Chapter 2 Maxwell’s Equations and Plane EM Waves
2-1 Dielectric and Conductor
→ → → → → →
Displacement vector: D = e 0 E+ P = e E = e 0 (1+ χ e )E = e 0e r E
n∆v
→ ∑ Pk Polarization vector: P = lim k =1 ∆v→0 ∆v
→ ∧ 1 P⋅ a 2 2 2 V = R dv', R 2 = (x − x') + (y − y') + (z − z') ∫∫∫v' 2 4πε 0 R
∧ ∧ ∂ ∧ ∂ ∧ ∂ 1 a ∇'= x + y + z ⇒ ∇' = R ∂x' ∂y' ∂z' R R 2
→ → 1 → 1 1 P ∇'⋅P ⇒ V = P⋅∇' dv' = ∇'⋅ dv'− dv' 4πε ∫∫∫v' R 4πε ∫∫∫v' R ∫∫∫v' R 0 0 → ∧ → 1 P⋅ a ' d − ∇'⋅P = n dS' + dv' 4πε ∫∫s' R ∫∫∫v' R 0
→ ∧
Surface charge density: ρps= P⋅ an .
→ Volume charge density: ρp= − ∇ ⋅ P
→ ∧ d → Total charge: Q= P⋅ an dS' + ∇ ⋅ P dv' = 0 ∫∫s' ∫∫∫v' → 1 → → ∇ ⋅ E = (ρ + ρ p ) ⇒ ∇ ⋅ε 0 E+ P = ρ ε 0
→ → → → → →
Define D = ε 0 E+ P ⇒ ∇ ⋅ D = ρ ⇔ D⋅ d S = Q ∫∫s
Dx ε11 ε12 ε13 Ex → ↔ → = ε ⋅ = ε ε ε ⋅ Note: Generally, D E or Dy 21 22 23 E y . Dz ε 31 ε 32 ε 33 Ez
ε 0 0 ↔ 11 ε = ε For biaxial dielectric, 0 22 0 0 0 ε 33 Dx 8 2 0 Ex = ε ⋅ Eg. For an anisotropic medium characterized by Dy 0 2 5 0 E y , Dz 0 0 9 Ez
find the value of the effective relative permittivity for (a) E = zˆE0 , (b)
E = E0 (xˆ − 2yˆ) , (c) E = E0 (2xˆ + yˆ).
Dx 8 2 0 0 0 0 = ε ⋅ = ε = ε ε (Sol.) (a) Dy 0 2 5 0 0E0 0 0E0 9 0 0E0 , r =9 Dz 0 0 9 1 9 1
Dx 8 2 0 1 4 1 = ε ⋅ − = ε − = ε − ε (b) Dy 0 2 5 0 2E0 0 8E0 4 0 2E0 , r =4 Dz 0 0 9 0 0 0
Dx 8 2 0 2 18 2 = ε ⋅ = ε = ε ε (c) Dy 0 2 5 0 1E0 0 9 E0 9 0 1E0 , r =9 Dz 0 0 9 0 0 0
Hall Effect: Current density: J = yˆJ 0 = Nqv
If the material is a conductor or an n-type semiconductor the charge carrier are electrons: q < 0 h h h Hall field: Eh = −v × B = −( yˆv0 ) × (zˆB0 ) = −xˆv0 B0
d Hall voltage: Vh = − Ehdx = v0 B0d ∫0
Ex 1 Hall coefficient: Ch = = < 0 J y Bz Nq If the material is a p-type semiconductor, the charge carries are holes: q > 0 h Hall field: Eh = xˆv0 B0
Hall voltage: Vh = −v0 B0d
Hall coefficient: Ch > 0 2-2 Boundary Conditions of Electromagnetic Fields Boundary conditions for electric fields:
Eg. Show that Et=0 on the conductor plane. (Proof) ∵ The E-field inside a conductor is zero,
→ → ∴ ⋅ = ∆ = ⇒ = ∫ E d l Et W 0 Et 0 → → ρ ∆S ρ ⋅ = ∆ = s ⇒ = s ∫∫ E d S En S En ε 0 ε 0
∧ → → Eg. Show that E1t= E2t and a ⋅ D − D = ρ on the interface between two n2 1 2 s dielectric.
→ →
(Proof) E⋅ d l = E1t ∆W − E2t ∆W = 0 , E1t=E2t ∫abcda → → → ∧ → ∧ ∧ ∆⋅ d S = ∆1⋅ an2 + ∆2 ⋅ an1 ∆S = an2 ⋅ (∆1 − ∆2 )∆S = ρ s ∆S ∫∫s ∧ → → a ⋅ D − D = ρ or D1n-D2n=ρs n2 1 2 s
If ρs=0, then D1n=D2n or ε1E1n=ε2E2n
Eg. Two dielectric media are separated by a charge free boundary. The electric
field intensity in media 1 at the point P1 has a magnitude E1 and makes an angle with the normal. Determine the magnitude and direction of the electric field
intensity at point P2 in medium 2. [交大電子所]
tana 2 ε 2 (Sol.) E2 sinα 2 = E1 sinα1 , ε 2 E2 cosα 2 = ε1E1 cosα1 ⇒ = tana1 ε1
2 2 2 2 E2 = E2t + E2n = (E2 sinα 2 ) + (E2 cosα 2 )
2 1/ 2 2 1/ 2 ε ε = (E sinα )2 + 1 E cosα = E sin 2 α + 1 cosα 1 1 ε 1 1 1 1 ε 1 2 2 Eg. Assume that z=0 plane separates two lossless dielectric regions with εr1=2 and
→ ∧ ∧ ∧ → →
εr2=3. If E1 in region 1 is x 2y − y 3x + z(5 + z), find E2 and D2 at z=0 in region 2.
