CHAPTER 3 MAGNETOSTATICS. 2 3.1BIOT-SAVART’S LAW 3.2AMPERE’S CIRCUITAL LAW 3.3MAGNETIC FLUX...

CHAPTER 3

MAGNETOSTATICS

2

MAGNETOSTATICS

3.1 BIOT-SAVART’S LAW

3.2 AMPERE’S CIRCUITAL LAW

3.3 MAGNETIC FLUX DENSITY

3.4 MAGNETIC FORCES

3.5 BOUNDARY CONDITIONS

3

INTRODUCTION

Magnetism and electricity were

considered distinct phenomena until

1820 when Hans Christian Oersted

introduced an experiment that showed

a compass needle deflecting when in

proximity to current carrying wire.

4

He used compass to show that current produces magnetic fields that loop around the conductor. The field grows weaker as it moves away from the source of current.

A represents current coming out of paper.

A represents current heading into the paper.

INTRODUCTION (Cont’d)

5

The principle of magnetism is widely used

in many applications:

Magnetic memory

Motors and generators

Microphones and speakers

Magnetically levitated high-speed

vehicle.


6


Magnetic fields can be easily visualized by sprinkling iron filings on a piece of paper suspended over a bar magnet.

7

The field lines are in terms of the magnetic

field intensity, H in units of amps per meter.

This is analogous to the volts per meter units

for electric field intensity, E.

Magnetic field will be introduced in a manner

paralleling our treatment to electric fields.


8

3.1 BIOT-SAVART’S LAW

Jean Baptiste Biot and Felix Savart arrived a

mathematical relation between the field and

current.

212

1211

4 R

aLH

dI

d

9

BIOT-SAVART’S LAW (Cont’d)

To get the total field resulting from a

current, sum the contributions from each

segment by integrating:

24 R

Id R

aL

H

10


Due to continuous current distributions:

Line current Surface current Volume current

11

In terms of distributed current sources, the

Biot-Savart’s Law becomes:

24 R

Id R

aL

H

24 R

dS R

aK

H

24 R

dV R

aJ

H

Line current

Surface current

Volume current


12

DERIVATION

Let’s apply

to determine the magnetic field, H

everywhere due to straight current carrying

filamentary conductor of a finite length AB .

24 R

Id R

aL

H

DERIVATION (Cont’d)

14

We assume that the conductor is along the z-

axis with its upper and lower ends

respectively subtending angles and at

point P where H is to be determined.

1 2

The field will be independent of z and φ and

only dependant on ρ.


15

The term dL is simply and the vector

from the source to the test point P is:

zdza

aaaR

zR zR

Where the magnitude is:

22 zR

And the unit vector:22

z

z zR

aaa


16

Combining these terms to have:

B

A

zz

R

z

zIdzR

Id

R

Id

2322

32

4

44

aaa

RLaLH


17

aaa zz zdzCross product of :

a

aaa

RL dz

z

dzdz

0

00

This yields to:

a H

B

A z

dzI23224


18

Trigonometry from figure,

z

tan So, cotz

Differentiate to get: decdz 2cos

a H

2

1

23222

22

cot

cos

4

decI


19


Remember!

)(cos)(cot1

)(cos)cot(

22

2

uecu

x

uuecu

x

20

Simplify the equation to become:

a

a

a H

12

33

22

coscos4

sin4

cos

cos

4

2

1

2

1

I

dI

ec

decI


21

Therefore,

aH 12 coscos4

I

This expression generally applicable for any

straight filamentary conductor of finite

length.

DERIVATION 1

22

As a special case, when the conductor is semifinite with respect to P,

,0,0,0,0

0,0,0

or at

at

B

A

The angle become: 02

01 0,90

So that,

aH4

I

DERIVATION 2

23

Another special case, when the conductor is infinite with respect to P,

,0,0

,0,0

at

at

B

A

The angle become: 02

01 0,180

So that,

aH2

I

DERIVATION 3

24

HOW TO FIND UNIT VECTOR aφ ?

From previous example, the vector H is in

direction of aφ, where it needs to be determine

by simple approach:

a

aaa l

Where,la unit vector along the line current

a unit vector perpendicular from the line current to the field point

25

EXAMPLE 1

The conducting triangular loop carries of 10A.

Find H at (0,0,5) due to side 1 of the loop.

26

SOLUTION TO EXAMPLE 1

• Side 1 lies on the x-y

plane and treated as a

straight conductor.

