Bài giảng Engineering electromagnetic - Chapter IX: The Steady Magnetic Field - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics The Steady Magnetic Field 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 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 2 The Steady Magnetic Field (1) 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Density 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 3 The Steady Magnetic Field (2) • The source of the steady magnetic field may be: – Permanent magnet – Electric field changing linearly with time – Direct current • Consider the field produced by a differential dc element in free space only The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 4 Biot – Savart Law (1) dL R12 Id L× a Id L× R 1 dH =R = P 4πR2 4 π R 3 aR12 I1 H: magnetic field intensity (A/m) The direction of H is determined by the right-hand rule I d L× a dH = 1 1R 12 2 π 2 4 R12 IdLa× Id La × dHH=R → = ∫ R 4πR2
4 π R 2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 5 Biot – Savart Law (2) I= Kb K b I= ∫ KdN I IdLK= dS IdLa× Ka × dS H =∫R = ∫ R
4πRR2S 4 π 2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 6 z dL 1 Biot – Savart Law (3) aR Id L× a dH = 1R 12 z’ a 2 π 2 z 4 R12 R12 = dL1 dz ' a z ρ − 2 aρ z ' a z =ρ − → = ρaρ R12 aρ z ' a z aR12 x y ρ2+ z ' 2 I Idz'a× (ρ a − z ') a ∞ Idz'a× (ρ a − z ') a → = zρ z → = zρ z dH2 H2 ∫ 4(π ρ 2+ z ') 2 3/2 −∞ 4(π ρ 2+ z ') 23/2 ×= ×= az aρ aa ϕ ; z a z 0 ∞ ρdz 'a Iρa ∞ → = I ϕ = ϕ dz ' H2 ∫ ∫ 4π −∞ (ρ 2+ z ' 2 ) 3/2 4π −∞ (ρ 2+ z ' 2 ) 3/2 z '=∞ Iρaϕ z ' I = = a π πρ ϕ 4 ρ2 ρ 2+ z ' 2 2 z '=−∞ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 7 z dL 1 Biot – Savart Law (4) aR z’ az I R z a H= a ϕ 12 z 2πρ aφ 2 x ρaρ y I z 0 y ρ aρ x I φ z =I α − α H(sin2 sin 1 ) a ϕ x α y 4πρ 2 α1 ρ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 8 Biot – Savart Law (5) I H= a ϕ 2πρ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 9 Biot – Savart Law (6) Id L× a H = ∫ R
4π R2 K× a dS H = ∫ R S 4π R2 J× a dV H = ∫ R V 4π R2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 10 z dH= d H + d H Ex. 1 Biot – Savart Law (7) 1 2 dH dH Given a circular hoop of radius a centered about the 2 1 P(0,0, z ) origin in the xy plane carries a constant current I. R2 Find MFI at P? R Id L× a 1 dH = 1R 1 0 φ a y 1 π 2 4 R1 I = ϕ x dL1 dL1 ad a ϕ = − + ϕ R1 a aρ z a z → =Iad + dH1 ( a az z a ρ ) 4π (z2+ a 23/2 ) R= z2 + a 2 1 ϕ −aaρ + z a =Iad − = z dH2 ( a az z a ρ ) aR1 4π (z2+ a 23/2 ) z2+ a 2 2Ia2 d ϕ π 2Ia2 d ϕ Ia 2 → = → = = a dH a z H∫ a z 2 23/2 z 4π (z2+ a 23/2 ) 0 4π (z2+ a 23/2 ) 2(z+ a ) The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 11 z Ex. 2 Biot – Savart Law (8) Find