field theory MAGNETOSTATICS

Dr. Thomas AfulloUkzn, Durban

Magnetostatic FieldsENEL2FT FIELD THEORY

1

REFERENCES 1. M.N. Sadiku: Elements of

Electromagnetics, Oxford University Press, 1995, ISBN 0-19-510368-8.

2. P. Lorrain, D. Corson: Electromagnetic Fields and Waves, W.H. Freeman & Co, 1970, ISBN: 0-7167-0330-0.

3. David T. Thomas: Engineering Electromagnetics, Pergamon Press, ISBN: 08-016778-0.


2

INTRODUCTION As we have noticed, an electrostatic field is produced

by static or stationary charges. If the charges are moving with a constant velocity, a

static magnetic field (or magnetostatic field) is produced.

There are two major laws governing magnetostatic fields:- The Biot-Savart Law- Ampere’s Circuital law.

Like Coulomb’s law, the Biot-Savart law is the general law of magnetostatics.

Just as in Gauss’s law, Ampere’s circuital law is a special case of Biot-Savart law and is easily applied in problems involving symmetrical current distribution.


3

MAGNETIC FORCES It is common experience that circuits carrying electric currents

exert forces on each other. For example, the force between two straight parallel wires carrying currents Ia and Ib is proportional to IaIb/, where is the distance between the wires.


4

Ib

Ia

dlb

dla

r

MAGNETIC FORCES The force is attractive if the currents flow in the same

direction, and it is repulsive if they flow in opposite directions.

For the more general case shown in the above figure, the force between the current-carrying conductors is given by:

This is the force exerted by current Ia on current Ib, and the line integrals are evaluated over the two circuits.

This is the magnetic force law. The vectors dla and dlb point in the direction of current flow, r is the distance between the two elements dla and dlb, and r1 is the unit vector pointing from dla to dlb.

The constant, o=4x10-7 H/m, is the permeability of free space.

a b

abbao

ab r

rxdlxdlIIF

2

1̂

4

Magnetostatic Fields ENEL2FT FIELD THEORY 5

MAGNETIC FORCES The force Fab can be expressed in symmetrical form by

expanding the triple vector product under the integral sign:

To show that the double integral of the first term on the right is zero, we note that:

This is the integral of dr/r2 around a closed curve, circuit b; which implies that the upper and lower limits of integration are identical. It is therefore zero.

2

1

2

1

2

1.ˆˆ.ˆ

r

dldlr

r

rdldl

r

rxdlxdlbabaab

a b

b

aa b

ba

r

rdldl

r

rdldl2

1

2

1ˆ.ˆ.


MAGNETIC FORCES We are thus left with the double integral of only the second

term for the triple vector product. Thus:

Despite the fact that the above integral for Fab is simpler and more symmetrical than that involving the triple vector product, it is not as useful.

This is because in the above integral, the force cannot be expressed as the interaction of the current b with the field in a.

Thus we use the earlier relationship to obtain:

a b

babao

ab r

dldlrIIF

2

1.ˆ

4

a

aao

a

babb

a b

abbao

ab

r

rxdlIB

BxdlIr

rxdlxdlIIF

2

1

2

1

ˆ

4

ˆ

4


MAGNETIC FORCES The vector Ba is called the magnetic induction due to

the circuit a at the position of the element dlb of circuit b.

Therefore the element of force, dF, on an element of wire of length dl carrying a current I in a region where the magnetic induction is B is given by:

If the current I is distributed in space with a (free) current density Jf, then, with dv’ the elemental volume, we have:

BxlIdFd

'

ˆ

4 '2

1dv

r

rxJB

v

fo


MAGNETIC FLUX As in electrostatics, where we use lines of force to describe

an electric field, we can describe a magnetic field by drawing lines of B that are everywhere tangent to the direction of B.

It is convenient to use the concept of flux, the flux of the magnetic induction B through a surface S being defined as the normal component of B integrated over S:

The flux is expressed in webers.

S

SdB

.


BIOT-SAVART’S LAW The Biot-Savart’s law states that the magnetic field

intensity, dH, produced at a point P shown in the figure below by the differential current element Idl is proportional to the product Idl and the sine of the angle between the element and the line joining P to the element, and is inversely proportional to the square of the distance R between P and the element.


10

R

ld

Current I

P

BIOT-SAVART’S LAW That is:

Here, k is the constant of proportionality. In SI units, the above equation becomes:

From the definition of cross product, it is seen that, in vector form,

Here, aR is the unit vector in the direction of vector R.


