gauss law in electrostatics; ∇. 𝐸⃗ = Or, ∇. 𝜖 𝐸⃗ = 𝜌 ∇. 𝐷⃗ = 𝜌 ………………………………………………….1 Where, 𝐷⃗ = 𝜖 𝐸⃗ is displacement current vector. 2. Gauss law in magnetostatics; ∇. 𝐵⃗ = 0 ……………………………………2 3. Faradays law of e.m. induction; Induced emf, 𝜖=− Or, ∫ 𝐸⃗. d 𝑙⃗ = − (∮ 𝐵⃗. 𝑑𝑆⃗ ) Or, ∮(∇ × 𝐸⃗ ). 𝑑𝑆⃗ = − (∮ 𝐵⃗. 𝑑𝑆⃗) ⃗ Or, ∇ × 𝐸⃗ = − ……………………………………3 4. Amperes circuital law; ∮ 𝐵⃗. 𝑑𝑙⃗ = 𝜇 𝐼 ∮ ∇ × 𝐵⃗. 𝑑𝑆⃗ = 𝜇 ∮ 𝐽⃗. 𝑑𝑆⃗ ∇ × 𝐵⃗ = 𝜇 𝐽⃗ ⃗ ∇× = 𝐽⃗ ∇ × 𝐻⃗ = 𝐽⃗ ………………………………………………………4 These four equations are called maxwells equations which are used in electrostatics and magnetostatics. Equation (3) & (4) also relates electric and magnetic fields. Among these four equations, first three are hold for both static and dynamic electric & magnetic field. But, fourth one is valid only for steady current. So, it does not hold for time varying field. 1 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Diaplacement current We have amperes circuital law, ∮ 𝐵⃗. 𝑑𝑙⃗ = 𝜇 𝐼 Here, we know, 𝐼 = ∮ 𝐽⃗. 𝑑𝑆⃗ and using stokes theorem, we get ∮ ∇ × 𝐵⃗. 𝑑𝑆⃗ = 𝜇 ∮ 𝐽⃗. 𝑑𝑆⃗ ∇ × 𝐵⃗ = 𝜇 𝐽⃗ ⃗ ∇× = 𝐽⃗ ∇ × 𝐻⃗ = 𝐽⃗ ………………………………………………………1 Taking Divergece on both sides, ∇. ∇ × 𝐻⃗ = ∇. 𝐽⃗ 0 = ∇. 𝐽⃗ ∇. 𝐽⃗ = 0 ………………………………………………………..2 But according to continuity equation, ∇. 𝐽⃗ = − ≠ 0 …………………………………………3 Thus, equation (2) & (3) are incompatible for time varying conditions. We must modify equation (1) to agree with equation (3). To do this, we add a term to equation (1), so that it becomes, ∇ × 𝐻⃗ = 𝐽⃗ +𝐽⃗ ………………………………………..4 Where, 𝐽⃗ is to be determined and defined. Now taking divergence on both sides, we get ∇. (∇ × 𝐻⃗ ) = ∇. (𝐽⃗ +𝐽⃗ ) 0 = ∇. 𝐽⃗ + ∇. 𝐽⃗ Or, − + ∇. 𝐽⃗ = 0 2 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet ∇. 𝐽⃗ = We know, ∇. 𝐷⃗ = 𝜌 , so, (∇. ⃗) ∇. 𝐽⃗ = ⃗ ∇. 𝐽⃗ = ∇. ⃗ Or, 𝐽⃗ = Substituiting this value in equation (4), we get ⃗ ∇ × 𝐻⃗ = 𝐽⃗ + ……………………………………………….5 ⃗ Where, 𝐽⃗ = is called displacement current density and 𝐽⃗ is conduction current density Here, equation (5) is maxwells fourth equation for time varying field. Now, on the basis of displacement current density, displacement current is defined as, ⃗ I = ∫ 𝐽⃗. d𝑆⃗ = ∫. d𝑆⃗ Hence, Displacement current is defined as the rate of change of the electric displacement field (𝑫⃗) in a magnetic field. In other words, this is a current produced in a magnetic field due to change in electric electric field with time. So this is the result of time varying electric field. The typical example of this current is the current through a capacitor when an alternating voltage source is applied to its plates. Significance of displacement current Displacement currents play a central role in the propagation of electromagnetic radiation, such as light and radio waves, through empty space. A traveling, varying magnetic field is everywhere associated with a periodically changing electric field that may be conceived in terms of a displacement current. Maxwell's equation in final form We have the maxwells equations; ∇. 𝐷⃗ = 𝜌 ∇. 𝐵⃗ = 0 ⃗ ∇ × 𝐸⃗ = − ⃗ ∇ × 𝐻⃗ = 𝐽⃗ + 3 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Proof: 1. ∇. 𝐷⃗ = 𝜌 We have according to Gauss law, ϕ = ∮ 𝐸⃗. 𝑑𝑆⃗ = q ∮ 𝐸⃗. 𝑑𝑆⃗ = q ∫ (∇. 𝐸⃗ ) 𝑑𝑣 = ∫ ρdv For arbitary volume, ∇. 𝐸⃗ = ∇. ϵ 𝐸⃗ = ρ ∇. 𝐷⃗ = ρ Which is first law of Maxwell. This states the relation between the electric field and the charges that produced it. this is the generalization of Coulombs law and valid for static charges as well as charges in uniform motion i.e. if electric field varies with time. 2. ∇. 𝐵⃗ = 0 Since the magnetic lines of force are either close or go of infinity then the number of magnetic force entering any arbitary close surface is exactly same as leaving it. It means the flux of magnetic induction vector B⃗ across any closed surfaces is always zero. i.e. ∫ B⃗. dS⃗ = 0 or, ∫ (∇.B⃗)dv= 0 For arbitary volume 𝑑𝑣, ∇.B⃗ = 0 Which is maxwells second law. ⃗ 3. ∇ × 𝐸⃗ = − According to faradays law of electromagnetic induction, e.m.f. induced in coil is, 𝐸 = − = − (∫ B⃗. dS⃗) …………………………………….1 Since, emf 𝐸 can be also calculated by calculating workdone in carying a unit charge completely around a close loop. i.e. 𝐸 = ∮ 𝐸⃗. d 𝑙⃗ ……………………………………..2 so, from equation (1) and (2). ∮ 𝐸⃗. d 𝑙⃗ = − (∫ B⃗. dS⃗) 4 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet ∮(∇ × 𝐸⃗ ). 𝑑𝑆⃗ = − (∮ 𝐵⃗. 𝑑𝑆⃗) ⃗ ∮(∇ × 𝐸⃗ ). 𝑑𝑆⃗ = −(∮. 𝑑𝑆⃗) ⃗ Or, ∇ × 𝐸⃗ = − This is maxwells third law. ⃗ 4. ∇ × 𝐻⃗ = 𝐽⃗ + We have amperes circuital law, ∮ 𝐵⃗. 𝑑𝑙⃗ = 𝜇 𝐼 Here, we know, 𝐼 = ∮ 𝐽⃗. 𝑑𝑆⃗ and using stokes theorem, we get ∮ ∇ × 𝐵⃗. 𝑑𝑆⃗ = 𝜇 ∮ 𝐽⃗. 𝑑𝑆⃗ ∇ × 𝐵⃗ = 𝜇 𝐽⃗ ⃗ ∇× = 𝐽⃗ ∇ × 𝐻⃗ = 𝐽⃗ …………………………………………………..1 Taking Divergece on both sides, ∇. ∇ × 𝐻⃗ = ∇. 𝐽⃗ 0 = ∇. 𝐽⃗ ∇. 𝐽⃗ = 0 ………………………………………………………..2 But according to continuity equation, ∇. 