MFI on the z axis? a 2 I = Ia d H z, red 2(z2+ a 23/2 ) 0 2 a y = Ia H z, blue I 2[(z− d )2 + a 23/2 ] x 2   → =Ia 1 + 1 Hz   2 (za2+ 23/2 ) [( zda − ) 2 + 23/2 ]  ∂H ∂2H z = 0 z = 2 0 ∂z = ∂z z d / 2 zd=/ 2, da = The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 12 z Ex. 3 Biot – Savart Law (9) P (0, 0, z) Find MFI on the z axis? L/2 2 dz' y = Ia z' H a z 2(z2+ a 23/2 ) –L/2 z= z − z', I = Kdz ' x a 2 K → = a Kdz ' dH z 2[(z− z ')2 + a 23/2 ] L /2 a2 Kdz ' →H = z ∫ −2 + 23/2 z'=− L /2 2[(z z ') a ] − + +  =K zL/2 + zL /2  = lim Hz K 22 22  →∞ 2 (zL− /2) + a ( zL + /2) + a  L The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 13 The Steady Magnetic Field 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Density 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 14 Ampere's Circuital Law (1)
∫ HL.d= I I The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 15 Ampere's Circuital Law (2) Ex. 1
∫ HL.d= I I z ρ dL aR dL = z’ az HHϕ a ϕ R12 dL=ρtan( d ϕ ) aϕ ≈ ρϕ d a ϕ π ρa 2 x ρ y →H.d L = Hϕ ρ d ϕ
∫ ∫ 0 I 2π = Hϕ ρ d ϕ ∫0 I H= a ϕ I πρ =Hϕ 2πρ = I →Hϕ = 2 2πρ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 16 Ampere's Circuital Law (3) Ex. 2 I I Hϕ = ρ 2πρ c I a I I Hϕ = ( a <ρ < b) b 2πρ ρ 2 ρ= ρ ρ= ρ ρ < = 2 1 1 a : Iρ I ρ ϕ= − ϕ ϕ= ϕ a2 →2πρ Hϕ = I 1 1 a2 IHρ= 2πρ ϕ ρ →Hϕ = I(ρ < a ) Hϕ 2π a2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 17 Ampere's Circuital Law (4) Ex. 2 I Hϕ = ( a <ρ < b) 2πρ ρ =ρ < Hϕ I( a ) c 2πa2 I a I b ρ > = + =−= c : IIenclosed inner conductor I outer conductor II 0 →Hϕ =0 (ρ > c ) The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 18 Ampere's Circuital Law (5) Ex. 2 I Hϕ = ( a <ρ < b) 2πρ ρ =ρ < Hϕ I( a ) c 2πa2 I a I b Hϕ =0 (ρ > c ) <ρ < b c : ρ22− 2 − ρ 2 = + =−b = c IIenclosed inner conductor I partial outer coductor II I cb22− cb 22 − Ienclosed Hϕ = 2πρ I c 2− ρ 2 →=Hϕ ( b <<ρ c ) 2πρ c2− b 2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 19 Ampere's Circuital Law (6) Ex. 2 ρ I HIϕ=(ρ < aH ); ϕ = ( a <ρ < b) 2π a2 2πρ I c 2− ρ 2 Hϕ=( bcH 0 ( ρ c ) c 2πρ c2− b 2 I a I b I 2πa 4a I 4π a a 3a 0 2a 3a= b 4a= c The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 20 J Ex. 3 Ampere's Circuital Law (7) b a A very long, hollow conductor of radii a & b is located along z axis, carrying a uniformly distributed current I in z direction. Find the magnetic field intensity at any point in space? ρ ≤a:H. d L = I =→= 0 H 0
∫ ρ ≤a a≤ρ ≤ b: H. dI L =
∫ a≤ρ ≤ b 2 2 = πρ Iρ − a ∫ H.d L 2 Hϕ →H = × I a ϕ
