1

MAGNETOSTATIC FIELD(STEADY MAGNETIC)

CHAPTER 7

7.1 INTRODUCTION - SOURCE OF MAGNETOSTATIC FIELD

7.2 ELECTRIC CURRENT CONFIGURATIONS

7.3 BIOT SAVART LAW

7.4 AMPERE’S CIRCUITAL LAW

7.5 CURL (IKAL)

7.6 STOKE’S THEOREM

7.7 MAGNETIC FLUX DENSITY

7.8 MAXWELL’S EQUATIONS

7.9 VECTOR MAGNETIC POTENTIAL

7.1 INTRODUCTION - SOURCE OF MAGNETOSTATIC FIELD

Originate from:

• constant current

• permanent magnet

• electric field changing linearly with time

Analogous between electrostatic and magnetostatic fields

Energy density

Scalar V withPotential

Related Maxwell equations

Factor

Field

Steady currentStatic chargeSource

MagnetostaticElectrostaticAttribute

D and E B and H

vD ∇

0∇ E JH

B

∇

0∇

VE 2

2

1Ewe 2

2

1Hwm

withVector A

AB

Two important laws – for solving magnetostatic field

• Biot Savart Law – general case

• Ampere’s Circuital Law – cases of symmetrical current distributions

7.2 ELECTRIC CURRENT CONFIGURATIONS

Three basic current configurations or distributions:

• Filamentary/Line current,

• Surface current,

• Volume current,

dlI

dsJ s

dvJ

(A.m) dvJdsJdlI s

Can be summarized:

7.3 BIOT SAVART LAW

A differential magnetic field strength, results from a differential current element, . The field varies inversely with the distance squared. The direction is given by cross product of

dHdlI

RadlI ˆ and

)Am( 4

ˆ x 1-

212

12112

R

adlIdH

R

Consider the diagram as shown:

)Am( 4

ˆ x 1-

22

l

R

R

adlIH

Total magnetic field can be obtained by integrating:

Similarly for surface current and volume current elements the magnetic field intensities can be written as:

)(Am 4

ˆ x

)(Am 4

ˆ x

1-22

1-212

12

121

s

R

Rs

R

dsaJH

R

dsaJdH

)(Am 4

ˆ x

)(Am 4

ˆ x

1-22

1-212

112

12

v

R

R

R

dvaJH

R

dvaJdH

Ex. 7.1: For a filamentary current distribution of finite length and along the z axis, find (a) and (b) when the current extends from - to +.

H H

Solution:

} 'dl zdz

1

2

z’-z

x rc y

(r c, ,z)

dH R

rc

(0,0,z’)

I

a

z z’

b

z

2/322

2

])'([4

)]'(ˆˆ[ x )'ˆ(= @

),,,',','(4

),,,',','(ˆ x '),,(

zzr

zzzrrdzzIdH

zrzrR

zrzradlIzrH

c

cc

b

a cc

ccRc

b

acc

b

a c

c

c

c

zzr

zz

r

I

zzr

dzIrH

zzr

dzrIdH

2/122

2/322

2/322

])'([

'

4

ˆ

])'([

'

4

ˆ=

])'([4

'ˆ=

ˆ x ˆ crz 0ˆ x ˆ zz=> ;

2/12222/322 ][ :Table Using

xcc

x

xc

dx

2/1222/122 ])([])([4

ˆ

zbr

za

zbr

zb

r

IH

ccc

Hence:

In terms of 1 and 2 :

A/m sinsin4

ˆ12

cr

IH

(r’, ’, z’)

(r, , z)

Hd

'ld

z

(b) When a = - and b = +, we see that 1 = /2, and 2 = /2

)(Am 2

ˆ1-

cr

IH

I

z Filamentary current

Flux H

H

Rl aa ˆˆˆ :r Unit vecto

The flux of in the direction and its density decrease with rc as

shown in the diagram.

H

Ex. 7.2: Find the expression for the field along the axis of the circular current loop carrying a current I.

H

d H

d a R

rddl ˆ

( 0 , 0 , z )

z R

( r c , , z )

C u r r e n t l o o p , I

I

Solution:

2/322 ][4

)ˆˆ( x )ˆ(=

zr

rrzzdrIdH c

Using Biot Savart Law

and

zr ˆ)ˆ( x ˆ

sinˆcosˆˆ x ˆ yxrz rz ˆˆ x ˆ

2/322

2

)(4z=

zr

dIrdH

rwhere the component was omitted due to symmetry

)(Am )(2

ˆ

; )(2

ˆ

)(4

ˆ

1-2/322

2

2/322

2

2

02/322

22

0

za

Iaz

ar zr

Irz

dzr

IrzdHH

Hence:

Ex. 7.3: Find the field along the axis of a s solenoid closely wound with a filamentary current carrying conductor as shown in Fig. 7.3.

