Prof. Sang-Yong Jungcontents.kocw.net/KOCW/document/2014/sungkyunkwan/... · 2016-09-09 ·...

L o g o www.emlab.kr

Prof. Sang-Yong Jung

( 23432B 031-299-4952 www.emlab.kr [email protected] )

Engineering Electromagnetics (Week 15)

L o g o

2

8.5_ The Nature of Magnetic Materials

8.6_ Magnetization and Permeability

8.7_ Magnetic Boundary Conditions

8.8_ The Magnetic Circuit

8.9_ Potential Energy and Forces on Magnetic Materials

8.10 _ Inductance and Mutual Inductance

Ch. 8 Magnetic Forces, Materials, and Inductance

전기기기연구실 (www.emlab.kr) Electric Machines LAB (www.emlab.kr)

3

□ Materials are characterized according to their configurations of -- and interactions

between -- atomic magnetic moments. On the atomic or molecular scale, magnetic

moments are associated with electron orbital configurations, electron spin, and

(to a lesser extent) nuclear spin.

□ The diagram shows contributions to the total magnetic moment of an atom that arise

from the electron orbit (morb) and electron spin (mspin), in which the latter vector may

add to or subtract from the orbital moment. In addition, electrons occurring in pairs

will have equal and opposite spin momenta.

8.5 THE NATURE OF MAGNETIC MATERIALS


4

□ Diamagnetisim

- The magnetic flux density, B0, is externally applied.

In this case, the summation of all orbital and spin magnetic moments associated with all

electrons in a single atom is precisely zero, with the atom in its unperturbed state.

- Application of a magnetic field leads to an outward radial force Fm on the orbiting

electron shown here, resulting in a decrease in its orbital velocity in order to maintain a

balance of Coulomb and magnetic forces.



5

- The velocity reduction will result in a lessening of the orbital magnetic moment, thus

producing a very small net magnetic moment in the direction of the spin moment;

i.e., in the opposite direction to the applied field.

So a net decrease in the magnetic flux density will occur.

- Examples of diamagnetic materials : metallic bismuth, copper, gold, silicon, germanium,

graphite, sulfur.



6

□ Paramagnetism

- In this case, the summation of all orbital and spin magnetic moments associated with

individual atoms is not zero, but is very small. Because of random orientations of

adjacent atoms, however, the average magnetic moment in the material is zero.

- Application of a magnetic field on the non-zero moments introduces a torque on the

ensemble which results in the partial alignment of the moments of adjacent atoms.

This strengthens the overall magnetic flux density.



7

- However, the diamagnetic effect, in addition to the tendency toward random orientations

due to molecular activity (increasing with temperature) reduces the net increase in B.

So a net increase in the B field will be seen, but this increase is very small.

- Examples of paramagnetic materials : potassium, tungsten, rare earth elements



8

□ Ferromagnetism

In ferromagnetic materials, fairly strong atomic or molecular magnetic moments exist.

Moments of adjacent molecules interact to cause partial alignment of moments within

small regions known as domains. Magnetic moment orientations from domain to

domain are random, thus resulting in a zero overall magnetic moment for the material.



9

- Application of a steady magnetic field produces greater torque on magnetic moments that

are closer to alignment with B0. Average alignment throughout the material thus increases,

but does so through the enlargement of the nearly-aligned domains, at the expense of

those that are not aligned with B0.

A significant increase in the overall magnetic

flux density occurs.

- Ferromagnetic elements :

iron, nickel, cobalt(room temperature), gadolinium,

dysprosium (low temperature).

- Another manifestation is found in superparamagnetic

materials, consisting of ferromagnetic particles suspended

in a non-ferromagnetic matrix (magnetic recording tape is

the most common example)



10

□ Antiferromagnetism

- In certain materials, and at low temperatures, adjacent magnetic moments are oriented

in opposite directions, this being the lowest energy state for the ensemble.

The application of an external field produces no change in the net flux density .