𝐽⃗ = − ≠ 0 …………………………………………3 Thus, equation (2) & (3) are incompatible for time varying conditions. We must modify equation (1) to agree with equation (3). To do this, we add a term to equation (1), so that it becomes, ∇ × 𝐻⃗ = 𝐽⃗ +𝐽⃗ ………………………………………..4 Where, 𝐽⃗ is to be determined and defined. Now, taking divergence on both sides, we get 5 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet ∇. (∇ × 𝐻⃗ ) = ∇. (𝐽⃗ +𝐽⃗ ) 0 = ∇. 𝐽⃗ + ∇. 𝐽⃗ Or, − + ∇. 𝐽⃗ = 0 ∇. 𝐽⃗ = We know, ∇. 𝐷⃗ = 𝜌 , so, (∇. ⃗) ∇. 𝐽⃗ = ⃗ ∇. 𝐽⃗ = ∇. ⃗ Or, 𝐽⃗ = Substituiting this value in equation (4), we get ⃗ ∇ × 𝐻⃗ = 𝐽⃗ + ……………………………………………….5 This is maxwells fourth equation. Electromagnetic wave(e.m.) Since maxwells third and fourth law shows that, when electric and magnetic fields are time dependent then they influenced each otherso they are coupled. Due to this coupling, the electric and magnetic fields transport energy over very large distances. These coupled fields produce travelling waves called electromagnetic waves. The e.m. wave includes, light rays, radio waves, T.V. signals etc. Charecteristics of e.m. wave The electromagnetic waves have following charecteristics, i. They travel at high velocity ii. They radiate radiation outward from a source. iii. Electric and magnetic fields are perpendicular to each other during propagation of e.m. wave. 6 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Plane electromagnetic waves in free space We have the maxwells equations, ∇. 𝐷⃗ = 𝜌 ∇. 𝐵⃗ = 0 ⃗ ∇ × 𝐸⃗ = − ⃗ ∇ × 𝐻⃗ = 𝐽⃗ + & 𝐵⃗ = 𝜇𝐻⃗ , 𝐷⃗ = 𝜖𝐸⃗ , 𝐽⃗ = 𝜎𝐸⃗ In free space, 𝜌 = 0, 𝜎 = 0, 𝜇 = 𝜇 and 𝜖 = 𝜖 so, On putting these values, we get ∇. 𝐸⃗ = 0 ………………………………………………………..1 ∇. 𝐻⃗ = 0 ………………………………………………………….2 ⃗ ∇ × 𝐸⃗ = −𝜇 …………………………………………..3 ⃗ ∇ × 𝐻⃗ = 𝜖 ………………………………………………………..4 Taking curl inboth side of equation (3) we get, (∇× ⃗ ) ∇ × ∇ × 𝐸⃗ = −𝜇 Since, ∇ × ∇ × 𝐸⃗ = ∇. ∇. 𝐸⃗ − ∇ 𝐸⃗ (∇× ⃗ ) So, ∇. ∇. 𝐸⃗ − ∇ 𝐸⃗ = −𝜇 Using equation (1) & (2), ⃗ 0 − ∇ 𝐸⃗ = −𝜇 (𝜖 ) ⃗ −∇ 𝐸⃗ = −𝜇 𝜖 ⃗ ∇ 𝐸⃗ − 𝜇 𝜖 = 0………………………………………….5 Similarly, after taking curl in equation (4), we get ⃗ ∇ 𝐻⃗ − 𝜇 𝜖 = 0………………………………………….6 7 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet These equations (5) & (6) gives the plane electromagnetic wave in free space. Again, the combined form of equation (4) & (5) is, ∂ 𝐴⃗ ∇ 𝐴⃗ − 𝜇 𝜖 =0 ∂t Or, 1 ∂ 𝐴⃗ ∇ 𝐴⃗ − =0 𝑣 ∂t Where, 1 =𝜇 𝜖 𝑣 So, 𝑣= = 3× 10 𝑚/𝑠 Hence, in free space, the velocity of electromagnetic wave is equal to the velocity of light. Poynting Vector It is defined as the energy per unit time per unit area transported by the field. If 𝑃⃗ is the poynting vector then it is given by, 𝑃⃗ = 𝐸⃗ × 𝐵⃗ Poynting theorem The electromagnetic