* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download EM wave in conductors (note11)
Survey
Document related concepts
Euler equations (fluid dynamics) wikipedia , lookup
Electrostatics wikipedia , lookup
Electromagnetism wikipedia , lookup
First observation of gravitational waves wikipedia , lookup
Lorentz force wikipedia , lookup
Equation of state wikipedia , lookup
Equations of motion wikipedia , lookup
Partial differential equation wikipedia , lookup
Diffraction wikipedia , lookup
Maxwell's equations wikipedia , lookup
Electrical resistivity and conductivity wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Transcript
§3. Electromagnetic Waves §3.4. EM fields (waves) in conductors The behaviour of EM waves in a conductor is quite different from that in a source-free medium. The conduction current in a conductor is the cause of the difference. We shall analyze the source terms in the Maxwell’s equations to simplify Maxwell’s equations in a conductor. From this set of equations, we can derive a diffusion equation and investigate the skin effects. §3.4.1 Skin Effects in Conductors A. Maxwell’s Equations in a Conductor Complete Maxwell’s equations: ρ − → ∇· E = ² − → ∂B − → ∇× E = − ∂t − → ∇· B = 0 − → ∂ E − → ∇× B = µ ~j + ² ∂t Fact 1: Conducting current dominates over the displacement current In a conductor, the electric field is the driving source for the conduction current. The collision is the source of impedance. The conduction current is governed by the Ohm’s law: − → − → j f = σE where σ (S/m) is the conductivity. The AC conductivity differs from the DC conductivity. Let ν be the collision frequency of electrons (current carrier in conductors) with ions and ω the frequency of the EM waves in the conductor. The equation of motion for electrons is: m → d− v − → → = −e E − mν − v dt → → Assume − v =− v 0 e−iωt and use ∂/∂t → −iω, we obtain − → → → −iωm− v = −e E − mν − v → Phys 463, E & M III, C. Xiao − → v = −e − → E m(ν − iω) 1 / note11 − → → Recall the current density is expressed by j = −en− v (n is the electron number density in the conductor), ne2 − → − → jf= E m(ν − iω) which gives the AC conductivity in conductor: σ(ω) = 1 ne2 ν − iω m In practical situations, ω ¿ ν ∼ 1014 (1/sec) (infrared range). So the DC conductivity ne2 mν σ= can be used. Let’s now compare the magnitude of conduction current with that of the displacement current. − → − → Assume E = E 0 e−iωt . Then ¯ → ¯¯ ¯− ¯ j f¯ σE σ ¯ − ¯= = ¯ ∂→ ¯ ²ωE ²ω ¯² E ¯ ∂t In copper, σ = 6 × 107 (S/m). The condition for jf ' ² ω= − → ∂E , ∂t or σ ²ω ' 1 leads to σ 6 × 107 = ∼ 7 × 1019 (rad/sec) ² 8.85 × 10−12 At frequencies ω < 1012 (rad/sec) (communication wave frequency), σ À 1 or ²ω ¯− ¯ ¯→ ¯ ¯ j f¯ À Ampere’s law in a conductor: ¯ − ¯ ¯ →¯ ¯ ∂E ¯ ¯² ¯ ¯ ¯ ¯ ∂t ¯ − → − → − → ∇× B = µ j f = µσ E Fact 2: No significant charge accumulation Because of the good conductivity, no significant charge accumulation in a conductor is expected (ρ ' 0). Phys 463, E & M III, C. Xiao 2 / note11 From the charge conservation and Gauss’s law ρ − → ∇· E = ² ∂ρf − → ∇· j f = − ∂t we obtain ∂ρf σ − → = −σ∇· E = − ρf ∂t ε So σ ρf = ρf (0)e− ε t The free charge ρf (0) dissipates in a characteristic time τ = σε , similar to the static case in which the charge will flow out to the edge of the conductor. If the transient phase is excluded, ρf = 0 can be assumed in a conductor. For simplicity, we shall consider “good” conductor case in which the displacement current can be ignored. Maxwell’s equations in a conductor: − → ∇· E − → ∇× E − → ∇· B − → ∇× B = 0 − → ∂B = − ∂t = 0 − → = µσ E It can be shown that the EM waves in conductor are also TEM (Transverse EM ) waves B. Diffusion Equation As we did before for waves in source-free media, let’s apply curl operator to the 2nd Maxwell’s equation: − → ³ ´ ∂B ∂ ∂ ³ − − → − → →´ ∇× ∇× E = −∇× = − (∇× B ) = − µσ E ∂t ∂t ∂t LHS of the equation So ³ ³ − − →´ →´ − → − → ∇× ∇× E = ∇ ∇· E − ∇2 E = −∇2 E − → ∂E ∇ E = µσ Diffussion equation ∂t Similarly, the magnetic field also satisfies the same diffusion equation: − → ∂B → 2− ∇ B = µσ Diffussion equation ∂t → 2− Phys 463, E & M III, C. Xiao 3 / note11 C. Skin Depth Suppose we have a plane wave. It comes from the −z direction and reaches a large − → conductor surface at z = 0. Outside of a conductor: E = E0 e−iωt~ex at z = 0. Assume the wave inside the conductor has the form − → − → E = E 0 ei(kz−ωt) where k is an unknown constant. Recall ∂ → −iω ∂t ∇ → ik, for the waves of the above type, we find from the diffusion equation − → − → (ik)2 E = −iωµσ E π k 2 = iωµσ = ωµσe 2 i → r ωµσ √ k = ±e ωµσ = ±(1 + i) 2 π i 4 Choose “+” sign to allow the electric field to damp (to “propagate”) in the +z direction. Separate the real and imaginary parts of k: k = k+ + ik− , k+ = k− = r ωµσ 2 − → − → E = E 0 e−k− z ei(k+ z−ωt) − → − → B = B 0 e−k− z ei(k+ z−ωt) (1) (2) If the “good” conductor assumption is not valid, the displacement current should be included in the 4th Maxwell’s equation. The solution for E and B are the same as above with s 1/2 r µ ¶ εµ σ 2 1+ ± 1 k± = ω 2 εω The equations (1) and (2) indicate that the amplitude of E and B fields decays to 1/e of their values at z = 0 in a distance: δ= 1 k− Phys 463, E & M III, C. Xiao 4 / note11 where δ is called skin depth For a good conductor (σ À εω): δ= s 2 (m) ωµσ Also, the wavelength is λ = 2π/k+ = 2πδ inqa good conductor. The wave decays significantly within one wavelength. Since δ ∝ 1/ωσ, deep penetration occurs for 1. Low frequency 2. poor conductor Example: skin depth at f = 60 Hz for copper. δ= s 2 = 8 × 10−3 m = 8 mm −7 7 2π × 60 × 4π × 10 × 6 × 10 Phys 463, E & M III, C. Xiao 5 / note11 There is no advantage to construct AC transmission lines using wires with a radius much larger than the skin depth because the current flows mainly in the outer part of the conductor. For a poor conductor (σ ¿ εω): 2 δ= σ s ε (m) µ independent of the frequency. Example: For sea water, µ = µ0 = 4π × 10−7 N/A2 , ε ' 70²0 = 6 × 10−10 C2 /N·m2 , and σ ' 5 (Ω·m)−1 . Sea water is a poor conductor for frequency ω σ f= À = 109 Hz 2π 2πε or λ ¿ 30 cm. The skin depth is s s 2 ε 2 70ε0 δ = = σ µ σ µ0 √ √ 2 70 2 70 = = ' 1 cm σZ 5 × 377 In the radioq frequency range (f ¿ 109 Hz) sea water is a good conductor, the skin depth δ = 2/(ωµσ) is quite short. To reach a depth δ = 10 m, for communication with submarines, ω 1 f= ' 500 Hz = 2π πµσδ 2 Phys 463, E & M III, C. Xiao 6 / note11 The wavelength in the air is about λ= c 3 × 108 = = 600 km f 500 The required λ/4 antenna would be gigantic. §3.4.2 Monochromatic plane waves in a Conductor A. Transverse waves The E and B fields in a conductor ~ t) = B ~ 0 e−k− z ei(k+ z−ωt) B(z, ~ t) = E ~ 0 e−k− z ei(k+ z−ωt) , E(z, can be rewritten as ~ t) = E ~ 0 ei(kz−ωt) , E(z, ~ t) = B ~ 0 ei(kz−ωt) B(z, They have the same functions as EM wave in vacuum, except that k is a complex number Following the same calculation for waves in vacuum, we can derive from ~ = 0, ∇·E ~ =0 ∇·B and the following results: ~k · E ~ = k~ez · E ~ =0 ~k · B ~ = k~ez · B ~ =0 ~ and B ~ are perpendicular to ~ez , the wave propagation direction. Both E Let’s assume ~ = E0 ei(kz−ωt)~ex = E0 e−k− z ei(k+ z−ωt)~ex E From ~ ~ = − ∂B ∇×E ∂t we obtain ~ = k E0 e−k− z ei(k+ z−ωt)~ey B ω |k| iφ = e E0 e−k− z ei(k+ z−ωt)~ey ω where à ! k− φ = tan , k+ √ For a good conductor φ = 450 , |k| = ωµσ −1 |k| = q 2 2 k+ + k− The B fields lags behind the electric fields Phys 463, E & M III, C. Xiao 7 / note11 §4.3.3 Reflection of EM Waves on a Conductor Surface We have seen that the EM waves do not penetrate the conductor deeply. Where do the waves go? Absorbed or reflected? Back to our example with a plane wave perpendicularly propagating to a conducting surface ~ = Hy~ey = H0 ei(z/δ−ωt) e−z/δ~ey H ~ = ~j = σEx~ex = σE0 ei(z/δ−ωt) e−z/δ~ex ∇×H ~ x = [∇ × (Hy~ey )]x = ∂Hz − ∂Hy = 1 − i Hy (∇ × H) ∂y ∂z δ | {z } =0 So 1−i Hy = σEx δ Phys 463, E & M III, C. Xiao 8 / note11 Characteristic impedance of the conductor Z = Ex 1−i 1−i = = Hy σδ σ 1−i = √ 2 r ωµ = σ s r ωµσ 2 −iωµ σ For 4mm microwave, f = 75 Ghz. If the conductor is aluminum (σAl = 2 × 107 S/m) r s ωµ 2π × 75 × 109 × 4π × 10−7 |ZAl | = = σ 2 × 107 = 0.17 Ω ¿ Zair = 377 Ω The reflectivity: Γ= ZAl − Zair −Zair ' = −1 ZAl + Zair Zair Almost complete reflection. Phys 463, E & M III, C. Xiao 9 / note11