- Example of antiferromagnetic materials : manganese oxide, nickel oxide, ferrous sulfide,

cobalt chloride



11

□ Ferrimagnetism

-In certain materials, the lowest energy configuration involves adjacent magnetic moments

that are oriented with opposing orientations, but with different magnitudes as shown. The

application of an external field produces a significant change in the net flux density, but

not as strong as in ferromagnetic materials. The most important materials in this class are

the ferrites, distinguished by having low conductivity.

- Example of ferrimagnetic materials : iron oxide magnetite, nickel-zinc ferrite, nickel

ferrite



12

□ Magnetic Material Summary



13

□ Magnetic Dipole Ensembles

- Consider a bound current, Ib, surrounding a differential area, dS.

The magnetic moment :

- if several such moments exist within a volume ∆𝑣 , then the total magnetic moment

within that volume will be the vector sum of the contributors, or:

(n is the dipole volume density – (dipoles per unit volume))

8.6 MAGNETIZATION AND PERMEABILITY

𝐦𝑡𝑜𝑡𝑎𝑙 = 𝐦𝑖

𝑛∆𝑣

𝑖=1

𝐦 = 𝐼𝑏𝑑𝐒


14

□ Magnetization

- The magnetization is the dipole moment per unit volume, in the limit as the volume

shrinks to a point:

(Units : dipole moment per unit volume, or Amps/m)

- If all dipoles are identical, each having moment m, and if they all have the same

orientation, then the magnetization simplifies to:


𝐌 = lim∆𝑣→0 1

∆𝑣 𝐦𝑖

𝑛∆𝑣

𝑖=1

𝐌 = 𝑛𝐦 = 𝑛𝐼𝑏𝑑𝐒


15

□ Bound Current Formulation

- Consider an arrangement of identical magnetic moments, oriented at angle 𝜃 to a large

closed path, along which they are arranged. The local path orientation is given by

differential length, dL, shown below for a small segment of the path. The closed path

defines a surface that intersects the charge orbits.



16

□ Bound Current Formulation

- The dipoles arranged around the loop will be contained within a tube whose volume

over a differential length will be

- The number of dipoles within that volume will therefore be

𝑁 = 𝑛𝑑𝑣 = 𝑛𝑑𝐒 ∙ 𝑑𝐋 (three are shown here)


𝑑𝑣 = 𝑑𝑆 cos 𝜃 𝑑𝐿 = 𝑑𝐒 ∙ 𝑑𝐋


17

- The differential bound current, crossing the surface along length dL will be:

𝑑𝐼𝐵 = 𝑛𝐼𝑏 𝑑𝐒 ∙ 𝑑𝐋 = 𝐌 ∙ 𝑑𝐋


Dipole current per unit volume

Differential volume, dv


18

□ Bound Current over a Closed Path

- The net bound current is found by integrating the differential current over the entire

closed path; i.e

- This equation says that if we go around a closed path and find dipole moments going our

way more often than not, there will be a corresponding current composed of,

for example, orbiting electrons crossing the interior surface.

- This equation has a decided resemblance to Ampere’s Circuital Law:

𝐼 = 𝐇 ∙ 𝑑𝐋


𝐼𝐵 = 𝐌 ∙ 𝑑𝐋


19

□ Relation Between B and H

- The total current in a general medium will consist of the sum of bound and free currents

- Defining B as the fundamental magnetic field quantity, Ampere’s Circuital Law

for the total current becomes:

where 𝐼𝑇 = 𝐼𝐵 + 𝐼


𝐁

𝜇0∙ 𝑑𝐋 = 𝐼𝑇

𝐇 ∙ 𝑑𝐋 = 𝐼 = 𝐼𝑇 − 𝐼𝐵 = (𝐁

𝜇0−𝐌) ∙ 𝑑𝐋

𝐇 =𝐁

𝜇0−𝐌

𝐁 = 𝜇0 (H + M)