• Join the point of interest

(0,0,5) to the beginning

and end of the line

current.

27

SOLUTION TO EXAMPLE 1 (Cont’d)

This will show how is

applied for any straight, thin, current carrying

conductor.

aH 12 coscos4

I

From figure, we know that 0cos90 10

1

and from trigonometry and 29

2cos 2 5

28


To determine by simple approach:a

xl aa zaa and so that,

yzxl aaaaaa

mAm

I

yy a a

aH

1.59029

2

54

10

coscos4 12

29

EXAMPLE 2

A ring of current with radius a lying in the x-y

plane with a current I in the direction. Find

an expression for the field at arbitrary point a

height h on z axis.

a

30

Can we use ?

aH 12 coscos4

I


Solve for each term in the Biot-Savart’s Law

31


We could find:

aL add

aaaR ahR zR

22 ah R

22 ah

ah zR

aa

a

32

It leads to:

2

02322

32

4

44

ah

ahIadR

Id

R

Id

z

R

aaa

RLaLH


The differential current element will give a field with:

a from

fromzazaa

aa

33

However, consider the symmetry of the problem:


The radial components cancel but the components adds, so:

za

2

02322

2

4d

ah

Ia zaH

34

This can be easily solved to get:

zah

IaaH

2322

2

2

At h=0 where at the center of the loop, this equation reduces to:

za

IaH

2


35


• For many problems involving surface current

densities and volume current densities, solving for

the magnetic field using Biot-Savart’s Law can be

quite cumbersome and require numerical

integration.

• There will be sufficient symmetry to be able to

solve for the fields using Ampere’s Circuital

Law.

36

3.2 AMPERE’S CIRCUITAL LAW

In magnetostatic problems with sufficient

symmetry, we can employ Ampere’s Circuital

Law more easily that the law of Biot-Savart.

The law says that the integration of H

around any closed path is equal to the net

current enclosed by that path. i.e.

encId LH

37

• The line integral of H around the path is termed the circulation of H.

• To solve for H in given symmetrical current

distribution, it is important to make a careful

selection of an Amperian Path (analogous to

gaussian surface) that is everywhere either

tangential or normal to H.

• The direction of the circulation is chosen such

that the right hand rule is satisfied.

AMPERE’S CIRCUITAL LAW (Cont’d)

38

DERIVATION 4

Find the magnetic field intensity

everywhere resulting from an infinite

length line of current situated on the

z-axis using Ampere’s Circuital Law.

39

DERIVATION 4 (Cont’d)

Select the best Amperian

path, where here are two

possible Amperian paths

around an infinite length line

of current.Choose path b which has a

constant value of Hφ

around the circle specified by the radius ρ

40


Using Ampere’s circuital law:

encId LH

We could find:

aL

aH

dd

H

So,

IdHId enc

2

0

aaLH

41

Solving for Hφ:

2

IH

Where we find that the field resulting from an infinite length line of current is the expected result:

aH

2

I Same as applying

Biot-Savart’s Law!


42

Use Ampere’s Circuital Law to find the

magnetic field intensity resulting from an

infinite extent sheet of current with current

sheet in the x-y plane.

DERIVATION 5

xxK aK

43


Rectangular amperian path of height Δh and

width Δw. According to right hand rule, perform

the circulation in order of a b c d a

44

We have:

a

d

d

c

c

b

b

aenc ddddId LHLHLHLHLH


From symmetry argument, there’s only Hy

component exists. So, Hz will be zero and thus

the expression reduces to:

d

c

b

aenc ddId LHLHLH

45

So, we have:

wH

dyHdyH

ddd

y

w

yyyw

yyy

d

c

b

a

20

0

aaaa

LHLHLH


46

The current enclosed by the path,

wKdyKKdSI x

w

x

0


This will give:

encId LH

wKwH xy 2

2x

yK

H

Or generally,

NaKH 2

1

47

EXAMPLE 3

An infinite sheet of current with exists

on the x-z plane at y = 0. Find H at P (3,2,5). m

Az aK 6

48


Use previous expression, that is:

NaKH 2

1

is a normal vector from the sheet to the test point P (3,4,5), where:Na

yN aa and zaK 6

So,

mA

xyz aaaH 362

1

49

Consider the infinite

length cylindrical

conductor carrying a

radially dependent current

Find H

everywhere.

zJ aJ 0

EXAMPLE 4

50

What components of H will be present?

