Bài giảng Engineering electromagnetic - Chapter X: Magnetic Forces & Inductance - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Engineering Electromagnetics Magnetic Forces & Inductance 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 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 2 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 3 Force on a Moving Charge (1) • In an electric field: F = QE • This force is in the same direction as the EFI (positive charge) • In a magnetic field: F = QvB • This force is perpendicular to both v & B • In an electromagnetic field: F = Q(E + vB) • (Lorentz force) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 4 Ex. 1 Force on a Moving Charge (2) 6 The point charge Q = 18 nC has a velocity of 5 10 m/s in the direction av = 0.04 ax – 0.05 ay + 0.2 az. Find the magnitude of the force exerted on the charge by the following fields: a) B = –3ax + 4 ay + 6 az mT; b) E = –3ax + 4 ay + 6 az kV/m; c) B & E acting together. = × FB Q v B a 0.04a− 0.05 a + 0.2 a v =v v =5 × 10 6 x y z 2 2 2 av 0.04+ 0.05 + 0.2 =×6 − + 5 10 (0.19ax 0.24 a y 0.95 a z ) m/s aaaxyz aaa xyz →=×= =×××−9 3 − FB Q v B Qvx v y v z 18 10 5 10 0.19 0.24 0.95 −3 4 6 Bx B y B z =− − + 0.47ax 0.36 a y 0.0036 a z mN →==2 + 2 + 2 = FBF B 0.47 0.36 0.0036 0.5928 mN Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 5 Ex. 1 Force on a Moving Charge (3) 6 The point charge Q = 18 nC has a velocity of 5 10 m/s in the direction av = 0.04 ax – 0.05 ay + 0.2 az. Find the magnitude of the force exerted on the charge by the following fields: a) B = –3ax + 4 ay + 6 az mT; b) E = –3ax + 4 ay + 6 az kV/m; c) B & E acting together. = =×−9 −+ + × 3 µ F = 0.5928 mN FE Q E 1810(3ax 4 a y 6)10 a z N B →==×−6 2 ++= 2 2 FEF E 18 10 3 4 6 0.1406mN = +× = + FEBQ( EvBFF ) E B =−6 −+++ 18.10 ( 3ax 4 a y 6 a z ) +− − + × −3 ( 0.47ax 0.36 a y 0.0036 a z ) 10 =− − + 0.53ax 0.29 a y 0.11 a z mN →==2 + 2 + 2 = FEBF EB 0.53 0.29 0.11 0.6141 mN Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 6 Ex. 2 Force on a Moving Charge (4) A test charge Q C, moving with a velocity v = ax + ay m/s, experiences no force in a region of electric & magnetic fields. If the magnetic flux density B = ax – 2az T, find E. F=Q( EvB +× ) = 0 →E =− v × B =− + × − (aaxy )( a x 2) a z = − + 2ax 2 a y a z V/m Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 7 y Ex. 3 Force on a Moving Charge (5) v –2 Given a magnetic flux density B = 10 ax T, find the force on an electron whose velocity is 10 7 m/s: 45 o a) In the x direction, y direction, & z direction. b) In the xy plane at 45 o to the x axis. 0 x =×=7 ×− 2 = FvBxQQ(10 a x 10 a x ) 0 =×=−×−19 7 × − 2 =× − 14 FvByQ 1.6 10 (10 aa yxz 10 ) 1.6 10 a N =×=−×−19 7 × − 2 =−× − 14 FvBzQ 1.6 10 (10 aa zx 10 ) 1.6 10 a y N =o + o 7 v(cos45 ax sin 45 a y )10 m/s =×=−×−19 o + o 72 × − FvBQ 1.6 10 (cos45 ax sin 45 a y )10 10 a x = × −14 1.13 10az N Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 8 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 9 Force on a Differential Current Element (1) • Force on a differential current element: dF = dQvB • If charges are in motion in a conductor, the force is transferred to the conductor • Consider only force on conductors carrying currents • If dQ = ρvdv (dv is an incremental volume) → dF = ρvdv vB • J = ρvv → dF = JBdv Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 10 Force on a Differential Current Element (2) dFJB= × dv JLdv= Id →dFLB = Id × →=FJBLBBL ×dv = Id ×=− I × d ∫V ∫ ∫ For a straight conductor in a uniform magnetic field: FLB=I × F= BIL sin θ Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 11 Ex. Force on a Differential Current Element (3) Find the total force on the loop. z = I = 10 10 A H a z az A/m 2π x 2π x y −6 − 10 2.10 (1, 0, 0) B=µ H =4 π × 10 7 a = a T (1, 2, 0) 0 2π x z x z −6 = − × =−×−3 2.10 × FI∫ B d L 5 10 ∫ az d L (3, 0, 0) 5 mA x x =−−8  3az ×+ 2 a z ×+ 1 a z ×+ 0 a z ×  10  dxax dy a y dx a x dy a y  ∫x=1x ∫ y = 03 ∫ x = 3 x ∫ y = 2 1  =−−8  3 +−+1 2 10 +−  = − × −8 10ln xay y ()ln a x xy a y () a x  1.33 10ax N  13 0 32  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 12 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 13 Force between Differential Current Elements (1) I d L× a dH = 1 1R 12 2 π 2 4 R12 = × → = × dF Id L B dd(F2 ) Id 22 L d B 2 = µ dBH2 0 d 2 I I →dd()F =µ 1 2 dd LLa ×× ( ) 2 0π 2 2 1R 12 4 R12 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 14 Ex.Force 1 between Differential Current Elements (2) z Given I1dL1 = – 3ay Am; I2dL2 = – 4az Am. I2dL2 Find the differential force on dL2. I I dd()F=µ 1 2 dd LLa × ( × ) R12 2 0π 2 2 1R 12 y 4 R12 π × −7 =4 10 × × Id2L 2( Id 1 L 1 a R 12 ) I dL 4π R2 1 1 12 x =− +− +− R12 (1 5) ax (6 2) a y (4 1) a z −4a + 4 a + 3 a =− + + →=x y z =++2 2 2 4ax 4 a y 3 a z aR12;R 12 443 42+ 4 2 + 3 2 π × −7 (3)(4−a ×− aaa + 4 + 3)  → =4 10 −× y xyz  d() d F2 (4) a z 4π (42+ 4 2 + 3) 23/2 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 15 Ex.Force 1 between Differential Current Elements (3) z Given I1dL1 = – 3ay Am; I2dL2 = – 4az Am. I2dL2 Find the differential force on dL2. I I dd()F=µ 1 2 dd LLa × ( × ) R12 2 0π 2 2 1R 12 y 4 R12 −7 − ×− + +  4π × 10 (3)(4ay aaa x 4 yz 3)  =( − 4a ) × I dL 4π z 2+ 2 + 23/2 1 1 (4 4 3) x ax a y a z × = A B Ax A y A z B B B x y z ax a y a z →− ×− + + = − = − + (3)(4ay aaa xyz 4 3) 0 30 3(3ax 4 a z ) −4 4 3 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 16 Ex.Force 1 between Differential Current Elements (4) z Given I1dL1 = – 3ay Am; I2dL2 = – 4az Am. I2dL2 Find the differential force on dL2. I I dd()F=µ 1 2 dd LLa × ( × ) R12 2 0π 2 2 1R 12 y 4 R12 −7 − ×− + +  4π × 10 (3)(4ay aaa x 4 yz 3)  =( − 4a ) × I dL 4π z 2+ 2 + 23/2 1 1 (4 4 3) x ax a y a z × = − ×− + + =− + A B Ax A y A z (3)(4ay aaa xyz 4 3) 3(3 aa xz 4) Bx B y B z ax a y a z →−×−×−++  = − = (4)(3)(4az a y aaa xyz 4 3)  0 0 4 36 a y −9 0 − 12 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 17 Ex.Force 1 between Differential Current Elements (5) z d( d F2 ) Given I1dL1 = – 3ay Am; I2dL2 = – 4az Am. I2dL2 Find the differential force on dL2. I I dd()F=µ 1 2 dd LLa × ( × ) R12 2 0π 2 2 1R 12 y 4 R12 −7 − ×− + +  4π × 10 (3)(4ay aaa x 4 yz 3)  =( − 4a ) × I dL 4π z 2+ 2 + 23/2 1 1 (4 4 3) x − ×− ×−+ +  = (4)az (3)(4 a y aaa xyz 4 3)  36 a y −7 → = 10 = × −8 d() d F2 36 a y 1.37 10a y N (42+ 4 2 + 3) 23/2 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 18 Ex.Force 2 between Differential Current Elements (6) z d( d F2 ) Given I1dL1 = – 3ay Am; I2dL2 = – 4az Am. I2dL2 Find the differential force on dL1. – 8 (d(dF2) = 1.37 10 ay N was found in Ex.1) R12 I I y dd()F=µ 1 2 dd LLa × ( × ) 2 0π 2 2 1R 12 4 R12 I I I1dL1 dd()F=µ 2 1 dd LLa × ( × ) 1 0π 2 1 2R 21 x 4 R21 d( d F1 ) − 4π × 10 7 =IdL ×( Id L × a ) π 2 1 1 2 2R 21 → =− × −8 4 R21 d( d F1 ) 1.83 10 a z =− +− +− R21 (5 1) ax (2 6) a y (1 4) a z Why d(dF2) ≠ d(dF1) ? Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 19 Force between Differential Current Elements (7) I I dd()F=µ 1 2 dd LLa × ( × ) 2 0π 2 2 112R 4 R12 I I d L× a  →FL =µ 12d × 112R  2 0π ∫ 2 ∫ 2 4 R12  I I d L× a  =µ 1 2 1R 12  × dL 0π ∫ ∫ 2 2 4 R12  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 20 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 21 Force & Torque on a Closed Circuit (1) • Force on a filamentary closed circuit: FBL= −I∫ × d • If B = const → FBL= −I × ∫ d • In an electrostatic field: ∫ dL = 0 • → the force on a closed filamentary circuit in a uniform magnetic field is zero • General : any real closed circuit carrying direct currents experiences a total vector force of zero in a uniform magnetic field Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 22 Force & Torque on a Closed Circuit (2) z z T T y y 0 0 R1 F1 F R2 R P P P 2 R 1 x x 12 F2 = – F1 = ×+ × TRF= × TRFRF1 1 2 2 = − × (RRF1 2 ) 1 = × RF21 1 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 23 Force & Torque on a Closed Circuit (3) = × y dT1 R 1 d F 1 I B = × = − dF1 Idx ax B 0 Idx( B0yza B 0 zy a ) 3 = − 1 dy x R1 dy a y 2 2 4 1 →=−dT dy a × Idx( B aa − B ) R 1 12 y 0 yz 0 zy = − 1 dxdyIB 0ya x dx 2 → + =− dT1 d T 3 dxdyIB 0 y a x = − 1 Similarly: dT3 dxdyIB 0 y a x 2 + = Similarly: dT2 d T 4 dxdyIB 0 x a y → = − = × dT dxdyI( B0xy a B 0 yx a ) dxdyI az B 0 =Id S × B Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 24 Force & Torque on a Closed Circuit (4) • The differential magnetic dipole moment: dm = Id S • Unit: Am2 • → dT = dmB • Holds for differential loops of any shape • In a uniform magnetic field: T = ISB = mB Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 25 Force & Torque on a Closed Circuit (5) Ex. 1 z Given B0 = –0.6 ay + 0.8 az T. Find the torque. y 0 T=I S × B 4 mA (1, 2, 0) →=×−3 × ×− + x T4 10 (1 2 az ) (0.6 aa y 0.8 z ) ax a y a z ax a y a z × = →× ×− + = = 1.2 a A B Ax A y A z 12az (0.6 a y 0.8) a z 0 0 2 x 0− 0.6 0.8 Bx B y B z → = × −3 T4.8 10 a x Nm Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 26 Ex. 2 Force & Torque on a Closed Circuit (6) z Find the force per meter between two infinite & parallel I I filamentary current carrying conductors that are separated d 1 2 & carry a current I in opposite direction. d y I 0 H= a ϕ 2πρ x I I I= I = I BHa==µ µ1 =− µ 1 a T 1 2 2020πρϕ 0 π x 2ρ=d , ϕ = π /2 2 d I I I1  = µ 1 2 dF= I d L × B =I( − dz a ) ×−µ a  0dz 2 a y 2 2 2 2 2z 0 2π d x  2πd 2 → = 1 µ I1 I 2 = µ I F2∫ = 0dz 2 a y 0 a y N/m z2 0 2π d 2π d Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 27 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 28 Magnetization & Permeability (1) • The magnetization is defined basing on the magnetic dipole moment m 2 • m = IbdS (unit: Am ) • Ib: the bound current circulates about a path enclosing dS n∆ v = • For ∆v, the