potential energy in electric and magnetic field are, U = ϵ E dv ……………………………..1 (Energy stored in electric field) U = ϵ H dv …………………………….2 (Energy stored in magneticfield) We have maxwells third and fourth equations are, ⃗ ∇ × 𝐸⃗ = − ……………………………………….3 ⃗ ∇ × 𝐻⃗ = 𝐽⃗ + ……………………………………….4 Taking dot product with 𝐻⃗ in equation (3) and by 𝐸⃗ with equation (4), we get, ⃗ 𝐻⃗. (∇ × 𝐸⃗ ) = −𝐻⃗. ……………………………………5 8 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet ⃗ & 𝐸⃗. ( ∇ × 𝐻⃗ ) = 𝐸⃗. 𝐽⃗ +𝐸⃗. ……………………………………6 Subtracting equation (5) by (6), we get ⃗ ⃗ 𝐻⃗. ∇ × 𝐸⃗ − 𝐸⃗. ( ∇ × 𝐻⃗ ) = −𝐻⃗. − 𝐸⃗. 𝐽⃗ −𝐸⃗. ⃗ ⃗ ∇. 𝐸⃗ × 𝐻⃗ = −(𝐻⃗. + 𝐸⃗. ) −𝐸⃗. 𝐽⃗ ……………………………………..7 Using, 𝐵⃗ = 𝜇 𝐻⃗ & 𝐷⃗ = ϵ 𝐸⃗ ⃗ ( ⃗) Then, 𝐸⃗. = 𝐸⃗. = ϵ ⃗ ( ⃗) & 𝐻⃗. = 𝐻⃗. = 𝜇 Then, ∇. 𝐸⃗ × 𝐻⃗ = − 𝜇 + ϵ − 𝐸⃗. 𝐽⃗ ∇. 𝐸⃗ × 𝐻⃗ =− ( 𝜇 𝐻 + 𝜇 𝐸 ) − 𝐸⃗. 𝐽⃗ On integrating both sides over volume, ′𝑉′ enclosed surface, ∫ ∇. 𝐸⃗ × 𝐻⃗ 𝑑𝑣 = − ∫ ( 𝜇 𝐻 + 𝜇 𝐸 )𝑑𝑣 − ∫ 𝐸⃗. 𝐽⃗ 𝑑𝑣 ∫ ∇. 𝐸⃗ × 𝐻⃗ 𝑑𝑣 = − ∫ 𝜇 𝐻 + 𝜇 𝐸 𝑑𝑣 − 0 ∫ ∇. 𝐸⃗ × 𝐻⃗ 𝑑𝑣 = − ∫ 𝜇 𝐻 + 𝜇 𝐸 𝑑𝑣 Here, the volume integral over the product of electric field intensity and current density is zero. Using gauss divergence theorem in left side, we get ∫ 𝐸⃗ × 𝐻⃗. 𝑑𝑆⃗ = − ∫ 𝜇 𝐻 + 𝜇 𝐸 𝑑𝑣 ∫ 𝑃⃗. 𝑑𝑆⃗ = − ∫ 𝜇 𝐻 + 𝜇 𝐸 𝑑𝑣…………………………………8 Where, 𝑃⃗ = 𝐸⃗ × 𝐻⃗ is called poynting vector. & 𝑊=∫ 𝜇 𝐻 + 𝜇 𝐸 𝑑𝑣 is total electromagnetic energy in vaccum space. Equation (8) is called Poynting theorem. 9 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Lossy dielectric: Lossy dielectric is a medium in which an electromagnetic waves losses power when it propagates through it due to poor conduction. In other words, a lossy dielectric is a partially conducting medium i.e. imperfect dielectric or conductor in which 𝜎 ≠ 0. Note: 1. In free space; 𝜎 = 0 , 𝜖 = 𝜖 & 𝜇 = 𝜇 2. In lossless dielectrics; 𝜎 = 0 , 𝜖 = 𝜖 𝜖 , 𝜇 = 𝜇 𝜇 & 𝜎 ≪ 𝜔𝜀 3. In lossy medium; 𝜎 = 0 , 𝜖 = 𝜖 𝜖 & 𝜇 = 𝜇 𝜇 4. In good conductor; 𝜎 ≅ ∞ , 𝜖 = 𝜖 𝜖 , 𝜇 = 𝜇 𝜇 & 𝜎 ≫ 𝜔𝜀 Wave propagating in a lossy dielectric medium Consider a lossy dielectic medium which is free from charge. So, 𝜌 = 0. Now, maxwells equation in lossy dielectric medium becomes, ∇. 𝐸⃗ = 0 …………………………………………………………………….1 ∇. 𝐻⃗ = 0 …………………………………………………………………….2 ⃗ ∇ × 𝐸⃗ = − ⃗ = −𝜇 …………………………………………………………………3 ⃗ And, ∇ × 𝐻⃗ = 𝐽⃗ + ⃗ = 𝜎𝐸⃗ + 𝜖 ………………………………………………………………..4 Here electric and magnetic fields are time varying field. So, can be expressed as, 𝐸⃗ = 𝐸 𝑒 …………………………………………5 & 𝐻⃗ = 𝐻 𝑒 ……………………………………………6 Using equation (5) & (6) in equation (3) & (4) we get, ∇ × 𝐸⃗ = − 𝜇𝑗𝜔𝐻⃗ ……………………………………..7 & ∇ × 𝐻⃗ = ( 𝜎 + 𝜖𝑗𝜔)𝐸⃗ ………………………………………8 10 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Taking curl in equation (7) we get, ∇ × ∇ × 𝐸⃗ = − 𝜇𝑗𝜔(∇ × 𝐻⃗ ) ∇. ∇. 