Chap 3 - Magnetostatics.pdf (2435k)

CHAPTER 3

MAGNETOSTATICS

3.1

BIOT-SAVART’S LAW

3.2

AMPERE’S CIRCUITAL LAW

3.3

MAGNETIC FLUX DENSITY

3.4

MAGNETIC FORCES

INTRODUCTION

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.





The principle of magnetism is widely used in

many applications:



Magnetic memory



Motors and generators



Microphones and speakers



Magnetically levitated high-speed

INTRODUCTION (Cont’d)

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.

3.1 BIOT-SAVART’S LAW

Jean Baptiste Biot and Felix Savart arrived a mathematical relation between the field and current.

2 12

12 1

1

4

R

a

L

H







I

d

BIOT-SAVART’S LAW (Cont’d)

To get the total field resulting from a

current, sum the contributions from each

segment by integrating:







2

4

R

Id

_R



a

L

BIOT-SAVART’S LAW (Cont’d)

Due to continuous current distributions:

In terms of distributed current sources, the Biot-Savart’s Law becomes:







2

4

R

Id



a

L

H







4

R

dS



a

K

H



dV

a

J

Line current

Surface current

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



a

L

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





The field will be independent of z and φ and only dependant on ρ.

The term dL is simply and the vector from the source to the test point P is:

z

dza





a

a

a

R











z

R

z

R

Where the magnitude is:

2 2







z

R

And the unit vector:

a





z

a





a



a



a



a





z



dz

Cross product of :

  





a

a

a

a

R

L

dz

z

dz

d









0

0

0

This yields to:







a

H





B

dz

I

Trigonometry from figure,

z







tan

z





cot



Therefore,









_



a

H

cos

₂

cos

₁

4





I

This expression generally applicable for any straight filamentary conductor of finite length.

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:





₉₀

_,





₀

So that,





a

H

4

I



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













,

0

,

0

,

0

,

0

at

at

B

A

The angle become:





₁₈₀

_,





₀

So that,

_I



HOW TO FIND UNIT VECTOR a

?

From previous example, the vector H is in direction of a_φ, where it needs to be determine by simple approach:



a





a

a

a



_l



Where,

l

a



a

EXAMPLE 1

SOLUTION TO EXAMPLE 1

• Side 1 lies on the x-y plane and treated as a straight conductor.

SOLUTION TO EXAMPLE 1 (Cont’d)

This will show how is

applied for any straight, thin, current carrying

conductor.











H cos ₂ cos ₁

4 

 I

From figure, we know that





₉₀



_cos





₀

29

2

SOLUTION TO EXAMPLE 1 (Cont’d)

To determine by simple approach:

_a

x l

a

a



z

a

a



and so that,

y z

x

l

a

a

a

a

a

a

















 





m

A

m

I

a

a

a

H

1

.

59

0

29

2

5

4

10

cos

cos

4

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

Can we use ?











H cos ₂ cos ₁

4 

 I

SOLUTION TO EXAMPLE 2

SOLUTION TO EXAMPLE 2 (Cont’d)

We could find:





a

L

ad

d





a

a

a

R



R



h



a



2 2

a

h





R

2 2

a

h

a

h

R







It leads to:







































4

4

4

a

h

a

h

Iad

R

Id

R

Id

a

a

a

R

L

a

L

H

SOLUTION TO EXAMPLE 2 (Cont’d)

The differential current element will give a field with: 

a

a

a

a



 



a

However, consider the symmetry of the problem:

SOLUTION TO EXAMPLE 2 (Cont’d)

The radial components cancel but the

components adds, so:z

a

















2

0 2

3 2 2

2

4

d

a

h

This can be easily solved to get:





a

h

Ia

a

H

2 2

2

2





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

z

a

I

a

H

2



BIOT-SAVART’S LAW (Cont’d)

• 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.

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.

enc

I

d





• 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

DERIVATION 4

Find

the

magnetic

field

intensity

everywhere resulting from an infinite

length line of current situated on the

DERIVATION 4 (Cont’d)

Select the best Amperian path, where here are two possible

Amperian paths around an infinite length line of current.

