Bài giảng Engineering electromagnetic - Chapter VI: Current & Conductors - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Current & Conductors Contents I. Introduction II. Vector Analysis III. Coulomb’s Law & Electric Field Intensity IV. Electric Flux Density, Gauss’ Law & Divergence V. Energy & Potential VI. Current & Conductors VII. Dielectrics & Capacitance VIII.Poisson’s & Laplace’s Equations IX. The Steady Magnetic Field X. Magnetic Forces & Inductance XI. Time – Varying Fields & Maxwell’s Equations XII. The Uniform Plane Wave XIII.Plane Wave Reflection & Dispersion XIV.Guided Waves & Radiation Current & Conductors - sites.google.com/site/ncpdhbkhn 2 Current & Conductors 1. Current & Current Density 2. Metallic Conductors 3. Conductor Properties & Boundary Conditions 4. The Method of Images 5. Semiconductors Current & Conductors - sites.google.com/site/ncpdhbkhn 3 Current & Current Density (1) • Current: dQ I = dt • Unit A (ampère) • Current is defined as the motion of positive charges Current & Conductors - sites.google.com/site/ncpdhbkhn 4 Current & Current Density (2) • Current: rate of movement of charge crossing a given reference plane (of one coulomb per second) • Current density: J (A/m2) • The increment of current ∆I crossing an incremental surface ∆S normal to the current density: ∆I = JN∆S • If the current density is not perpendicular to the surface: ∆I = J. ∆S • Total current: I= J. d S ∫S Current & Conductors - sites.google.com/site/ncpdhbkhn 5 Current & Current Density (3) ∆ =ρ ∆ z Qv v ∆ =ρ ∆ =ρ ∆ ∆ Qv v v SL ∆Q =ρ ∆ S ∆ x v ∆x ∆S ∆Q → ∆I =ρ ∆ S y ∆ = v ∆ I t ∆∆L ∆t x x =ρ ∆Sv v x → = ρ Jx v v x ∆ = ∆ I Jx S = ρ Jv v Current & Conductors - sites.google.com/site/ncpdhbkhn 6 Current & Current Density (4) Ex. 1 z 2 2 z = 2 Given J = 2ρzaρ + 7 zsin φaφ mA/m . Find the total current leaving the circular band. z = 1 = = ρ = 3 I∫ J. d S ∫ Jρ =3 .d S S S x y J=2 × 3z a + 7 z sin 2 ϕ a ρ =3 ρ ϕ z dρ =6za + 7 z sin 2 ϕ a ρ ϕ dz =ρ ϕ = ϕ z+dz dS d dz aρ3 d dz a ρ z 0 φ x φ+d φ ρ ρdφ Current & Conductors - sites.google.com/site/ncpdhbkhnρ+d ρ 7 Current & Current Density (5) Ex. 1 z 2 2 z = 2 Given J = 2ρzaρ + 7 zsin φaφ mA/m . Find the total current leaving the circular band. z = 1 = = ρ = 3 I∫ J. d S ∫ Jρ =3 .d S S S x y J=2 × 3z a + 7 z sin 2 ϕ a ρ =3 ρ ϕ =6za + 7 z sin 2 ϕ a ρ ϕ → = ϕ Jρ =3 .d S 18 zd dz dS=ρ d ϕ dz aρ = 3 d ϕ dz a ρ z=2ϕ = 2 π z=2 →I = 18 zdϕ dz =2π × 18 zdz = 169 mA ∫z=1 ∫ ϕ = 0 ∫z=1 Current & Conductors - sites.google.com/site/ncpdhbkhn 8 Current & Current Density (6) The current leaving a closed surface: I= J. d S
∫S The total charge in the surface: Qi The law of conservation of charge dQ →=I∫ J. d S =− i
S dt • in circuit analysis, I = dQ/dt because this is an entering current • in electromagnetism, I = – dQ/dt because this is a leaving one Current & Conductors - sites.google.com/site/ncpdhbkhn 9 Current & Current Density (7) dQ I=J. d S = − i
∫ dQi S dt →∫ ( ∇ .J ) dv =− V dt
∫J.Sd= ∫ ( ∇ . J ) dv (div. theo.) S V Q= ρ dv i∫V v d ∂ρ →( ∇ .J ) dv =− ρ dv = − v ∫ ∫ v ∫ dv Vdt V V ∂t ∂ρ ∂ρ →( ∇.J ) ∆v = −v ∆ v → ∇.J = − v ∂t ∂t Current & Conductors - sites.google.com/site/ncpdhbkhn 10 Ex. 2 Current & Current Density (9) e−t Consider the current density J = a A/m2. r r ∂ρ e−t  −v = ∇ = ∇ .J . a r  ∂t r  ∂ ∂ ∂ 12 1 1 Dϕ ∇=.D (r D ) + (sin)θ Dθ + rrr2 ∂r sinθθ ∂ r sin θϕ ∂ ∂ρ ∂ −t  − t −t − t →−=v 1 2 e = e →=−ρ e + =+ e 2r  2 v ∫ 2dtKr() 2 Kr () ∂trr ∂  r  r r r Suppose ρv → 0 as t → ∞, then K(r) = 0 e−t J e−t  e − t  →ρ = C/ m3 →==vr = r m/ s v 2 r ρ  2  r v r  r  Current & Conductors - sites.google.com/site/ncpdhbkhn 11 Current & Conductors 1. Current & Current Density 2. Metallic Conductors 3. Conductor Properties & Boundary Conditions 4. The Method of Images 5. Semiconductors Current & Conductors - sites.google.com/site/ncpdhbkhn 12 Metallic Conductors (1) • The quantum theory • Valence band, conduction band, energy gap • Metallic conductors: no energy gap between valence & conduction bands • In metallic conductors: F = – eE Current & Conductors - sites.google.com/site/ncpdhbkhn 13 Metallic Conductors (2) F = – eE • In free space, the electron will accelerate • In conductors, the electron will soon obtain a constant average velocity: vd = – µeE 2 • µe: the mobility of an electron, m /Vs, positive • Ex.: Al: 0.0012; Cu: 0.0032; Ag: 0.0056 • J = ρvv • → J = – ρe µeE Current & Conductors - sites.google.com/site/ncpdhbkhn 14 Metallic Conductors (3) J = – ρe µeE • ρe : free-electron charge density, negative • J is in the same direction as E J = σE • σ : conductivity, S/m • Ex.: Al: 3.82
10 7; Cu: 5.80
10 7; Ag: 6.17
10 7 σ = – ρe µe Current & Conductors - sites.google.com/site/ncpdhbkhn 15 Metallic Conductors (4) Uniform E E J S I σ I= J. d S = JS →J = ∫S S Uniform J a L V= − E. d L ab ∫b a V I L = − E.∫ d L →σ =→=V I b L Sσ S = −E.L = E.L →V = RI ba ab L R = (Ohm’s law) →V = EL V σ S →J = σ J= σ E L − a V ∫ E.d L R =ab = b I σE.d S ∫S Current & Conductors - sites.google.com/site/ncpdhbkhn 16 Metallic Conductors (5) dL R = i i σ iS i N N dL th →RR = = i i cell dL i ∑i ∑σ i=1 i = 1 iS i dL →R = ∫ σ S Current & Conductors - sites.google.com/site/ncpdhbkhn 17 Ex. Metallic Conductors (6) A material with conductivity σ = m/ρ + k, where m & k are constants, fills the space between two concentric, cylindrical conductors of radii a & b. L is the length of each conductor. Find the resistance of the material? dL R = i i σ iS i = ρ dLi d b dρ 1 kb+ m →R =∫ = ln m a (kρ+ m )2 π L 2 π Lk ka + m σ = + k i ρ = πρ Si 2 L Current & Conductors - sites.google.com/site/ncpdhbkhn 18 Current & Conductors 1. Current & Current Density 2. Metallic Conductors 3. Conductor Properties & Boundary Conditions 4. The Method of Images 5. Semiconductors Current & Conductors - sites.google.com/site/ncpdhbkhn 19 Conductor Properties & Boundary Conditions (1) • Given some electrons in the interior of a conductor • They will begin to accelerate away from each other, until they reach the surface of the conductor • Characteristic 1: the charge density inside a conductor is zero, the exterior surface has a surface charge density • Within a conductor: no charge → no current → no electric field intensity (Ohm) • Characteristic 2: the electric field intensity within the conductor is zero Current & Conductors - sites.google.com/site/ncpdhbkhn 20 Conductor Properties & Boundary Conditions (2) EN E D ∆S a ∆w b ∆h ∆h ∆h D c
∫ E.d L = 0 N d ∆w Et Dt Conductor b c d a → + + + = 0 ∫a ∫ b ∫ c ∫ d ∆h  ∆ h  →∆−()EwE  ++0 E  = 0 t Nb,at2  Na ,at 2  Ewithin conductor = 0 ∆h → 0 → ∆ = → = → =ε = → = = Et w 0 Et 0 Dt0 E t 0 Dt E t 0 D.d S = Q →+∫ ∫ + ∫ = Q
∫S topbottom sides → ∆= =ρ ∆ DSQN S S = ∆ = = ∫ DN S ; ∫ 0; ∫ 0 → =ρ = ε top bottom sides DN S0 E N Current & Conductors - sites.google.com/site/ncpdhbkhn 21 Conductor Properties & Boundary Conditions (3) EN E D ∆S a ∆w b = = ∆h Dt E t 0 ∆h ∆h D N d ∆w c Et =ε = ρ DN0 E N S Dt Conductor x V= −E. d L = 0 xy ∫y Characteristics of conductors in static field: 1. The static EFI inside a conductor is zero 2. The static EFI at the surface of a conductor is everywhere directed normal to that surface 3. The conductor surface is an equipotential surface Current & Conductors - sites.google.com/site/ncpdhbkhn 