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CHAPTER 3

MAGNETOSTATICS

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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

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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.

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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)

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The principle of magnetism is widely used in

many applications:

Magnetic memory Motors and generators

Microphones and speakers

Magnetically levitated high-speed

vehicle.


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Magnetic fields can be easily visualized by

sprinkling iron filings on a piece of paper

suspended over a bar magnet.

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The field lines are in terms of themagnetic field

intensity,H in units of amps per meter.

This is analogous to the volts per meter units

forelectric field intensity,E.

Magnetic field will be introduced in a manner

paralleling our treatment to electric fields.


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3.1BIOT-SAVART’S LAW

Jean Baptiste Biot and Felix Savart arrived a

mathematical relation between the field and

current.

2

12

1211

4 R

aLH

π

×= d I

d

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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

π

aLH

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Due to continuous current distributions:

Line current Surface current Volume current

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In terms of distributed current sources, the

Biot-Savart’s Law becomes:

∫ ×= 24 R Id

Rπ

aL

H

∫ ×

=2

4 R

dS R

π

aK H

∫ ×

=24 R

dV R

π

aJH

Line current

Surface current

Volume current


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DERIVATION

Let’s apply

to determine the magnetic field, H everywhere

due to straight current carrying filamentary

conductor of a finite length AB .

∫

×=

2

4 R

Id R

π

aLH

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DERIVATION (Cont’d)

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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

α 2

α

The field will be independent of z and φ and

only dependant on ρ.


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The term dL is simply and the vectorfrom the source to the test pointP is:

z dz a

ρ ρ aaaR +−==→

z R z R

Where the magnitude is:

22 ρ += z R

And the unit vector:22 ρ

ρ ρ

++−=

z

z z R

aaa


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Combining these terms to have:

( )

( )∫

∫ ∫

+

+−×=

×=×= B

A

z z

R

z

z Idz

R Id

R Id

2322

32

4

44

ρ π

ρ

π π

ρ aaa

R LaLH


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( ) ρ ρ aaa +−× z z z dz Cross product of :

φ

φ ρ

ρ ρ

a

aaa

R L dz z

dz d

z

=−

=×

→

000

This yields to:

( ) φ

ρ π

ρ aH ∫

+= B

A z

dz I 2322

4


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Trigonometry from figure,

z

ρ α =tan So, α ρ cot= z

Differentiate to get: α α ρ d ecdz 2cos−=

( ) φ α

α α ρ ρ

α α ρ

π aH

∫ +−=∴2

1

23222

22

cot

cos

4

d ec I


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Remember!

( ) ( ))(cos)(cot1

)(cos)cot(

22

2

uecu

xuuecu

x

=+

∂∂−=∂∂

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Simplify the equation to become:

( ) φ

φ

α

α

φ

α

α

α α πρ

α α πρ

α ρ

α α ρ

π

a

a

aH

12

33

22

coscos4

sin4

cos

cos

4

2

1

2

1

−=

−=

−=

∫ ∫

I

d I

ec

d ec I


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Therefore,

( ) φ

α α πρ

aH12

coscos

4

−= I

This expression generally applicable for any

straight filamentary conductor of finite length.

DERIVATION 1

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As a special case, when the conductor is semifinite

with respect toP,

( )( ) ( )−∞∞ ,0,0,0,0

0,0,0orat

at

B A

The angle become: 0

20

1 0,90 == α α

So that,

φ πρ

aH4

I =

DERIVATION 2

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Another special case, when the conductor is

infinite with respect toP,

( )( )∞−∞,0,0

,0,0at

at

B A

The angle become: 0

20

1 0,180 == α α

So that,

φ πρ

aH2

I =

DERIVATION 3

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HOW TO FIND UNIT VECTOR aφ ?

From previous example, the vectorH is in

direction ofaφ, where it needs to be determine

by simple approach:

φ a

ρ φ aaa ×= l Where,

l a

unit vector along the line current

ρ aunit vector perpendicular from theline current to the field point

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EXAMPLE 1

The conducting triangular loop carries of 10A.FindH at (0,0,5) due to side 1 of the loop.

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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.

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SOLUTION TO EXAMPLE 1 (Cont’d)

This will show how is

applied for any straight, thin, current carrying

conductor.