total magnetic dipole moment: mtotal ∑ mi i=1 • n: number of magnetic dipole in a unit volume n∆ v = 1 • Definition of the magnetization: Mlim ∑ mi ∆v →0 ∆ v i=1 • M: the (total) magnetic dipole moment per unit volume Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 29 Magnetization & Permeability (2) n∆ v = 1 Mlim∑ mi : the (total) magnetic dipole moment per u nit volume ∆v →0 ∆ v i=1 dI= I ndSLSL. d = nI d . d b b b →dI = nm. d L = b mIb d S → = → = dIb ML. d Ib ∫ ML. d Surface defined by closed path m= Id S θ dL dS Ib Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 30 Magnetization & Permeability (3) = ∫ H.d L I T = B H   µ →=−=B − 0 IIIT b ∫ M  . d L = + µ  IT I b I 0 B =µ + = Redefine: HM= − BHM0 ( ) Ib ∫ M. d L µ 0 (if M = 0 then B = µ0H) M Definition of the magnetic susceptibility: χ = m H µ= + χ Defintion of the relative permeability: R1 m µ= µ µ Definition of the permeability: 0 R BH=µ Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 31 Magnetization & Permeability (4) I= J. d S b∫S b I= J. d S T∫S T I= J. d S ∫S = ∇× = Ib ∫ ML. d M J b   B = → ∇× = J Ib∫ J b . d S µ T S  0  ML.d= ( ∇ × MS ). d ∇×H = J ∫ ∫ S Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 32 I Ex. Magnetization & Permeability (5) a A line current I of infinite extent is within a cylinder of radius a that has permeability µ, the cylinder is surrounded by free space. ρ Find B, H, & M everywhere, & the current density? µ µ I 0 I= HL. d = Hϕ 2πρ →Hϕ = ∫ 2πρ  µI µHϕ =, 0 < ρ < a  2πρ →B =  ϕ µ µ=0I ρ >  0Hϕ , a  2πρ Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 33 I Ex. Magnetization & Permeability (6) a A line current I of infinite extent is within a cylinder of radius a that has permeability µ, the cylinder is surrounded by free space. ρ Find B, H, & M everywhere, & the current density? B µ µ =µ + →MH = − 0 BHM0( ) µ 0 µ  ( µ− µ ) I −=1H0 , 0 <<ρ a → = µ  ϕ πρµ Mϕ 0  2 0 0, ρ > a ∂Mϕ 1 ∂ JM=∇×=− aρ +(ρM ϕ ) a = 0, 0 << ρ a b ∂z ρ ∂ ρ z Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 34 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 35 Magnetic Boundary Conditions (1) ∆L Ht1 BN1 ∆S ∫ B.d S = 0 S µ 1 Ht2 → ∆− ∆= BN1 SB N 2 S 0 BN2 aN12 µ2 → = BN2 B N 1 µ µ χ µ →HH = 1 →=MHHMχ = χ 1 = m2 1 N2µ N 1 NmNm2222µ N 1 χ µ N 1 2 2m 1 2 = → ∆− ∆=∆ ∫ H.d L I Ht1 LH t 2 L KL B B χ → − = →t1 − t 2 = K →M =m2 M − χ K Ht1 H t 2 K µ µ t2χ t 1 m 2 1 2 m1 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 36 Magnetic Boundary Conditions (2) ∆L Ht1 BN1 ∆S (H− H ) × a = K 1 2N 12 µ 1 Ht2 (H− H ) = a × K BN2 aN12 t1 t 2 N 12 µ2 Normal Tangential µ H= 1 H − = N2µ N 1 Ht1 H t 2 K 2 B B t1− t 2 = K B= B µ µ N2 N 1 1 2 χ µ χ MM= m2 1 M=m2 M − χ K N2χ µ N 1 t2χ t 1 m 2 m1 2 m1 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 37 Ex. 1 Magnetic Boundary Conditions (3) Where z > 0 (region 1), µ = µ1 = 4 µH/m; where z < 0 (region 2), µ2 = 7 µH/m; at z = 0, given a surface current K = 80 ax A/m. In region 1 there is a magnetic field B1 = 2 ax – 3ay + az mT. Find B2. = =−+−−= BN1( B.aa 1 NN 12 ) 12 [(2 aaa xyzzzz 3 ).