Magnetostatic Fields

Dr. Talal Skaik Islamic University of Gaza – Palestine

2012

2

Introduction In chapters 4 to 6, static electric fields characterized by E or D

(D=εE) were discussed.

This chapter considers static magnetic fields, characterised by H

or B (B=μH).

As we have noticed, a distribution of static or stationary charges

produces static electric field.

If the charges move at a constant rate (direct current- DC), a

static magnetic field is produced (magnetostatic field).

Static magnetic field are also produced by stationary permanent

magnets.

3

Magnet and Magnetic Field

A compass needle is deflected by the direct current flowing in a conductor

Magnetic Field Around Current Carrying Wires

The iron filings form circles around the wire along the magnetic field

6

Electrostatic Fields have dual equations for magnetostatic fields

Biot-Savart’s law states that the magnetic field intensity dH produced at a point P by a differential current element Idl is proportional to the product Idl and the sine of the angle α 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 and its direction can be obtained by right handed screw rule.

Biot-Savart’s Law

7

2

2

2

sin

sin = , 1/ 4

sin=

4

I dldH

R

kI dlor dH k

R

I dldH

R

8

2

2 3

R

sin=

4

l l RH

4 4

where =|R| and a =R/

R

I dldH

R

I d a I dd

R R

R R

The direction of dH can be determined by the right-hand rule or right-handed screw rule.

The direction of magnetic field intensity H (or current I) can be represented by a small circle with a dot or cross sign depending on whether H (or I) is out of, or into the page.

9

Biot-Savart’s Law

For different current distributions:

10

2

2

2

l (line current)

4

K S (surface current) (K: surface current density)

4

J (volume current) (J: volume current density)

4

R

L

R

S

R

v

I d aH

R

d aH

R

dv aH

R

Biot-Savart’s Law

Current distributions: (a) line current, (b) surface current, (c) volume current.

Consider conductor of finite length AB, carrying current from point A to point B.

Magnetic Field of straight Conductor

11

3

z

3/22 2

2

2 2

3

Consider the contribution H at

l Rdue to l at (0,0,z): H ,

4

But l and R= a a , ,

l R a , Hence H= a4

Letting z= cot , cosec ,

cosecH

4

z

d P

I dd d

R

d dz a z so

I dzd dz

z

dz d

I d

2 2

3

1 1

2 1

sincosec 4

H= cos cos4

Ia a d

Ior a

tan /

cot

z

z

2 21+cot x=cosec x

2 1

o o

1 2

H= cos cos4

When the conductor is semi-infinite, so that point A is now at O(0,0,0)

while B is at (0,0, ), =90 , =0 :

Ia

o o

1 2

H= 4

When the conductor is of infinite length, point A is at (0,0, ) while

B is at (0,0, ) , =180 , =0 :

H= 2

A simple appr

Ia

Ia

oach to determine :

where is is a unit vector along the line current, and is a unit vector

along the perpendicular line from the li

l

l

a

a a a

a a

ne current to the field point. 12

2 1H= cos cos4

Ia

13 α1:outside, α2: inside

The conducting triangular loop in the figure carries a current of 10 A. Find H at (0,0,5) due to side 1 of the loop. 14

Example 7.1

(a) conducting triangular loop, (b) side 1 of the loop.

2 1

1 2

o

1 2

Applying H= cos cos , that is valid for any straight, 4

thin, current carrying conductor. , , and are found as follows:

2cos cos90 0, cos , =5

29

, but , and l l x

Ia

a

a a a a a

1 2 1

1

1

, Hence

H cos cos4

10 2H 0 ( )

4 (5) 29

H 59.1 mA/m

z

x z y

y

y

a a

so a a a a

Ia

a

a

15

Example 7.1 Solution

16

Example 7.2 Find H at (-3,4,0) due to the current filament shown in the figure.

(a) current filament along semi-infinite x- and z-axes, aℓ and a for H2 only; (b) determining a for H2.

1 2 1

2

2 2 1

1/2

o o

1 2

Let H=H +H at P(-3,4,0), where H is due

to current filament along x-axis, and H is due to current filament

along z-axis.

H cos cos4

At P(-3,4,0), =(9+16) 5,

=90 , =0

l

Ia

a a a

,

but ,

3 4and

5 5

3 4 4 3

5 5 5 5

l z

x y

z x y x y

a a

a a a

a a a a a a

17

Example 7.2 Solution

2

2

2

or alternatively, from figure b,

4 3sin cos

5 5

4 33Thus, H 1 0

4 (5) 5

H 38.2 28.65 mA/m

in this case is the negative of the regular of

3cylindrical coordinates: H 1 0 (

4 (5)

x y x y

x y

x y

a a a a a

a a

a a

a a

a

1

2 1

1

1 2

) 47.75 mA/m

To find H at P:

=4, 0 , cos 3 / 5, and . Hence,

3 3H 1 23.88 mA/m

4 (4) 5

Thus H=H +H 38.2 28.65 23.88 mA/m

or H= 47.75

o

l x y z

z z

x y z

a

a a a a a a

a a

a a a

a

23.88 mA/mza 18

A circular loop located at X2+y2=9, z=0 carries a direct current of 10 A along aФ. Determine H at (0,0,4) and (0,0,-4).