( ) φ α α πρ

aH 12 coscos4

−= I

From figure, we know that 0cos90 10

1 =∴= α α

and from trigonometry and29

2

cos 2 =α 5= ρ

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To determine by simple approach:φ a

xl aa = z aa = ρ and so that,

y z xl aaaaaa −=×=×= ρ φ

( )

( ) ( ) m Am

I

y y aa

aH

1.59029

2

54

10

coscos4

12

−=−

−=

−=∴

π

α α πρ

φ

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EXAMPLE 2

A ring of current with radiusa lying in the x-y

plane with a currentI in the direction. Find

an expression for the field at arbitrary point a

heighth on z axis.

φ a+

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Can we use ?( ) φ α α πρ

aH 12 coscos4

−= I


Solve for each term in

the Biot-Savart’s Law

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We could find:

φ φ aL ad d =

ρ aaaR ah R z R −==

→

22 ah +=R

22 ah

ah z R

+

−=∴ ρ

aa

a

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It leads to:

( )( )∫

∫ ∫

= +−×=

×=

×=

π

φ

ρ φ

π φ

π π 2

02322

32

4

44

ahah Iad

R

Id

R

Id

z

R

aaa

R LaLH


The differential current element will give a field

with: ρ a from

from z a

z aa ×φ ( ) ρ φ aa −×

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However, consider the symmetry of the problem:


The radial components

cancel but the

components adds, so: z a

( ) ∫ =+=

π

φ

φ π

2

02322

2

4

d ah

Ia z aH

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This can be easily solved to get:

( ) z

ah

IaaH

2322

2

2 +=

At h=0 where at the center of the loop, this

equation reduces to:

z a

I aH

2=


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• For many problems involvingsurface current

densities andvolume 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 tosolve for the fields using Ampere’s Circuital Law.

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3.2AMPERE’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 ofH

around any closed path is equal to the net

current enclosed by that path. i.e.

enc I d =•∫ LH

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• The line integral ofH around the path is termedthe circulation ofH.

• To solve forH in given symmetrical current

distribution, it is important to make a carefulselection of an Amperian Path (analogous to

gaussian surface) that is everywhere either

tangential or normal toH.

• The direction of the circulation is chosen such

that the right hand rule is satisfied.

AMPERE’S CIRCUITAL LAW (Cont’d)

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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.

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DERIVATION 4 (Cont’d)

Select the best Amperian path,

where here are two possible

Amperian paths around an

infinite length line of current.

Choose pathb which has a

constant value of Hφ

around the circle specified

by the radiusρ

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Using Ampere’s circuital law:

enc I d =•∫ LH

We could find:

φ

φ φ

φ ρ aL

aH

d d

H

=

=

So,

I d H I d enc =•==• ∫ ∫ =

π

φ

φ φ φ φ ρ 2

0

aaLH

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Solving for Hφ:

πρ φ

2

I H =

Where we find that the field resulting from an

infinite length line of current is the expected

result:

φ πρ

aH2

I = Same as applyingBiot-Savart’s Law!


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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

x x K aK =

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Rectangularamperian path of height ∆h and width∆w. According to right hand rule, perform the

circulation in order of a b c d a

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We have:

∫ ∫ ∫ ∫ ∫ •+•+•+•==•a

d

d

c

c

b

b

a

enc d d d d I d LHLHLHLHLH


From symmetry argument, there’s only H y

component exists. So, Hz will be zero and thus the

expression reduces to:

∫ ∫ ∫ •+•==•d

c

b

a

enc d d I d LHLHLH

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So, we have:

( )

w H

dy H dy H

d d d

y

w

y y y

w

y y y

d

c

b

a

∆=

•+•−=

•+•=•

∫ ∫

∫ ∫ ∫ ∆

∆

2

0

0

aaaa

LHLHLH


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The current enclosed by the path,

w K dy K KdS I x

w

x ∆=== ∫ ∫ ∆

0


This will give:

enc I d =•∫ LH

w K w H x y ∆=∆2

2

x y

K H =

Or generally,

N aK H ×= 21

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EXAMPLE 3

An infinite sheet of current with exists

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

A z aK 6=

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Use previous expression, that is:

N aK H ×=2

1

is a normal vector from the sheet to the testpoint P (3,4,5), where: N a

y N aa = and z aK 6=

So,

m A

x y z aaaH 362

1−=×=

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Consider the infinite length

cylindrical conductor

carrying a radially

dependent current

FindH everywhere. z J aJ ρ 0=

EXAMPLE 4

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What components of H will be present?

Finding the field at

some pointP, the

field has both

and components. ρ a

φ a


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The field from the

second line currentresults in a

cancellation of the

components

ρ a


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To calculateH everywhere, two amperian paths

are required:

Path #1 is for

Path #2 is for

a≤ ρ

a> ρ


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Evaluating the left side of Ampere’s law:

φ

π

φ φ φ πρ φ ρ H d H d 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

ρ π φ ρ ρ

φ ρ ρ ρ

ρ

ρ

π

φ

J d d J

d d J d I z z

==

•=•=

∫ ∫

∫ ∫

= =

aaSJ


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Solving to get Hφ:

3

2

0 ρ φ

J H = Or φ

ρ aH

3

2

0 J = for a≤ ρ

The current enclosed for the path #2:

3

2 30

0

2

0

20

a J d d J d I

a π φ ρ ρ

ρ

π

φ

==•= ∫ ∫ ∫ = =

SJ

Solving to get Hφ:

φ ρ

aH3

3

0a J = for a> ρ


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EXAMPLE 5

FindH everywhere

for coaxial cable as

shown.

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Even current

distributions are

assumed in theinner and outer

conductor.

Consider fouramperian paths.


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It will be four amperian paths:

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

a≤ ρ

ba ≤

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As previous example, only Hφ component is

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

circuital law:

φ

π

φ φ φ πρ φ ρ H d H d 22

0

=•=• ∫ ∫ aaLH

For the path #1:

∫ •= SJ d I enc


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We need to find current density, J for innerconductor because the problem assumes an event

current distribution (ρ

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So, z z

a

I

dS

I aaJ

2π ==

We therefore have:

2

2

2

0 02

a

I

d d a

I d I z z enc

ρ

ρ φ ρ π

π

φ

ρ

ρ

=

•=•= ∫ ∫ ∫ = =

aaSJ


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Equating both sides to get:

2

2

2 22 a

I

a

I H

π

ρ

πρ

ρ φ == for a≤ ρ

For the path #2:

The current enclosed is justI,

Therefore:

I I enc =

I I H d enc ===•

∫ φ πρ 2LH

πρ φ

2

I H = for ba ≤

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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 anevent current distribution (a

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We therefore have (for AP#3):

( ) ( )

22

22

2

022

bc

b I

d d bc

I d

b

z z

−−−=

•−−

=• ∫ ∫ ∫ = =

ρ

ρ φ ρ π

π

φ

ρ

ρ

aaSJ

But, the total current enclosed is:

22

22

22

22

bc

c I

bc

b I I

d I I enc

−−

=

−−

−+=

•+= ∫

ρ ρ

SJ


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So we can solve for path #3:

22

22

2bc

c I I H d enc −

−===•∫ ρ

πρ φ LH

−−=

22

22

2 bc

c I H

ρ

πρ φ

for cb ≤ ρ This shows the shieldingability by coaxial cable!!

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Summarize the results to have:

>≤

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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.


( d)

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The Ampere’s Circuital Law can be rewritten in

terms of a current density, as:

∫ ∫ •=•SJLH d d

Use the point form of Ampere’s Circuital Law to

replace J, yielding:

( )∫ ∫ •×∇=• SHLH d d This is known asStoke’s Theorem.


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3.3MAGNETIC 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 µ µ µ µ 0== HB Where μ is the permeability with:

m H 70

104 −×= π µ

MAGNETICFLUXDENSITY(C ’d)

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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:

∫ •=Φ SB d

MAGNETICFLUXDENSITY(C t’d)

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Fundamental features of magnetic fields:

• The field lines form a

closed loops. It’s differentfrom electric field lines,

where it starts on positive

charge and terminates on

negative charge



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• The magnet cannot be

divided in two parts, but it

results in two magnets. The magnetic pole cannot

be isolated.


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The net magnetic flux passing through a

gaussian surface must be zero, to getGauss’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

EXAMPLE6

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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 theaz

direction.

Try to sketch this!

SOLUTIONTOEXAMPLE6

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The relation betweenB andH is:

φ

πρ

µ µ aHB

2

00

I ==

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

∫ •=ΦSB d

where dS is in the aφ direction.