( a )]( aa ) mT → = = BN2 B N 1 a z mT = + → = − BBB1N 1 t 1 BBBt1 1 N 1 →= −+− = − Bt1 (2 aaa xyz 3 )()2 a z aa xy 3 mT −3 B (2a− 3 a )10 →==Ht1 x y =−500 a 750 a A/m t1 µ −6 x y 1 4.10 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 38 Ex. 1 Magnetic Boundary Conditions (4) Where z > 0 (region 1), µ = µ1 = 4 µH/m; where z < 0 (region 2), µ2 = 7 µH/m; at z = 0, given a surface current K = 80 ax A/m. In region 1 there is a magnetic field B1 = 2 ax – 3ay + az mT. Find B2. = − Ht1 500 a x 750 a y A/m − = × (Ht1 H t 2 ) a N 12 K →=− ×= − −−× HHaKttN2 1 12 500 a x 750 a yzz ( a ) 80 a ax a y a z × = A B Ax A y A z Bx B y B z →= − += − Ht2 500 a x 750 aa yy 80 500 a x 670 a y A/m Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 39 Ex. 1 Magnetic Boundary Conditions (5) Where z > 0 (region 1), µ = µ1 = 4 µH/m; where z < 0 (region 2), µ2 = 7 µH/m; at z = 0, given a surface current K = 80 ax A/m. In region 1 there is a magnetic field B1 = 2 ax – 3ay + az mT. Find B2. = − Ht2 500 a x 670 a y A/m →=µ =×−6 − =− BHtt2 2 2 7 10 (500 aaaa xyxy 670 ) 3.5 4.69 mT = + BBB2N 2 t 2 = BN2 a z mT →=+= − + BBB2N 2 tt 2 3.5 a x 4.69 aa y z mT Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 40 Ex. 2 Magnetic Boundary Conditions (6) µ= 1000 µ µ= 1000 µ A uniform magnetic field of strength B = 1.2 T 0 Air 0 exists within an iron core. If an air gap is cut 30 o with the orientation shown, find the magnitude B and direction of B in the gap. = =o =× = BN2 B N 1 Bcos30 1.2 0.866 1.0 T 1 2 B B HH= →t1 = t 2 t1 t 2 µ µ 0 µ 1 →=B0 B =××1.2 sin 30o = 0.06 mT t2µ t 1 10000 1000 →=22 += 2 + ×− 32 ≈ B2 BN 2 B t 2 1.0 (0.06 10 ) 1 T Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 41 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 42 The Magnetic Circuit (1) = −∇ E = −∇V H Vm B B V= E. d L V= H. d L AB ∫A mAB ∫A JE= σ BH= µ I= J. d S Φ = B.d S ∫S ∫S = Φ V= IR Vm R m d = d R = Rm σ S µS = = ∫ E.d L 0 ∫ H.d L I total Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 43 The Magnetic Circuit (2) 1 FNI= = ∫ H. d L 2 =H.d L + H. d L + H. d L F = NI ∫1 ∫ 2 ∫ 3 3 =ℓ + ℓ + ℓ HHH11 22 33 Rm1 R1 E F Ф Rm2 I R2 Rm3 R3 Φ+ Φ+ Φ= + + = Rm1 R m 2 R m 3 F RI1 RI 2 RI 3 E Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 44 The Magnetic Circuit (3) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 45 The Magnetic Circuit (4) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 46 Ex. 1 The Magnetic Circuit (5) The iron core has an average length of 0.44 m & a cross-section of 0.02 0.02 m2. The air gap is 2 mm. It is wound with 400 turns. Find the current producing ℓair a magnetic flux of 0.141 mWb in the air gap? − Φ0.141 × 10 3 B = = = 0.35T iron −2 − 2 Siron (2× 10 )(2 × 10 ) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 47 The Magnetic Circuit (6) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 48 Ex. 1 The Magnetic Circuit (5) The iron core has an average length of 0.44 m & a cross-section of 0.02 0.02 m2. The air gap is 2 mm. It is wound with 400 turns. Find the current producing ℓair a magnetic flux of 0.141 mWb in the air gap? − Φ0.141 × 10 3 B = = = 0.35T iron −2 − 2 Siron (2× 10 )(2 × 10 ) → = Hiron 850 A/m = +2 =× − 4 2 Sair (0.02 0.002) 4.84 10 m − Φ0.141 × 10 3 H == =×2.32 105 A/m air µ −7 − 4 airS air ()()4π × 10 4.84 × 10 =ℓ + ℓ =×+×××=5− 3 F Hiron iron H air air 850 0.44 2.32 10 2 10 838 A F 838 →=I = = 2.09 A N 400 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 49 Ex. 2 The