πρ 2 2 ρ 2− 2 2 b− a = a I≤ρ ≤ I a b b2− a 2 I ρ ≥b : H = a ϕ 2πρ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 21 Ampere's Circuital Law (8) + − = 3 z HLHx1 x 2 ( L ) KL y → − = 1 3’ y Hx1 H x 2 K y x − = 1’ Hx3 H x 2 K y → = K = Kyay HHx3 x 1 2  1 2’ H= K( z > 0) L  x2 y →  z 1 H= − K( z < 0)  x2 y h → =1 × H K a N 2 K = –Kyay H=× K a (0 <<z h ) 0  N H =0 (z h ) K = Kyay The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 22 The Steady Magnetic Field 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Density 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 23 Curl (1) H = H0 = Hx0ax + Hy0ay + Hz0az HL.d= I 4 3
∫ z ∆x ∆ = ∆ (H. L ) 12−Hy ,12 − y ∂ 1∆ 2 H y 1  y H− ≈ H + ∆ x  y,1 2 y 0 ∂x 2  x y ∂H  →∆ ≈ +1 y ∆∆ (H. L ) 1− 2Hy 0 x  y 2 ∂x  ∂  ∆ ≈ −∆≈− +1 H x ∆∆ ()H. L 23−Hx ,23 − () xH x 0 yx  2 ∂y  ∂H  ∆ ≈− −1 y ∆∆ (H. L ) 3− 4Hy 0 x  y 2 ∂x  ∂  ∆ ≈ −1 Hx ∆∆ (H. L ) 4− 1Hx 0 y  x 2 ∂y  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 24 Curl (2) H = H0 = Hx0ax + Hy0ay + Hz0az HL.d= I 4 3
∫ z ∆x ∂H  ∆ ≈ +1 y ∆∆ (H. L ) 1− 2Hy 0 x  y 1∆ 2 2 ∂x  y y ∂  x ∆ ≈− +1 H x ∆∆ (H. L ) 2− 3Hx 0 y  x 2 ∂y  1 ∂H  ∂H ∂  ∆ ≈− −y ∆∆ → ≈y −H x ∆∆ (H. L ) 3− 4Hy 0 x  y
∫ H.d L   x y 2 ∂x  ∂x ∂ y  ∂  ∆ ≈ −1 H x ∆∆ (H. L ) 4− 1Hx 0 y  x 2 ∂y  ≈ ∆ + ∆
∫ HL.d ( H.L )(12− H.L ) 23 − +∆ +∆ (H. L )34− ( H. L ) 41 − The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 25 Curl (3) H = H0 = Hx0ax + Hy0ay + Hz0az ∂H ∂  ≈y −Hx ∆∆ 4 3
∫ H.d L   x y z ∆ ∂x ∂ y  x
∫ H.d L = ∆ I 1∆y 2 ∆I ≈ J ∆∆ xy z x y ∂H ∂  → ≈y −Hx ∆∆≈∆∆
∫ H.d L   xyJxyz ∂x ∂ y  H.d L ∂H ∂ H.d L ∂H ∂ →
∫ ≈y −≈Hx →
∫ =−=y H x Jz lim J z ∆∆xy ∂ x ∂ y ∆x, ∆ y → 0 ∆∆xy ∂ x ∂ y H.d L ∂ ∂H
∫ =H z −y = lim J x ∆y, ∆ z → 0 ∆∆yz ∂ y ∂ z H.d L ∂ ∂
∫ =HHx − z = lim J y ∆z, ∆ x → 0 ∆∆zx ∂ z ∂ x The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 26 Curl (4) H.d L ∂H ∂
∫ =y −H x = lim J z ∆x, ∆ y → 0 ∆∆xy ∂ x ∂ y H.d L ∂ ∂H
∫ =Hz −y = lim J x ∆y, ∆ z → 0 ∆∆yz ∂ y ∂ z H.d L ∂ ∂
∫ =H x −H z = lim J y ∆z, ∆ x → 0 ∆∆zx ∂ z ∂ x ∫ H.d L Define ()rotH = lim
N ∆S → 0 ∆ N SN - SN : planar area enclosed by the closed line integral - (rot H)N : the component (of rot H) perpendicular to SN The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 27 Curl (5) ∫ H.d L ()rotH = lim
N ∆ → ∆ SN 0 SN ∂∂H∂ ∂  ∂ H ∂ =−Hzy +−Hx H z  +−y H x rot H ax  a y  a z ∂∂yz ∂∂ zx   ∂∂ xy ax a y a z ∂ ∂ ∂ rot H = ∂x ∂ y ∂ z Hx H y H z rot HH= ∇× The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 28 Curl (6) ∂∂H∂ ∂  ∂ H ∂ =∇×=Hz −y +Hx − H z  +y − H x rot HH ax  a y  a z ∂∂yz ∂∂ zx   ∂∂ xy ∂∂∂HH  ∂  ∂∂(ρ HH )  ∇×=1Hz −ϕρ +− H z + 1 ϕρ − 1 H  aaρ  ϕ  a z ρϕ∂∂z  ∂∂ z ρ  ρρρϕ ∂ ∂  ∂(sin)Hθ ∂ ∂ ∂ () rH  ∇×=1ϕ −Hθ + 1 1 Hr − ϕ H  ar  a θ rsinθ∂∂ θ ϕ  r sin θϕ ∂∂ r  ∂ ∂ 1 (rHθ ) Hr  + −  aϕ r∂ r ∂ θ  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 29 Ex. Curl (7) Find the curl of the following vectors: =2 + 2 + a) A xy ax yz a y xy a z z sin ϕ b)B=ρ cos ϕ a + a z ρ ρ θ ϕ =2 θ ϕ + cos sin c)C r sin cos ar a θ r2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 30 Curl (8) ∂∂HH ∂∂  ∂ ∂ =∇×=HHHHz −y + xz −  +y − x Curl: rot HH ax  a y  a z ∂∂yz ∂∂ zx   ∂∂ xy ∂V ∂ V ∂ V Gradient: ∇=V a + a + a ∂xx ∂ y y ∂ z z ∂D ∂D ∂D Divergence: ∇=.D x +y + z ∂x ∂ y ∂ z The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 31 Curl (9) ∂∂HH∂ ∂  ∂ ∂ =∇×=HHz −y +HHx − z  +y − x rot HH ax  a y  a z ∂∂yz ∂∂ zx   ∂∂ xy The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 32 Curl (10) ∂∂H∂ ∂  ∂ H ∂ ∇×=Hz −y +Hx − H z  +y − H x rot H= H ax  a y  a z ∂∂yz ∂∂ zx   ∂∂ xy H.d L ∂H ∂
∫ =y −H x = lim J z ∆x, ∆ y → 0 ∆∆xy ∂ x ∂ y H.d L ∂ ∂H
∫ =Hz −y = lim J x ∆y, ∆ z → 0 ∆∆yz ∂ y ∂ z H.d L ∂ ∂
∫ =HHx − z = lim J y ∆z, ∆ x → 0 ∆∆zx ∂ z ∂ x ∇×H = J (the second of Maxwell’s four equations) The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 33 The Steady Magnetic Field 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Density 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 34 Stokes' Theorem (1) aN ≈ IN ∆S J N ∫ H.d L ∆S ∆S →J ≈
N ∆S ∆ = S ∆ IN
∫ H.d L ∆ S S = ∇× JHN( ) N ∆S ∫ H.d L ∆S →
≈∇×(H ) =∇× ( H ). a ∆S N N S → ≈∇× ∆=∇× ∆
∫ HL.d∆S ( Ha ). N S ( HS ). →HL.d = ( ∇× HS ). d
∫ ∫ S The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 35 z Ex. 1 Stokes' Theorem (2) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 ∫HL.d= ∫ ( ∇× HS ). d z
S 2 φ = 0.25 π r = 5 y dr x y rd θ = +θ + θ ϕ x dLa drr rd aθ rsin d a ϕ rsin θdφ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 36 z Ex. 1 Stokes' Theorem (3) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 = +θ + θ ϕ dLa drr rd aθ rsin d a ϕ φ = 0.25 π r = 5 → = ++θ θ ϕ HLHa.d .( drr rd aθ r sin d a ϕ )
∫
∫ y =Hdr + Hrdθ + Hrsin θ d ϕ
∫r
∫θ
∫ ϕ x H dr= H dr + H dr + H dr
∫r ∫1 r ∫ 2 r ∫ 3 r → = ∫ Hr dr 0 = → =
1,2,3:r 5 dr 1,2,3 0 θ
∫ Hθ rd →θ =
∫ Hθ rd 0 Hθ = 0 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 37 z Ex. 1 Stokes' Theorem (4) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 = +θ + θ ϕ
∫H.d L
∫ Hdrr
∫ Hrdθ
∫ Hr ϕ sin d φ = 0.25 π r = 5 =
∫ Hr dr 0 y θ =
∫ Hθ rd 0 x Hrϕsinθϕ d= Hr ϕ sin θϕ d + Hr ϕ sin θϕ d + Hr ϕ sin θϕ d
∫ ∫1 ∫ 2 ∫ 3 1,3:ϕ= const →d ϕ = 0 1,3 →Hrϕsinθϕ d = Hr ϕ sin θϕ d
∫ ∫ 2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 38 z Ex. 1 Stokes' Theorem (5) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 = +θ + θ ϕ
∫H.d L
∫ Hdrr
∫ Hrdθ
∫ Hr ϕ sin d φ = 0.25 π r = 5 =
∫ Hr dr 0 y θ =
∫ Hθ rd 0 x Hrϕsinθϕ d= Hr ϕ sin θϕ d
∫ ∫ 2 0.25 π 0.25 π →H.d L = Hrϕ sin θ d ϕ = Hϕ rsin θ d ϕ = Hϕ 5sin(0.22π ) d ϕ