𝐸⃗ − ∇ 𝐸⃗ = − 𝜇𝑗𝜔.( 𝜎 + 𝜖𝑗𝜔)𝐸⃗ {using equation (8)} 0−∇ 𝐸⃗ = − 𝜇𝑗𝜔.( 𝜎 + 𝜖𝑗𝜔)𝐸⃗ ∇ 𝐸⃗ − 𝜇𝑗𝜔( 𝜎 + 𝜖𝑗𝜔)𝐸⃗ = 0 ∇ 𝐸⃗ − 𝛾 𝐸⃗ = 0…………………………………………9 Where, 𝛾 = 𝜇𝑗𝜔( 𝜎 + 𝜖𝑗𝜔) ……………………….10 is called propagation constant in dielectric medium. Similarly, in case of magnetic field, wave equation becomes, ∇ 𝐻⃗ − 𝛾 𝐻⃗ = 0…………………………………11 Equation (9) & (11) are called homogeneous equation or vector wave equation. Here, 𝛾 is complex quantity which can be expressed as, 𝛾 = 𝛼 + 𝑗𝛽 ⇒ 𝛾 = (𝛼 + 𝑗𝛽) ……………………………………………….12 From equation (10) & (12), we get 𝜇𝑗𝜔( 𝜎 + 𝜖𝑗𝜔) = (𝛼 + 𝑗𝛽) 𝜇𝑗𝜔𝜎 + 𝜖𝜇𝑗 𝜔 = 𝛼 − 𝛽 + 𝑗𝛼𝛽 𝜇𝑗𝜔𝜎 − 𝜖𝜇𝜔 = 𝛼 − 𝛽 + 𝑗𝛼𝛽 Equating real and imaginary part on both sides, we get 𝛼 𝛽 = 𝜇𝜔𝜎 ……………………………………………….13 & 𝛼 − 𝛽 = − 𝜖𝜇𝜔 ……………………………………14 On solving this equation, we get, 𝜇𝜖 𝜎 𝛼=𝜔 ( 1+( ) −1 2 𝜔𝜖 & 𝛽=𝜔 ( 1+ +1 11 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Again, without loss of originality, Let us suppose that, the wave is propagated along z- axis and it has only amplitude along x-axis then, We can expressed 𝐸⃗ & 𝐻⃗ as, 𝐸⃗ = 𝐸 (𝑧)𝑎 …………………………………………………….15 & 𝑍⃗ = 𝐻 (𝑧)𝑎 On substituiting these values in equation (9) we get, ( ) ( ) ( ) + + − 𝛾 𝐸 (𝑧) = 0 Here, wave is only along z-direction. So, first and second term is zero. Hence, ( ) − 𝛾 𝐸 (𝑧) = 0…………………………………………16 This is scalar wave equation with solution, 𝐸 (𝑧) = 𝐸 𝑒 +𝐸 𝑒 For incident wave For reflected wave Where, 𝐸 & 𝐸 are constant that represents amplitude of wave. Since wave is only propagate along z- axis (i.e. no reflection component) So, 𝐸 (𝑧) = 𝐸 𝑒 Now, inserting the time factor in solution of equation (16) we get, 𝐸 (𝑧, 𝑡) = 𝐸 𝑒 𝑒 With this value, we can write, 𝐸⃗ (𝑧, 𝑡) = 𝑅𝑒{𝐸 (𝑧, 𝑡)𝑎 } = 𝑅𝑒{𝐸 𝑒 𝑒 𝑎 } = 𝑅𝑒{𝐸 𝑒 𝑎 } 12 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Since, 𝛾 = 𝛼 + 𝑗𝛽 So, 𝐸⃗ (𝑧, 𝑡) = 𝑅𝑒{𝐸 𝑒 ( ) 𝑎 } 𝐸⃗ (𝑧, 𝑡) = 𝑅𝑒{𝐸 𝑒 𝑒 ( ) 𝑎 } 𝐸⃗ (𝑧, 𝑡) = 𝑅𝑒{𝐸 𝑒 [cos(𝜔𝑡 − 𝛽𝑧) + 𝑗𝑠𝑖𝑛(𝜔𝑡 − 𝛽𝑧)]𝑎 } 𝐸⃗ (𝑧, 𝑡) = 𝐸 𝑒 cos(𝜔𝑡 − 𝛽𝑧) 𝑎 { Taking real part only} Similarly, in case of magnetic field, 𝐻⃗ (𝑧, 𝑡) = 𝐻 𝑒 cos(𝜔𝑡 − 𝛽𝑧) 𝑎 Here, if 𝐸⃗ is along x-axis, then 𝐻⃗ is along y-axis because 𝐸⃗ & 𝐻⃗ are in perpendicular to eachother. It is note that, 𝐻 = Where, 𝜂 is intrinsic ipedence and is given by, 𝜂= = |𝜂|e Equation (17) & (18) represents the 𝐸⃗ & 𝐻⃗ field propagate