11

22

sinsin

R

kIdldH

R

IdldH

24

sin

R

IdldH

32 44

ˆ

R

RxlId

R

axlIdHd R

BIOT-SAVART’S LAW Just as we have different charge configurations, we can

have different current distributions: the line current, the surface current, and the volume current, as shown below.

If we define K as the surface current density (A/m2), and J as the volume current density (A/m3), then we have:


12

J dvJ

IdlI

K

dSK

Line current density

Surface current density

Volume current density

dvJdSKdlI

BIOT-SAVART’S LAW Thus in terms of the distributed source currents , the

Biot-Savart law becomes:


13

)(

4

ˆ

)(4

ˆ

)(4

ˆ

2

2

2

currentvolumeR

axdvJH

currentsurfaceR

axSKdH

currentlineR

axlIdH

v

R

S

R

L

R

FIELD DUE TO A STRAIGHT CURRENT-CARRYING CONDUCTOR

Consider the field due to a straight current-carrying filamentary conductor of finite length AB, as shown above.


14

A

B

z

dl

RI

P

1

2

z

O


Let us assume the conductor is along the z-axis, with its upper and lower ends respectively subtending angles 1 and 2 at P, the point at which the magnetic field strength, H, is to be determined.

If we consider the contribution dH at P due to an element dl at (0,0,z), we have, from Biot-Savart’s law:


15

ˆ4

ˆ

ˆˆ;ˆ

4

2/322

3

z

dzIH

dzRxld

zzRzdzld

R

RxlIdHd


Let us make the following substitutions:

Note that H is always along the unit vector , irrespective of the length of the wire or the point of interest P.


16

ˆcoscos4

ˆsin4

1ˆ

cos

cos

4

coscot

12

33

22

2

2

1

2

1

IH

dec

decIH

decdzzLet

FIELD DUE TO A STRAIGHT CURRENT-CARRYING CONDUCTOR – EXAMPLE

The conducting triangular loop below carries a current of 10 A. Find the magnetic filed intensity H at (0,0,5) due to side 1 of the loop.


17

y

x

1

1 2

10A

23

1


Solution: The problem can be solved using the following figure:

R

lId


x

y

1

P(0,0,5)

z


From the Biot-Savart law, we obtain:

dzdx

zz

d

dz

d

dx

zxz

x

xz

yIzdx

xz

ydxI

xz

zxxxIdxHd

xzx

z

xz

x

xz

z

zxadxxld

R

axlIdHd

R

R

2

22

2/3222222

2/3222/322

2

sec

seccos

tan

tantan

4

ˆ

4

ˆsin

4

sinˆcosˆˆ

tantan

cos;sin

sinˆcosˆˆ;ˆ

4

ˆ



Therefore we obtain:

mAy

z

IyH

zz

IyH

z

Iyd

z

IyH

dz

Iy

z

dzIy

xz

yIzdxHd

zzzzxz

zz

/ˆ0591.05/2tansin4

ˆ

/2tansin4

ˆ

sin4

ˆcos4

ˆ

cos4

ˆsec

sec

4ˆ

4

ˆ

sectan1tan

1)5,0,0(

1

/2tan

0

/2tan

0

33

22

2/322

332/3232/32222/322

11



1. Find H due to side 3 of the rectangular loop

mAyxH

Ans

/ˆ03063.0ˆ03063.0

:



As a special case, when the conductor is semi-infinite (with respect to P), point A is now at O(0,0,0), while point B is at (0,0,); then 1=90o, 2=0o. Then we have:

Another special case is when the conductor is infinite in length.

For this case, point A is at (0,0,-),while B is at (0,0,); then 1=180o, 2=0o. Then we have:


22

ˆ

4;ˆ

4

IHB

IH

ˆ

2;ˆ

2

IB

IH

FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORS

Consider two long parallel conductors, separated by distance , carrying current in the same direction as shown in the figure below.


23

Ia Ib

Ba

dF

dlb

FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORS

The current Ia produces a magnetic induction Ba, as shown, at the position of current Ib. The force acting on an element Idlb of this current is:

The last expression is the force per unit length of the wire.

22

2ˆ

ˆ2

ˆˆ2

ˆ2

babba

bba

abb

abb

aa

abb

II

dl

dFdlIIdF

dlIIFd

IxdlzI

IxldIFd

IB

BxldIFd


FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORS-EXAMPLE

Consider a current-carrying conductor of finite length L placed a distance d from another current-carrying conductor of infinite length.

Determine the magnetic force per unit length acting on the finite conductor.