SOLUTIONTOEXAMPLE6(C t’d)

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So, φ ρ aS dz d d =

Therefore,

Wb I

dz d I

d

z

aa

SB

60

2

0

05.0

01.0

0

1061.101.0

05.0ln

2

2

2

−

= =

×==

•=

•=Φ

∫ ∫ ∫

π

µ

ρ πρ

µ φ φ

ρ


34MAGNETICFORCES

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3.4MAGNETIC FORCES

Upon application of a magnetic field, the wire is

deflected in a direction normal to both the field and the

direction of current.

MAGNETICFORCES(Cont’d)

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MAGNETIC FORCES (Cont’d)

The force is actually acting on the individual

charges moving in the conductor, given by:

BuF

×= qmBy the definition of electric field intensity, the

electric forceFe acting on a chargeq within an

electric field is:

EF qe =


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A total force on a charge is given byLorentz force

equation:

( )BuEF ×+= q


The force is related to acceleration by the

equation from introductory physics,

aF m=


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To find a force on a current element, consider a

line conducting current in the presence of

magnetic field with differential segmentdQ of

charge moving with velocityu:

BuF ×= dQd

dt d Lu =

But,


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BLF ×=∴ Id d

So,BLF ×= d

dt

dQd

Since corresponds to the currentI in

the line,

dt dQ


We can find the force from a collection of

current elements

12212 ∫ ×= BLF d I


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

axis. For current elementa,

z aa Idz Id aL

=But, the field cannot exert magnetic force

on the element producing it. From field of

second elementb, the cross product will

be zero sinceIdL andaR in same

direction.

MAGNETIC FORCES (Contd)

EXAMPLE7

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EXAMPLE 7

If there is a field from a

second line of current

parallel to the first, what

will be the total force?

SOLUTIONTOEXAMPLE7

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The force from the magnetic field of line 1 acting

on a differential section of line 2 is:

12212 BLF ×= d I d Where,

φ πρ

µ aB

2

101

I =

By inspection from figure,

x y aa −== φ ρ , Why?!?!


SOLUTIONTOEXAMPLE7

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( ) ( )

( )∫ −=

−=−×=

0210

12

21010212

2

22

L

y

y x z

dz y

I I

dz y

I I

y

I dz I d

aF

aaaF

π

µ

π

µ

π

µ

z dz d aL =2Consider , then:

y y

L I I aF

π

µ

2

210

12 =∴



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Generally, ( )∫ ∫

××=

212

121212

012

4 R

d d I I

aLLF

π

µ

• 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 fieldB1 by Biot-Savart’s

law, then use to findF12 .∫ ×= 12212 BLF d I

MAGNETIC FORCES (Contd)

EXAMPLE8

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EXAMPLE 8

The magnetic flux density in a region of free space

is given byB= −3xax+ 5yay− 2zaz T. Find the

total force on the rectangular loop shown which

lies in the planez= 0 and is bounded byx= 1,x=

3,y= 2, andy= 5, all dimensions in cm.

Try to sketch this!

SOLUTIONTOEXAMPLE8

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87

The figure is as shown.


SOLUTIONTOEXAMPLE8(Cont’d)

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SOLUTION TO EXAMPLE 8 (Contd)

BL F x Id loo∫ =

A I 30=

First, note that in the planez= 0, thezcomponent

of the given field is zero, so will not contribute to the

force. We use:

Which in our case becomes with,

z y x z y x aaaB 253 −+−=and

SOLUTIONTOEXAMPLE8(Cont’d)

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( )

( )

( )

( )∫

∫

∫

∫

+−

++−

++−×

++−×=

=

=

=

=

02.0

05.001.0

01.0

03.0

05.0

05.0

02.003.0

03.0

01.002.0

5330

5330

5330

5330

y x x y

y y x x

y x x y

y y x x

y x xdy

y x xdx

y xdy

y xdx

aaa

aaa

aaa

aaaFSo,

SOLUTION TO EXAMPLE 8 (Contd)

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35BOUNDARYCONDITIONS

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91

3.5BOUNDARY 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 enc I d =•∫ LH

BOUNDARYCONDITIONS(Cont’d)

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92

BOUNDARY CONDITIONS (Contd)

Consider,


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93

A pair of magnetic media separated by a sheet

current densityK. Choose a rectangular Amperian

path of width ∆w and height ∆h centered at the

interface. The current enclosed by the path is:

∫ ∆== w K KdW I enc



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∫ ∫ ∫ ∫ ∫ ∆=•+++=•b

a

c

b

d

c

a

d

w K d d )( 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,


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95

( )221

21

11

2

0

0

2/

0

h H H

dL H dL H d

cb

w H dL H d

ba

N N

h

N N N N N

h

N

c

b

b

a

w

! ! ! !