in lossy medium. From Equation (17) & (18), it is seen that, the wave propagating alonh z-axis is decreasing its amplitude exponentially by factor 𝑒. Hence, 𝛼 is called attenuation constant or attenuation coefficient of medium. It measured the rate of decay of wave in medium and measured in the unit of Nepers per meter (Np/m). Where, 1Np/m = 8.68 𝑑𝐵(𝑑𝑒𝑐𝑖𝑏𝑒𝑙) a. Plane wave in lossless dielectric medium The medium which has zero conductivity means no transfer of electron is called lossless dielectric. In lossless dielectric medium, 𝜎 = 0 , 𝜖 = 𝜖 𝜖 , 𝜇 = 𝜇 𝜇 & 𝜎 ≪ 𝜔𝜀 Using this value in value of 𝛼 & 𝛽 as, 𝜇𝜖 𝜎 𝛼=𝜔 ( 1+( ) −1 2 𝜔𝜖 13 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet & 𝛽=𝜔 ( 1+ + 1 we get, 𝛼=0 & 𝛽 = 𝜔√𝜇𝜖 Then, Velocity of e.m. wave is, 𝑢= = , √ And, 𝜆 = Also, fot this lossless dielectrics, 𝜂= Becomes, 𝜂= (Use 𝜎 = 0) This gives the relation for intrinsic impedence in lossless dielectric. b. Plan wave in free medium In free medium, the conductivity in medium is zero, but permitivity and permeability have lowest value. i.e. = 0 , 𝜖 = 𝜖 & 𝜇 = 𝜇 Using this value in value of & 𝛽 , we get, 𝛼=0 & 𝛽= 𝜔 𝜇 𝜖 Now, Velocity of e.m. wave is, 𝑢 = Or, 𝑢= = 𝑐 = 3 × 10 𝑚/𝑠 This shows that speed of e.m. wave in vaccum is equal to velocity of light, 3 × 10 𝑚/𝑠. Also, intrinsic impedence of free space is, 𝜂= is, 14 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet 𝜂 = 𝜂 = = 120𝜋 = 377Ω Thus, total resistance offer by e.m. wave in free space is 377Ω. c. Plane wave in good conductor A perfect or good conductor is one in which 𝜎 ≫ 𝜔𝜀 so that, ≫ ∞ then, 𝛼= 𝛽= = = 𝜋𝑓𝜇𝜎 Now, Velocity of e.m. wave is, 𝑢 = = = = & 𝜆= Also, intrinsic impedence of free space is, 𝜂= is, 𝜂= ………………………….………………..1 Again in conducting medium, 𝐸⃗ (𝑧, 𝑡) = 𝐸 𝑒 cos(𝜔𝑡 − 𝛽𝑧) 𝑎 ……………………………. ………………….2 & 𝐻⃗ (𝑧, 𝑡) = 𝐻 𝑒 cos(𝜔𝑡 − 𝛽𝑧) 𝑎 = 𝑒 cos(𝜔𝑡 − 𝛽𝑧) 𝑎 𝐻⃗ (𝑧, 𝑡) = 𝑒 cos(𝜔𝑡 − 𝛽𝑧) 𝑎 ………………………..3 Equation (2) and (3) shows that, 𝑤ℎ𝑒𝑛 𝐸⃗ & 𝐻⃗ travel in conducting medium, its amplitude is attenuted by the factor 𝑒. The distance ar which amplitude of e.m. wave is decrease by factor 𝑒 (i.e. about 37% of its original value) is called skin depth or penetration depth of medium. 15 Jagat P Panday (Assistant Professor) Electromagnetism Mid-West University, Surkhet Hence, at 𝑧 = 𝛿 = , 𝐸 𝑒 =𝐸 𝑒 =𝐸 𝑒 = 37% 𝑜𝑓 𝐸 The skin depth is a measure of depth to which electromagnetic waves can penetrate the medium. The skin depth is denoted by 𝛿 and given by, 𝛿= = For conducting medium. 16