25

d

II

Ba

dF

dlbInfinitely long wire

Finite wire of length L

z

FORCE BETWEEN TWO LONG PARALLEL CURRENT-CARRYING CONDUCTORS-EXAMPLE

THE Solution is as follows:

mNd

I

L

F

Ld

Idz

d

IF

d

I

dz

dFdz

d

IFd

d

IdzxzIFd

d

IB

L

/2

ˆ

2ˆ

2ˆ

22ˆ

ˆ2

ˆ

ˆ2

2

2

0

2

22


FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP Consider the circular loop shown below, with the loop

having radius .

The magnetic field intensity dH at point P(0,0,h) contributed by current element Idl is given by Biot-Savart’s law:


27

R

x

y

z

h

ld

P

1ld 1R

FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP

Further, from the diagram, we note that:

Therefore we determine the elemental magnetic field intensity due to current element Idl to be:


28

34 R

RxlIdHd

zdhdzhxdRxld

zhR

dld

ˆˆˆˆˆ

ˆˆ

;ˆ

2

zdhdh

I

R

RxlIdHd ˆˆ

442

2/3223


Similarly we determine the elemental magnetic field dH1 intensity due to current element Idl1 to be:

Therefore, by symmetry, the H contributions along add up to zero because the radial components produced by pairs of current elements 180o apart cancel each other. Thus:


29

zdHdHzdhdh

I

R

RxlIdHd

zdhdzhxdRxld

dldzhR

z ˆˆˆˆ44

ˆˆˆˆˆ

ˆ;ˆˆ

22/3223

1

111

211

1

2/322

2

2/322

22

0

22/322 2

ˆ4

2ˆˆ

4ˆ

h

Iz

h

Izzd

h

IdHzH z


The magnetic induction is thus given by:

Thus the magnetic induction is maximum in the plane of the current-carrying loop (h=0), and it drops off as h∞ or h».

2/322

2

2ˆ

h

IzHB


FIELD DUE TO A CIRCULAR CURRENT-CARRYING LOOP – EXAMPLE

A circular loop located on x2+y2=9, z=0, carries a direct current of 10 A. Determine the magnetic field intensity at:

A) (0,0,4) B) (0,0,-4)


31


SOLUTION: For the circular loop, the radius =3. therefore, at a height h

above the x-y plane, the field strength is given by:

mAzzH

mAzzH

h

IzH

/ˆ36.0432

3)10(ˆ

/ˆ36.0432

3)10(ˆ

2ˆ

2/322

2

)4,0,0(

2/322

2

)4,0,0(

2/322

2



A thin loop of radius 5 cm is placed on the plane z=1 cm so that its centre is at (0,0,1 cm). The loop carries a 0.05A current.

Determine H at: A) (0,0, -1 cm) B) (0,0, 10 cm)

Answer:

mAzHB

mAzHA

/ˆ0573.0)

/ˆ4.0)


FIELD DUE TO A SOLENOID Consider a solenoid of length L, consisting of N turns of wire

carrying a current I, whose cross-section is shown in the figure below.

The number of turns per unit length, n, is given by n=N/L


L

dz

1

2

z

P

Cross-section of a solenoid

z

FIELD DUE TO A SOLENOID For a single loop centered about the z-axis, the value of H a

distance z is:

For an arbitrary point z along the length of the solenoid, the incremental electric field intensity for small length dz is:

2/322

2

2ˆ

z

IzH

2/322

2

2ˆ

z

ndzIzHd


FIELD DUE TO A SOLENOID From the sketch, we obtain:

In the middle of a very long solenoid, 1=0o, 2=180o; therefore:

12

12

33

22

2/322

2

222222

22

coscos2

ˆ

coscos2

ˆˆsin2

ˆsin2

cos

cos

2ˆ

2ˆ

coscot1

coscos

cottan/

2

1

L

NIzH

nIzzd

nIH

zdnI

Hd

ec

decnIz

z

ndzIzHd

ecz

decdzecd

dz

z


nIzL

NIzH ˆˆ

FIELD DUE TO A TORROID For a toroidal coil of radius R, with N coils of wire

carrying a current I, the total values of B and H are obtained from the figure below.


