Magnetic Circuits Outline • Ampere’s Law Revisited • Review of Last Time: Magnetic Materials • Magnetic Circuits • Examples 1 Electric Fields Magnetic Fields oE · dA = ρdV B · dA =0 S V S = Qenclosed GAUSS GAUSS FARADAY AMPERE d E · dl = − B· dA H · dl dt C S C d = J · dA + E · dA dt dλ S S emf = v = dt 2 Ampere’s Law Revisited In the case of the magnetic field we can see that ‘our old’ Ampere’s law can not be the whole story. Here is an example in which current does not gives rise to the magnetic field: B B =0?? Side View II I I B B Consider the case of charging up a capacitor C which is connected to very long wires. The charging current is I. From the symmetry it is easy to see that an application of Ampere’s law will produce B fields which go in circles around the wire and whose magnitude is B(r) = μoI/(2πr). But there is no charge flow in the gap across the capacitor plates and according to Ampere’s law the B field in the plane parallel to the capacitor plates and going through the capacitor gap should be zero! This seems unphysical. 3 Ampere’s Law Revisited (cont.) If instead we drew the Amperian surface as sketched below, we would have concluded that B in non-zero ! B B Side View I I II B Maxwell resolved this problem by adding a term to the Ampere’s Law. In equivalence to Faraday’s Law, the changing electric field can generate the magnetic field: d H · dl = J · dA + E · dA COMPLETE dt AMPERE’S LAW C S S 4 Faraday’s Law and Motional emf What is the emf over the resistor ? dΦ vΔt emf = − mag dt L v B In a short time Δt the bar moves a distance Δx = v*Δt, and the flux increases by ΔФmag = B (L v*Δt) There is an increase in flux through the circuit ΔΦmag emf = = BLv as the bar of length L moves to the right Δt (orthogonal to magnetic field H) at velocity, v. from Chabay and Sherwood, Ch 22 5 from Chabay and Sherwood, Ch 22 Faraday’s Law for a Coil The induced emf in a coil of N turns is equal to N times the rate of change of the magnetic flux on one loop of the coil. Moving a magnet towards a coil produces a time-varying magnetic field inside the coil Rotating a bar of magnet (or the coil) produces a time-varying magnetic field Will the current run inside the coil CLOCKWISE or ANTICLOCKWISE ? 6 Complex Magnetic Systems DC Brushless Stepper Motor Reluctance Motor Induction Motor f = q v × B H · dl = Ienclosed B · dA =0 C S We need better (more powerful) tools… Magnetic Circuits: Reduce Maxwell to (scalar) circuit problem Energy Method: Look at change in stored energy to calculate force 7 Magnetic Flux Φ [Wb] (Webers) due to macroscopic & microscopic Magnetic Flux Density B [Wb/m2] = T (Teslas) Magnetic Field Intensity H [Amp-turn/m] due to macroscopic currents B = μo H + M = μo H + χmH = μoμrH Faraday’s Law dΦ emf = − mag dt emf = ENC · dl and Φmag = B · ndˆ A 8 Example: Magnetic Write Head Ring Inductive Write Head Shield GMR Read Head Recording Medium Horizontal Magnetized Bits Bit density is limited by how well the field can be localized in write head 9 Review: Ferromagnetic Materials B Bs B B,J r Initial Magnetization Curve H −H 0 C HC H>0 −B rhysteresis −Bs H 0 Slope = μ i C Behavior of an initially unmagnetized H r : coercive magnetic field strength material. B S : remanence flux density Domain configuration during several stages B : saturation flux density of magnetization. 