DERIVATION 4 (Cont’d)

Using Ampere’s circuital law:

enc

I

d







H

L

We could find:

  





a

L

a

H

d

d

H





I

d

H

I

d









Solving for H_φ:





2

I

H



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



a

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 a

DERIVATION 5 (Cont’d)

Rectangular amperian path of height Δh and width

We have:































d d

c c

b b

a

enc

d

d

d

d

I

d

L

H

L

H

L

H

L

H

L

H

DERIVATION 5 (Cont’d)

From symmetry argument, there’s only H_y component exists. So, H_z will be zero and thus the expression reduces to:



















c b

a

enc

d

d

I

d

L

H

L

H

L

So, we have:

 

w

H

dy

H

dy

H

d

d

d



































2

a

a

a

a

L

H

L

H

L

H

The current enclosed by the path,

w

K

dy

K

KdS

I

x









_

_

0

DERIVATION 5 (Cont’d)

This will give:

enc

I

d







H

L

w

K

w

H







2

2

K

H



a

K

H





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

A

z

SOLUTION TO EXAMPLE 3

Use previous expression, that is:

N

a

K

H





2

1

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

a

y N

a

a



K



6

a

m

A

x y

z

a

a

a

H

6

3

2

Consider the infinite length

cylindrical conductor

carrying a radially

dependent current

Find

H

everywhere.

What components of H will be present?

Finding the field at

some point P, the

field has both

and components. 

a



a

The field from the second line current results in a

cancellation of the components



a

To calculate H everywhere, two amperian paths are required:

Path #1 is for

Path #2 is for

a





a





Evaluating the left side of Ampere’s law:

 

 





d





H

H

d

2









_



H

L

a

a

This is true for both amperian path. The current enclosed for the path #1:

2

2 2 0



















K

d

d

K

d

d

K

d

I













 





a

a

S

K

Solving to get H_φ: 3 2 0



H  _Or H  a_

3

2 0

K

 for





_a

The current enclosed for the path #2:

3 2 ₀ 3

0 2 0 2 0 a K d d K d I a











 

Solving to get H_φ:

  a H 3 3 0a K

 for





_a

EXAMPLE 5

Even current

distributions are assumed in the inner and outer conductor.

Consider four amperian paths.

It will be four amperian paths:

   

Therefore, the magnetic field intensity, H will

a





b

a







c

b











c

As previous example, only H_φ component is present, and we have the left side of ampere’s circuital law:

 

 





d





H

H

d

2

2

0









_



H

L

a

a

 For the path #1:







K d

S

I

We need to find current density, K for inner

conductor because the problem assumes an event current distribution (ρ<a is a solid volume where current distributed uniformly).

z

dS

I

a

K



Where,

2 2

,

S

d

d

a

d

d

dS

a





















_{ }

So, z z

a

I

dS

I

a

a

K







We therefore have:

2 2 2 0 0 2

a

I

d

d

a

I

d

I























 

a

a

S

K

Equating both sides to get:

2 2

2

2

2

_a

I

a

I

H



















a

 For the path #2:

The current enclosed is just I, Therefore:

I

I



I

I

H

d











H

L

2



I

_

_

_

 For the path #3:

SOLUTION TO EXAMPLE 5 (Cont’d)

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





_

_





b

c

I

dS

I

a

a

K











We therefore have (for AP#3):









 













But, the total current enclosed is:

d I I_enc     







SOLUTION TO EXAMPLE 5 (Cont’d)

So we can solve for path #3:

2 2 2 2

2

b

c

c

I

I

H

d















H

L























2

_c

_b

c

I

H





 for

b







c

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

0





H





c

SOLUTION TO EXAMPLE 5 (Cont’d)

Summarize the results to have:

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

J

H







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

some particular point.

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





H



d

L



J



d

S

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









H



d

L







H



d

S

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



H

B

Where μ is the permeability with:

m

H

7

0

4

10



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:







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

MAGNETIC FLUX DENSITY (Cont’d)

MAGNETIC FLUX DENSITY (Cont’d)

The net magnetic flux passing through a

gaussian surface must be zero, to get Gauss’s Law for magnetic fields:

0







B d

S

By applying divergence theorem, the point form of Gauss’s Law for static magnetic fields:

0





EXAMPLE 6

Find the flux crossing the portion of the

plane

φ

=

π

/4 defined by 0.01m <

ρ

< 0.05m

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

filament of 2.5A is along the z axis in the

a

direction.

SOLUTION TO EXAMPLE 6

The relation between B and H is:









H

a

B

2

0 0

I





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









B d

S

So,

d

S



d



dz

a

I

dz

d

I

d

a

a

S

B

05

.

