Lecture 2 Review 2 More on Maxwell Wave Equations Boundary Conditions Poynting Vector Transmission Line A. Nassiri - ANL Maxwell’s equations in differential form ∇.D = ρ Gauss' law for electrostatics ∇.B = 0 Gauss' law for magnetostatics ∂D ∇ × H = J + Ampere' s law dt ∂B ∇ × E = − Faraday' s law dt ∂ρ ∇.J = − Equation of continuity ∂t D = εE Varying E and H fields are coupled B = µH Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 2 Electromagnetic waves in lossless media - Maxwell’s equations Constitutive relations Maxwell ∂D D = εE = ε ε E ∇× H = J + r o dt B = µH = µ µ H ∂B r o ∇×E = − dt J = σE SI Units ∇ = ρ • J Amp/ metre2 .D • D Coulomb/metre2 • H Amps/metre • B Tesla ∇.B = 0 Weber/metre2 Volt-Second/metre2 Equation of continuity • E Volt/metre • ε Farad/metre • µ Henry/metre ∂ρ • σ Siemen/metre ∇.J = − ∂t Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 3 Wave equations in free space • In free space – σ=0 ⇒J=0 – Hence: ∂D ∂D ∇× H = J + = dt dt ∂B ∇×E = − dt – Taking curl of both sides of latter equation: ∂B ∂ ∂ ∇×∇× E = −∇× = − ∇× B = −µ ∇× H ∂t ∂t o ∂t ∂ ∂D = −µo ∂t ∂t ∂ 2E ∇×∇× E = −µoε ∂t 2 Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 4 Wave equations in free space cont. ∂ 2E ∇×∇×E = −µoε ∂t 2 • It has been shown (last week) that for any vector A where is the Laplacian operator 2 Thus: ∇×∇× A = ∇∇.A − ∇ A ∂ 2 ∂ 2 ∂ 2 ∇2 = + + ∂x2 ∂y 2 ∂z 2 2 2 ∂ E ∇∇.E − ∇ E = −µoε ∂t 2 There are no free charges in free space so ∇.E=ρ=0 and we get 2 2 ∂ E ∇ E = µoε ∂t 2 A three dimensional wave equation Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 5 Wave equations in free space cont. • Both E and H obey second order partial differential wave equations: ∂ 2E ∇2E = µ ε o ∂ 2 t2 2 ∂ H ∇ H = µoε ∂t 2 What does this mean – dimensional analysis ? Volts/metre Volts/metre = µoε metre2 seconds2 -2 – µοε has units of velocity – Why is this a wave with velocity 1/ √µοε ? Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 6 Uniform plane waves - transverse relation of E and H • Consider a uniform plane wave, propagating in the z direction. E is independent of x and y ∂E ∂E = 0 = 0 ∂x ∂y In a source free region, ∇.D=ρ =0 (Gauss’ law) : ∂E ∂E ∂E ∇.E = x + y + z = 0 ∂x ∂y ∂z E is independent of x and y, so ∂E ∂E ∂E x = 0, y = 0 ⇒ z = 0 ⇒ E = 0 (E = const is not a wave) ∂x ∂y ∂z z z So for a plane wave, E has no component in the direction of propagation. Similarly for H. Plane waves have only transverse E and H components. Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 7 Orthogonal relationship between E and H: • For a plane z-directed wave there are no variations along x and y: ∂ ∂A ∂ Az y H y ∂H x ∇× A = a x − + ∇× H = −a + a ∂y ∂z x ∂z y ∂z ∂Ax ∂Az a y − + ∂D ∂z ∂x = ∂A ∂A ∂t a y − x z ∂x ∂y ∂ ∂Ex Ey ∂Ez = εa + a + a ∂D x ∂t y ∂t y ∂t ∇× H = J + dt Equating terms: and likewise for : ∇×E = −µo ∂H ∂t ∂H ∂E ∂E ∂H − y = ε x y = µ x ∂z ∂t ∂z o ∂t ∂H ∂E ∂E ∂H x = ε y x = µ y ∂z ∂t ∂z o ∂t Spatial rate of change of H is proportionate to the temporal rate of change of the orthogonal component of E & v.v. at the same point in space Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 8 Orthogonal and phase relationship between E and H: • Consider a linearly polarised wave that has a transverse component in (say) the y direction only: ∂ H y ∂Ex = − − = ε Ey Eo f (z vt) ∂z ∂t ∂H x ∂E = ∂H ∂E ⇒ ε y = −εvE f ′(z − vt) ∂z x = ε y ∂t o ∂z ∂t ⇒ = −ε ′ − + = −ε − H x vEo ∫ f (z vt)dz const vEo f (z vt) = −εvEy ε H x = − Ey µo Similarly ∂ E y ∂H x = µo ε ∂z ∂t ∂ H y = Ex ∂Ex H y µ = µo o ∂z ∂t H and E are in phase and orthogonal Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 9 ε