→ ∧ ∧ ∧ → → ∧ ∧
(Sol.) E1 = x 2y − y 3x + z 5, E1t (z = 0) = E2t (z = 0) = x 2y − y 3x ,
→ → → →
D1n (z = 0) = D2n (z = 0) ⇒ 2 E1n (z = 0) = 3E2n (z = 0) → 2 ∧ ∧ 10 → ∧ ∧ ∧ 10 E (z = 0) = z 5 = z , ∴ E (z = 0) = x 2y − y 3x + z 2n 3 3 2 3 → ∧ ∧ ∧ 10 D (z = 0) = x 2y − y 3x + z 3ε 2 3 0
Eg. A lucite sheet (εr=3.2) is introduced perpendicularly in a uniform electric
→ ∧ → → → field E0 = x E0 in free space. Determine Ei , Di and Pi inside the lucite. [中 央地球物理所]
→ ∧ ∧ ∧
(Sol.) Di = x Di = x D0 = xε 0 E0
→ → → ∧ 1 1 E0 Ei = Di = Di = x ε ε 0ε r 3.2 → → → ∧ 1 ∧ P = D − ε E = x1− ε E = x 0.6875ε E (C / m) i i 0 i 3.2 0 0 0 0
Eg. Dielectric lenses can be used to collimate electromagnetic fields. The left surface of the lens is that of a circular cylinder, and right surface is a plane. If
→ ∧ ∧
E1 at point P(r0,45°,z) in region 1 is ar 5 − aφ 3, what must be the dielectric
→ constant of the lens in order that E3 in region 3 is parallel to the x-axis?
→ ∧ ∧
(Sol.) Assume E2 = ar E2r + aφ E2φ , ∵ E1t = E2t = Eφ ⇒ E2φ = −3
→ →
For E3 // x − axis ⇒ E2 // x − axis ⇒ E2φ = −E2r ⇒ E2r = 3
∧ → ∧ → 5 a ⋅ D = a ⋅ D ⇒ ε E = ε E , ε 5 = ε ε 3 ⇒ ε = n 1 n 2 1 r1 2 r 2 0 0 r 2 r 2 3
Eg. A positive point charge Q is at the center of a spherical dielectric shell of an inner radius Ri and an outer radius Ro. The dielectric constant of the shell is εr.
→ → → Determine E , V , D , and P as functions of the radial distance R. [高考]
→ → → →
(Sol.) P = D− ε 0 E = ε 0 (ε r −1)E
R>Ro: Q Q = = E aˆ R 2 , V 4πε 0 R 4πε 0 R Q D = aˆ and P = 0 R 4πR 2
Ri Q Q Q 1 Q = ˆ = ˆ = ˆ = ˆ − E aR 2 aR 2 , D aR 2 , P aR 1 2 4πε 0ε r R 4πεR 4πR ε r 4πR Ro Q R Q Q 1 1 1 V = − dR − dR = 1− + ∫∞ 2 ∫ 2 Ro 4πε 0 R 4πε 0ε r R 4πε 0 ε r Ro ε r R Q Q = = = R R Q Q 1 1 1 1 1 V = V − dR = 1− − 1− + ∫ 2 = Ri R Ri 4πε 0 R 4πε 0 ε r Ro ε r Ri R Boundary conditions for magnetic fields: Eg. Show that μ1H1n=μ2H2n and aˆn2 × (H 1 − H 2 ) = J . → → ⋅ = ⇒ ∆ − ∆ = (Proof) ∫∫ B d S 0 B1n S B2n S 0 , B1n=B2n ⇒ μ1H1n=μ2H2n ⋅ = ⇒ ⋅ = ⋅ ∆ + ⋅ −∆ = ∆ ∫ H dl I ∫ H dl H1 w H 2 ( w) J sw w abcda ⇒ H1t − H 2t = J sw ⇒ aˆn2 × (H 1 − H 2 ) = J If J=0, then H1t=H2t Eg. Two magnetic media with permeabilities μ1 and μ2 have a common boundary. The magnetic field intensity in medium 1 at the point P1 has a magnitude H1 and makes an angle α1 with the normal. Determine the magnitude and the direction of the magnetic field intensity at point P2 in medium 2. µ H cosa = µ H cosa tana µ (Sol.) 2 2 2 1 1 1 ⇒ 2 = 2 H 2 sina2 = H1 sina1 tana1 µ1 −1 µ2 ⇒ a2 = tan ( tana1) µ1 1 2 2 2 2 2 2 µ1 2 H 2 = H 2t + H 2n = (H 2 sinα2 ) + (H 2 cosα2 ) = H1 sin α1 + ( cosα1) µ2 Eg. Consider a plane boundary (y=0) between air (region 1, μr1=1) and iron (region 2, μr2=5000). (a) Assuming B1 = 0.5xˆ −10yˆ (mT), find B2 and the angle that B2 makes with the interface. (b) Assuming B2 = 10xˆ + 0.5yˆ (mT), find B1 and the angle that B1 makes with the normal to the interface. (Sol.) B2x 0.5 (a) B1 = 0.5xˆ −10yˆ , B2 = B2x xˆ + B2 y yˆ , H 2x = = H1x = ⇒ B2x = 2500 5000µo µo µ 2 B1x B2 y = B1y = −10 ⇒ B2 = 2500xˆ −10yˆ , tana 2 = tana1 = 500 = 250 µ1 B1y B1x B2x (b) B2 = 10xˆ + 0.5yˆ , B1 = B1x xˆ + B1y yˆ , H1x = = H 2x = µ1 µ 2 1 10 ⇒ B1x = B2x = = 0.002 , B1y = B2 y = 0.5 , ∴ B1 = 0.002xˆ + 0.5yˆ , µ BE 5000 B1x 0.002 tana1 = = = 0.004 B1y 0.5 Magnetic flux lines round a cylindrical bar magnet: Eg. A very large slab of material of thickness d lies perpendicularly to uniform magnetic field intensity H 0 = zˆH 0 . Ignoring edge effect, determine the magnetic field intensity in the slab: (a) if the slab material has a permeability μ, (b) if the slab permanent magnet having a magnetization vector M i = zˆM i . [台大物研] (Sol.) µo (a) B2 = µ H 2 , B = B ⇒ µ H = µ H , H 2 = zˆH = zˆ H 2 2z 1z 2 o o 2 µ o (b) B 2 = µo (H 2 + M i ) , B2z = B1z ⇒ µo (H 2 + M i ) = µo H o ⇒ H 2 = zˆ(H o − M i ) Eg. Assume that N turns of wire are wound around a toroidal core of a ferromagnetic material with permeability μ. The core has a mean radius r0, a circular cross section of radius a (a << r0), and a narrow air gap of length lg, as shown in Figure. A steady current I0 flows in the wire. Determine (a) the magnetic flux density Bf in the ferromagnetic core; (b) the magnetic field intensity Hf in the core; and, (c) the magnetic field intensity Hg in the gap. [台大 電研] (Sol.) B B ⋅ = = = f π − + f = ∫ H d l NI o , B f B g aˆf B f , (2 ro l g ) l g NI o C µ µo µ o µNI o µ o NI o B f = ⇒ H f = aˆ f µ o (2πro − l g ) + µl g µ o (2πro − l g ) + µl g µNI o H g = aˆφ µo (2πro − l g ) + µl g 2-3 Steady-state Currents ∆Q Nqv ⋅ aˆ ∆s∆t Differential current: ∆I = = n = Nqv ⋅ ∆s ∆t ∆t d Current density: J = Nqv = ρv (A/m2), I = ∫∫ J ⋅ dS (A) s Let v = µE, J = ρv = −µρE = σE m : mobility σ : conductivity σ = −ρeue + ρ huh electrons holes Eg. An emf V is applied across a parallel-plate capacitor of area S. The space between the conducting plates is filled with two different lossy dielectrics of thicknesses d1 and d2, permittivities ε1 and ε2, and conductivities σ1 and σ2, respectively. Determine (a) the current density between the plates, (b) the electric field intensities in both dielectrics. [高考] (Sol.) d d V = (R + R )I = 1 + 2 I 1 2 σ σ 1S 2 S I V σ σ V J = = = 1 2 S (d1 σ 1 )+ (d 2 σ 2 ) σ 2 d1 + σ 1d 2 σ 2V σ 1V V = E1d1 + E2 d 2 , J = σ 1E1 = σ 2 E2 , E1 = , E2 = σ 2 d1 + σ 1d 2 σ 2 d1 + σ 1d 2 Eg. Assume a rectangular conducting sheet of conductivity σ, width a, and height b. A potential difference is applied to the side edges. Find (a) the potential distribution, (b) the current density everywhere within the sheet. [台科大電子所] (Sol.) Vo (a) V(x)=Cx, V(a)=Ca=V0 ⇒ V(x)= x a Vo σVo (b) E = −∇V (x) = −xˆ ⇒ J = σ E = −xˆ a a dQ d ∂ρ Equation of continuity: I = ∫∫ J ⋅ dS = − = − ∫∫∫ ρdv = ∫∫∫∇ ⋅ Jdv ⇒ ∇ ⋅ J + = 0 s dt dt v′ ∂t σ ∂p ρ ∂ρ − t If J = σE, σ∇ ⋅ E + = σ + = 0, ρ= ρ e e ∂t ε ∂t 0 Eg. Lightning strikes a lossy dielectric sphere εr=1.2, σ=10S/m, of radius 0.1m at time t=0, depositing uniformly in the sphere a total charge 1mC. Determine for all t for (a) the electric field intensity both inside and outside the sphere, (b) the current density in the sphere, (c) calculate the time it takes for the charge density in the sphere to diminish to 1% of its initial value, (d) calculate the charge in the electrostatic energy stored in the sphere as the charge density diminished from the initial value to 1% of the value. What happens to this energy? (e) determine the electrostatic energy stored in the space outside the sphere. Does this energy charge with time? σ Q − t (Sol.) ρ = = 0.239(C m3 ), ρ = ρ e e 0 4 0 πb3 3 4π 3 R ρ σ ρ R − t 10 < = ˆ 3 = ˆ 0 e = ˆ × 9 ⋅ −7.42×10 t (a) R b : Ei a R 2 a R e a R 7.5 10 R e (V m) 4πeR 3e Q Q > = ˆ = ˆ × 6 ....R b : E0 a R 2 a R 2 10 (V m) 4πe 0 R R 10 −7.42×1010 t (b) R < b : J i = σEi = aˆ R 7.5×10 Re ....R > b : J 0 = 0 s − t ρ n100 −12 (c) e e = = 0.01⇒ t = = 4.88 ×10 (s) ρ 0 (s e ) ξ 2 e −2 t W = 2 ' ∝ e ⇒ i = = −4 (d ) Wi ∫∫∫ Ei dv e (0.01) 10 2 v' (Wi )0 ∞ e 0 2 2 3 (e) W0 = E0 4πR dR = 4.5 ×10 (J ) 2 ∫0 Boundary conditions for current densities: Governing Equations for Steady Current Density Differential Form Integral Form ⋅ = ∇ ⋅ J = 0 ∫s J ds 0 J 1 ∇ × = 0 ⋅ = ∫c J d 0 σ σ ∇ ⋅ J = 0 ⇒ J1n = J 2n J J1t σ 1 ∇ × = 0 ⇒ = σ σ J 2t 2 Eg. Two lossy dielectric media with permittivities and conductivities (ε1, σ1) and (ε2, σ2) are in contact. An electric field with a magnitude E1 is incident from medium 1 upon the interface at