20

□ Current and Related Current Densities

- IT = I + IB

1. Bound Current :

2. Conduction Current :

3. Total Current :


𝐼𝐵 = J𝐵 ∙

𝑆

𝑑𝐒

𝐼𝑇 = J𝑇 ∙

𝑆

𝑑𝐒

𝐼 = J ∙

𝑆

𝑑𝐒


21

□ Magnetic Susceptibility and Relative Permeability

- The increase in magnetization with increasing applied H is governed by:

- Values of 𝜇𝑟 are negative for diamagnetic materials: positive for all the rest.

permeability :


𝜒𝑚 : magnetic susceptibility function(frequency domain) 𝐌 = 𝜒𝑚𝐇

𝐁 = 𝜇0 𝐇 + 𝜒𝑚𝐇 = 𝜇0𝜇𝑟𝐇

𝜇 = 𝜇0𝜇𝑟

𝐁 = 𝜇 𝐇

(the relative permeability is: 𝜇𝑟 = 1 + 𝜒𝑚 )


22

□ Anisotropic Media

- In certain crystalline materials, structural constraints may result in the induced

magnetization vector oriented in a different direction than the applied H field.

- In such cases, the B field is constructed according to a tensor relation, expressed

in the following matrix form :

𝐵𝑥 = 𝜇𝑥𝑥𝐻𝑥 + 𝜇𝑥𝑦𝐻𝑦+𝜇𝑥𝑧𝐻𝑧

𝐵𝑦 = 𝜇𝑦𝑥𝐻𝑥 + 𝜇𝑦𝑦𝐻𝑦+𝜇𝑦𝑧𝐻𝑧

𝐵𝑧 = 𝜇𝑧𝑥𝐻𝑥 + 𝜇𝑧𝑦𝐻𝑦+𝜇𝑧𝑧𝐻𝑧



23

□ Magnetic Field Boundary Conditions

- Consider an interface between two media that have different permeabilities. We want to

find the relationship between magnetic fields at the boundary, on either side. This is

done by considering separately the normal and tangential components of the field.

8.7 MAGNETIC BOUNDARY CONDITIONS


24

□ Boundary Condition for the Normal Component of B

- In this case, we use Gauss’ Law for the magnetic flux density:

apply this to the right circular cylinder as shown:

𝐵𝑁1 = 𝐵𝑁2

The normal component of B is continuous across a boundary.


𝐁 ∙ 𝑑𝐒 = 0

𝑠

𝐵𝑁1∆𝑆 − 𝐵𝑁2∆𝑆 = 0


25

□ Boundary Condition for the Tangential Component of H

- In this case, we allow for the possibility of a surface current density, K, at the boundary,

Ampere’s Circuital Law states:

over the indicated path, this becomes:

if no surface current exists, then the two tangential fields are equal at the boundary!


𝐇 ∙ 𝑑𝐋 = 𝐼

𝐻𝑡1∆𝐿 − 𝐻𝑡2∆𝐿 = 𝐾∆𝐿

𝐻𝑡1 − 𝐻𝑡2 = 𝐾


26

□ Electric field & electric potential relations:

𝐄 = −𝛻𝑉

𝑉𝐴𝐵 = 𝐄 ∙ 𝑑𝐋𝐵

𝐴

□ Magnetic field & scalar magnetic potential relations:

𝐇 = −𝛻𝑉𝑚

8.8 THE MAGNETIC CIRCUIT

𝑉𝑚𝐴𝐵 = 𝐇 ∙ 𝑑𝐋𝐵

𝐴

(Vm : the magnetomotive force, or mmf)


27

□ Point form of Ohm’s Law for the electric field:

J = 𝜌𝐄

□ Current flux:

𝐼 = J ∙ 𝑑𝐒

𝑆

□ Analogy to Ohm’s Law for the magnetic field:

𝐁 = 𝜇0𝐇 (B assumes the role of current density)

□ Magnetic “current” flux:

Φ = 𝐁 ∙ 𝑑𝐒

𝑠



28

□ In large scale form, Ohm’s Law :

- A similar rule can be constructed that relates mmf to magnetic flux:

This is the analogous equation for Ohm’s Law in a magnetic circuit.