H

surface current

flux

Fig. 7.3: (a) Closely wound solenoid (b) Cross section (c) surface current, NI (A).

• Total surface current = NI Ampere

• Surface current density, Js = NI / l Am-1

• View the dz length as a thin current loop that

carries a current of Jsdz = (NI / l )dz

Solution:

)(Am )(2

ˆ 1-2/322

2

za

IazH

surface current

Solution from Ex. 7.2:

2/322

2

)(2ˆ

za

adzNI

zHd

Therefore at the center of the solenoid:Hd

2/122

2/

2/-2/322

2

)4(ˆ

)(2ˆ ∫

a

NIz

za

dzNIazH

Hence:

If >> a :

a

at the center of the solenoidsJz

NIzH ˆˆ

(Am-1)

Field at one end of the solenoid is obtained by integrating from 0 to :

2/122 )(2ˆ

a

NIzH

If >> a :

2ˆ

2ˆ sJ

zNI

zH

which is one half the value at the center.

x

z

y

3A

3A

to ∞

(-3,4,0)

to ∞

Ex. 7.4: Find at point (-3,4,0) due to the filamentary current as shown in the Fig. below.

H

Solution:

zx HHH

Total magnetic field intensity is given by :

yxyxz ˆ5

3ˆ

5

4ˆ

5

4ˆ

5

3-ˆ-

12 sinsin4

ˆ

c

z r

IH

5

ˆ4ˆ3

169

ˆ4ˆ3ˆ

yxyx

R

RaR

Unit vector:

To find :zH

Rl aa ˆˆˆ

yx

yx

r

IH

cz

ˆ65.28ˆ2.38

1054

3ˆ

5

3ˆ

5

4

sinsin4

ˆ12

Hence:

1 = /2 ; 2 = 0

x

zto ∞

(-3,4,0)

y

3A

Ra

-3

4

To find :xH

Unit vector:

12 sinsin4

ˆ

c

x r

IH

Rl aa ˆˆˆ

zyx ˆˆˆ

x

z

to ∞

2 = /2

(-3,4,0)

y

Ra

3 A

-3

sin 1 = -3/5

yy

R

RaR ˆ

16

ˆ4ˆ

z

z

r

IH

cx

ˆ88.23

5

3-1

44

3ˆ

sinsin4

ˆ12

Hence:

A/m ˆ88.23ˆ65.28ˆ2.38 zyxHHH zx Hence:

7.4 AMPERE’S CIRCUITAL LAW

• Solving magnetostaic problems for cases of symmetrical current distributions.

The line integral of the tangential component of the magnetic field strength around a closed path is equal to the current enclosed by the path :

Definition:

enIdlH

Graphical display for Ampere’s Circuital Law interpretation of Ien

Path (loop) (a) and (b) enclose the total current I , path c encloses only part of the current I and path d encloses zero current.

I

(a) (b) (c) (d)

en

l

IdlH

I I I 0

Ex. 7.5: Using Ampere’s circuital law, find field for the filamentary current I of infinite length as shown in Fig. 7.6.

H

I x

Amperian loop

rddl ˆ

H

d

r

Filamentary current of infinite length

z

dl} y

Solution:

Construct a closed concentric loop as shown in Fig. 7.6a.

Fig. 7.6a

z

y

x

to +

to -

Fig. 7.6

I

(A/m) ˆ2

)2(ˆˆ2

0

r

IH

IrHIdrHIrdHIldHll

enc

(similar to Ex. 7.1(b) using Biot Savart)

Ex. 7.5: Find inside and outside an infinite length conductor of infinite cross section that carries a current I A uniformly distributed over its cross section and then plot its magnitude.

