Magnetic Circuits. Outline. Ampere s Law Revisited Review of Last Time: Magnetic Materials Magnetic Circuits Examples

Transcription

1 Magnetic Circuits Outline Ampere s Law Revisited Review of Last Time: Magnetic Materials Magnetic Circuits Examples 1

2 Electric Fields Magnetic Fields S ɛ o E da = ρdv B V = Q enclosed S da =0 GAUSS GAUSS C FARADAY AMPERE ( ) d E dl = B da H dl dt S C = J da + d dt dλ S S ɛe da emf = v = dt 2

3 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 I I 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) = μ o I/(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

4 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 I I 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: C d H d l = J da + dt S S ɛ E d A COMPLETE AMPERE S LAW 4

5 Faraday s Law and Motional emf What is the emf over the resistor? dφ emf = dt mag vδt L v B In a short time Δt the bar moves a distance Δx = v*δt, and the flux increases by ΔФ mag = B (L v*δt) emf = ΔΦ mag Δt = BLv There is an increase in flux through the circuit as the bar of length L moves to the right (orthogonal to magnetic field H) at velocity, v. 5 from Chabay and Sherwood, Ch 22

6 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 inside the coil Will the current run CLOCKWISE or ANTICLOCKWISE? 6

7 Complex Magnetic Systems DC Brushless Stepper Motor Reluctance Motor Induction Motor C H dl = Ienclosed S B da =0 f = q ( v B ) 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

8 Magnetic Flux Φ [Wb] (Webers) Magnetic Flux Density B [Wb/m 2 ] = T (Teslas) Magnetic Field Intensity H [Amp-turn/m] due to macroscopic & microscopic B = μ o ( ) ( ) H + M = μ o H + χ m H due to macroscopic currents = μ o μ r H Faraday s Law emf = emf = dφ dt mag E NC dl and Φ mag = B ˆndA 8

9 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

10 B Review: Ferromagnetic Materials H C B r B s 0 HC H B,J Initial Magnetization Curve H>0 B r hysteresis B s C H r : coercive magnetic field strength B S : remanence flux density B : saturation flux density H 0 Slope = μ Behavior of an initially unmagnetized i material. Domain configuration during several stages of magnetization. 10

11 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)? C H dl = S J da 11

12 Electrical Circuit Analogy + V i Charge is conserved i Flux is conserved i S B d A =0 Electrical EQUIVALENT CIRCUITS Ф Magnetic i φ V R I R 12

13 Electrical Circuit Analogy Electromotive force (charge push)= v = E d l i i + V Electrical i EQUIVALENT CIRCUITS Magneto-motive force (flux push)= C i H d l = I enclosed Magnetic φ Ф V R I R 13

14 Electrical Circuit Analogy Material properties and geometry determine flow push relationship i φ V R I R B = μ OHM s LAW J = σe oμrh DC B Recovering macroscopic variables: I = V J d A = σ E da = σ H l A l ρl V = I = I = IR σa A Ni =ΦR 14

15 Reluctance of Magnetic Bar Magnetic OHM s LAW l Ni =ΦR A φ l R = μa 15

16 Flux Density in a Toroidal Core Core centerline H R B = 2R i N turn coil (of an N-turn coil) μni 2πR μni =2πRB = lb lb mmf = Ni = =Φ l μ μa mmf =ΦR 16

17 Electrical Circuit Analogy Electrical Voltage v Current i Resistance R Conductivity 1/ρ Current Density J Electric Field E Magnetic Magnetomotive Force I = Ni Magnetic Flux φ Reluctance R Permeability μ Magnetic Flux Density B Magnetic Field Intensity H 17

18 Toroid with Air Gap Magnetic flux Electric current Why is the flux confined mainly to the core? A = cross-section area Can the reluctance ever be infinite (magnetic insulator)? Why does the flux not leak out further in the gap? 18

19 C H dl = S J da = I enclosed Fields from a Toroid Magnetic flux Electric current A = cross-section area H = Ni 2πR ( ) B = μ o H + M Ni B = μ 2πR μa Φ=BA = Ni 2πR 19

20 Scaling Magnetic Flux Ni =ΦR Magnetic flux & R = l μa R = 2πR μa Electric current μa Φ=BA = Ni 2πR A = cross-section area Same answer as Ampere s Law (slide 9) 20

21 Magnetic Circuit for Write Head Core Thickness = 3cm 8cm 2cm 2cm l R = μa 0.5cm R core φ R gap i N=500 A = cross-section area I + - Φ 21

22 Parallel Magnetic Circuits 10cm 10cm 1cm Gap a i Gap b 0.5cm A = cross-section area 22

23 A Magnetic Circuit with Reluctances in Series and Parallel Shell Type Transformer Magnetic Circuit N 1 turns + + v v2 N 2 turns φ a R 1 φ b R 3 + R 2 N 1 i φ c Depth A l 2 l 1 l=l 1 +l 2 N 2 i 2 - φ c l1 R1 = R 2 = l 2 μa μa 23

24 Faraday Law and Magnetic Circuits Ф i 1 i sinusoidal v 1 N 1 N 2 v Primary Secondary Load Laminated Iron Core A = cross-section area Flux linkage λ = NΦ dλ v = dt Step 1: Estimate voltage v 1 due to time-varying flux Step 2: Estimate voltage v 2 due to time-varying flux v 2 v 1 = 24

25 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

26 Stored Energy in Inductors In the absence of mechanical displacement W S = P elec dt = ivdt = dλ i = dt i (λ) dλ For a linear inductor: λ i (λ) = L Stored energy W S = λ λ λ 2 dλ = 0 L 2L 26

27 Relating Stored Energy to Force Lets use chain rule W S (Φ,r) W S dφ W S dr = + dt Φ dt r dt This looks familiar dw S dr = i v f r dt dt di dr = il f r dt dt Comparing similar terms suggests W f r = r S 27

28 Energy Balance heat i v electrical dw S dt dr f r dt mechanical dw S (λ, r) W S dλ W S dr = + neglect heat dt λ dt r dt For magnetostatic system, dλ=0 no electrical power flow dw S dt = dr f r dt 28

29 x Linear Machines: Solenoid Actuator 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 f r = WS 1 λ 2 r λ W S (λ, r) = 2 L 29

30 KEY TAKEAWAYS COMPLETE AMPERE S LAW C H d l = S J d A + d dt S ɛ E d A Electrical Voltage v Current i Resistance R Conductivity 1/ρ Current Density J Electric Field E Magnetic Magnetomotive Force I = Ni Magnetic Flux φ Reluctance R Permeability μ Magnetic Flux Density B Magnetic Field Intensity H RELUCTANCE R = l μa 30

31 MIT OpenCourseWare Electromagnetic Energy: From Motors to Lasers Spring 2011 For information about citing these materials or our Terms of Use, visit: