L6-1MAGNETIC FORCES ON FLAT SURFACESLorentz Force Law:WI1I2sF [N/m]F [N/m] = Nqv11×μooH2=I1×μ H12v∫∫Hd•=sCAw∫J•da=I21≅ 2H2WF =μzˆo1I I2/ 2W [N/m]Currents exert no net force on themselvesTotal H method:v∫∫Hd•=s •aCAw∫Jd=Let I12 = I = I = HoWF1o [N/m] = I1×μ <H>H= zzˆˆI1oμo = μoI1I2/ 2W [N/m]21P = F/W = zˆμ H22oo [N/m ]2"Magnetic pressure"0 for C1I for C2zyxIWC1HC2s << Wconductor<H> = Ho/2magnetic pressureI1HoH0=zJyF1cfq m=×vμNoH []L6-2ROTARY WIRE MOTORSingle wire loop spinning in uniform H:Axial forces from wires at ends cancelF [Nm-1] =×I μoHT =×v∫r F ds = r2WIμoH zˆ [Nm]cT = IAμoH zˆ (A is loop area, N = 1 turn)T = NIAμoH zˆ for N-turn coil fNSWrθIT,HdsTorque = f(θ):T[Nm]With commutatorNIμoH(θ)A0π 2πθCommutators:Switch currents to maximize torqueCan have N coils and 2N brushesCθspring-loadedcarbon brusheszˆL6-3MOTOR BACK VOLTAGEForce on electron inside moving wire:Open-circuit wire:Open-circuit voltage:( )feee= −+Ev×μoH=0⇒=Eve−×μoH insideΦ=EW=eovμHW [V]Force balance−ev ×μoHH-ev−eEe+-ΦHWv,f,xˆfeMechanical power output, N turns:Pm= ωT = ωNIAμoH [W] I = (V - Φ)/RΦ = 2NvμoHW = 2NωrμoHW = NAμoHωPm= ωN(V - NAμoHω)AμoH/R = ωK1– ω2K2 RI+ +VΦ- -PpPm(ω)0ωpωωmaxmotorgeneratorωmax= V/NAμoH= 2ωpΦp= V/2L6-4RELUCTANCE MOTOR FIELDS2-Pole Reluctance Motor:()μμ≅o4∇•B0= ⇒ BμμBgap ⇒ H≅ Hgap<<Hgap [ > ~10]μμo∫∫J •=da NI =vvH•ds =∫Hgap+Hμ•ds ≅2bHAccgap (if Hμv∫ds << 2bHcgap)NITherefore H ≅gap [A /m]2bμμrotorstatorgap bθDIN turns+ V -TN turnsI+ V -μμrotorstatorDdsArea AL6-5RELUCTANCE MOTOR TORQUEμμrotorstatorgap bθDN turnsI+ V -TSet V = dΛ/dt = 0:We power coil until overlap is maximum, then coast until it is zeroΛ= N∫∫B•da = NμoHAgap gapA = N2μoIAgap/2b (Agap= RDθ)Λ⇒ = LI L = N2μoRDθ/2b112Λ2wm = LI = 22LFields:NIHgap=2bMagnetic Flux Linkage Λ:dwmΛΛ2-dL1 22bT = - = - = I222 [ ∝ ], Λμ=N Hd2θθd2N μ RDθogapRθDo1 = μoH2dVgap2bD R = Wolumegap [Nm] Torque2dθMagnetic pressure = Energy density [J/m3= N/m2]L6-6¾-POLE RELUCTANCE MOTORWinding Excitation Plan:First excite windings A and B, pulling pole 1 into pole B. Pole area A = constant, temporarily.When Δθ = π/3, excite B and C.When Δθ = 2π/3, excite C and A. Repeating this cycle results in nearly constant clockwise torque.To go counter-clockwise, excite BC, then AB, then CA.Torque:Only one pole is being pulled in here; the other excited winding has either one rotor pole fully in, or one entering and one leaving that cancel. Many pole combinations are used (more poles, more torque).AB1μ4θ2μCABCL6-7ELECTRIC AND MAGNETIC PRESSUREElectric and magnetic pressures equal the field energy densities, J/m3Both field types only pull along their length, and only push laterallyThe net pressure is the difference between two sides of any boundaryσ = ∞σ = ∞Area AArea BElectric pressure Pe= [N/m2] or [J/m2]1||2ε2oEForce fe= BPeForce fe= APeμ = ∞μ = ∞Area AArea BMagnetic pressure Pm= [N/m2] or [J/m2]1||2μ2oHForce fm= APmσ = ∞Force fm= BPmArea AForce fm= APmL6-8FORCES ON NEUTRAL MATTERKelvin polarization force density:Kelvin magnetization force density:xzy•+++++++ε > εo-------+-+-+V-fx/2fx/2dEIf ∇×E =0 = ∇•E, then:Field gradiants ⊥⇒E E is curvedCurved E pulls electric dipoles into stronger field regions for ε>εoIf H∇× =0 = ∇•B, then:Field gradiants ⊥⇒H H is curvedμ > μofx2fx1וμoμ > μoInduced current loopBB1B2ICurved H pulls current loops into stronger field regions for μ > μoB1> B2MIT OpenCourseWare http://ocw.mit.edu 6.013 Electromagnetics and Applications Spring 2009 For information about citing these materials or our Terms of Use, visit:
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