37

FIELD DUE TO A TORROID Note that is the magnetic permeability of the material

around which the current-carrying coil is wound. It is assumed that the coil lies on the x-y plane. Note that

with the length, L, of the torroid, the B and H fields are given by:

It is assumed that the coil lies on the x-y plane. If the coil has a circular cross-section with radius ,

and if <<R, then the total flux inside the toroid is:

ˆ2

ˆ2

ˆ

R

NIB

R

NI

L

NIH

R

NI

R

NIBA

2)(

2

22


FORCE ON A POINT CHARGE MOVING IN A MAGNETIC FIELD – THE LORENTZ FORCE

Let us determine the force F on a single charge Q moving at a velocity v in a magnetic field B.

The force on a current element Idl is:

If the cross-sectional area of the wire is da, the current I is given by:

Here n is the number of carriers per unit volume, v is their average drift velocity, and Q is the charge per carrier.

BxlIdFd

QdavnI


FORCE ON A POINT CHARGE MOVING IN A MAGNETIC FIELD – THE LORENTZ FORCE

Therefore the total charge flowing per second is the charge on the carriers that are contained in a length v of the wire.

Then the force on the element dl becomes:

The force on a single charge Q moving at a velocity v in a field B is:

More generally, if there is also an electric field E, the force is:

This is the Lorentz force.

BxvndadlQBxlIdFd

BxvQF


BxvEQF

AMPERE’S CIRCUITAL LAW – MAXWELL’S EQUATION

Ampere’s circuital law states that the line integral of the tangential component of H around a closed path is the same as the net current enclosed by the path.

In other words:

Ampere’s circuital law is similar to Gauss’s law, and is easily applied to determine H when the current distribution is symmetrical.

Ampere’s circuital law is a special case of the Biot-Savart law.


41

enclosedIldH

.

AMPERE’S CIRCUITAL LAW – MAXWELL’S EQUATION

By applying Stoke’s theorem, we obtain:

We can further simplify this, with J the current density, to obtain:

This is the third Maxwell’s equation, which is Ampere’s law in differential form.

SdHxldHI L Sencl

..

Sencl JHxSdJI

.


APPLICATION OF AMPERE’S CIRCUITAL LAW – INFINITE LINE CURRENT

Consider an infinitely long filamentary current I along the z-axis, as shown below.

This path, on which ampere’s law is to be applied, is known as the Amperian path (analogous to the Gaussian surface).


43

z

y

x

ld

Amperian path

APPLICATION OF AMPERE’S CIRCUITAL LAW – INFINITE LINE CURRENT

We choose a concentric circle as the Amperian path in view of the previous considerations.

Since this path encloses the entire current I, according to Ampere’s law,

This is the result expected, from the application of the Biot-Savart law.


44

aI

H

HdHadaHI

ˆ2

2.ˆ.ˆ

AMPERE’S CIRCUITAL LAW – INFINITELY LONG COAXIAL TRANSMISSION LINE

Consider an infinitely long transmission line consisting of two concentric cylinders having their axes along the z-axis.

The cross-section of the line is shown below.


45

t

b

aAmperian path (one of 4)


The inner conductor has radius a and carries a current I, while the outer conductor has inner radius b and thickness t, and carries return current –I.

We want to determine H everywhere, assuming the current is uniformly distributed in both conductors.

Since the current distribution is symmetrical, we apply Ampere’s law along the Amperian path for each of the four possible regions:

Region 1: 0<<a (This is the only one shown in the figure) Region 2: a<<b Region 3: b<<b+t Region 4: >b+t


46


For Region 1, we apply Ampere’s law, giving:

Since the current is uniformly distributed over the cross-section, we have:

1Re

..g

encSdJIldH

2

2

2

1Re

2

2

2

2

2

2.

.

ˆ;ˆ

a

IH

a

IHldH

a

Idd

a

ISdJI

zddSdza

IJ

g

enc



For Region 2: a<<b, using the Amperian path (not shown in the figure) gives:

For Region 3, b<<b+t, we have:

2

2

.2Re

IH

IH

IIldHg

enc

zttb

IJ

SdJIIldHg g

enc

ˆ

..

22

3Re 3Re



Thus we obtain:

For Region 4, >b+t, we obtain:

btt

bIH

btt

bII

dttb

IISdJII

enc

bgenc

21

2

21

.

2

22

2

22

2

022

3Re

02

12 2

22

bbtt

bIH



Therefore from the equations we have, for a coaxial cable:

Thus Ampere’s law can only be used to find H due to symmetric current distributions for which it is possible to find a closed path over which H is constant in magnitude.

tb

tbbbtt

bI

baI

aa

I

H

0

ˆ2

12

ˆ2

0ˆ2

2

22

2


MAGNETIC VECTOR POTENTIAL The magnetic flux through a surface S is given by:

However, unlike electric flux lines, magnetic flux lines always close upon themselves. This is due to the fact that it is not possible to have isolated magnetic poles or magnetic charges.