□ For a straight wire of uniform conductivity,

- a similar medium of uniform permeability,


ℜ =𝑑

𝜇𝑠

𝑉𝑚 = Φℜ ( ℜ : the reluctance of the medium, in direct analogy to resistance)

𝑉 = 𝐼𝑅

𝑅 =𝑑

𝜎𝑆


29

□ In an electric circuit, Kirchoff’s Voltage Law :

𝐄 ∙ 𝑑𝐋 = 0

□ The magnetic analogy is expressed as Ampere’s Circuital Law :

𝐇 ∙ 𝑑𝐋 = 𝐼𝑡𝑜𝑡𝑎𝑙

- Generally, the closed path integral of H may include N turns of wire :

𝐇 ∙ 𝑑𝐋 = 𝑁𝐼

This quantity is the mmf around a closed path, which we use as Vm in our magnetic

circuit equation.



30

□ Including the Nonlinear Relation between B and H

- In ferromagnetic materials, B increases with increasing H, but in a nonlinear manner as

shown in the typical curve below

- So we really don’t know what the actual permeability is until we have definite

knowledge of B or H.



31

□ Hysteresis

Domain wall shifting in ferromagnetic materials introduces semi-permanent

magnetization states that are slow to respond to changes in applied magnetic fields. The

resulting magnetization curve demonstrates the hysteresis phenomenon as shown here.



32

Coercive field in transitioning B from negative

to positive values

Increasing H to high positive values lines up

all magnetic moments, and a single domain is

left (in the extreme case).

The core is thus in saturation. Further increase

in H leads to an increase in B through the free

space permeability



33

Decreasing the applied H field to zero leaves

many dipoles still aligned, and we have the

remanent magnetic flux density, Br .

The material has become a permanent magnet.

The remanent flux density is reduced to zero by

applying an opposing magnetic field strength, -

Hc known as the coercive field (or coercive

force).



34

Remanent magnetic flux density, for increasing

H field from negative to positive values

Increasing H to high negative values again

leads to saturation



35

□ Energy and Energy Density in the Magnetic Field

- In a previous lecture it was demonstrated that the energy in the electric field within

volume v is given by:

𝑊𝐸 =1

2 𝐃 ∙ 𝐄𝑑𝑣

𝑣𝑜𝑙

𝐽

- It would seem logical that the magnetic field energy within volume v is:

- Both results are true, but are restricted to linear media.

(permittivity and permeability are constant with field strength)

8.9 POTENTIAL ENERGY AND FORCES ON MAGNETIC MATERIALS

𝑊𝐻 =1

2 𝐁 ∙ 𝐇𝑑𝑣

𝑣𝑜𝑙

𝐽

𝑊𝐻 =1

2 𝝁𝐻2𝑑𝑣

𝑣𝑜𝑙

=1

2 𝐵2

𝝁𝑑𝑣

𝑣𝑜𝑙

𝐽


36

□ Magnetic Flux and Flux Linkage

- Consider a solenoid of length d, which carries current I in each of N windings.

The turns have individual areas Si which may be different from each other.

The B field may also evaluate differently in each turn.

- The Flux Linkage : the sum of the magnetic fluxes through

all turns, where the flux evaluated at each surface is the total

flux there, generated by all turns acting together.

Units : flux linkage are Weber-turns [Wb-t]

- If all turns (and B through each) are equal, then the above simplifies to:

8.10 INDUCTANCE AND MUTUAL INDUCTANCE

𝜆 = Φ𝑖

𝑁

𝑖=1

𝜆 = 𝑁Φ = 𝑁 𝐁 ∙ 𝑑𝐒

𝑆

(Φ𝑖 = 𝐁𝑖 ∙ 𝑑𝐒𝑖)

𝑆𝑖


37

□ Inductance Definition

The flux linkage:

The inductance of the device is defined as the flux linkage

per unit current,

(if all turns are identical, the last equality applies)

Units : Weber-turns per Ampere, [ Wb-t/A ]

𝜆 = Φ𝑖

𝑁

𝑖=1

(Φ𝑖 = 𝐁𝑖 ∙ 𝑑𝐒𝑖

𝑆𝑖)


𝐿 ≡𝜆

𝐼=𝑁Φ 𝐼


38

□ Example: Inductance of a Coaxial Line

A length d of coax, as shown here. The magnetic field strength between conductors is:

The magnetic flux is now the integral of B over the

flat surface between radii a and b, and of length d along z.

