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PPT No. 5 E&M Waves Mechancial waves require media capable of storing kinetic and potential energies. Electromagentic waves require media capable of storing electric (capacitive) and magnetic (inductive) energies. LC transmission line Consider a transmission line having an inductance per unit length L / l (H/m) and capacitance per unit length C / l (F/m). Kirchhoff's voltage theorem L ∂i ∂V L ∂i V ( x ) = V ( x + Δx ) + Δ x = V ( x ) + Δ x + Δx l ∂t ∂x l ∂t ∂V L ∂i =− ∂x l ∂t Current theorem C ∂V i ( x ) = i ( x + Δx ) + Δx l ∂x ∂i C ∂V =− ∂x l ∂t ∂ 2V L ∂ 2i C L ∂ 2V =− = ∂x 2 l ∂t∂x l l ∂t 2 ∂ 2V 1 ∂ 2V = ∂t 2 C L ∂x 2 l l 1 Wave velocity c = CL l l In vacuum, L / l = μ0 , C / l = ε 0 . Then 1 c= = 3.0 × 108 m/sec ε 0 μ0 Impedance Let V ( x, t ) =V0sin ( kx − ωt ) , i ( x, t ) =i0sin ( kx − ωt ) ∂V L ∂i L in =− . kV0 = − ( −ω )i0 ∂x l ∂t l V0 L ω L/l μ0 = = = = 377 Ohms in vacuum. i0 l k C /l ε0 Examples: μ0 1 ⎛ b ⎞ Coaxial cable Z = (Ω ) ε 2π ⎜ a ⎟ ln ⎝ ⎠ μ0 1 ⎛ d ⎞ Parallel wire lines Z = ⎟ (Ω ) ε0 π ⎜ a ⎠ ln ⎝ b ε,μ 2a wire radius a d L/l 1C 2 1L 2 V = Zi = i→ V = i C /l 2 l 2 l In electromagnetic wave, electric energy = magnetic energy In mechanical wave, potential energy = kinetic energy C 2 Total energy density = V l Energy transfer rate = wave velocity × energy density C 2 V2 Power = c V = (W) l Z Example: A dc volatge source 6V is suddenly connected to a 50 Ω coaxial cable filled with insulator of permittivity ε = 2ε 0 . Assuming no reflection, find the current wave. V 6V The current is i = = = 0.12 A. The wave velocity Z 50 Ω is 1/ εμ0 = c / 2 = 0.71c. The power delivered by the V2 source is P = Vi = = 0.72 W. Z Example Discuss how the voltage and current waves develop after the switch is closed. 50 After the switch is closed, a voltage pulse of × 10 = 5 V propagtes toward the 20Ω load. 50+50 20-50 After τ = l / c, reflected voltage is × 5 = −2.14 V and the line voltage is 5-2.14=2.86 V. 20+50 There is no further reflection at the source, Therefore, the final voltage is 2.86 V and final current is 2.86 V/20 Ω = 0.143 A. 10 V DC theory gives i = = 0.143 A. 70 Ω Quarter wavelength transformer for impedance matching To achieve impedance matching between two impedances Z1 and Z 2 , insert Z 3 = Z1 Z 2 quarter wavelength long (or thick). The reflection at the first boundary is Z − Z1 Γ1 = 3 Z 3 + Z1 and that at the second boundary is Z − Z3 Γ2 = 2 Z3 + Z2 λ λ The second reflected wave travels additional distance of ×2 = and the total reflection is 4 2 Z 3 − Z1 Z 2 − Z 3 − Z 3 + Z1 Z 3 + Z 2 Z 3 − Z1 Z 2 − Z 3 The condition for no refelction is − = 0 or Z 3 = Z1 Z 2 Z 3 + Z1 Z 3 + Z 2 Anti-reflection coating: Coating material should have impedance of 377 × 251 = 307Ω 1 1 λair or index of refaction of n = 1.5 = 1.22. Coating thickness is λcoat = . If λair = 550 nm 4 4 1.22 1 550 (green), coating thickness should be × = 112 nm. Coating material like MgO_2 has 4 1.22 the desired permittivity. http://physics.usask.ca/%7Ehirose/ep225/emref.htm Example: For impedance matching between 50 Ω coaxial cable and 300 Ω coaxial cable can be achieved if a third cable having an impedance of 50 × 300 = 122 Ω and λ /4 long is inserted. 1 Example: A resistive plate with thickness d having a conductiviy σ = (S/m) is placed at Z0d a distance λ /4 in front of a conducting plate. Explain how reflection can be avoided. Wave reflection by a conductor plate is almost complete. At λ /4, the voltage is maximum and current is minimum → open circuit. Then the incient wave sees an admittance of 1 1 Y = ×d = or impedance of Z 0 and impedance matching is achieved. Such antireflection Z0d Z0 technique has important military application. resistive plate metal λ/4 Brewster's angle sin θ Oblique incidence at angle θ . Snell's law = ng sin θ ' Electric field in the incidence plane and magnetic field out of the page. Continuity of H-field H i + H r = H t Continuity of tangential components of E field Ei cos θ − Er cos θ = Et cos θ ' Z0 Ei = Z 0 H i , Er = Z 0 H r , Et = Ht ng 1 1 ( H i − H r ) cos θ = H t cos θ ' = ( H i + H r ) cos θ ' ng ng tan (θ − θ ' ) Hr = Hi tan (θ + θ ' ) If θ + θ ' = π / 2, tan (θ + θ ' ) = ∞ and reflection becomes zero. Brewster's angle θ B = tan −1 ( ng ) . If ng = 1.5, θ =56.3° . Example: Reflection at air-glass boundary. nglass = 1.5 μ0 μ0 1 Impedance of air Z a = = 377 Ω. Glass Z g = = × 377 = 251 Ω ε0 ε 1.5 Electric field reflection is 251 − 377 = −0.2 251 + 377 Energy (power) reflection is ( −0.2 ) = 4%. 