an angle α1 and measured from the common normal, as in Figure. (a) Find the magnitude and direction of E 2 in medium 2. (b) Find the surface charge density at the interface. (Sol.) (a) E1t = E 2t => E1 sinα 1 = E 2 sinα 2 , J 1n = J 2n => s 1E1 cosα 1 = s 2 E 2 cosα 2 2 s 1 2 E 2 = E1 sin a + ( cosa ) 1 s 1 => 2 s 2 a 2 = a tan tan 1 s 1 σ 1 (b) D2n − D1n = ρs => ε 2E 2n −ε 1E1n = ρs , ρs= ε 2 − ε 1E1 coσα σ 2 Eg. Two conducting media with conductivities σ1 and σ2 are separated by an interface. The steady current density in medium 1 at point P1 has a magnitude J1 and makes an angle α1 with the normal. Determine the magnitude and direction of the current density at point P 2 in Medium 2. [台大電研] (Sol.) tana 2 s 2 J1 cosa1 = J 2 cosa 2 , s 2 J1 sina1 = s 1 J 2 sina 2 ⇒ = tana1 s 1 1 2 2 2 2 2 2 s J = J + J = (J sinα ) + (J cosα ) = 2 J sinα + (J cosα )2 2 2t 2n 2 2 2 2 s 1 1 1 1 1 J = J ⇒ s E = s E s s 1n 2n 1 1n 2 2n 2 1 ⇒ ρ s = ε1 − ε 2 E2n = ε1 − ε 2 E1n . If D1n − D2n = ρ s ⇒ ε1E1n − ε 2 E2n = ρ s s 1 s 2 s 2 >> s 1 ⇒ ρ s = ε1E1n = D1n . 2-4 Maxwell’s Equations and Plane EM Waves ∂D Note: is equivalent to a current density, called the displacement current density. ∂t Eg. A voltage source V0sin(ωt), is connected across a parallel-plate capacitor C. Find the displacement current in the capacitor. dv A (Sol.) i = C C = CV ω cosωt = ε V ω cosωt C dt 0 d 0 V ∂D d A = ε ⇒ = ε 0 ω = ⋅ = ε ω ω = D E D sin t , iD ∫∫ dS V0 cos t iC d A ∂t d ∂V Lorentz condition: ∇ ⋅ A + µε =0 ∂t ∂D ∂ ∂A ∂V ∂ 2 A ∇× B = µJ + µ = ∇×∇× A = µJ + µε (−∇V − ) = ∇(∇ ⋅ A) − ∇ 2 A = µJ − ∇(µε ) − µε ∂t ∂t ∂t ∂t ∂t 2 ∂ 2 A ∂V ⇒ ∇ 2 A − µε = −µJ + ∇(∇ ⋅ A + µε ) ∂t 2 ∂t ∂ 2 A If Lorentz Condition holds, we have ∇2 A − µε = −µJ ∂t 2 ∂A ∂ ∂ ∂V ρ ∵ ∇ ⋅ D = ρ = ∇ ⋅ εE = ρ = ∇ ⋅ ε (−∇V − ) ⇒ ∇ 2V + (∇ ⋅ A) = ∇ 2V + (−µε ) = − ∂t ∂t ∂t ∂t ε ∂ 2V ρ ∴ ∇ 2V − µε = − ∂t 2 ε Effective permittivity: ∂E σ ∇ × H = J + ε = σE + jωεE = jω(ε + )E = jωε E ∂t jω C σ ⇒ ε = ε − j = ε'− jε''⇒ σ = ωε '' . Similarly, µ = µ'− jµ'' C ω ε '' σ Loss tangent: tanδ = = C ε ' ωε Eg. A sinusoidal electric intensity of amplitude 250V/m and frequency 1GHz exists in a lossy dielectric medium that has a relative permittivity of 2.5 and loss tangent of 0.001. Find the average power dissipated in the medium per cubic meter. (Sol.) σ tanδ e = 0.001 = , ωe 0e r −9 9 10 −4 σ = 0.001(2π10 )( )(2.5) = 1.39×10 (S / m) 36π 1 1 1 p = JE = σE 2 = × (1.39×10−4 ) × 2502 = 4.34(W / m3 ) 2 2 2 Maxwell’s Equations in the source-free regions: ∂H ∂E ∇ × E = −µ , ∇ × H = ε , ∇ ⋅ E = 0, ∇ ⋅ H = 0 ∂t ∂t − β +θ 3 4 − β +θ Phasor representations: Eg. xˆAe j( z ) , (xˆ + yˆ )e j( z ) , etc. 5 5 Instantaneous representations: Eg. xˆAcos(ωt − βz +θ ) = Re[xˆAe− j(βz+θ ) ⋅ e jωt ] , etc. ∂H In case E and H are proportional to ejωt, we have ∇ × E = −µ = − jωµH and ∂t ∂E ∇ × H = ε = jωεE . ∂t Eg. Given that H = yˆ2cos(15πx)sin(6π109 t − βz) in air, find E and β. − jβz 2 2 2 2 (Sol.) Phasor: H = yˆ2cos(15πx)e , (15π ) + β = ω µ0ε 0 = 400π ⇒ β = 13.2π 1 E = ∇ × H = [xˆ158π cos(15πx) + zˆj180π sin(15πx)]e − jβz jωe 0 E(x, z,t) = Re[E(x, z)e jωt ] Eg. Given that E = yˆ0.1cos(10πx)sin(6π109 t − βz) in air, find H and β. − jβz 2 2 2 2 (Sol.) Phasor: E = yˆ0.1cos(10πx)e , (10π ) + β = ω µ0ε 0 = 400π ⇒ β = 10 3π 1 j − β H = − ∇ × E = [xˆ0.1β cos(10πx) + zˆ0.1(10π )cos(10πx)]e j z jωµ0 ωµ0 H (x, z,t) = Re[H (x, z)e jωt ] Eg. The electric field intensity of a spherical wave in free space is E E = aˆ 0 sinθ cos(ωt − kR) . Determine the magnetic field intensity. R R E (Sol.) Phasor: E = aˆ 0 sinθ ⋅ e− jkR R R 1 ∂ E0 − jkR E0 e 0 − jkR − jωµ0 H = ∇ × E = aˆφ (RE0 ) = aˆφ (− jk) sinθ ⋅ e ⇒ H = aˆφ sinθ ⋅ e R ∂R R R µ0 Plane EM waves excited by a current sheet: Given J (t) = −xˆJ (t) at z=0, the field components of the EM plane wave excited by z 1 z the current density are E(z,t) = xˆ J (t ) and H (z,t) = ± yˆ J (t ) , 2 v p 2 v p respectively. If it is a sinusoidal EM plane wave, J (t) = −xˆJ 0 cos(ωt) at z=0. J J We have E(z,t) = xˆ 0 cos(ωt kz) , H (z,t) = ± yˆ 0 cos(ωt kz) . 