Thus the total flux through a closed surface in a magnetic field must be zero; that is:

s

SdB

.

0. SdB


MAGNETIC VECTOR POTENTIAL This is Gauss’s law for magnetostatic fields. By applying the divergence theorem, we obtain:

This is the fourth Maxwell’s equation. In order to satisfy the above equation, we define the vector

magnetic potential, A, such that:

0.

0..

B

dvBSdBV

AxBB

0.Magnetostatic Fields ENEL2FT FIELD THEORY 52

MAGNETIC VECTOR POTENTIAL Just as we defined the electric scalar potential, V as:

We can define:

r

dQV

o4

currentvolumer

dvJA

currentsurfacer

SKdA

currentliner

lIdA

o

o

o

o

o

o

4

4

4


MAGNETIC VECTOR POTENTIAL Let us apply Stoke’s theorem to the flux equation:

L

Ls

s

s

ldA

ldASdAx

SdAxAxB

SdB

.

..

.

.


MAGNETIC VECTOR POTENTIAL - EXAMPLE:

Given the magnetic vector potential,

Calculate: The total magnetic flux density The total magnetic flux crossing the surface: /2,

1<<2 m, 0<z<5 m.

zA ˆ4/2


MAGNETIC VECTOR POTENTIAL - EXAMPLE: SOLUTION

Wb

dzdSdB

dzdSd

SdB

AAxB

z

z

75.3

4

5

2

1.

ˆ

.

ˆ2

ˆ

5

0

2

1

2

1

2


MAGNETIC VECTOR POTENTIAL EXAMPLE: A current distribution gives rise to a vector magnetic

potential:

Calculate: A) B at (-1, 2, 5) B) The flux through the surface defined by z=1, 0<x<1, -

1<y<4.

Ans:

zxyzyxyxyxA ˆ4ˆˆ 22

Wb

zyxB

20

ˆ3ˆ40ˆ20


MAGNETIC VECTOR POTENTIAL Let us consider the curl of the magnetic flux density:

However, for a static magnetic field, we have:

JAA

AAAxxBx

JBx

o

o

2

2

.

.

zoz

yoy

xox

o

JA

JA

JA

JAA

2

2

2

20.


MAGNETIC BOUNDARY CONDITIONS The magnetic boundary conditions are defined as the

conditions that B and H must satisfy at the boundary between different media.

To derive these conditions, we make use of Gauss’s law for magnetic fields, namely:

We also use Ampere’s circuital law, namely:

0. SdB

IldH

.


MAGNETIC BOUNDARY CONDITIONS Consider the boundary between two magnetic media 1

and 2, characterized respectively by 1 and 2, as shown below.


60

B1 B1n

B1t

B2

B2n

B2t

Medium 1,

Medium 2,

h

S

MAGNETIC BOUNDARY CONDITIONS Applying Gauss’s law to the pillbox and allowing h0, we

obtain:

The above equation shows that the normal component of B is continuous at the boundary.

It also shows that the normal component of H is discontinuous at the interface; that is, H undergoes some change at the interface.

nn

nn

nn

HH

BB

SBSB

2211

21

210


MAGNETIC BOUNDARY CONDITIONS Similarly, we apply Ampere’s circuital law to the closed

path abcda shown in the figure below, where surface current K on the boundary is assumed normal to the path.


62

H1 H1n

H1t

H2

H2n

H2t

Medium 1,

Medium 2,

h

w

a b

cd

K

MAGNETIC BOUNDARY CONDITIONS We obtain:

As h-0, the above equation becomes:

This shows that the tangential component of H is discontinuous. The above equation may be written in terms of B as:

2222 122211

hH

hHwH

hH

hHwHwK

nntnnt

KHHtt

21


KBB

tt 2

2

1

1

MAGNETIC BOUNDARY CONDITIONS If the boundary is free of current or the media are not

conductors, K=0, and we have the tangential components of H being equal:

If the fields make an angle with the normal to the interface, then, from the normal and tangential components of B (with K=0) we have:

2

2

1

1

21;

tt

tt

BBHH

2

1

2

1

2

2

2

211

1

1

222111

tan

tan

sinsin

coscos

B

HHB

BBBB

tt

nn


MAGNETIC CIRCUITS Consider a current-carrying conductor formed into a coil

of N turns around a doughnut-shaped iron core (magnetic material), the electrically-induced magnetic flux lines will be largely concentrated inside the iron core.