Finding the field at

some point P, the

field has both

and

components.

a

a


51

The field from the

second line current

results in a

cancellation of the

components

a


52

To calculate H everywhere, two amperian paths are required:

Path #1 is for

Path #2 is for

a

a


53

Evaluating the left side of Ampere’s law:

HdHd 22

0

aaLH

This is true for both amperian path.

The current enclosed for the path #1:

3

2 30

0

2

0

20

0

JddJ

ddJdI zz

aaSJ


54

Solving to get Hφ:

3

20

J

H Or

aH3

20J for a

The current enclosed for the path #2:

3

2 30

0

2

0

20

aJddJdI

a

SJ

Solving to get Hφ:

aH3

30aJ

for a


55

EXAMPLE 5

Find H

everywhere for

coaxial cable as

shown.

56

Even current

distributions are

assumed in the

inner and outer

conductor.

Consider four

amperian paths.


57

It will be four amperian paths:

Therefore, the magnetic field intensity, H will be determined for each amperian paths.

aba cb

c


58

As previous example, only Hφ component is

present, and we have the left side of ampere’s circuital law:

HdHd 22

0

aaLH

For the path #1:

SJ dIenc


59

We need to find current density, J for inner conductor because the problem assumes an event current distribution (ρ<a is a solid volume where current distributed uniformly).

zdS

IaJ

Where,

22

0 0

, addSdddSa


60

So,zz

a

I

dS

IaaJ

2

We therefore have:

2

2

2

0 02

a

I

dda

IdI zzenc

aaSJ


61

Equating both sides to get:

2

2

2 22 a

I

a

IH

for a

For the path #2:

The current enclosed is just I,

Therefore:

IIenc

IIHd enc 2LH

2

IH for ba


62

For the path #3:


For total current enclosed by path 3, we need to find the current density, J in the outer conductor because the problem assumes an event current distribution (a<ρ<b is a solid volume where current distributed uniformly) given by:

zzbc

I

dS

IaaJ

22

63

We therefore have (for AP#3):

22

22

2

022

bc

bI

ddbc

Id

bzz

aaSJ

But, the total current enclosed is:

22

22

22

22

bc

cI

bc

bII

dIIenc

SJ


64


So we can solve for path #3:

22

22

2bc

cIIHd enc

LH

22

22

2 bc

cIH

for cb

For the path #4, the total current is zero. So,

0H for c This shows the shielding ability by coaxial cable!!

65


Summarize the results to have:

c

cbbc

cI

baI

aa

I

02

2

2

22

22

2

a

a

a

H

66

Expression for curl by applying Ampere’s Circuital Law might be too lengthy to derive, but it can be described as:

JH The expression is also called the point form of Ampere’s Circuital Law, since it occurs at some particular point.


67

The Ampere’s Circuital Law can be rewritten in terms of a current density, as:

SJLH dd

Use the point form of Ampere’s Circuital Law to replace J, yielding:

SHLH dd

This is known as Stoke’s Theorem.


68

3.3 MAGNETIC FLUX DENSITY

In electrostatics, it is convenient to think in

terms of electric flux intensity and electric flux

density. So too in magnetostatics, where

magnetic flux density, B is related to magnetic

field intensity by:r 0 HB

Where μ is the permeability with:

mH 70 104

69

MAGNETIC FLUX DENSITY (Cont’d)

The amount of magnetic flux, φ in webers

from magnetic field passing through a

surface is found in a manner analogous to

finding electric flux:

SB d

70

Fundamental features of magnetic fields:

• The field lines form a

closed loops. It’s different

from electric field lines,

where it starts on positive

charge and terminates on

negative charge


71


• The magnet cannot be

divided in two parts, but it

results in two magnets.

The magnetic pole cannot

be isolated.

72


The net magnetic flux passing through a

gaussian surface must be zero, to get

Gauss’s Law for magnetic fields:

0 SB d

By applying divergence theorem, the point

form of Gauss’s Law for static magnetic fields:

0 B

73

EXAMPLE 6

Find the flux crossing the portion of the

plane φ=π/4 defined by 0.01m < r <

0.05m and 0 < z < 2m in free space. A

current filament of 2.5A is along the z axis

in the az direction.

Try to sketch this!

74


The relation between B and H is:

aHB200I

To find flux crossing the portion, we need to use:

SB d

where dS is in the aφ direction.

75

So, aS dzdd Therefore,

WbI

dzdI

d

z

aa

SB

60

2

0

05.0

01.0

0

1061.101.0

05.0ln

2

2

2


76

3.4 MAGNETIC FORCES

Upon application of a magnetic field, the wire is

deflected in a direction normal to both the field and

the direction of current.