10 Thin Film Write Head Recording Current Magnetic Head Coil Magnetic Head Core Recording Magnetic Field How do we apply Ampere’s Law to this geometry (low symmetry) ? H · dl = J · dA C S 11 Electrical Circuit Analogy Charge is conserved… Flux is ‘conserved’… B · dA =0 i i S + V i Ф EQUIVALENT CIRCUITS Electrical Magnetic i φ V R 12 Electrical Circuit Analogy Electromotive force (charge push)= Magneto-motive force (flux push)= v = E · dl H · dl = Ienclosed C i i i + V Ф EQUIVALENT CIRCUITS Electrical Magnetic i φ V R 13 Electrical Circuit Analogy Material properties and geometry determine flow – push relationship i φ V R B = μoμrH OHM’s LAW J = σEDC B Recovering macroscopic variables: V I = J · dAE = σ · dA = σ A H l l ρl V = I = I = IR σA A Ni =Φ 14 Reluctance of Magnetic Bar Magnetic “OHM’s LAW” l Ni =Φ φ l A = μA 15 Flux Density in a Toroidal Core Core centerline H μNi R B = 2πR 2R μNi =2πRB = lB N turn lB l coil mmf = Ni = =Φ μ μA (of an N-turn coil) i mmf =Φ 16 Electrical Circuit Analogy Electrical Magnetic Voltage v Magnetomotive Force = Ni Current i Magnetic Flux φ Resistance R Reluctance Conductivity 1/ρ Permeability μ Current Density J Magnetic Flux Density B Electric Field E Magnetic Field Intensity H 17 Toroid with Air Gap Magnetic flux Electric current A = cross-section area Why is the flux confined mainly to the core ? Can the reluctance ever be infinite (magnetic insulator) ? Why does the flux not leak out further in the gap ? 18 Fields from a Toroid Magnetic flux H · dl = J · dA C S Electric = Ienclosed current A = cross-section area B = μ H+ M Ni o Ni H = B = μ 2πR 2πR μA Φ=BA = Ni 2πR 19 Scaling Magnetic Flux Ni =Φ Magnetic flux & l 2πR = = Electric μA μA current A = cross-section area μA Φ=BA = Ni 2πR Same answer as Ampere’s Law (slide 9) 20 Magnetic Circuit for ‘Write Head’ Core Thickness = 3cm 2cm l = μA 8cm 2cm 0.5cm core φ gap N=500 i + - A = cross-section area Φ ≈ 21 Parallel Magnetic Circuits 10cm 10cm i 1cm Gap a Gap b 0.5cm A = cross-section area 22 A Magnetic Circuit with Reluctances in Series and Parallel “Shell Type” Transformer Magnetic Circuit 1 + + φa φb N1 turns v v2 N2 turns 1 3 + 2 - - N1i1 - + φc N i Depth A 2 2 - φ l2 l1 l=l1+l2 c l l = 1 = 2 1 μA 2 μA 23 Faraday Law and Magnetic Circuits Ф i1 i2 + + + Load sinusoidal v N N v 1 1 2 2 - - - Primary Secondary Laminated Iron Core A = cross-section area dλ λ = NΦ v = Flux linkage dt Step 1: Estimate voltage v1 due to time-varying flux… v2 Step 2: Estimate voltage v2 due to time-varying flux… = v1 24 Complex Magnetic Systems DC Brushless Stepper Motor Reluctance Motor Induction Motor Powerful tools… Magnetic Circuits: Reduce Maxwell to (scalar) circuit problem Makes it easy to calculate B, H, λ Energy Method: Look at change in stored energy to calculate force 25 Stored Energy in Inductors In the absence of mechanical displacement… dλ W = P dt = ivdt = i = i (λ) dλ S elec dt For a linear inductor: λ i (λ)= L Stored energy… λ 2 λ λ WS = dλ = 0 L 2L 26 Relating Stored Energy to Force Lets use chain rule … W (Φ,r) ∂W dΦ ∂W dr S =S + S dt ∂Φ dt ∂r dt This looks familiar … dW dr S = i · v − f dt r dt di dr = iL − f dt r dt Comparing similar terms suggests … ∂W f = − S r ∂r 27 Energy Balance heat i · v dr −fr dWS dt electrical dt mechanical dW (λ, r) ∂W dλ ∂W dr S =S + S dt ∂λ dt ∂r dt neglect heat For magnetostatic system, dλ=0 no electrical power flow… dW dr S = −f dt r dt 28 Linear Machines: Solenoid Actuator x l Coil attached to cone If we can find the stored energy, we can immediately compute the force… …lets take all the things we know to put this together… ∂W 1 λ2 f = − S | W (λ, r)= r ∂r λ S 2 L 29 KEY TAKEAWAYS d H · dl = J · dA + E · dA COMPLETE AMPERE’S LAW dt C S S Electrical Magnetic Voltage v Magnetomotive Force = Ni Current i Magnetic Flux φ RELUCTANCE Resistance R Reluctance l 1/ρ μ = Conductivity Permeability μA Current Density J Magnetic Flux Density B Electric Field E Magnetic Field Intensity H 30 MIT OpenCourseWare http://ocw.mit.edu 6.007 Electromagnetic Energy: From Motors to Lasers Spring 2011 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. .