H

C2

I a

xC1

r

yconductor

Amperian path

Solution:

For r a (C2) :

mAr

IH

IrH

IldH enc

c

/ ˆ2

)2(2

H(a)= I/2πa

a

H1

H2

H(r)

r

mAa

IrH

a

rIrH

IldH enc

c

/ ˆ2

)2(

2

2

2

1

For r ≤ a (C1) :

Ex. 7.6: Find field above and below a surface current distribution of infinite extent with a surface current density

H.Am ˆ -1yJJ ys

Solution:

lJIldHldHldHldHldH yen

l

1

2

2

'2

'2

'1

'1

1

x

y

z

.Am ˆ -1yJJ ys

1

1’

2

2’

3

3’

Amperian path 1-1’-2’-2-1

1

2

Graphical display for finding and using Ampere’s circuital law:H

x

12

z

2dH

1dH

1dH

2dH

rdH

rdH

Filamentary current

yx

z

Surface current

From the construction, we can see that above and below the surface current will be in the and directions, respectively.

Hx x

lJdxxxHdxxxH yxx ˆ)ˆ(ˆˆ2

'2

2

'1

1

1

x

y

z

.Am ˆ -1yJJ ys

1

1’

2

2’

3

3’

Amperian path 1-1’-2’-2-1

1

2

1

2

'2

'1

0 and

where

since is perpendicular to dlH

lJlHlH yxx 21

Hence:

Similarly if we takes on the path 3-3'-2'-2-3, the equation becomes:

lJlHlH yxx 23

xxx HHH 31

Therefore:

HAnd we deduce that above and below the surface current are equal, its becomes:

lJlHlH yxx

0z ; 2

1

0z ; 2

1

yx

yx

JH

JH

)ˆ( 2

1

ˆ 2

1

xJH

xJH

y

y

In vector form:

naJH ˆ2

1

yx

naz ˆ

It can be shown for two parallel plate with separation h, carrying equal current density flowing in opposite direction the field is given by: H

) 0 z andh z ( ; 0

)h z0 ( ; ˆ

naJH

xxx

h)z ( ; 0 z

0

h

yx

naz ˆ

ˆnaJH

) 0 z andh z ( ; 0

)h z0 ( ; ˆ

naJH

) 0 (z ; 0

7.5 CURL (IKAL)

The curl of a vector field, is another vector field.H

y

H

x

Hz

x

H

z

Hy

z

H

y

HxH xyzxyz

∂

∂-

∂

∂ˆ

∂

∂-

∂

∂ˆ

∂

∂-

∂

∂ˆ∇

For example in Cartesian coordinate, combining the three components, curl can be written as:H

zyx HHHzyx

zyx

H∂

∂

∂

∂

∂

∂ˆˆˆ

∇

And can be simplified as:

z

HrH

rrr

H

z

Hr

z

HH

rH rzrz ˆ

∂

∂-

∂

∂1ˆ∂

∂-

∂

∂ˆ

∂

∂-

∂

∂1∇

ˆ∂

∂-

∂

∂1ˆ∂

∂-

∂

∂

sin

11

ˆ∂

∂-

∂

sin∂

sin

1∇

rr H

r

rH

rr

rHH

r

rHH

rH

Expression for curl in cyclindrical and spherical coordinates:

cyclindrical

spherical

7.5.1 RELATIONSHIP OF ANDH J

JH ∇

Meaning that if is known throughout a region, then H

will produce for that region.J

JH ∇

Ex. 7.7: Find for given field as the following.H x H

cr

IH

2ˆ(a) for a filamentary current

22ˆ

a

IrH c

(b) in an infinite current carrying

conductor with radius a meter

2ˆ sJxH (c) for infinite sheet of uniformly

surface current Js

(d) in outer conductor of coaxial cable

22

22

2ˆ

bc

rc

r

IH c

c

Solution:

02

ˆˆ x

I

rr

zHr

rr

zH

ccc

cc

cr

IH

2ˆ(a) 0 ,

2ˆˆ

zr

c

HHr

IHH

c =>

Cyclindrical coordinate

z

HrH

rrr

H

z

Hr

z

HH

rH rzrz ˆ

∂

∂-

∂

∂1ˆ∂

∂-

∂

∂ˆ

∂

∂-

∂

∂1∇

= 0 = 0 = 0

Hence:

22ˆ

a

IrH c

Solution:

(b)

)(Am ˆˆ

2

2ˆ

2∂

∂ˆ x

2-2

22

Jza

Iz

a

Ir

r

z

a

Irr

rr

zH c

c

cc

cc

z

HrH

rrr

H

z

Hr

z

HH

rH rzrz ˆ

∂

∂-

∂

∂1ˆ∂

∂-

∂

∂ˆ

∂

∂-

∂

∂1∇

= 0 = 0 = 0

Cyclindrical coordinate

Hence:

Solution:

(c)

Cartesian coordinate

Hence:

2ˆ sJxH

0 x H

y

H

x

Hz

x

H

z

Hy

z

H

y

HxH xyzxyz

∂

∂-

∂

∂ˆ

∂

∂-

∂

∂ˆ

∂

∂-

∂

∂ˆ∇

= 0= 0 = 0

because Hx = constant and Hy = Hz = 0.