∫ ∫ 2 ∫0 ∫0 2 0.25 π = 3.19 Hϕ d ϕ ∫0 2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 39 z Ex. 1 Stokes' Theorem (6) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 = +θ + θ ϕ
∫H.d L
∫ Hdrr
∫ Hrdθ
∫ Hr ϕ sin d φ = 0.25 π r = 5 0.25 π = 3.19 Hϕ d ϕ y ∫0 2 x Hϕ =18 × 5sin(0.22π )cos ϕ = 57.37cos ϕ 2 0.25 π 0.25 π →H.d L = 3.19 × 57.37cos ϕ d ϕ = 182.84cos ϕd ϕ
∫ ∫ 0 ∫0 =ϕ0.25 π = π = 182.84sin0 182.84sin(0.25 ) 129.27 A The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 40 z Ex. 1 Stokes' Theorem (7) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 =
∫ H.d L 129.27A φ = 0.25 π r = 5 (∇× H ). d S y ∫S ∂(H sinθ ) ∂  ∇×=1 ϕ − Hθ x H  a r r sin θ∂ θ ∂ ϕ  ∂∂  ∂ ∂ 1 1Hr (rH ϕ ) 1 (rH θ ) H r  + −  aθ + −  a ϕ rsin θ∂∂ ϕ rrr   ∂∂ θ  1 1 1  =()36r sinθθϕ cos cosa + 6 r cos ϕ − 36 r sin θϕ cos  a θ rsinθr r  sin θ  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 41 z Ex. 1 Stokes' Theorem (8) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 =
∫ H.d L 129.27A φ = 0.25 π r = 5 (∇× H ). d S y ∫S x     =1()θθϕ + 1 1 ϕ − θϕ ∫  36r sin cos cosar  6 rrd cos 36 sin cos  aθ  S S rsinθ r  sin θ       =()θϕ +1 ϕ − θϕ ∫  36cos cosar  6cos 36sin cos  aθ  d S S  sin θ   The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 42 z Ex. 1 Stokes' Theorem (9) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 ∫ (∇× H ). d S z S 2 φ = 0.25 π r = 5 y dr x y rd θ = 2 θ θ ϕ x dS rsin dd a r rsin θdφ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 43 z Ex. 1 Stokes' Theorem (10) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 =
∫ H.d L 129.27A φ = 0.25 π r = 5 y x     ∇× =()θϕ +1 ϕ − θϕ ∫ (H ). d S ∫  36cos cosar  6cos 36sin cos  aθ  d S S S  sin θ   = 2 θ θ ϕ dS rsin dd a r →( ∇×HS ).d = 36cosθ cos ϕ aS d = (36cosθ cos ϕ )(5)2 sin θθϕd d ∫S ∫ S r ∫S The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 44 z Ex. 1 Stokes' Theorem (11) 1 Given H = 6 rsin φar + 18 rsin θcos φaφ A/m. Verify Stokes’ theorem. 3 HL.d= ( ∇× HS ). d
∫ ∫ S 2 =
∫ H.d L 129.27A φ = 0.25 π r = 5 2 (∇×H ).d S = (36cosθ cos ϕ )(5) sin θθϕ d d y ∫S ∫ S 0.25π 0.22 π x = (36cosθ cos ϕ )(5)2 sin θθϕd d ∫0 ∫ 0 0.22 π π 0.25 π 0.25 1 2  = 900 sinθ  cos ϕd ϕ = 182.84cos ϕd ϕ ∫0 ∫0 2  0 0.25 π 0.25 π = 182.84cos ϕd ϕ =182.84sinϕ = 129.27 A ∫0 0 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 45 Ex. 2 Stokes' Theorem (12) Extract
∫ H . d L = I from ∇×H = J ∇×H = J →∇×(H.S ) d = J. d S →( ∇×H.S ) d = J.S d = I ∫S ∫ S HL.d= ( ∇× HS ). d
∫ ∫ S →
∫ HL.d = I The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 46 z Ex. 3 Stokes' Theorem (13) Given A = –yax + xay – zaz = ρaφ – zaz. Verify Stokes’ theorem for the circular bounding contour in the xy plane; check the result for: a) The flat circular surface in the xy plane b) The hemispherical surface bounded by the contour c) The cylindrical surface bounded by the contour 0 R y AL.d= ( ∇ × AS ). d x
