Magnetic Circuits

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

1

Electric Fields Magnetic Fields

∮� · �dA =

∫� �εoE ρdV B

S V

∮· dA = 0

S

= Qenclosed

GAUSS GAUSS

FARADAY AMPERE

∮d�E · �dl = − � � � �B dA H dl

C dt

(∫S

·) ∮

C

·

� · �=

∫d

J dA+

∫�εE · �dA

dtdλ 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:

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.

Side View

�B

�B�B

�B = 0??

I

II I

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 !

�� BB Side View

I

I I I

�BMaxwell 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� � �l

C

· COMPLETE H d =

∫AMPERE’S LAW

S

· �J dA+dt

∫S

ε �E · d �A4

Faraday’s Law and Motional emf What is the emf over the resistor ?

dΦemf = − mag

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 as the bar of length L moves to the right

(orthogonal to magnetic field H) at velocity, v.

from Chabay and Sherwood, Ch 22

dt

emf =ΔΦmag

Δt= BLv

L

vΔt

v�B

5

Faraday’s Law for a Coil

from Chabay and Sherwood, Ch 22

Will the current run CLOCKWISE or ANTICLOCKWISE ?

Rotating a bar of magnet (or the coil) produces a time-varying magnetic field

inside the coil

Moving a magnet towards a coil produces a time-varying magnetic field inside the coil

6

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.

Complex Magnetic Systems

DC Brushless Stepper Motor Reluctance Motor Induction Motor

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

∫�H

C

· �dl = Ienclosed

∫�B

S

· � �dA = 0 f = q(�v × �B

)

7

Magnetic Flux Φ [Wb] (Webers)

Magnetic Flux Density B [Wb/m2] = T (Tesla

Magnetic Field Intensity H [Amp-turn/m]

due to macroscopic & microscopic

s)

due to macroscopic currents

Faraday’s Law

�(� � �H

)�B = μo +M

)= μo

(�H + χmH = μoμrH

dΦemf = − mag

dt

emf =

∮�ENC · �dl and Φmag =

∫� · �B n̂dA

8

Example: Magnetic Write Head

Bit density is limited by how well the field can be localized in write head

Horizontal Magnetized Bits

Ring Inductive Write Head

Shield

GMR Read Head

Recording Medium

9

Review: Ferromagnetic Materials

hysteresis

0

B

B,J

H

C

r H : coercive magnetic field strength S B : remanence flux density

B : saturation flux density

H 0

Initial Magnetization

Curve

Slope = i Behavior of an initially unmagnetized

material. Domain configuration during several stages of magnetization.

�Bs�Br

− �HC�HC

− �Br

− �Bsμ

H > 0

10

Thin Film Write Head

How do we apply Ampere’s Law to this geometry (low symmetry) ?

Recording Current

Magnetic Head Coil

Magnetic Head Core

Recording Magnetic Field

∫�H

C

· �dl =

∫�J

S

· �dA

11

Electrical Circuit Analogy

Charge is conserved… Flux is ‘conserved’…

Electrical

Ф EQUIVALENT

CIRCUITS

Magnetic

i

ii

+ V

∫S

�B · d �A = 0

V

i φ

��R

12

Electrical Circuit Analogy

Electromotive force (charge push)= Magneto-motive force (flux push)=

Electrical

Ф

Magnetic

EQUIVALENT CIRCUITS

V

i

i

φ

��

∮C

�H · d�l = Ienclosedv =

∫�E · d�l

+ V

ii

R

13

Electrical Circuit Analogy

Material properties and geometry determine flow – push relationship

� �B = μOHM’s LAW � � oμJ = σE rHDC

�B

Recovering macroscopic variables:

�I =

∫V�J · �d = σ

∫� �A E · dA = σ H

lA

l ρlV = I = I = IR

σA A Ni = Φ�

V

i φ

��R

14

Reluctance of Magnetic Bar

Magnetic “OHM’s LAW”