2 Example: Brewster's angle π If θ + θ ' = in refraction and if the electric field is in the incidence plane, reflection can be avoided. 2 θ θ π/2 θ' Ei Er θ θ n=1 ng θ' Et Electric field in the incident plane Magnetic field parallel to the boundary π In this case, reflection vanishes if θ + θ ' = or 2 θ = arctan ( ng ) : Brewster's angle If ng = 1.5, θ = 56.3 y 0.20 0.15 0.10 0.05 0.00 0 10 20 30 40 50 60 70 80 x If the electric field is in the incident plane, tan (θ − θ ' ) r= In the figure r 2 is plotted. tan (θ + θ ' ) The Brewster's angle is at θ = tan −1 ( n ) = 56.3 if n = 1.5. http://physics.usask.ca/%7Ehirose/ep225/emwave.htm Polarization In plane waves propagating in z direction, E x and H y (e.g., laser light) Sun and most light sources emit randomly polarized light. Scattered and reflected light is polarized to some dgree. Reflection at Brewster's angle → linearly polarized normal to the incident plane. Example: Find the intensity of unpolarized light when it goes through 3 polarizer plates at angles 0 , θ , and 90 . 1 The first plate cuts y component. → intensity I0 2 1 After the second plate, I 0 cos2 θ 2 1 After the third plate, I 0 cos2 θ sin 2 θ 2 1 If θ = 45 , the intensity becomes I 0 8 Doppler shift Doppler effect in E&M wave is fundamentally different from that in sound waves. cs − Vo Sound wave f '= f0 cs − Vs 1− β2 V E&M wave f ' = f 0 where β = r 1 − β cos θ c Vr = relative velocity between the source and observer Example: A distant nebula is receding from the Earth at a velocity 0.4c. 1 − ( 0.4 ) 2 The frequency is Doppler shifted by = 0.655 and the wavelength 1 + 0.4 1 by = 1.523. 0.655 Example: Police radar emits 4 GHz microwave and detects reflection from a moving car. If the car is moving at 150 km/hr toward the radar, what is the frequency change? 1 V 150000 / 3600 The car detects f ' = f 0 where β = = = 1.4 × 10−7 1− β c 3 × 108 1 The radar detects f '' = f = (1 + 2 β ) f 0 (1 − β ) 2 0 Δf = 2 β f 0 = 1120 Hz Radiation of E&M waves Electric field profile of accelerated charge The charge is accelerated for a short duration after which undergoes drift motion. The kink in the field line has a transverse field component which is the radiation field. For radiation of electromagnetic waves, charge must be accelerated (or decelerated). Radiation electric field Charge q is accelerated for a duration Δt after which it travels at a constant velocity v = aΔt. Coulomb field q q EC = = 4πε 0 r 4πε 0 ( ct ) 2 2 ER vt sin θ From = EC c Δt we find the radiation field qa ER = sin θ 4πε 0c r 2 Radiation power 2 ER 2 1 ⎛ qa ⎞ Poynting flux = ⎜ sin θ ⎟ Z0 Z 0 ⎝ 4πε 0 c 2 r ⎠ Radiation power is π ER 2 1 2 q2 a 2 P=r ∫ 2π sin θ dθ = 2 3 (W) 0 Z0 4πε 0 3 c 3 where π 4 ∫0 sin 3 θ dθ = 3 Example: An electron having an energy of 1 keV undergoes cyclotron motion in a magnetic field of 2 T. Find the acceleration and radiation power. Lorentz force = evB 2E 2000 × 1.6 × 10−19 The velocity is v = = −31 =1.87 × 107 m/sec me 9.1 × 10 Force is evB = 1.6 × 10−19 × 1.87 × 107 × 2=5.98 × 10-12 N F Acceleration is a = =6.6 × 1018 m/sec 2 me 1 2 e2 a 2 Radiation power is P = = 2.46 W=1.54 × 103 eV/sec 4πε 0 3 c 3 Cyclotron Radiation (Non-relativistic) Bending magnet e-beam B into page Synchrotron Radiation Radaition beam (relativistic) Radiation power increases as γ4, frequency as γ2 R orbit curvature raidus Wiggler Free electron laser Wavelength Spatial period/γ2 Radiation by Antenna Radiation by antenna is caused by oscillatory acceleration of electrons. If electron is under oscillatory motion at frequency ω , the acceleration is a = ωv The radiation power per one electron is 1 2 e 2ω 2 v 2 P= 4πε 0 3 c 3 If there are many electrons in an antenna l (m) long, Il = qv and 1 2 I 2 l 2ω 2 Z 0 P= = ( kl ) I 2 (W) 2 4πε 0 3 c 3 6π Z0 R= ( kl ) is called radiation resistance. 2 6π The radiation resistance of λ /2 antenna is R = 73.1 Ω. Other radiation mechanisms Laser and maser (QM mechanism) Ionized gas = plasma (QM) Cherenkov radiation by energetic particles in matter Bremsstrahlung (braking radiation): collision between unlike charges Transition radiation due to passage of charges at a boundary of dielectrics

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Lecture Presentations, Sound Waves, Quantum Effects, Engineering Physics, University of Saskatchewan, Department of Physics, Physics and Engineering

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posted: | 3/29/2010 |

language: | Norwegian |

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