∫ ∫ S π 2 2 2 2 dL= Rd ϕ a ϕ →A.d L = R d ϕ →A.d L = Rdϕ = 2 π R
∫ ∫ 0 ∂∂A ∂∂  ∂ A ∂ ∇×=−Azy +− AA xz  +− y A x = A ax  a y  aa zz2 ∂∂yz ∂∂ zx   ∂∂ xy (∇×=AS ).d (2 a ). dS a = 2 dS = 2 π R 2 ∫flat ∫ flatz z ∫ flat The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 47 z Ex. 3 Stokes' Theorem (14) Given A = –yax + xay – zaz = ρaφ – zaz. Verify Stokes’ theorem for the circular bounding contour in the xy plane; check the result for: a) The flat circular surface in the xy plane b) The hemispherical surface bounded by the contour c) The cylindrical surface bounded by the contour 0 R y 2π R2 =AL . d = ( ∇× AS ). d x
∫ ∫ S (∇ ×AS ).d = (2 a ).( Rdd2 sinθ θ ϕ ) a ∫hemi ∫ hemi z r = θ az .a r cos π / 2 π/ 2 2 π cos2 θ →∇×(A ).d S = R2 sin 2 θ dd θ ϕ = − 2π R2 = 2π R2 ∫hemi ∫θ=0 ∫ ϕ = 0 2 θ =0 (∇ ×A ).d S = ? ∫cy The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 48 The Steady Magnetic Field 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Density 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 49 Magnetic Flux & Magnetic Flux Density (1) • The magnetic flux density B is defined in free space: B = µ0H • Unit: Wb/m2 or T or G (1T = 10000G) π –7 • Permeability µ0 = 4
10 H/m • Definition of magnetic flux: Φ = BS.d ∫S • Electric flux: Ψ =DS.d = Q
∫S The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 50 Magnetic Flux & Magnetic Flux Density (2) • Gauss’ law for the magnetic field: B.d S = 0
∫S • The last of Maxwell’s four equations: ∇.B = 0 D.d S = Q = ρ dv • Maxwell’s four equations:
∫S ∫ V v ∇ = ρ E.d L = 0 .D v
∫ ∇× = E 0 H.Ld= I = J.S d
∫
∫ S ∇×H = J B.d S = 0 ∇.B = 0
∫S The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 51 Ex. 1 Magnetic Flux & Magnetic Flux Density (3) 2 2 Given the magnetic field B = 3 xy az Wb/m . Determine the magnetic flux crossing the portion of the xy plane lying between x = 0, x = 1, y = 0, & y = 1. Φ = B.d S ∫S 1 1 2 = 2 → Φ = ∫ ∫ 3xy dxdy = 0.5 Wb B3xy a z x=0 y = 0 = dS dxdy a z The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 52 Ex. 2 Magnetic Flux & Magnetic Flux Density (4) Find the flux between the conductors of the coaxial line. I Hϕ = ( a <ρ < b) 2πρ µ 0I d →B =µ H = a ϕ 0 2πρ c Φ = ∫ B.d S I a S I b dS= dρ dz a ϕ d b µ µ → Φ = 0I ρ 0Id b ∫ ∫ aϕ.d dz a ϕ = ln 0 a 2πρ 2π a The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 53 Ex. 3 Magnetic Flux & Magnetic Flux Density (5) Find the total flux through the rectangular circuit. I dρ a L Φ = B.d S ρ ∫S µ I B=µ H = 0 a 0 πρ ϕ 2 b dS= Ld ρ a ϕ ρ =b µ I µ IL b → Φ = 0 Ld ρ = 0 ln ∫ρ =a 2πρ 2π a