2 2 Electromagnetic wave spectrum: 2-5 Plane EM waves in a simple, nonconducting and source-free region In a simple, nonconducting and source-free region: ∂H ∂E ∇ × E = −µ , ∇ × H = ε , ∇ ⋅ E = 0, ∇ ⋅ H = 0 ∂t ∂t ∂ ∂ 2 E ∂ 2 E ∇ × ∇ × E = −µ (∇ × H ) = −µε = ∇(∇ ⋅ E) − ∇ 2 E = −∇ 2 E ⇒ ∇ 2 E − µε = 0 . ∂t ∂t 2 ∂t 2 1 Velocity of the plane EM wave: v= µε -7 1 -9 1 8 In vacuum, μ0=4π×10 , ε0= ×10 ⇒ c = ≈ 3×10 (m / s) . π 36 m 0ε 0 2π 2π Wave number: k=ω/v=ω µε = = v / f λ Assume E ∝ e jωt ⇒ ∇2 E + k 2 E = 0 (drop ejωt factor) d 2 E(z) Suppose E = E(z) ⇒ + k 2 E = 0 ⇒ E(z) = E + e − jkz + E −e jkz dz 2 0 0 Traveling wave in +z-direction: + + − jkz jωt + E0 (z,t) = Re[E0 e ⋅ e ] = E0 cos(ωt − kz) dz ω Let ωt-kz=constant ⇒ Phase velocity: vp= = dt k + + + + If E = xˆE x (z),∇ × E = − jωµ(xˆH x + yˆH y + zˆH z ) 1 1 ωµ µ ⇒ H + = H + = 0, H + (z) = − ⋅ (− jk)E + (z) = E + (z) , where η= = , x z y jωµ x η x k ε and η0=120π ≈377Ω in free space. TEM waves (Transverse electromagnetic waves): E and H ⊥ direction of propagation ( aˆn ) − − − jkx x jk y y jkz z − jk ⋅R − jkaˆn ⋅R E(R) = E(x, y, z) = E0e = E0e = E0e , where R = xˆx + yˆy + zˆz , 2 2 2 2 k = aˆn k , and k x + k y + k z = ω µε − jkaˆn ⋅R − jkaˆn ⋅R − jkaˆn ⋅R − jkaˆn ⋅R ∇ ⋅ E = 0 = ∇ ⋅ (E0e ) = e ∇ ⋅ E0 + E0 ⋅ (∇e ) = E0 ⋅ (∇e ) − jkaˆn ⋅R − jkaˆn ⋅R = E0 ⋅ − j(xˆk x + yˆk y + zˆk z )e = − jk(E0 ⋅ aˆn )e aˆn ⋅ E0 = 0 ⇒ E0 ⊥ aˆn (TE). Similarly, ∇ ⋅ H = 0 ⇒ H 0 ⊥ aˆn (TM) Relation between E-field and H-field of the plane EM wave: 1 1 ωµ µ E(R) = ∇ × H (R) = (− jk)aˆ × H (R) ⇒ E(R) = −aˆ × H (R) , where η= = jωε jωε n n k ε 1 1 1 H (R) = − ∇ × E(R) = aˆ × E(R) ⇒ H (R) = aˆ × E(R) ⇒ H ⊥ aˆ jωµ n n n Eg. The instantaneous expression for the magnetic field intensity of a uniform plane wave propagating in the +y direction in air is given by −6 7 π H = zˆ4×10 cos(10 πt − k y + ) A/m. (a) Determine k0 and the location where 0 4 H z vanishes at t=3ms. (b) Write the instantaneous expression for E . ω 107 π π (Sol.) ω = 107 π ⇒ k = = = , aˆ = yˆ 0 c 3×108 30 n π π π 2n +1 1 (a) cos[(2n +1) ] =0⇒ 107 π × 3×10−3 − y + = π ⇒ y = 30(3×104 − − n) 2 30 4 2 4 π π (b) E(z,t) = − aˆ × H (z,t) , E(z,t) = −xˆ480π10−6 cos(107 πt − y + ) 0 n 30 4 Eg. A 100MHz uniform plane wave E = xˆEx propagates in the +z direction. -4 Suppose εr=4, μr=1, σ=0, and it has a maximum value of 10 V/m at t =0 and z=0.125m. (a) Write the instantaneous expressions for E and H . (b) Determine the location where E is a positive maximum when t=10-8sec. 4π µ0 µr (Sol.) k = ω µ0 µrε 0ε r = , aˆn = zˆ , η = = 60π 3 ε 0ε r −4 8 (a) E(z,t) = xˆE x = xˆ10 cos(2π ×10 t − kz + θ ) has the maximum in case of π 4π π 2π ×108 t − kz + θ = 0 ⇒ θ = ⇒ E(z,t) = xˆ10−4 cos(2π108 t − z + ) , 6 3 6 η 1 η 10−4 4π π H (z,t) = aˆ × E(z,t) = yˆ cos(2π108 t − z + ) η n 60π 3 6 4π π 13 3n (b) cos(2nπ ) = 1, 2π108 (10 −8 ) − z + = 2nπ ⇒ z = ± 3 max 6 max 8 2 Polarization of the EM wave: The direction of electric field of the EM wave. In the following text, we assume all EM waves to be z-propagated if we do not specify them. Linear polarizations in the x and the y-direction, − j(kz+θ ) − j(kz+θ ) respectively: E = xˆEx e , E = yˆE y e Linear polarization in general case: − j(kz+θ ) − j(kz+θ ) E = xˆEx e + yˆE y e , where Ex and Ey are in phase (we can assume the both to be real). Right–hand circular polarization: − j(kz+θ ) − j(kz+θ ) E = xˆE0e − yˆjE0e Left–hand circular