As we have only one turn (N = 1), the result is also the flux linkage:

with d = 1, the inductance per unit length :

𝐁 = 𝜇0𝐇 =𝜇0𝐼

2𝜋𝜌𝑎∅

𝜆 = Φ = 𝐁 ∙ 𝑑𝐒 =

𝑆

𝜇0𝐼

2𝜋𝜌𝐚∅

𝑏

𝑎

∙𝑑

0

𝑑𝜌𝑑𝑧 𝐚∅ =𝜇0𝐼𝑑

2𝜋 ln𝑏

𝑎

𝐿 =𝜆

𝐼=𝜇02𝜋ln( 𝑏

𝑎) 𝐻/𝑚


𝐻∅ =𝐼

2𝜋𝜌 (a < ρ < 𝑏)


39

□ Example: Inductance of a toroidal

In the problem of a toroidal coil of N turns and a current I,

If the dimensions of the cross section are small compared

with the mean radius of the toroid 𝜌0, then the total flux is

Multiplying the total flux by N, and dividing by I, we have the inductance

𝐁∅ =𝜇0𝑁𝐼

2𝜋𝜌

Φ =𝜇0𝑁𝐼𝑆

2𝜋𝜌0

𝐿 =𝜇0𝑁2𝑆

2𝜋𝜌0 𝐻/𝑚


(S : the cross-sectional area)


40

□ Suppose that our toroid has an appreciable spacing between turns, a short part of which

might look like Figure.

- The flux linkages are no longer the product of the flux at the mean radius times the total

number of turns. In order to obtain the total flux linkages we must look at the coil on a

turn-by-turn basis.



41

- In reality, flux density generated by each turn may not link the entire coil. Such fringing

fields, shown here, may link only one or two turns.

( i : the flux linking the ith turn)

- Rather than doing this, we usually rely on experience and empirical quantities called

winding factors and pitch factors to adjust the basic formula to apply to the real physical

world.

- An equivalent definition for inductance may be made using an energy point of view,

(𝑁∅)𝑡𝑜𝑡𝑎𝑙= ∅1 + ∅2 + ∙∙∙ +∅𝑖 ∙∙∙ +∅𝑁 =

∅𝑖

𝑁

𝑖=1

𝐿 =2𝑊𝐻

𝐼2


(I : the total current flowing in the closed path)

(𝑊𝐻 : the energy in the magnetic field produced by the current)


42

- After using the equation to obtain several other general expressions for inductance,

- express the potential energy 𝑊𝐻 in terms of the magnetic fields,

𝐿 =2𝑊𝐻

𝐼2=𝑁Φ 𝐼

𝐿 = 𝐁 ∙ 𝐇 𝑑𝑣

𝑣𝑜𝑙

𝐼2

𝐿 = 1

𝐼2 𝐇 ∙ (𝛻 × 𝐀) 𝑑𝑣

𝑣𝑜𝑙


(replace B by ∇ ×A)

→


43

- The vector identity

may be proved by expansion in rectangular coordinates. The inductance is

- After applying the divergence theorem to the first integral and letting ∇ ×H = J in the

second integral,

∇ · (A × H) ≡ H· (∇ × A) − A· (∇ × H)

𝐿 = 1

𝐼2[ 𝛻 ∙ 𝐀 × 𝐇 𝑑𝑣 + 𝐀 ∙ 𝛻 × 𝐇 𝑑𝑣 ]

𝑣𝑜𝑙

𝑣𝑜𝑙

𝐿 = 1

𝐼2[ 𝐀 × 𝐇 ∙ 𝑑𝑠

𝑆

+ 𝐀 ∙ J𝑑𝑣 ]

𝑣𝑜𝑙



44

- The surface integral is zero, as the surface encloses the volume containing all the

magnetic energy, and this requires that A and H be zero on the bounding surface.