The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 54 The Steady Magnetic Field 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Densit y 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 55 Magnetic Potential (1) • The scalar magnetic potential Vm is defined by: = −∇ H Vm ∇× = → ∇ × = = ∇ × −∇ • H J H J (Vm ) • Curl of gradient of a scalar field is zero, therefore: =−∇ = HVm ( J 0) ∇ = →∇ =µ ∇ = .B 0 .B0 .H 0 →µ ∇−∇ = →∇2 = = 0 .(Vm ) 0 Vm 0 (J 0) a V= − H. d L m, ab ∫b The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 56 y Magnetic Potential (2) P(ρ, π/4, 0) a <ρ <b :J = 0 I φ x = I <ρ < c H a ϕ (a b) I 1 ∂V a b 2πρ → =−∇V =− m πρm ϕ ρ∂ ϕ = −∇ = 2 HVm ( J 0) ∂V I I →m = − →V = − ϕ ∂ϕ2 π m 2π I 1  AssumeV = 0 →=V2 n −  π ( n =±± 0, 1, 2, ...) m ϕ =0 mP 2π  4  1  =I n −  ( n =±± 0, 1, 2, ...) 8  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 57 Magnetic Potential (3) • Definition of a vector magnetic potential A: BA= ∇× • Unit: Wb/m 1 1 1 • HBA= = ∇× →∇×HJA = = ∇×∇× µ µ µ 0 0 0 • A may be determined by: dL R P µ aR = 0Id L I A ∫ µ Id L
4π R dA = 0 4π R The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 58 Magnetic Potential (4) R=ρ2 + z 2 µ z = 0Id L P(ρ, φ, z) dA Id L = Idz az 4π R µ Idz → = 0 y dL= dz a dA a z z 4π ρ 2+ z 2 R=ρ 2 + z 2 x φ ρ z µ →=0Idz az == dAAAz dϕ0 d ρ 0 π ρ 2+ 2 1 4 z →dHA = ∇× d 1 µ HA= ∇× 0 µ 1 ∂dA  0 = − z a µ∂ ρ  ϕ 0   Idz ρ →dH = a ϕ 4π (ρ 2+ z 23/2 ) The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 59 Magnetic Potential (5) µ Id L dA = 0 4π R • In the case of current flow throughout a volume with a density J, then: Id L = Jdv µ Jdv →A = 0 ∫V 4π R The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 60 Ex. 1 Magnetic Potential (6) A very long, straight conductor lies along the z axis, carrying a uniform current I in the z direction. Find the magnetic potential difference between two points in space? a V= − H. d L m, ab ∫b =ρ + ρ ϕ + a I I dLa dρ d a ϕ dz a z →V = − d ϕ =(ϕ − ϕ ) m, ab ∫b 2π 2π b a I H= a ϕ 2πρ The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 61 z L Ex. 2 Magnetic Potential (7) Id L R Find the vector magnetic potential in the plane bisecting a straight piece of thin wire of finite length 2 L in free space. 0 ρ ρ I P( ,0,0) µ Id L dA = 0 −L 4π R = IdL Idz ' a z R=ρ 2 + ( z ') 2 L µ µ 2+ρ 2 + →ρ = 0Idz ' = 0I L L A( ,0,0) ∫ a z ln az 2 2 4π 2 2 z'=− L 4π ρ + (z ') L+ρ − L The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 62 The Steady Magnetic Field 1. Biot – Savart Law 2. Ampere’s Circuital Law 3. Curl 4. Stokes’ Theorem 5. Magnetic Flux & Magnetic Flux Density 6. Magnetic Potential 7. Derivation of the Steady – Magnetic – Field Law The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 63 (1) • Use formulae/definitions: × =Id L a R =µ =∇× HBHBA∫ 0