polarization: − j(kz+θ ) − j(kz+θ ) E = xˆE0e + yˆjE0e Right–hand elliptical polarization: − j(kz+θ ) − j(kz+θ ) E = xˆE10e − yˆjE20e ( E10 ≠ E20 ) Left–hand elliptical polarization: − j(kz+θ ) − j(kz+θ ) E = xˆE10e + yˆjE20e ( E10 ≠ E20 ) We can receive/transmit linearly-polarized EM waves by a linear dipole antenna. We can receive/transmit circularly-polarized EM waves by a circular reflector antenna. Instantaneous Expression for E of right–hand elliptical–polarization (drop phase factor e-jθ): E(z,t) = Re{[xˆE e − jkz − yˆjE e − jkz ]e jωt }= xˆE cos(ωt − kz) + yˆE sin(ωt − kz) 10 20 10 20 = xˆE1 (z,t) + yˆE2 (z,t) E (0,t) E (0,t) E (0,t) E (0,t) E (0,t) ⇒ cos(ωt) = 1 , sin(ωt) = 2 ⇒ [ 1 ]2 + [ 2 ]2 = 1, ωt = tan −1 2 E10 E20 E10 E20 E1 (0,t) 1 1 1. xˆE = (xˆE − yˆjE ) + (xˆE + yˆjE ) : A linearly polarized plane wave can be x 2 x y 2 x y resolved into a right –hand and left–hand elliptically- or circularly-polarized waves. E + E E − E E − E E + E 2. xˆE − yˆjE = (xˆ .0 1 − yˆj 1 0 ) + (xˆ 0 1 + yˆj 0 1 ) : 0 0 2 2 2 2 A circularly–polarized plane wave can be resolved into two opposite elliptically–polarized waves. E + E E + E E − E E − E 3. xˆE − yˆjE = (xˆ .1 2 − yˆj 1 2 ) + (xˆ 1 2 + yˆj 1 2 ) : 1 2 2 2 2 2 An elliptically–polarized plane wave can be resolved into two opposite circularly–polarized waves. Eg. The E field of a uniform plane wave propagating in a dielectric medium is z z given by E(t, z) = xˆ2cos(108 t − ) − yˆ sin(108 t − ) V/m. (a) Determine the 3 3 frequency and wavelength of the wave. (b) What is the dielectric constant of the medium? (c) Describe the polarization of the wave. (d) Find the corresponding H field. (Sol.) Phasor: E = xˆ2e − jz / 3 + yˆje − jz / 3 1 2π (a) ω = 108 ⇒ f = 1.59×107 Hz , k = ⇒ λ = = 2 3π 3 k ω (b) v = = 3 ×108 = 1/ µ ε ε ⇒ ε = 3 β 0 0 r r (c) It is the left–hand elliptically-polarized wave propagating along +z direction. µ0 120π (d) η = = , aˆn = zˆ ε 0ε r 3 η 1 η 1 3 H = aˆ × E = zˆ × (xˆ2e − jz / 3 + yˆje − jz / 3 ) = (yˆ2e − jz / 3 − xˆje − jz / 3 ) η n η 120π 3 zz ⇒Hzt( , ) = Re[ Hze ( )jtω ] = [ xˆ sin(1088 t −+ ) yˆ 2cos(10 t − )] 120π 33 Eg. Write down the instantaneous expression for the electric- and magnetic-field intensities of sinusoidal time-varying uniform plane wave propagating in free space and having the following characteristics: (1) f=10GHz; (2) direction of propagation is the +z direction; (3) left-hand circular polarization; (4) the initial condition is the electric field in the z=0 plane and t=0 having an x-component equal to E0 and a y-component equal to √3E0. [台大電研] ω 2π (Sol.) ω = 2π ×1010 , v = = c = 3×108 ⇒ k = ×102 k 3 Phasor: E = xˆAe − j(kz+θ ) + yˆjAe − j(kz+θ ) for the left-hand circular polarization ⇒ E(z,t) = Re{[xˆAe −( jkz+θ ) + yˆjAe − j(kz+θ ) ]e jωt }= xˆAcos(ωt − kz + θ ) − yˆAsin(ωt − kz + θ ) -1 z=0 and t=0, E(0,0) = xˆAcos(θ ) − yˆAsin(θ ) = xˆE0 + yˆ 3E0 ⇒ θ=tan (- √3), A=2E0 2π 2π E(z,t) = xˆ2E cos[2π1010 t − 10 2 z + tan −1 (− 3)] − yˆ2E sin(2π1010 t − 10 2 z + tan −1 (− 3)] 0 3 0 3 η 1 η 1 − j(kz+θ ) − j(kz+θ ) 2E0 − j(kz+θ ) − j(kz+θ ) H = aˆn × E = zˆ ×[xˆAe − yˆjAe ] = [yˆe + xˆje ] η0 η0 120π ⇒ H (z,t) = Re[H (z)e jωt ] Application of polarization: Liquid Crystal Display (LCD) The polarizations of incident lights are synchronized by the rotations of molecules of liquid crystal, which were controlled by an AC voltage. And then the output polarizer can block the orthogonally-polarized lights to control the output optical intensities. Poynting vector: P = E × H ∂B ∂D ∂B ∂D ∇ × E = − , ∇ × H = J + ⇒ ∇ ⋅ (E × H ) = H ⋅ (∇ × E) − E ⋅ (∇ × H ) = −H ⋅ − E ⋅ − E ⋅ J ∂t ∂t ∂t ∂t ∂(µH ) ∂(εE) ∂ 1 2 ∂ 1 2 2 = −H ⋅ − E ⋅ − E ⋅ J = − ( µ H ) − ( ε E ) −σ E ∂t ∂t ∂t 2 ∂t 2 d ∂ ε 2 µ 2 2 ∴ ∫∫(E × H ) ⋅ dS = ∫∫∫∇ ⋅ (E × H )dv = − ∫∫∫( E + H )dv − ∫∫∫s E dv s v ∂t v 2 2 v ⇒ P = E × H is the electromagnetic power flow per unit area. Instantaneous power density: P(z,t) = Re[E(z)e jωt ]× Re[H (z)e jωt ] η η 1 E − j(βz+θ ) Set E(z) = xˆE (z) = xˆE e−(α + jβ )z ⇒ H (z) = [αˆ × E(z)] = yˆ 0 e−αz ⋅ e η , x 0 η n η jωt −αz ∴ E(z,t) = Re[E(z)e ] = xˆE0e cos(ωt − βz) η E and H (z,t) = Re[H (z)e jωt ] = yˆ 0 e −αz cos(ωt − βz −θ ) η η ⇒ P(z,t) = E(z,t) × H (z,t) = Re[E(z)e jωt ]× Re[H (z)e jωt ] 2 E0 −2αz 2 = zˆ e [cosθη + cos(2ωt − 2βz −θη )] ∝ E 2η 0 1 Average power density: P = Re(E × H * ) av 2 2 T η 1 η E0 −2az Pav = P(z,t)dt = zˆ e cosθη , where T is the period. And it can be proved that T ∫0 2η 1 P = Re(E × H * ) . av 2 Eg. Show that P(z,t) of a circularly–polarized plane wave propagating in a lossless medium is a constant. (Sol.) Assuming right–hand circularly–polarized plane wave, aˆn = zˆ E(z,t) = E0 [xˆ cos(ωt − βz) + yˆ sin(ωt − βz)] η 1 η E H (z,t) = (aˆ × E) = 0 [−xˆ sin(ωt − βz) + yˆ cos(ωt − βz)] η n η η η η E 2 P(z,t) = E(z,t) × H (z,t) = zˆ 0 η Eg. The radiation electric field intensity of an antenna system is E = aˆθ Eθ + aˆφ Eφ , find the expression for the average outward power flow per unit area. η 1 η Eφ E (Sol.) aˆ = aˆ , H = (aˆ × E) = (−aˆ + aˆ θ ) n r η n θ η φ η * η 1 η η * 1 Eφ * Eθ 1 2 2 P = Re(E × H ) = Re[(aˆθ Eθ + aˆφ Eφ ) × (−aˆθ + aˆφ )] = aˆ ( Eθ + Eφ ) av 2 2 η η 2η r Eg. Find P on the surface of a long, straight conducting wire of radius b and conductivity σ that carries a direct current I. Verify Poynting’s theorem. I J I I I 2 (Sol.) J = zˆ ⇒ E = = zˆ , H = aˆφ ⇒ P = E × H = −aˆ πb 2 σ σπb 2 2πb r 2σπ 2b3 2 d I 2 2 − P ⋅ dS = − P ⋅ aˆ dS = ⋅ 2πb = I ( ) = I R ∫∫ ∫∫ r 2 2 2 s s 2sπ b sπb 2-6 Plane EM Wave in a Lossy Media σ σ ∇ × H = J + jωεE = σE + jωεE = jω(ε − j )E = jωε E ,ε = ε − j = ε ' − jε '' . ω c c ω ' '' Similarly, µc = µ − jµ σ Complex wave number: k = ω µε . Loss tangent: tanδ ≈ ε '' ε ' = c c c ωε σ 1 Propagation constant: γ = jk = jω µε = α + jβ = jω µε (1+ ) 2 c c jωε −γ − −α − β E ∝ e z = e jkc z = e z ⋅ e j z If the medium is lossless, α=0; else if the medium is lossy, α>0. 2π Phase constant: β = λ µε σ 1 µε σ 1 ⇒ α = ω [ 1+ ( ) 2 −1] 2 , β = ω [ 1+ ( ) 2 +1] 2 2 ωε 2 ωε σ σ µ 1 σ Case 1 Low-loss Dielectric: << 1 ⇒ α ≈ , β ≈ ω µε [(1+ ( ) 2 ] ωε 2 ε 8 ωε µ σ Intrinsic impedance: η ≈ (1+ j ) c ε 2ωε ω 1 1 1 σ Phase velocity: v = = ≈ [1− ( ) 2 ] p β ωε µε c µε 8 σ ωµσ Case 2 Good Conductor: >> 1 ⇒ α = β ≈ = πfµσ , ωε 2 µ jωµ πfµ α and ηc = ≈ ( ) = (1+ j) = (1+ j) ε c σ σ σ ω 2ω Phase velocity: v = ≈ p β µσ 1 1 Skin Depth (depth of penetration): δ = = . α πfµσ 1 1 λ For a good conductor, δ = ≈ = α β 2π 7 Eg. E(t, z) = xˆ100cos(10 πt) V/m at z=0 in seawater: εr=72, μr=1, σ=4S/m. (a) Determine α, β, vp, and ηc. (b) Find the distance at which the amplitude of E is 1% of its value at z=0. (c) Write E(z,t) and H(z,t) at z=0.8m, suppose it propagates in the +z direction. 7 6 (Sol.) ω = 10 π , f=5×10 Hz, σ/ωε0εr=200>>1, ∴ Seawater is a good conductor in this case. πfµ (a) α = πfµσ = 8.89Nπ / µ = β , η = (1+ j) c σ ω 2π 1 v = = 3.53×106 m / s , λ = = 0.707m , δ = = 0.112m p β β α 1 (b) e −αz = 0.01 ⇒ z = ln(100) = 0.518m α (c) E(z,t) = Re[E(z)e jωt ] = xˆ100e−αz cos(ωt − βz) z = 0.8m ⇒ E(0.8,t) = xˆ100e −0.8α cos(ωt − 0.8β ) = xˆ0.082cos(107 πt − 7.11) 1 Ex (0.8) jωt 7 H (0.8,t) = aˆ n × E(0.8,t) , H (0.8,t) = yˆ Re[ e ] = yˆ0.026cos(10 πt −1.61) ηc Eg. The magnetic field intensity of a linearly polarized uniform plane wave propagating in the +y direction in seawater εr=80, μr=1, σ=4S/m is π H = xˆ0.1sin(1010 πt − ) A/m. (a) Determine the attenuation constant, the phase 3 constant, the intrinsic