Magnetic force 2018b mk

Nguyễn Công Phương Engineering Electromagnetics Magnetic Forces & Inductance Contents I II III IV V VI VII VIII IX X XI XII XIII XIV XV Introduction Vector Analysis Coulomb’s Law & Electric Field Intensity Electric Flux Density, Gauss’ Law & Divergence Energy & Potential Current & Conductors Dielectrics & Capacitance Poisson’s & Laplace’s Equations The Steady Magnetic Field Magnetic Forces & Inductance Time – Varying Fields & Maxwell’s Equations Transmission Lines The Uniform Plane Wave Plane Wave Reflection & Dispersion Guided Waves & Radiation Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential Current Elements Force & Torque on a Closed Circuit Magnetization & Permeability Magnetic Boundary Conditions The Magnetic Circuit Potential Energy of Magnetic Fields Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 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 = Qv B • This force is perpendicular to both v & B • In an electromagnetic field: F = Q(E + v B) • (Lorentz force) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Force on a Moving Charge (2) Ex The point charge Q = 18 nC has a velocity of 106 m/s in the direction av = 0.04ax – 0.05ay + 0.2az Find the magnitude of the force exerted on the charge by the following fields: a) B = –3ax + 4ay + 6az mT; b) E = –3ax + 4ay + 6az kV/m; c) B & E acting together FB = Qv × B av 0.04a x − 0.05a y + 0.2a z v=v = × 10 av 0.042 + 0.052 + 0.22 = × 106 (0.19a x − 0.24a y + 0.95a z ) m/s ax ay → FB = Qv × B = Q vx vy Bx By az ax ay az v z = 18 × 10−9 × × 103 0.19 −0.24 0.95 −3 Bz = −0.47a x − 0.36a y + 0.0036a z mN → FB = FB = 0.47 + 0.36 + 0.0036 = 0.5928 mN Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Ex Force on a Moving Charge (3) The point charge Q = 18 nC has a velocity of 106 m/s in the direction av = 0.04ax – 0.05ay + 0.2az Find the magnitude of the force exerted on the charge by the following fields: a) B = –3ax + 4ay + 6az mT; b) E = –3ax + 4ay + 6az kV/m; c) B & E acting together FE = QE = 18 × 10−9 ( −3a x + 4a y + 6a z ) ì 103 N FB = 0.5928 mN → FE = FE = 18 × 10 −6 32 + 42 + 62 = 0.1406 mN FEB = Q (E + v × B ) = FE + FB = 18.10 −6 ( −3a x + 4a y + 6a z ) + + ( −0.47a x − 0.36a y + 0.0036a z ) ×10−3 = −0.53a x − 0.29a y + 0.11a z mN → FEB = FEB = 0.532 + 0.292 + 0.112 = 0.6141 mN Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Ex 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 ( E + v × B) = → E = −v × B = −(a x + a y ) × ( a x − 2az ) = 2a x − 2a y + az V/m Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Ex y Force on a Moving Charge (5) v 10–2 Given a magnetic flux density B = ax T, find the force on an electron whose velocity is 107 m/s: a) In the x direction, y direction, & z direction b) In the xy plane at 45o to the x axis 45o x Fx = Qv × B = Q(107 a x × 10−2 a x ) = Fy = Qv × B = −1.6 × 10−19 (10 a y × 10 −2 a x ) = 1.6 × 10 −14 a z N Fz = Qv × B = −1.6 × 10−19 (107 a z × 10−2 a x ) = −1.6 × 10−14 a y N v = (cos 45o a x + sin 45o a y )107 m/s F = Qv × B = − 1.6 × 10 −19 (cos 45o a x + sin 45o a y )10 × 10 −2 a x = 1.13 × 10 −14 a z N Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential Current Elements Force & Torque on a Closed Circuit Magnetization & Permeability Magnetic Boundary Conditions The Magnetic Circuit Potential Energy of Magnetic Fields Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Force on a Differential Current Element (1) • Force on a differential current element: dF = dQv B • 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 = ρvdvv B • J = ρvv → dF = J Bdv Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 10 The Magnetic Circuit (6) Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 51 The Magnetic Circuit (5) Ex The iron core has an average length of 0.44 m & a cross-section of 0.02 0.02 m2 The air gap is mm It is wound with 400 turns Find the current producing a magnetic flux of 0.141 mWb in the air gap? Biron ℓair Φ 0.141 × 10−3 = = = 0.35T −2 −2 Siron (2 × 10 )(2 ×10 ) → H