4π R2 • to show that µ Jdv A = 0 ∫V 4π R µ Jdv Id L× a AH=∫0 → = ∫ R V 4π R
4π R2 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 64 (2) µ Jdv Id L× a AH=∫0 → = ∫ R V 4π R
4π R2 µ J dv Current element at ( x , y , z ), A at (x , y , z ) →A = 0 1 1 1 1 1 2 2 2 2 ∫V π 4 R12 BH= µ  BA∇× 0  →H = = = ∇× µ µ BA 0 0 ∇ ×A ∇ µ J dv 1 J dv 1 J  →=H 2 2 =× 2 0 1 1 = ∇ × 1 1 = ∇ × 1  dv 2 µ µ∫V π π ∫V 2 π ∫V 2 1 0 04 R 12 4 R12 4 R12  ∇×()()SV =∇ S × V + S ( ∇× V ) 1 1  1  →=H  ∇  ×+J() ∇× J  dv 2π ∫V 2 1 211 4 R12  R 12  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 65 (3) µ Jdv Id L× a AH=∫0 → = ∫ R V 4π R
4π R2 Current element at ( x1, y1, z1), A at (x2, y2, z2) 1 1  1  →=H  ∇  ×+J() ∇× J  dv 2π ∫V 2 1 211 4 R12  R 12  ∇ × = 2J 1 0 1 1   →=H ∇  × J  dv 2π ∫V 2 11 4 R12   1 R a R=( xx − )(2 +− yy )( 2 +− zz ) 2 →∇ =−12 =− R 12 12 21 21 21 2 3 2 R12 R12 R 12 1 a× J 1 J× a →H = R12 1 dv = 1R 12 dv 2π ∫V 2 1 π ∫V 2 1 4 R12 4 R12 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 66 (4) µ Jdv Id L× a AH=∫0 → = ∫ R V 4π R
4π R2 Current element at ( x1, y1, z1), A at (x2, y2, z2) 1 J× a →H = 1R 12 dv 2π ∫V 2 1 4 R12 = J11dv I 1 d L 1 I d L× a →H = 1 1R 12 2
∫ π 2 4 R12 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 67 (5) ∇×H = J ∇×H = J B 1 BH= µ →∇×HA =∇× = ∇×∇× 0 µ µ 0 0 BA= ∇× ∇×∇×A = ∇( ∇ .A ) − ∇2 A ∇22 = ∇ + ∇ 2 + ∇ 2 AAAAxx a yy a zz a ∇( ∇.A ) − ∇2 A →∇×H = µ 0 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 68 (6) ∇×H = J ∇( ∇.A ) − ∇2 A →∇×H = µ 0 µ J dv A = 0 1 1 2 ∫V π 4 R12 ∇.()SA = A. () ∇ S + S ( ∇ .A ) µ 1  1  → ∇.A =0  J . ∇  +( ∇ .J )  dv 22π ∫V 12 211 4 RR12  12  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 69 (7) ∇×H = J µ 1  1  → ∇.A =0  J . ∇  +( ∇ .J )  dv 22π ∫V 12 211 4 RR12  12  1 (∇.J )dv = 0 ∫V 2 1 1 R12 ∇1 =R12 = −∇ 1 13 2 RR12R12 12 µ 1   → ∇.A =0  − J .  ∇   dv 22π ∫V 11 1 4 R12   The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 70 (8) ∇×H = J µ 1   → ∇.A =0  − J .  ∇   dv 22π ∫V 11 1 4 R12   ∇.()SA = A. () ∇ S + S ( ∇ .A ) µ 1 J  → ∇.A =0() ∇ .J − ∇ . ( 1 )  dv 22π ∫V 111 1 4 RR12 12  The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 71 (9) ∇×H = J µ 1 J  → ∇.A =0() ∇ .J − ∇ . ( 1 )  dv 22π ∫V 111 1 4 RR12 12  ∂ρ ∇.J =−v = 0 1 1 ∂t J.d S= ∇ .J dv
∫S ∫ V µ J →∇.A =−0 1 d S = 0 2 2π
∫S 1 4 1 R12 The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 72 (10) ∇×H = J ∇( ∇.A ) − ∇2 A →∇×H = µ 0 µ J dv ∇.A = 0 A = 0 x x ∫V π ∇2 = − µ 4 R Ax0 J x ρ dv  V = v →∇ 2 A =− µ J →∇2A =− µ J ∫V πε y0 y 0 4 0 R  ρ ∇2 A = − µ J ∇2V = − v z0 z ε 0 →∇×H = J The Steady Magnetic Field - sites.google.com/site/ncpdhbkhn 73