Magnetic Circuit (6) ℓ1 ℓ3 F Ф 1 ℓ Ф Ф2 2 3 −ℓ = ℓ = ℓ FHHH11 22 33 Φ = Φ + Φ 1 2 3 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 50 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 51 Potential Energy of Magnetic Fields (1) 1 W= B.H dv H 2 ∫V 1 = µH2 dv 2 ∫V 1 B2 = dv 2 ∫V µ Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 52 Ex. Potential Energy of Magnetic Fields (2) I Find the magnetic energy associated with unit length of an a infinitely long straight wire of radius a carrying a current I. = I ρ Hinside a ϕ 2π a2 1 W= µ Hdv2 inside 2 ∫V 2 1 a I 2  µI =µ ρ2  ()1 × 2 πρρ × d = ∫ 2 4  π 2 0 4π a  16 I H= a ϕ outside 2πρ 1 1 ∞ I 2  = µ 2 =µ()1 × 2 πρ × d ρ = ∞ Woutside ∫ 0 Hdv ∫ 0 2 2  2 V 2 0 4π ρ  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 53 Magnetic Forces & Inductance 1. Force on a Moving Charge 2. Force on a Differential Current Element 3. Force between Differential Current Elements 4. Force & Torque on a Closed Circuit 5. Magnetization & Permeability 6. Magnetic Boundary Conditions 7. The Magnetic Circuit 8. Potential Energy of Magnetic Fields 9. Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 54 Inductance & Mutual Inductance (1) NΦ • Define inductance of a coil as: L = • A.k.a self-inductance I • Φ: magnetic flux • N: number of turns • I: current flowing in the coil • H (henry) ↔ weber-turn/A • Stands for linear inductor Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 55 Inductance & Mutual Inductance (2) µ Id b Φ = 0 ln 2π a d Φ c L = I a I I b µ d b →L = 0 ln H 2π a µ b →per-meter inductance: L = 0 ln H/m 2π a Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 56 Inductance & Mutual Inductance (3) µ NI µ NIS B = 0 → Φ = 0 φ πρ πρ ρ 2 2 0 0 NΦ L = I µ N2 S →L = 0 πρ 2 0 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 57 Inductance & Mutual Inductance (4) 2W L = H I 2 1 W= B.H dv H 2 ∫V ∫ B.Hdv →L = V I 2 = ∇× N BA =Φ+Φ+ +Φ+ +Φ = Φ Total 1 2 ...i ... N∑ i i 1 →=L∫ H.( ∇× A ) dv I 2 V Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 58 Inductance & Mutual Inductance (5) 1 L=∫ H.( ∇× A ) dv I 2 V ∇.( A × H )( = H. ∇× A )( − A. ∇× H ) 1 →=L∫ ∇×.() A H dv + ∫ A. () ∇× H dv  I 2 V V  DSD.d= ∇ . dV ∫S ∫ V ∇×H = J 1 →=L∫(A × H ). d S + ∫ A.J dv  I 2 S V  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 59 Inductance & Mutual Inductance (6) 1 L=(A × H ). d S + A.J dv  2 ∫ ∫  1 I S V →L = A.J dv 2 ∫V (A× H ).d S = 0 I ∫S µJ A = dv ∫V 4π R 1 µJ  →L = ∫ ∫ dvdv  .J 1 µId L  I 2 V V 4π R  →L = ∫ ∫  .L Id I 2 4π R  JLdv≈ Id µ dL  =   .Ld 4π ∫ ∫ R  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 60 Inductance & Mutual Inductance (7) 1 µIdLL  µ  d L=∫∫ .L.L Id = ∫∫  d I 2 4πRR  4 π  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 61 Inductance & Mutual Inductance (8) 2W NΦ L=H ↔ L = I 2 I 1 L= ∫ A.J dv 1 I 2 V →L = A. d L I ∫ 1 JLdv≈ Id →L =( ∇× A ). d S I ∫S Stokes' theorem: AL.d= ( ∇ × AS ). d BA= ∇× ∫ ∫ S → = 1 L∫ B. d S Φ NΦ I S →L = With N turns: L = I I Φ = B.d S ∫S Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 62 Inductance & Mutual Inductance (9) Φ = N2 12 • Definition of mutual inductance : M12 I1 • Φ12 : flux linking I1 & I2 • N2: number of turns in circuit 2 • Unit: H Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 63