Magnetic Circuits

Outline

• ’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 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 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

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 on one loop of the coil.

Moving a 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 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 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 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 : flux density Domain configuration during several stages B : 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 ’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 = Ni Current i Magnetic Flux φ Resistance R Reluctance Conductivity 1/ρ Permeability μ J Magnetic Flux Density B Electric Field E Magnetic Field Intensity H

17 Toroid with Air Gap Magnetic flux

Electric current

A = cross-section Why is the flux confined mainly to the core ?

Can the reluctance ever be infinite (magnetic ) ?

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 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” 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 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

In the absence of mechanical displacement…     dλ W = P dt = ivdt = i = i (λ) dλ S elec dt

For a linear : 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 flow… dW dr S = −f dt r dt

28 Linear Machines: 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

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