The inductance :

- The vector magnetic potential A due to J is given by Equation, Chapter 7,

- the inductance may therefore be expressed more basically as a rather formidable

double volume integral,

𝐿 = 1

𝐼2 𝐀 ∙ J𝑑𝑣

𝑣𝑜𝑙

𝐴 = 𝜇J

4𝜋𝑅 𝑑𝑣

𝑣𝑜𝑙

𝐿 = 1

𝐼2 (

𝜇J

4𝜋𝑅 𝑑𝑣

𝑣𝑜𝑙

) ∙ J𝑑𝑣

𝑣𝑜𝑙



45

- A slightly simpler integral expression is obtained by restricting our attention to current

filaments of small cross section for which J dν may be replaced by I dL and the volume

integral by a closed line integral along the axis of the filament,

- The inductance is a function of the distribution of the current in space or the geometry of

the conductor configuration.

𝐿 = 1

𝐼2 (

𝜇𝐼𝑑𝑳

4𝜋𝑅 ) ∙ 𝐼 𝑑𝐿

=𝜇

4𝜋 (

𝑑𝑳

𝑅 ) ∙ 𝑑𝐿



46

- To obtain our original definition of inductance , let us hypothesize a uniform current

distribution in a filamentary conductor of small cross section so that J dν becomes I dL,

- For a small cross section, dL may be taken along the center of the filament.

Apply Stokes’ theorem

- If we now let the filament make N identical turns about the total flux, an idealization

that may be closely realized in some types of inductors, the closed line integral

must consist of N laps about this common path,

𝐿 = 1

𝐼2 𝐀 ∙ J𝑑𝑣

𝑣𝑜𝑙

= 1

𝐼 𝐀 ∙ 𝑑𝐋

𝐿 = 1

𝐼 (∇ × A) ∙ 𝑑𝐒

𝑆

= 1

𝐼 B ∙ 𝑑𝐒

𝑆

= Φ 𝐼

𝐿 =𝑁Φ 𝐼



47

- The interior of any conductor contains magnetic flux, and this flux links a variable

fraction of the total current, depending on its location. These flux linkages lead to an

internal inductance, which must be combined with the external inductance to obtain the

total inductance.

- In Chapter 11, we will see that the current distribution in a conductor at high frequencies

tends to be concentrated near the surface. The internal flux is reduced, and it is usually

sufficient to consider only the external inductance. At lower frequencies, however,

internal inductance may become an appreciable part of the total inductance.



48

- The mutual inductance between circuits 1 and 2, 𝑀12 , in terms of mutual flux linkages,

(Φ 12 : the flux produced by 𝐼1 which links the path of the filamentary current 𝐼2 , and

𝑁2 is the number of turns in circuit 2)

- The mutual inductance, therefore, depends on the magnetic interaction between two

currents.

𝑀12 =𝑁2Φ 12𝐼1



49

- In terms of a mutual energy,

𝐁1 : the field resulting from 𝐼1 (with 𝐼2 = 0)

𝐇2 : the field arising from 𝐼2 (with 𝐼1 = 0)

𝑀12 =𝑁2Φ 12𝐼1=1

𝐼1𝐼2 𝐁1 ∙ 𝐇2 𝑑𝑣

𝑣𝑜𝑙

=1

𝐼1𝐼2 𝜇𝐇1 ∙ 𝐇2 𝑑𝑣

𝑣𝑜𝑙

𝑀12 = 𝑀21


∴

Date post:	08-Sep-2018
Category:	Documents
Upload:	truongtuong
View:	214 times
Download:	0 times

Prof. Sang-Yong Jungcontents.kocw.net/KOCW/document/2014/sungkyunkwan/... · 2016-09-09 ·...

Documents