0

2

2

















 











3.4 MAGNETIC FORCES

Upon application of a magnetic field, the wire is

MAGNETIC FORCES (Cont’d)

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

B

u

F

_m



q



By the definition of electric field intensity, the electric force F_e acting on a charge q within an electric field is:

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



E

u

B



F



q





MAGNETIC FORCES (Cont’d)

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

a

MAGNETIC FORCES (Cont’d)

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:

B

u

F



dQ



d

dL

u



B

L

F







d

Id

So,

B

L

F



d



dt

dQ

d

Since corresponds to the current I in the line,

dt

dQ

MAGNETIC FORCES (Cont’d)

We can find the force from a collection of current elements

1 2

2

12





L



B

Consider a line of current in +a_z direction on the z axis. For current element a,

z a a

Idz

Id

L



a

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 a_R in same

EXAMPLE 7

If there is a field from a

second line of current

The force from the magnetic field of line 1 acting on a differential section of line 2 is:

1

2

2

12

L

B

F



I

d



d

Where,







a

B

2

1 0 1

I



By inspection from figure,





 

 



_













2

2

2

dz

y

I

I

dz

y

I

I

y

I

dz

I

d

a

F

a

a

a

F













dz

d

L



a

Consider , then:

y

y

L

I

I

a

F





2





Generally,













2 12

12 1

2 1

2 0 12

4

_R

d

d

I

I

L

L

a

F





• 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 B by Biot-Savart’s

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.

The figure is as shown.

SOLUTION TO EXAMPLE 8 (Cont’d)

B

L

F

Id

x

loop





A

I



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,

z y

x

y

z

x

a

a

a

B





3



5



2















































5

3

30

5

3

30

5

3

30

y

x

x

dx

y

x

dy

y

x

dx

a

a

a

a

a

a

a

a

a

F

Simplifying these becomes:













N

dy

dx

dy

dx

a

a

a

a

a

F

027

.

0

150

.

0

081

.

0

06

.

0

)

01

.

0

)(

3

(

30

)

05

.

0

)(

5

(

30

)

03

.

0

)(

3

(

30

)

02

.

0

)(

5

(

30

































mN

z

a

F





36



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







B d

S

enc

I

d





BOUNDARY CONDITIONS (Cont’d)

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:









KdW

K

w

I

_enc

   



















a

c

b

d

c

a

d

w

K

d

d

L

(

H

L

)

H

BOUNDARY CONDITIONS (Cont’d)

The sheet current is heading into the page and

2 1 1 1 2 0 0 2 / 0 dL H dL H d c b w H dL H d b a h N N N N N h N c b b a w T T T T            











For first and second integral,





h

H

H

dL

H

dL

H

d

a

d

w

H

dL

H

d

d

c











































a

a

a

a

L

H

a

a

L

H

BOUNDARY CONDITIONS (Cont’d)

K

H

H





2 1

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



H



H





K



₁

₂

21

a

In more general case,

BOUNDARY CONDITIONS (Cont’d)

BOUNDARY CONDITIONS (Cont’d)

The Gauss’s Law,

Where,























top

d

d

d

d

S

B

S

B

S

B

S

B

The pillbox is short enough, so the flux out of

0







B d

S

We have









0

2 1 2 1



























S

B

B

dS

B

dS

B

d

a

a

a

a

S

B

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

2

1

N

N

B

B



EXAMPLE 9

The magnetic field intensity is given as:

In a medium with µ

=6000 that exist for z

< 0. Find

H

in a medium with µ

=3000 for

z>0.

m

A

x

y

x

a

a

a

Recall that, for a conductor-dielectric interface:

0



T

E

D

_N





_S

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

If medium 2 is superconductor, the equations for magnetostatic fields become:

0



N

B

K

H

a

_N



₁



 

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

PRACTICAL APPLICATION



Loudspeakers



Maglev (Magnetically Levitated

• 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

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

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:

12 12

12 a

R  R

2 12 12 1 1

4

R

a

L

H







I

d

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

SUMMARY (2)







2

4

R

dS



a

K

H







4

R

dv



a

J

H

Surface current

Volume current

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

SUMMARY (3)

and from a current sheet of extent it is:

N

a

K

H





2

1

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

I

d





SUMMARY (4)

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

J

H















H



d

L







H



d

S

• A closed line integral is related to surface integral by

Stoke’s Theorem:

r









SUMMARY (5)

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

m

H

7

0

4

10











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









B d

S

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

0





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:



E

u

B



F



q





The force F₁₂ from a magnetic field B₁ on a current carrying line I₂ is:

1 2

2

12





L



B

SUMMARY (7)

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



H



H





K



21

a

Where a₂₁ is unit vector normal from medium 2 to medium 1, and:

2

1 N

N

B

VERY IMPORTANT!