iron = 850 A/m S air = (0.02 + 0.002) = 4.84 × 10 −4 m H air Φ 0.141 × 10 −3 = = = 2.32 × 10 A/m − − àair Sair ì 10 4.84 ì 10 ( )( ) F = H ironℓ iron + H air ℓ air = 850 × 0.44 + 2.32 × 105 × × 10−3 = 838 A F 838 →I = = = 2.09 A N 400 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 52 The Magnetic Circuit (6) Ex F ℓ1 Ф1 Ф2 ℓ3 ℓ2 Ф3 F − H1ℓ1 = H 2ℓ = H 3ℓ Φ1 = Φ2 + Φ Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 53 Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential Current Elements Force & Torque on a Closed Circuit Magnetization & Permeability Magnetic Boundary Conditions The Magnetic Circuit Potential Energy of Magnetic Fields Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 54 Potential Energy of Magnetic Fields (1) WH =  B.Hdv V =  µ H dv V B =  dv V µ Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 55 Potential Energy of Magnetic Fields (2) Ex Find the magnetic energy associated with unit length of an infinitely long straight wire of radius a carrying a current I Hinside = Winside 2π a ρ aϕ =  µ H 2dv V  µ I a I2 =   µ ρ  (1 × 2πρ × d ρ ) =  16π  4π a  Houtside = Woutside I I a I 2πρ aϕ   ∞ I =  µ0 H 2dv =   µ0 2  (1 × 2πρ × d ρ ) = ∞  4π ρ  V Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 56 Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential Current Elements Force & Torque on a Closed Circuit Magnetization & Permeability Magnetic Boundary Conditions The Magnetic Circuit Potential Energy of Magnetic Fields Inductance & Mutual Inductance Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 57 Inductance & Mutual Inductance (1) Φ =  B.dS =  µr µ0 H.dS S S →Φ∼ I  H.dL = I Φ L= I Φ L= N I Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 58 Inductance & Mutual Inductance (2) Φ = N  B.d S = N  S S Φ L=N I µ r µ0 H.dS = N µr µ0 HS Φ a d N b I  H.dL = NI NI → Hd = NI → H = d I https://www.stlfinder.com/model/continuous -rectangular-coil/1975752 I → Φ = N µ r µ0 S d I N µr µ S N S d = µ µ →L= r I d Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 59 Inductance & Mutual Inductance (3) µ0 Id b Φ= ln 2π a Φ L= I d I I c a b µ0 d b →L= ln H 2π a µ0 b → per-meter inductance: L = ln H/m 2π a Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 60 Inductance & Mutual Inductance (4) Φ = N  B.dS a S = N  µr µ 0H.d S r S  H.dL =  HdL NI = ri I = H (2π r ) NI →H = 2π r I https://3dwarehouse.sketchup.com/model/ec8884f9 04c69cbb92e83e251d26ee96/Toroidal-Inductor-Coil b N Ib  NI  = µ µ ln → Φ = N  µr µ  r  (bdr ) 2π ri r π   ri Φ N 2b → L = = µr µ0 ln I 2π ri Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 61 Inductance & Mutual Inductance (5) z H = H1 + H x B1 I I = az + az 2π y 2π (d − y) Φ =  µ0H.dS l I S d  I  I =  µ0  az + a z  dS S 2π (d − x)   2π x µ Il = 2π d − r0  r0 y 1 µ Il d − r0   +  dy = 2π ln r  y d−y →L= I r0 B2 Φ µ0l d − r0 = ln I 2π r0 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 62 Inductance & Mutual Inductance (6) L= L=  I2 V A.J dv I2 →L= I Jdv ≈ Id L Stokes' theorem: 2WH  A.dL  A.dL = S → L =  B.dS I S Φ =  B.dS NΦ ↔L= I → L =  (∇ × A ).dS I S (∇ × A ).dS B = ∇× A Φ →L= I NΦ With N turns: L = I S Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 63 Inductance & Mutual Inductance (7) • Definition of mutual inductance: M12 N 2Φ12 = I1 • Φ12: flux linking I1 & I2 • N2: number of turns in circuit • Unit: H Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 64 F= Q Q1Q2 4πε R aR W = −Q  E.d L I= dQ dt H= I 2πρ R= aϕ E= Q 4πε R aR V = −  E.d L D =εE Q C= V V I B = µH Φ =  B.d S L= Φ I F = − I  B × dL Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 65 ... 1.13 × 10 −14 a z N Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential... 10−8 a x N   −8 Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 12 Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential...  × dL  12  Magnetic Forces & Inductance - sites.google.com/site/ncpdhbkhn 20 Magnetic Forces & Inductance Force on a Moving Charge Force on a Differential Current Element Force between Differential