19

Example 7.3

(a) circular current loop, (b) flux lines due to the current loop.

3

The magnetic field intensity dH at point P(0,0,h)

contributed by current element Idl is :

l RH

4

l , R=(0,0, ) ( , ,0)

R , and

l R 0 0

z

z

I dd

R

where d d a h x y

a ha

a a a

d d

2

2

3/22 2

0

Hence,

H= 4

By symmetry, the contributions along add up to zero H 0

z

z z z

h d a d a

h

Id h d a d a dH a dH a

h

a

20

Example 7.3 - solution

2 2 2

3/2 3/22 2 2 2

0

2

3/22 2

2

3/2

2 Thus, H=

4 4

or H=

2

(a) Substituting I=10 A, =3, h=4 gives

10(3) H 0,0,4 = 0.36 A/m

2 9 16

(b) replacing by

z zz z

z

zz

I d a I adH a

h h

Ia

h

aa

h

2

- ,

l R

(z-component remains the same)

Hence, H 0,0,4 H 0,0, 4 0.36 A/m

z

z

h

d h d a d a

a

21

Example 7.3 - solution

AMPERE'S CIRCUIT LAW - MAXWELL'S EQUATION

22

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

enc

enc

enc

enc

In other words, the circulation of H equals ; that is,

H l= (integral form of 3rd Maxwell's equation)

Apply Stoke's theorem,

H l H S

But, J S

H J (differential

L S

S

I

d I

I d d

I d

form of third Maxwell's equation)

AMPERE'S CIRCUIT LAW - MAXWELL'S EQUATION

23

enc

2

enc enc

H l apply to Amperian path ( ).

current enclosed by amperian path.

is found if J(A/m ) is known: J S

or if K(A/m) is known for surface current density.

L

S

d L

I

I I d

encH l=d I

Consider infinite filamentary current I along the z-axis. To determine H at point P, allow Amperian path passing through P, such that H is constant provided ρ is constant.

APPLICATIONS OF AMPER'S LAW

24

A. Infinite Line Current

2

0

The whole current is enclosed by the path,

according to Amper's law:

a a

= (2 )

or H= a2

I

I H d

I H d H

I

encH l=d I

25

B. Infinite Sheet of Current

y y

enc

Consider an infinite current sheet in the z=0 plane with uniform

current desnity K= a A/m. Applying Ampere's law to the

rectangular closed 1-2-3-4-1 path gives: H l

Consid

y

K

d I K b

er the sheet as comprising of filaments

Consider H above and below the sheet due to a pair of

filamentary currents.

d

26

0

0

2 3 4 1

1 2 3 4

The resultant H has only an x-component.

H on one side is the negative of that on the other side.

0H=

0

Evaluating the line integral of H along the closed path:

H. l

x

x

d

H a z

H a z

d

enc

0 0 0 0

n n

H. l=

10( )+( )( ) 0( ) ( )=2

2

1 0

2H=

10

2

In general, for an infinite sheet of current density K A/m,

1H= K a where a is a unit normal vector dire

2

y

y y

y x

y x

d I K b

a H b a H b H b K b H K

K a z

K a z

cted from

the current sheet to the point of interest

27

Magnetic field of Infinite Sheet of Current

n

1H= K a

2

28

Infinitely Long Coaxial Transmission Line

Consider infinitely long coaxial transmission line of two concentric cylinders

The inner conductor has radius and carries current , the outer conductor has

inner radius b and thickness t and carries

a I

return current .

To determine H everywhere, apply Ampere's law in four possible regions:

0 , a , ,

I

a b b b t b t

29

Infinitely Long Coaxial Transmission Line

30

1

1

enc

2

2

2

enc 2 2

0 0

For region 0 : apply Ampere's law to path L

H l J S

Since the current is uniformly distributed over

the cross section, J= , S=

J S

L

z z

a

d I d

Ia d d d a

a

I II d d d

a a

1

2

2

2

2

2 2

2

enc 2

2 or 2

For region a : Apply Ampere's law to path L

H l (since the whole current is enclosed by L )

2 or (same as infinite straight2

L

L

I

a

I IH dl H H

a a

b

d I I

IH I H

filamentary current)

31

3

3

enc

enc 2 2

2 2 2

enc 22 20

2 2

2

For region :

Apply Ampere's law to path L

H l 2

where

J S, J=

, 12

12 2

L

z

b

b b t

d H I

II I d a

b t b

I bThus I I d d I

t btb t b

I bH

t bt

4

4For region : Apply Ampere's law to path L

H. l 0 or 0L

b t

d I I H

2

2 2

2

a 02

a aPutting all equations 2H=

together:

1 a 2 2

0

Ia

a

Ib

I bH b b t

t bt

b t

Infinitely Long Coaxial Transmission Line

32

Infinitely Long Coaxial Transmission Line

33

Planes z=0 and z=4 carry current K= -10 ax A/m and K=10 ax A/m, respectively. Determine H at (a) (1,1,1) (b) (0,-3,10)

34

Example 7.5

0 4 0 4

0 n

1 n

Let H=H +H , where H +H are

the contributions due to current

sheets z=0 and z=4,

(a) At (1,1,1) , which is between the plates (0 ( 1) 4),

H (1/ 2)K a (1/ 2)( 10a ) a 5a A/m

H (1/ 2)K a (1/ 2)(10a ) (

x z y

x

z

0 n

1 n

a ) 5a A/m

Hence, H=10a A/m

(b) At (0,-3,10) , which is above the sheets ( 10 4 0),

H (1/ 2)K a (1/ 2)( 10a ) a 5a A/m

H (1/ 2)K a (1/ 2)(10a ) a 5a A/m

Hence, H= 0 A/m

z y

y

x z y

x z y

z

35

Example 7.5

A toroid whose dimensions are shown in the figure has N turns and carries current I. Determine H inside and outside the toroid. 36

Example 7.6

37

enc

0 0

0

The net current enclosed by the Amperian path is . Hence,

H l= (2 )

or H= , for 2

where is is the mean radius of the toroid.

Outside the toroid, the current enclos

NI

d I H NI

NIa a

ed by an Amperian path is

- 0 and Hence =0NI NI H

Example 7.6

38

Magnetic Field of a Toroid

•Electric flux density and Electric field intensity are related by D=ε0 E

in free space.

•Similarly, the magnetic flux density B is related to the magnetic

field intensity H by:

•Where μ0 is known as the permeability of free space.

•The magnetic flux through a surface is given by

•Where Ψ is in webers (Wb), B is in (Wb/m2) or teslas (T).

Magnetic Flux Density

39

0B H

7

0 4 10 H/m

= B SS

d

•Magnetic flux line is a path to which B is tangential at every point on the line. •Each flux line is closed and has no beginning or end.

Magnetic Flux Lines

40

•Unlike electric flux lines, magnetic flux lines always close upon themselves. •This is because it is not possible to have isolated magnetic poles (or magnetic charges).

41

•In an electrostatic field the flux crossing a closed surface is the same as the charge enclosed.

•So it is possible to have an isolated electric charge and the flux lines produced by it need not be closed.

D SS

d Q

B S 0S

d

42

It is not possible to isolate the north and south poles of a magnet.

43

Broken Magnet

Gauss’s Law for magnetostatic fields

44

The total flux through a closed surface in a magnetic field is zero. Applying divergence theorem, Or This is the Maxwell’s fourth equation. It states that magnetostatic fields have no sources or sinks.

B S B 0S v

d dv

B S 0d

B 0

Maxwell’s Equations for Static Fields

45

*In electrostatics electric field intensity and potential are related by:

*Similar to this we can relate magnetic field intensity with two magnetic potentials: •Magnetic scalar potential (Vm) •Magnetic vector potential (A)

*Magnetic scalar potential Vm is related to H by the relation:

Magnetic Scalar and Vector Potentials

46

E V

m

2

J= H= ( ) 0 (since for any scalar, ( ) 0)

so the magnetic scalar potential V is only defined in the region where J=0.

satisfies Laplace's equation 0 (J=0)

m

m m

V V

V V

H if J=0mV

47

Since for a magnetostatic field B=0

For any vector ( A)=0

we can define A such that

B= A

In many EM problems it is more convenient to first find A and then

find B from it.

magnetic vector potential

A

0

0

0

l for line current

4

for surface current4

J for volume current

4

L

S

v

Id

R

KdSA

R

dvA

R

Magnetic Vector Potential

Magnetic flux from vector potential

48

The magnetic fluc through a given area can be found from

= B S

Applying Stoke's theorem, we obtain,

= B S ( A) S A l

A l

S

S S L

L

d

d d d

d

49

Example 7.7 Given the magnetic vector potential A=-ρ2/4 az Wb/m, calculate the total magnetic flux crossing the surface Ф=π/2, 1≤ρ≤2 m, 0≤z≤5 m.

25 2

2

10 1

Method 1

B A= a a , S= a2

Hence,

1 1 15B S= (5)

2 4 4

3.75 Wb

z

z

Ad d dz

d d dz

Example 7.7 Solution

50

1 2 3 4

1 3

5 02 2

2 4

0 5

Method 2

A l=

where is the path bounding surface .

Since A has only z-component,

0

11 (2)

4

1 15(1 4)(5) 3.75 Wb

4 4

L

d

L S

dz dz

Example 7.7 Solution Continued A=-ρ2/4 az