ε H = − E H = E x µ y y x o µo • The ratio of the magnetic to electric fields strengths is: 2 2 Ex + E y E µ = = o = η Note: 2 2 H ε H x + H y E E 1 = = = c which has units of impedance µ B o H µoεo Volts / metre = Ω amps / metre and the impedance of free space is: µ 4π ×10−7 o = = 120π = 377Ω ε 1 − o ×10 9 36π Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 10 Orientation of E and H • For any medium the intrinsic impedance is denoted by η E E η = − y = x and taking the scalar product H x H y E.H = Ex H x + E y H y = ηH y H x −ηH x H y = 0 so E and H are mutually orthogonal Taking the cross product of E and H we get the direction of wave propagation E× H = a z (Ex H y − E y H x ) A×B = a x (Ay Bz − Az By )+ 2 2 = a z (ηH y −ηH x ) a y (Az Bx − Ax Bz )+ 2 E× H = a zηH a z (Ax By − Ay Bx ) Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 11 A ‘horizontally’ polarised wave • Sinusoidal variation of E and H • E and H in phase and orthogonal ε H y = Ex µo Hy Ex E× H Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 12 A block of space containing an EM plane wave • Every point in 3D space is characterised by – Ex, Ey, Ez – Which determine • Hx, Hy, Hz • and vice versa – 3 degrees of freedom λ Ex ε H y = Ex µo E× H ε H x = − E y Hy µo Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 13 Power flow of EM radiation 2 • Energy stored in the EM field in the thin box is: εE u = E 2 dU = dU E + dU H = (uE + uH )Adx µ H 2 2 2 o εE µ H uH = = + o dU Adx 2 2 2 2 ε = εE Adx H = E y µ x Power transmitted through the box is dU/dt=dU/(dx/c).... o λ dx Ex E× H Hy Area A Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 14 Power flow of EM radiation cont. dU = εE 2 Adx dU εE 2 ε E 2 S = = Adx = = W/m2 Adt A(dx c) µo η • This is the instantaneous power flow – Half is contained in the electric component – Half is contained in the magnetic component • E varies sinusoidal, so the average value of S is obtained as: 2π E = E sin (z − vt) o λ E 2 sin 2 (z − vt) S = o η E 2 E 2 S = o RMS(E 2 sin 2 (z − vt))= o η o 2η S is the Poynting vector and indicates the direction and magnitude of power flow in the EM field. Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 15 Example • The door of a microwave oven is left open – estimate the peak E and H strengths in the aperture of the door. – Which plane contains both E and H vectors ? – What parameters and equations are required? • Power-750 W • Area of aperture - 0.3 x 0.2 m • impedance of free space - 377 Ω • Poynting vector: E 2 S = = ηH 2 W/m2 η Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 16 E 2 Power = SA = A = ηH 2 A Watts η Power 750 E = η = 377 = 2,171V/m A 0.3.0.2 E 2170 H = = = 5.75A/m η 377 −7 B = µo H = 4π ×10 ×5.75 = 7.2μTesla Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 17 Constitutive relations -12 • permittivity of free space ε0=8.85 x 10 F/m -7 • permeability of free space µo=4πx10 H/m D = εE = ε rεoE Normally ε (dielectric constant) and µ • r r = µ = µ µ – vary with material B H r oH – are frequency dependant J = σE • For non-magnetic materials mr ~1 and for Fe is ~200,000 • er is normally a few ~2.25 for glass at optical frequencies – are normally simple scalars (i.e. for isotropic materials) so that D and E are parallel and B and H are parallel • For ferroelectrics and ferromagnetics er and mr depend on the relative orientation of the material and the applied field: µ µ µ Bx xx xy xz H x µ − jκ 0 At By = µ yx µ yy µ yz H y µ = jκ µ 0 microwave ij B µ µ µ H z zx zy zz z frequencies: 0 0 µo Massachusetts Institute of Technology RF Cavities and Components for Accelerators USPAS 2010 18 Constitutive relations cont..