77

MAGNETIC FORCES (Cont’d)

The force is actually acting on the individual charges moving in the conductor, given by:

BuF qm

By the definition of electric field intensity, the

electric force Fe acting on a charge q within an

electric field is:

EF qe

78

A total force on a charge is given by Lorentz force equation:

BuEF q


The force is related to acceleration by the equation from introductory physics,

aF m

79


To find a force on a current element, consider a line conducting current in the presence of magnetic field with differential segment dQ of charge moving with velocity u:

BuF dQd

dt

dLu

But,

80

BLF Idd

So,BLF d

dt

dQd

Since corresponds to the current I in the line,

dtdQ


We can find the force from a collection of current elements

12212 BLF dI

81

Consider a line of current in +az direction on the z

axis. For current element a,

zaa IdzId aL

But, the field cannot exert magnetic force on the element producing it. From field of second element b, the cross

product will be zero since IdL and aR in

same direction.


82

EXAMPLE 7

If there is a field from a

second line of current

parallel to the first, what

will be the total force?

83

The force from the magnetic field of line 1

acting on a differential section of line 2 is:

12212 BLF dId

Where,

aB

210

1I

By inspection from figure,

xy aa , Why?!?!


84

0210

12

21010212

2

22

Ly

yxz

dzy

II

dzy

II

y

IdzId

aF

aaaF

zdzd aL 2Consider , then:

yy

LIIaF

2

21012


85

Generally,

212

121212

012 4 R

ddII

aLLF

• Ampere’s law of force between a pair of current-

carrying circuits.

• General case is applicable for two lines that are not

parallel, or not straight.

• It is easier to find magnetic field B1 by Biot-Savart’s

law, then use to find F12 . 12212 BLF dI


86

EXAMPLE 8

The magnetic flux density in a region of free

space is given by B = −3x ax + 5y ay − 2z az T.

Find the total force on the rectangular loop

shown which lies in the plane z = 0 and is

bounded by x = 1, x = 3, y = 2, and y = 5, all

dimensions in cm.Try to sketch this!

87

The figure is as shown.


88


B L F xIdloop

AI 30

First, note that in the plane z = 0, the z

component of the given field is zero, so will not

contribute to the force. We use:

Which in our case becomes with,

zyx zyx aaaB 253 and

89

02.0

05.001.0

01.0

03.005.0

05.0

02.003.0

03.0

01.002.0

5330

5330

5330

5330

yxxy

yyxx

yxxy

yyxx

yxxdy

yxxdx

yxdy

yxdx

aa a

aa a

aa a

aa aFSo,


90

Simplifying these becomes:

N

dydx

dydx

z

zz

zz

a

aa

aaF

027.0150.0081.006.0

)01.0)(3(30)05.0)(5(30

)03.0)(3(30)02.0)(5(30

02.0

05.0

01.0

03.0

05.0

02.0

03.0

01.0

mNz aF 36


91

3.5 BOUNDARY CONDITIONS

We could see how the fields behave at the

boundary between a pair of magnetic materials

which derived using Ampere’s Circuital Law and

Gauss’s Law for magnetostatic fields:

0 SB d

encId LH

92

BOUNDARY CONDITIONS (Cont’d)

Consider,

93

A pair of magnetic media separated by a sheet

current density K. Choose a rectangular

Amperian path of width Δw and height Δh

centered at the interface. The current enclosed

by the path is:

wKKdWIenc


94

b

a

c

b

d

c

a

d

wKdd )( LHLH


The sheet current is heading into the page and

use the right hand rule to determine the

direction of integration around the loop. So,

95

221

21

11

2

0

0

2/

0

hHH

dLHdLHd

cb

wHdLHd

ba

NN

h

NNNNNh

N

c

b

b

a

w

TTTT

aaaaLH

aaLH

For first and second integral,


96

221

12

22

2

0

0

2

0

hHH

dLHdLHd

ad

wHdLHd

dc

NN

h

NNNh

NNN

a

d

wTTTT

d

c

aaaaLH

aaLH


For third and fourth integral,

97

KHH TT 21

Combining the result, we get the first boundary condition for magnetostatic field,

KHH 2121a

In more general case,


Where a21 is unit vector normal from medium 2 to medium 1

98


Second boundary condition can be determined by applying Gauss’s Law over a small pillbox shaped Gaussian surface,

99

The Gauss’s Law,

Where,

sidebottomtop

dddd SBSBSBSB

The pillbox is short enough, so the flux out of

the side is negligible.