This is the example of a simple magnetic circuit.


65

Flux

I

I

N-turn coil, with current I

Cross-sectional area A, mean length L

Iron core

MAGNETIC CIRCUITS The flux in this magnetic circuit is analogous to the

electric current in an electric circuit. In the figure, there are N turns, each having a current I;

thus the cause of the induced flux is the current flow NI.

The analogy in an electric circuit is the fact that a voltage potential difference is the cause of the flow of the current carriers (that is, V causes I).

The quantity NI is called the magnetomotive force (MMF). It is the driving force behind the existence of the magnetic flux .

Thus: turnsAmpereNIMMFFm


MAGNETIC CIRCUITS If the same magnetomotive force (MMF) is

applied to similar iron cores, each with a different mean length Li, the resulting magnetic flux density for the i-th coil is:

One therefore expects that stronger magnetic fields will result in iron cores with shorter length L.

Therefore the other quantity of interest in magnetic circuits is the magnetizing force, or magnetic field intensity:


67

BAL

NIB

i

;

i

m

i L

F

L

NIBH

MAGNETIC CIRCUITS: RELUCTANCE When the current I or the number of turns N is increased

in the simple magnetic circuit, the magnetomotive force, Fm, is increased, resulting in a higher flux in the magnetic core.

Thus, we have:

Based on the analogies previously established, the electric voltage, E or V is analogous to the magnetomotive force Fm(=NI), and the electric current I is analogous to the magnetic flux.

The equation relating Fm to is thus similar to Ohm’s law for electric circuits: V=RI.

kF

F

m

m


MAGNETIC CIRCUITS: RELUCTANCE Thus the constant of proportionality, k, is actually a

measure of the opposition to the establishment of magnetic flux.

This quantity is called the reluctance, R of the magnetic circuit, and is analogous to the resistance R of an electric circuit.

Hence Ohm’s law for the magnetic circuit can be expressed as: WbA

FRRF m

m /


MAGNETIC CIRCUITS: PERMEABILITY It is easier to establish or set up the magnetic flux

lines in some materials (e.g. iron) than it is in other materials (e.g. air).

The magnetic lines of force, like electric current, always try to follow the path of least resistance.

Permeability is the property of materials that measures its ability to permit the establishment of magnetic lines of force. It is analogous to conductivity in electric circuits.

Air is taken as the reference material, with its permeability called o. The permeability of any other material is given by: ro


MAGNETIC CIRCUITS: PERMEABILITY Where r is called the relative permeability. Non-

magnetic materials (e.g. air, glass, copper and aluminum) are characterized by their r which is approximately unity. Magnetic materials such as iron, steel, cobalt, nickel, and their alloys are called ferromagnetic materials, as characterized by high values of r(100 to 100,000 or more)

From the definitions of reluctance, R of the magnetic circuit, and permeability of the material, it is clear that one is the opposite of the other.

Thus, we have the relationship between reluctance and permeability:

A

L

ALNI

NI

BA

NIFR m


MAGNETIC CIRCUITS: EXAMPLES

1. The simple magnetic circuit with one coil would around a doughnut-shaped iron core has a cross-sectional area of 50cm2 and a mean length of 2 m. The relative permeability of the magnetic material of the core is 80. If the current coil has 150 turns and the resulting flux is 80 Wb, determine:

a) The reluctance of the magnetic circuit (Ans: 3.98x106 A/Wb)

b) The value of the current flowing in the coil (Ans: 2.123 A)

2. The same magnetic circuit in example 1 has a magnetomotive force of 200 A. If the length of the coil is 40cm, and the permeability of the core is 6x10-4 Wb/m2, determine:

a) The magnetizing force, H (Ans: 500 A/m)b) The flux density B in the iron core (Ans: 0.3 Wb/m2)


72

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW

The figure below shows an example of a simple series magnetic circuit, made up of three different types of materials, including the air gap.

Since there is only one path for the magnetic flux lines , it must be the same in all parts of this series magnetic circuit. This is similar to a series electric circuit where the current is the same in all series components.

I

1

I

2

Material 1

Material 2

Air Gap

N

1

N

2



As E and V are analogous to Fm (=NI) and R or HL, a law similar to KVL (algebraic sum of voltage rise and drop in a closed loop =0) also applies to the closed-loop series magnetic circuit. This is Ampere’s circuital law.

Ampere’s circuital law states that the sum of the magnetomotive force (MMF or Fm) rises equals the sum of the MMF drops around any closed path of a magnetic circuit.