Aφ

l Ni = Φ�

l� =μA

15

Flux Density in a Toroidal Core

N turn coil

(of an N-turn coil) i

Core centerline

H

μNiR B =2R

2πR

μNi = 2πRB = lB

lBmmf = Ni =

μ= Φ

l

μA

mmf = Φ�

16

Electrical Circuit Analogy

Electrical Magnetic

Voltage v Magnetomotive Force � = Ni

Current i Magnetic Flux φResistance R Reluctance �

1/ρConductivity Permeability μCurrent Density J Magnetic Flux Density BElectric Field E Magnetic Field Intensity H

17

Toroid with Air Gap

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 ?

A = cross-section area

Electric current

Magnetic flux

18

Fields from a Toroid

A = cross-section area

Electric current

Magnetic flux

∫�H

C

· �dl =

∫�J

S

· �dA

= Ienclosed

NiH =

�B = μo

(� �H +M

)Ni

B = μ2πR 2πR

μAΦ = BA = Ni

2πR

19

Scaling Magnetic Flux

&

Same answer as Ampere’s Law (slide 9)

A = cross-section area

Electric current

Ni = Φ� Magnetic flux

l� =2πR

=μA

�μA

μAΦ = BA = Ni

2πR

20

i N=500

2cm 8cm

2cm

0.5cm

Core Thickness = 3cm

Magnetic Circuit for ‘Write Head’

A = cross-section area

+ -

�core �gap

�

φ

l� =μA

Φ ≈21

Parallel Magnetic Circuits

A = cross-section area

10cm 10cm

0.5cm 1cm i

Gap a Gap b

22

A Magnetic Circuit with Reluctances in Series and Parallel

“Shell Type” Transformer

+ + N1 turns vv 2 N2 turns

1 - -

Depth A l2 l1 l=l1+l2

N1i1

N2i2

φa φb

φc

φc

Magnetic Circuit

�1

�3+ �2

- +

-

l� 11 =

μA�2 =

l2μA

23

Faraday Law and Magnetic Circuits

Flux linkage

Step 2: Estimate voltage v2 due to time-varying flux…

+ + sinusoidal v1 v2

- -

A = cross-section area

Step 1: Estimate voltage v1 due to time-varying flux…

Ф

N1 N2

+ -

i1 i2

Load

Primary Secondary

Laminated Iron Core

dλλ = NΦ v =

dt

v2v1

=

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…

For a linear inductor:

Stored energy…

dλWS =

∫Pelecdt =

∫ivdt =

∫i =dt

∫i (λ) dλ

λi (λ) =

Lλ

WS =

∫λ′ λ2

dλ′ =0 L 2L

26

Relating Stored Energy to Force

Lets use chain rule …

This looks familiar …

Comparing similar terms suggests … ∂Wfr = − S

∂r

WS (Φ, r) ∂WS dΦ ∂WS dr= +

dt ∂Φ dt ∂r dt

dWS dr= i

dt· v − fr

dtdi dr

= iLdt

− frdt

27

Energy Balance

l dWS

dt

heat

electrica mechanical

For magnetostatic system, dλ=0 no electrical power flow…

dWS dr=

dt−fr

dt

i · v dr−frdt

dWS (λ, r) ∂WS dλ ∂WS dr= + neglect heat

dt ∂λ dt ∂r dt

28

Linear Machines: Solenoid Actuator

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− S 1 λ2

∂r|λ WS (λ, r) =

2 Lfr =

x

l

29

KEY TAKEAWAYS COMPLETE AMPERE’S LAW

RELUCTANCE

Electrical Magnetic

Voltage Magnetomotive Force Current Magnetic Flux Resistance Reluctance Conductivity Permeability Current Density Magnetic Flux Density Electric Field Magnetic Field Intensity

R

i

1/ρ

J

E

v � = Ni

φ

�μ

B

H

∫C

�H · d�l =∫S

�J · d �A+d

dt

∫S

ε �E · d �A

� =l

μA

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.