impedance, the phase velocity, the wavelength, and the skin depth. (b) Find the location at which the amplitude of H is 0.01 A/m. (c) Write the expressions for E(y,t) and H(y,t) at y=0.5m as function of t. (Sol.) (a) σ/ωε=0.18<<1, ∴ Seawater is a low-loss dielectric in this case. σ µ µ σ ⇒ α ≈ = 83.96Np / m η ≈ (1+ j ) = 41.8e j0.0283π 2 ε c ε 2ωε 1 σ ω 1 β ≈ ω µε [(1+ ( ) 2 ] = 300π , v = = 3.33×107 m / s , δ = = 1.19×10−2 m , 8 ωε p β α 2π λ = = 6.67 ×10−3 m β 0.01 1 (b) e −αy = ⇒ y = ln10 = 2.74×10−2 m 0.1 α π (c) H (y,t) = xˆ0.1e −αy sin(1010 πt − βy − ) , y = 0.5, β = 300π 3 π ⇒ H (0.5,t) = xˆ5.75×10−20 sin(1010 πt − ) 3 η η π aˆ = yˆ ⇒ E(0.5,t) = −η aˆ × H (0.5,t) = zˆ2.41×10−18 sin(1010πt − + 0.0283π ) n c n 3 Eg. Given that the skin depth for graphite at 100 MHz is 0.16mm, determine (a) the conductivity of graphite, and (b) the distance that a 1GHz wave travels in graphite such that its field intensity is reduced by 30dB. 1 (Sol.) (a) δ = = 0.16×10−3 ⇒ σ = 0.99×105 S / m πfmσ (b) At f=109Hz, α = pfmσ = 1.98×104 Np / m −αz 1.5 −4 − 30(dB) = 20log10 e ⇒ z = = 1.75×10 m α log10 e Eg. Determine and compare the intrinsic impedance, attenuation constant, and 7 7 skin depth of copper σcu=5.8×10 S/m, silver σag=6.15×10 S/m, and brass 7 σbr=1.59×10 S/m at following frequencies: 60Hz and 1GHz. 1 α (Sol.) α = πfµσ , δ = , f ↑⇒ δ ↓ , η = (1+ j) α c σ −6 2 −3 Copper: 60Hz⇒ ηc = 2.02(1+ j) ×10 Ω , α = 1.17 ×10 Np / m , δ = 8.53×10 m −3 5 −6 1GHz⇒ ηc = 8.25(1+ j) ×10 Ω , α = 4.79×10 Np / m , δ = 2.09×10 m dω 1 Group velocity: v = = g dβ dβ / dω E(t, z) = E0 cos[(ω + ∆ω)t − (β + ∆β )z] + E0 cos[(ω − ∆ω)t − (β − ∆β )z] = 2E0 cos(t∆ω − z∆β )cos(ωt − βz) dz ∆ω 1 dω 1 Let t∆ω − z∆β =constant ⇒ v = = = = = g dt ∆β ∆β / ∆ω dβ dβ / dω dv dv Eg. Show that v = v + β p and v = v − λ p g p dβ g p dλ ω dω dv p (Proof) v = , ω = v β , v = = v + β p β p g dβ p dβ 2π β λ dv p ∵ β = , βλ = 2π , λdβ + βdλ = 0 ⇒ = − , v = v − λ λ dβ dλ g p dλ An example of longitudinal vp>0 but longitudinal vg=0 in barber’s pole. Eg. A 3GHz, y-polarized uniform plane wave propagates in the +x direction in a nonmagnetic medium having a dielectric constant 2.5 and a loss tangent 10-2. (a) Determine the distance over which the amplitude of the propagating wave will be cut in half. (b) Determine the intrinsic impedance, the wavelength, the phase velocity, and the group velocity of the wave in the medium. (c) Assuming π E = yˆ50sin(6π109 t + ) V/m at x=0, write the instantaneous expression for H 3 for all t and x. −2 σ −2 9 1 −9 −3 (Sol.) 10 = << 1 ⇒ σ = 10 × 2π × 3×10 × ×10 × 2.5 = 4.166×10 ωε 36π 1 σ It is a low–loss dielectric material: β = ω mε [1+ ( ) 2 ] = 99.34rad / m 8 ωε µ σ η ≈ (1+ j ) = 238∠0.29° Ω c ε 2ωε σ µ 1 (a) α ≈ =0.497, e −0.497d = ⇒ d = 1.395m 2 ε 2 ω 8 dω 1 8 (b) v = = 1.8973×10 m / s , v = = = 1.8975×10 m / s p β g dβ (dβ / dω) π π η j η η j( −0.0016π ) −0.497 x 3 1 −0.497 x 3 (c) E = yˆ50e ⋅ e ⇒ H = aˆn × E = zˆ0.21e ⋅ e ηt ⇒ H (x, t) = zˆ0.21e −0.497 x ⋅sin(6π10 9 t − 31.6πx + 0.332π ) A/m Plasma: Ionized gasses with equal electron and ion densities. Ionosphere: 50~500 Km in altitude Simple model of plasma: An electron of charge –e, mass m, position x 2 2 d x 2 e − e − eE = m = −mω x ⇒ x = E ⇒ Electric dipole p = −ex = E dt 2 mω 2 mω 2 Ne 2 ∴ Total electric dipole moment: P = Np = − E mω 2 2 2 2 Ne ω p Nε = e + = e − = e − ω = D 0 E P 0 (1 2 )E 0 (1 2 )E , where p is the plasma mω e 0 ω mε 0 angular frequency. ω 2 f 2 ε = ε (1− p ) = ε (1− p ) . 0 ω 2 0 f 2 f 2 Propagation constant: γ = jω µε ⋅ 1− ( p ) 0 f 2 η0 Intrinsic impedance of the plasma: ηc = where η 0= 120π (Ω) f 1− ( p ) 2 f Case 1 f Case 2 f>fp: γ is pure imaginary, ηc is real ⇒ EM wave can propagate through the plasma.