In general, Ampere’s circuital law can be stated as follows:

Here, is the same amount of magnetic flux in the series magnetic circuit, R1 is the reluctance of part 1 of the circuit (similarly for R2 and R3), H1 is the magnetic field intensity of part 1 of the circuit (similarly for H2 and H3), and L1 is the length of this part of the circuit (similarly for L2 and L3).

...

..'lg

332211

321

LHLHLH

RRRRsMMFappliedofsumebraicA T



From the above equation, one can easily see that in a series magnetic circuit,

That is, the net algebraic sum of the applied MMFs in the assumed positive direction of is the sum of times the sum of the reluctances of each part of the series circuit, consisting of materials 1 and 2 and the air gap (material 3).

The above form of Ampere’s circuital law is used if the dimensions and the permeabilities of each portion of the circuit are known, so that the reluctances can be calculated.

R To ta l ser ies re luc ce R R RT tan 1 2 3


MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW In the magnetic circuit shown in the figure below, the

flux induced by the MMF (NI) can be split into two parts, since there are two different paths of the magnetic flux lines, in either branch a (a) or branch b (b).


76

I

N

a

b

b


The sum of the two magnetic fluxes must be the same as the total flux, :

This is the law of conservation of flux. It is similar to KCL at a node of an electric circuit. If the reluctance of branch a is Ra, and the reluctance

of branch b is Rb, then the equivalent reluctance of branches a and b is:


77

a b

RR R

R Reqa b

a b

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW – EXAMPLE 1

Consider the magnetic circuit for a relay shown below. The average length of the iron core is 40 cm, and the length of the air gap is 0.2 cm, while the average cross-sectional area is 2.5 m2. The number of turns of the coil is 50, and r for iron is 200. Determine the current required to produce a flux density of 0.1 Wb/m2 in the air gap.


78

I

N

Air Gap

Iron

MAGNETIC CIRCUITS – AMPERE’S CIRCUITAL LAW – EXAMPLE 1

Solution:

o

iron o r

ironiron

iron

a ir gapo

a ir gap

T iron a ir gap

X X W b Am

X X X W b Am

RL

A X X XX A W b

RL

A X X XX A W b

R R R X A

4 1 0 1 2 5 7 1 0

1 2 5 7 1 0 2 0 0 2 5 1 3 1 0

1 0 4

2 5 1 3 1 0 2 5 1 06 3 6 6 1 0

1 0 0 0 2

1 2 5 7 1 0 2 5 1 06 3 6 4 1 0

1 2 7 3 1 0

7 6

6 4

4 4

6

6 4

6

7

. /

. . /

.

. .. /

.

. .. /

.

/

( . )( . ) . .

.

.

Wb

MMF F R BAR X X A

MMF N I A I

I A

m T T

0 1 2 5 1 0 1 2 7 3 1 0 3 1 8 2 4

3 1 8 2 4 5 0

6 3 6 5

4 7


ELECTROMAGNETIC INDUCTION A current-carrying conductor produces a magnetic field. The next question is: can a magnetic field result in a

current flow, or induced voltage? In other words, is the electromagnetic induction process reversible?

Faraday observed that when the magnetic lines of force (flux, ) linking a conductor are changed, a voltage will be induced across the terminals of the conductor.

The magnetic flux linkage can be changed by either moving the conductor or the magnetic filed itself in such a way that the conductor cuts across the magnetic lines of force.

The induced voltage and the resulting induced current are produced only if the cutting action is exhibited.


80

ELECTROMAGNETIC INDUCTION In the figure below, orig is the original magnetic flux of

the stationary horse-shoe magnet. If the conductor is moved perpendicular to the magnetic

lines of force, the galvanometer’s pointer deflects, indicating an induced current flow (iind) resulting from the induced voltage, vind.


81

GalvanometerN

S

conductorInduced current

ELECTROMAGNETIC INDUCTION – THE TRANSFORMER

The transformer is essentially just two (or more) inductors, sharing a common magnetic path.

Any two inductors placed reasonably close to each other will work as a transformer, and the more closely they are coupled magnetically, the more efficient they become.

When a changing magnetic field is in the vicinity of a coil of wire (an inductor), a voltage is induced into the coil which is in sympathy with the applied magnetic field.

A static magnetic field has no effect, and generates no output. Many of the same principles apply to generators, alternators, electric motors and loudspeakers.


ELECTROMAGNETIC INDUCTION – THE TRANSFORMER The figure shows the basics of all transformers. A coil (the

primary) is connected to an AC voltage source - typically the mains for power transformers. The flux induced into the core is coupled through to the secondary, a voltage is induced into the winding, and a current is produced through the load.