22 Conductor Properties & Boundary Conditions (4) Ex. Given V = x2 – 10 yz V & P(2, 1, 2) lies on a conductor – free space boundary. Find V, E, D, ρS at P, & the equation of the conductor surface. =2 − ×× =− →− =2 − VP 2 10 1 2 16 V 16x 10 yz = −∇ = −∇2 − =− + + E V (x 10 yz ) 2xax 10 z a y 10 y a z V/m → =− + + EP( 2x a x 10 z a y 10 y a z ) x=2, y = 1, z = 2 =− + + 40ax 20 a y 10 a z V/m =ε = ×−12 −+ + 2 DEPP0 8.854 10 ( 40 aaa xyz 20 10 ) nC/ m ρ = S, PD N == ×−12 2 ++= 2 2 2 DN, PD P 8.854 10 40 20 10 406 pC/m →ρ = 2 S, P 406 pC/m Current & Conductors - sites.google.com/site/ncpdhbkhn 23 Current & Conductors 1. Current & Current Density 2. Metallic Conductors 3. Conductor Properties & Boundary Conditions 4. The Method of Images 5. Semiconductors Current & Conductors - sites.google.com/site/ncpdhbkhn 24 The Method of Images (1) + Q + Q Equipotential surface, V = 0 Equipotential surface, V = 0 – Q • Dipole: the plane between the 2 charges is zero potential • That plane can be represented by a vanishingly thin conducting plane, infinite in extent • → the dipole can be substituted for a system of a charge and a conducting plane, & then the fields above the conducting plane obtain equivalence Current & Conductors - sites.google.com/site/ncpdhbkhn 25 The Method of Images (2) + Q + Q Equipotential surface, V = 0 Equipotential surface, V = 0 – Q + Q + Q Equipotential surface, V = 0 Equipotential surface, V = 0 – Q Current & Conductors - sites.google.com/site/ncpdhbkhn 26 Ex. 1 The Method of Images (3) ρL ρL + 1 + 1 – 5 – 5 Equipotential surface, V = 0 Equipotential? surface, V = 0 1. Coulomb’ s law + 5 2. Gauss’ law – 1 3. Laplace’s equation & E = –sV – ρL Current & Conductors - sites.google.com/site/ncpdhbkhn 27 P(x, y, z) Ex. 2 The Method of Images (4) Q Given Q at (0, 0, d). Find the potential & EFI at P ? d =Q = Q V+Q 4πε R πε 2+ 2 + − 2 Equipotential surface, V = 0 0 1 40 x y ( zd ) − − =Q = Q P(x, y, z) V−Q R 4πε R πε 2+ 2 + + 2 1 0 2 40 x y ( zd ) Q d R2   Q 1 1 d V = −  4πε 22 222 2 –Q 0  xy++−() zd xy +++ () zd  Q xx  yy  zdzd+ −   E=−∇=−V  −  aa + −  + −  a  πε 33 x 33  y 33  z 4 0 RR21  RR 21  RR 21   Current & Conductors - sites.google.com/site/ncpdhbkhn 28 y Ex. 3 The Method of Images (5) 5 nC 2 Find the potential at P ? 1 P =2 + 2 = R1 1 1 1.41 0 1 2 x =2 + 2 = y R2 3 1 3.16 –5 nC 5 nC 2 R1 R =2 + 2 = 1 2 P R3 3 3 4.24 0 R3 1 2 x R =12 + 3 2 = 3.16 4 R4 –5 nC − 510× 9  1 1 1 1  V = −+−= 14.03 V 5 nC P πε   4 0R 1 R 2 R 3 R 4  Current & Conductors - sites.google.com/site/ncpdhbkhn 29 z Ex. 4 The Method of Images (6) Q A point charge Q at a distance d from a center of a grounded conducting sphere of radius a. Find the image charge? y Problem: find q & b x =−θ2 + θ 2 =+− 22 θ R1 ( dR cos ) ( R sin ) RdRd 2 cos =θ −+2 θ 2 =+− 22 θ z R2 (cos R bR ) (sin) RbRb 2 cos Q R1 P Q q Q mQQ1 m  V =− =− = − R P πε πε πε πε πε   d q 2 4R1 4 R 2 4 R 1 4 R 2 4  RR 12  R c b θ  a a m= → q =− Q 0 x 1 m  d d a R= a → V = 0 → − = 0 →  P R R a2 1 2 b =  d Current & Conductors - sites.google.com/site/ncpdhbkhn 30 Current & Conductors 1. Current & Current Density 2. Metallic Conductors 3. Conductor Properties & Boundary Conditions 4. The Method of Images 5. Semiconductors Current & Conductors - sites.google.com/site/ncpdhbkhn 31 Semiconductors • Germani, silicon – – • Conductivity of conductors: J – – E – – σ = – ρe µe – – • Conductivity of semiconductors: σ = – ρe µe + ρh µh • h: hole – J – E – • At 300K: – – 2 2 – µe, Germani : 0,36 m /Vs; µh, Germani : 0,17 m /Vs 2 2 – µe, Silicon : 0,12 m /Vs; µh, Silicon : 0,025 m /Vs Current & Conductors - sites.google.com/site/ncpdhbkhn 32