∆+−=

•+•=⋅

→∆=•=•

→

∫ ∫ ∫

∫ ∫

∆−

∆

∆

aaaaLH

aaLH

For first and second integral,


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98


Second boundary condition can be determined byapplying Gauss’s Law over a small pillbox shaped

Gaussian surface,


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The Gauss’s Law,

Where,

∫ ∫ ∫ ∫ •+•+•=• "#debottomto

d d d d SBSBSBSB

The pillbox is short enough, so the flux out ofthe side is negligible.

0=•∫ SB d



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We have

( )

( ) 02121

=∆−=

−⋅+⋅=⋅∫ ∫ ∫

S B B

dS BdS Bd

N N

N N N N N N

aaaaSB

Since ∆S can be chosen unequal to zero, it follows

that:

21 N N BB =


EXAMPLE 9

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101

9

The magnetic field intensity is given as:

In a medium with µr1=6000 that exist for z

< 0. FindH2 in a medium with µr2=3000 for

z>0.

m A x y x aaaH 3261 ++=


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102

BOUNDARY CONDITIONS (Cont’d)

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103

Recall that, for a conductor-dielectric interface:

0=! $ S N % ρ =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:

0=B

( )

BOUNDARY CONDITIONS (Cont’d)

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104

If medium 2 is superconductor, the equations for

magnetostatic fields become:

0= N B

K Ha

=× 1 N ( )1( )2

The second expression is logical since the magnetic field

lines must form closed loops and cannot suddenly

terminate even on a superconductor.

( )

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CHAPTER 3

END

PRACTICAL APPLICATION

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Loudspeakers

Maglev (Magnetically Levitated

Trains)

LOUDSPEAKERS

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• Paper or plastic coneaffixed 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

MAGLEV

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108

MAGLEV(Cont’d)

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109

• Interaction betweenelectromagnets in the train and

the current carrying coils in the

guide rail provide levitation.

• By sending waves along theguide 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.

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SUMMARY (2)

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

∫ ×

=24 R

dS R

π

aK H

∫ ×

=2

4 R

d& R

π

aJH

Surface current

Volume current

•The magnetic field intensity resulting from an

infinite length line of current is:

φ πρ

aH2

I =

SUMMARY (3)

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and from a current sheet of extent it is:

N aK H ×=2

1 WhereaN 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 ofH is equal to the

net current enclosed by the circulation path

enc I d =•∫ LH

SUMMARY (4)

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113

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

JH =×∇

( )∫ ∫ •×∇=• SHLH d d

• 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 µ =

SUMMARY (5)

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r µ µ µ 0=Material permeability µ can be written as:and the free space permeability is:

m H 7

0 104 −×= π µ

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

∫ •=Φ SB d Since magnetic flux forms closed loops, we have

Gauss’s Law for static magnetic fields:

0=•∫ SB d

SUMMARY (6)

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• The total force vectorF acting on a chargeq movingthrough magnetic and electric fields with velocityu is

given byLorentz Force equation:

( )BuEF

×+= q The forceF12 from a magnetic fieldB1 on a current

carrying lineI2 is:

12212 ∫ ×= BLF d I

SUMMARY (7)

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• The magnetic fields at the boundary between

different materials are given by:

( ) K H H =−× 2121a

Wherea21 is unit vector normal from medium 2

to medium 1, and:

21 N N BB =

VERY IMPORTANT!

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From electrostatics and magnetostatics, we cannow present all four of Maxwell’s equation for

static fields:

enc

enc

I d

d

d Qd

LH

LE

SB

SD

∫ ∫

∫ ∫

=•

=•

=•=•

0

0

JH

E

B

D

=×∇=×∇=•∇=•∇

0

0

& ρ

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