83


Note that the time-varying current ip flowing in the primary coil, generates a time-varying flux p in the space surrounding the coil.

Part of this flux, ps, will link with the secondary coil. This part is called the mutual flux.

Another part of this flux, pp, will not link with the secondary coil. This last part is called the leakage flux of the primary coil.

The primary and the induced secondary voltages, vs and vp, are:

dt

dNv

dt

dNv

psss

ppp


ELECTROMAGNETIC INDUCTION – THE TRANSFORMER Note that the flux is proportional to the current producing it.

Thus:

Here, Rp is the reluctance of the magnetic path of p, and Rps is the reluctance of the magnetic path of ps.

Thus the expressions for vp and vs become:


85

ps

sps

p

pp

ps

pps

p

ppp

A

lR

A

lR

R

iN

R

iN

;

;

dt

diM

dt

di

R

NN

R

iN

dt

dN

dt

dNv

dt

di

R

N

R

iN

dt

dN

dt

dNv

pp

ps

sp

ps

pps

psss

p

p

p

p

ppp

ppp

2


The constants of proportionalities are the inductance, Lp of the primary coil, and the mutual inductance, M, between the two coils:

Both M and Lp have the same physical units (the Henry), and both are constants, depending on the physical parameters and dimensions of the magnetic flux paths, related through the coupling coefficient, k. By exciting the secondary coil and determining the induced voltage and current in the primary side, we find that:


86

ps

sp

p

pp R

NNM

R

NL ;

2

10;

;2

kkLLkM

R

NN

R

NNM

R

NL

p

ssps

p

pssp

sp

sp

ps

sp

s

ss


The first and most important characteristic of an ideal transformer is that its primary and secondary fluxes have no leakage components.

Thus all the flux produced due to the flow of ip links with the secondary coil, while all the flux produced due to the flow of is links with the primary coil.

That is:

s

p

s

pps

psss

ppp

spsp

s

sp

p

ps

spspsp

N

N

v

v

dt

dN

dt

dNv

dt

dNv

LLLLkM

k

;

1

;,



From KVL, we have, for a signal with frequency :

sspp

p

s

p

s

s

p

p

s

s

p

s

p

sps

ss

p

pp

s

ss

p

pp

p

s

s

p

ss

pp

s

p

sssppp

IVIV

N

N

N

N

N

N

L

L

N

N

I

I

RRA

lR

A

lR

R

NL

R

NL

L

L

N

N

Lv

Lv

I

I

ILjvILjv

2

2

22

;;;

/

/

;


Consider vector A. Then the divergence of vector A at a point P is the outward flux per unit volume as the volume shrinks about P. That is:

The divergence theorem states that, for a closed surface S, which closes a volume v:

The curl of vector A is defined as the circulation per unit area. That is:

Here, the area S is bounded by the curve L, and a is the unit vector normal to the surface S. Then Stoke’s theorem states that:


89

v

SdAAAdiv S

v

.

. lim0

vS

dvASdA

..

as

ldAAxAcurl L

sˆ

.lim

0

SL

SdAxldA

..

1. Conservation of Electrostatic Field: Consider an electrostatic field which produces a field E. Then,

But, from definition of potential difference between point 1 and point 2, we have:

For a closed loop, the potential difference would be V2-V2=0 Thus:

This is Maxwell’s curl equation for an electrostatic filed.


90

SL

SdExldE

..

12

2

1. VVldE

L

L

0

0.. 11

Ex

VVSdExldE SL

2. Maxwell’s Curl Equation for Magnetostatic Field Similarly, from Ampere’s circuital law:

3. Divergence Theorem: Maxwell’s Scalar equation for D. From Gauss’s law, we have:

If we assume that Q is the total charge in a volume enclosed by surface S, with volume charge density v; and applying the divergence theorem, we have:


91

JHx

SdJSdHxldHI SL Sencl

...

S SdDQ

.

v

V v

VS

D

dvQBut

dvDSdDQ

.

..

4. Divergence Theorem: Maxwell’s Scalar equation for B

For the magnetic field intensity, the total flux enclosed by a surface S is given by:

Applying the divergence theorem, and noting that there is no magnetic charge density in a given volume – that is, it is not possible to have isolated magnetic poles, since magnetic flux lines always close upon themselves, we have:

This is Maxwell’s fourth equation for static fields.


92

HB

SdBS

.

0.

0..

B

dvBSdB VS

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