0 SB d


100

We have

0

21

21

SBB

dSBdSBd

NN

NNNNNN

aaaaSB

Since ΔS can be chosen unequal to zero, it follows that:

21 NN BB


101

EXAMPLE 9

The magnetic field intensity is given as:

In a medium with µr1=6000 that exist for z

< 0. Find H2 in a medium with µr2=3000

for z>0.

mAxyx aaaH 3261

102


103

Recall that, for a conductor-dielectric interface:

0TE SND Generally, it is not exist for magnetostatic fields. If one of the media is superconductor, where the magnetic field rapidly attenuates away from the surface, such that:

0B


104

If medium 2 is superconductor, the equations

for magnetostatic fields become:

0NB

KHa 1N 1 2

The second expression is logical since the magnetic field lines must form closed loops and cannot suddenly terminate even on a superconductor.


CHAPTER 3

END

106

PRACTICAL APPLICATION

Loudspeakers

Maglev (Magnetically

Levitated

Trains)

107

• Paper or plastic cone affixed to a voice coil (electromagnet) suspended in a magnetic field.

•AC Signals to the voice coil moves back and forth, resulting vibration of the cone and producing sound waves of the same frequency as the AC signal

LOUDSPEAKERS

108

MAGLEV

109

MAGLEV (Cont’d)

• Interaction between electromagnets in the train and the current carrying coils in the guide rail provide levitation.

• By sending waves along the guide rail coils, the train magnet pushed/pulled in the direction of travel. The train is guided by magnet on the side of guide rail.

• Computer algorithms maintain the separation distance.

110

SUMMARY (1)

•For a differential current element I1dL1 at point 1,

the magnetic field intensity H at point 2 is given by the law of Biot-Savart,

Where is a vector from the source element at point 1 to the location where the field is desired at point 2. By summing all the current elements, it can rewritten as:

121212 aR R

212

1211

4 R

aLH

dI

d

24 R

Id R

aL

H

111

•The Biot-Savart law can be written in terms of surface and volume current densities:

SUMMARY (2)

24 R

dS R

aK

H

24 R

dv R

aJ

H

Surface current

Volume current

•The magnetic field intensity resulting from an infinite length line of current is:

aH

2

I

112

SUMMARY (3)

and from a current sheet of extent it is:

NaKH 2

1 Where aN is a unit vector normal from

the current sheet to the test point.

•An easy way to solve the magnetic field intensity in problems with sufficient current distribution symmetry is to use Ampere’s Circuital Law, which says that the circulation of H is equal to the net current enclosed by the circulation path

encId LH

113

SUMMARY (4)

• The point or differential form of Ampere’s circuital Law is:

JH

SHLH dd

• A closed line integral is related to surface integral by Stoke’s Theorem:

• Magnetic flux density, B in Wb/m2 or T, is related to the magnetic field intensity by

HB

114

r 0

SUMMARY (5)

Material permeability µ can be written as:and the free space permeability is:

mH 70 104

• The amount of magnetic flux Φ in webers through a surface is:

SB d

Since magnetic flux forms closed loops, we have Gauss’s Law for static magnetic fields:

0 SB d

115

SUMMARY (6)

• The total force vector F acting on a charge q moving through magnetic and electric fields with velocity u is given by Lorentz Force equation:

BuEF q

The force F12 from a magnetic field B1 on a current

carrying line I2 is:

12212 BLF dI

116

SUMMARY (7)

• The magnetic fields at the boundary between different materials are given by:

KHH 2121a

Where a21 is unit vector normal from medium 2

to medium 1, and:

21 NN BB

117

VERY IMPORTANT!

From electrostatics and magnetostatics, we can now present all four of Maxwell’s equation for static fields:

enc

enc

Id

d

d

Qd

LH

LE

SB

SD

0

0

Integral Form

JH

E

B

D

0

0v

Differential Form

CHAPTER 3 MAGNETOSTATICS. 2 3.1BIOT-SAVART’S LAW 3.2AMPERE’S CIRCUITAL LAW 3.3MAGNETIC FLUX...

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Transcript of CHAPTER 3 MAGNETOSTATICS. 2 3.1BIOT-SAVART’S LAW 3.2AMPERE’S CIRCUITAL LAW 3.3MAGNETIC FLUX...