Solution:

(d)

Cyclindrical coordinate

Hence:

22

22

2ˆ

bc

rc

r

IH c

c

z

HrH

rrr

H

z

Hr

z

HH

rH rzrz ˆ

∂

∂-

∂

∂1ˆ∂

∂-

∂

∂ˆ

∂

∂-

∂

∂1∇

= 0 = 0 = 0

)(Am )(

ˆ

2

2

ˆ

2∂

∂ˆ x

2-22

22

22

22

Jbc

Iz

bc

rI

r

z

bc

rc

r

Ir

rr

zH

c

c

c

cc

cc

7.6 STOKE’S THEOREM

Stoke’s theorem states that the integral of the tangential component of a vector field around l is equal to the integral of the normal component of curl over S.H

H

sdHldHsl

In other word Stokes’s theorem relates closed loop line integral

to the surface integral

ldH sdH

surface S

Hna

s

unit vector normal to s

Path l

Consider an open surface S whose boundary is a closed surface l

It can be shown as follow:

naHs

dlHˆ∇∫

dls

Hna

∫ dlH naH ˆ ∇ s sH ∇

≈∫∑lk

k

m

k

dlH 1

k

m

k

sH

∇∑1

dsHdlHSl

∫ ∇

From the diagram it can be seen that the total integral of the surface enclosed by the loop inside the open surface S is zero since the adjacent loop is in the opposite direction. Therefore the total integral on the left side equation is the perimeter of the open surface S.

If therefore

s

0Δ →ks ∞=m

Hence:

surface S

Hna

Path l

s k

dlk

where loop l is the path that enclosed surface S and this equation is called Stoke’s Theorem.

Ex. 7.8: was found in an infinite conductor of

radius a meter. Evaluate both side of Stoke’s theorem to find the current flow in the conductor.

)/(2

ˆ2

mAa

IrH

II

rdrdza

Izd

I

rdrdza

Irad

a

Ir

sdHldH

a

arl

sl

ˆˆ2

ˆ2

ˆˆ2

ˆ

)(

0

2

02

2

0

22

Solution:

222ˆ

2

2ˆ

2∂

∂ˆ x

a

Iz

a

Ir

r

z

a

Irr

rr

zH c

c

cc

cc

7.7 MAGNETIC FLUX DENSITY

Magnetic field intensity : Teslas HB o )/( 2mWb

H/mo 104 7- where permeability of free space

∫s

m dsB Magnetic flux : that passes through the surface S.

ds

nadsds ˆ

s

B

S

mdΨ B= ds cos

dsB =

s

m sdBΨHence:

)( 0 WbsdBΨs

m

I

H

In magnetics, magnet poles have not been isolated:

0

0

B

dvBsdBvs

4th. Maxwell’s equation for static fields.

rH 310Ex. 7.9: For (Am-1), find the m that passes through a plane

surface by, ( = /2), (2 r 4), and (0 z 2).

Wb10 x 8.150

121010

ˆ10ˆ

4-

32

0

4

2

3

2

0

4

2

3

oo

o

s

m

rdrdz

drdzrdsBΨ

Solution:

7.8 MAXWELL’S EQUATIONS

POINT FORM

INTEGRAL FORM

vD encQv

dvvs

sdDv

dvD

0 x E 0 x l

ldEs

sdE

JH x encIs

sdJl

ldHs

sdH x

0 B 0 s

sdBv

dvB

ED

HB

Electrostatic fields :

Magnetostatic fields:

7.9 VECTOR MAGNETIC POTENTIAL

0s

sdB magnet poles have not been isolated

To define vector magnetic potential, we start with:

0∇∫ v

dvBdsB

Using divergence theorem:

0∇∇ A

From vector identity:

0∇ B

Therefore from Maxwell and identity vector, we can defined if is a vector magnetic potential, hence:

<=>

AB ∇

A

where is any vector.A

=>

SUMMARY

en

s sl

IsdHsdJldH Stoke’s theorem

Ampere’s circuital law

Maxwell’s equations

s v

m dvBsdB 00

Divergence theorem

Gauss’s law

Magnetic flux lines close on themselves (Magnet poles cannot be isolated)

Maxwell’s equations