magnetostatic field in free space - guc - …eee.guc.edu.eg/courses/communications/comm402... ·...

ELECTROMAGNETIC PROF. A.M.ALLAM

MAGNETOSTATIC FIELDIN FREE SPACE

EMF

Jean-Baptiste Biot1774-1862

Felix Savart1791-1841


�

The magnetostatic field is produced by a constant current flow; charges move with constant velocity

Like the electrostatic field is produced by

static or stationary charges


1-Ampere’s law of force

II11 II22

CC11 CC22

RaRR

11dI ��

22dI ��

The magnetic interaction of currents is experimentally established in vacuum

by Ampere. The magnetic force between two current-carrying elements is:

� �××=

2C 1C221

112221

ˆII

4 Radd

F Ro ��

��

�

πµ

µµµµµµµµoo is the permeability of free space( ) [H/m] 104 7−×= πµo

Is the force exerted on loop (2) due to current passing in loop (1), 1221 FF��

−=

Currents in the same direction

Ra11dI ��

22dI ��××××××××

21Fd�

(Attraction force)(Attraction force)

12Fd�

Currents in the opposite direction

Ra11 ��

dI 22 ��

dI

21Fd�

(Repulsion force)(Repulsion force)

12Fd�

Currents

021 =Fd�

(No Force)(No Force)2211 II �

��

dd ⊥

Ra11 ��

dI 22 ��

dI××××××××

××××××××

=12Fd�

Ra

RR11dI ��

22dI ��

221

112221

ˆII

4 Radd

Fd Ro ××= ��

��

�

πµ

For small elements of the loops:


Similar to the electrostatic, the definition of an electric field was developed as

the force acting on a unit charge, an analogous definition can be found for

the stationary magnetic field which is the force exerted on current carrying

element��

��

� ××= 221

112221

ˆI

4I

Rad

dFd Ro ��

��

πµ

122I Bdd�

��

×=

Ra

RR��

IdPP

221

111

ˆI

4 Rad

Bd Ro ×= ��

�

πµ

�×=

c2

ˆ I

4 Rad

B Ro ��

�

πµ

×××××××× Bd�

B�

Is the magnetic flux density (magnetic induction)

[ Web/m2 ]� ×=2C

12221 I BdF�

��

2-Biot – Savart law

,

B�

In practice, one does not always deal with current flowing in thin conductors and hence it is necessary to generalize the defining equation for in terms of volumetric current distribution :

dV ) . ( ).( I JdSdJdSdJd�

��

��

��

===

ˆ )r(

4)(

v2 vd

RaJ

rB Ro ′×′= �

′

��

πµ [ Web/m2 ] �d

σσσσ

dSdSJ�II

Jean-Baptiste Biot

1774-1862Felix Savart1791-1841

]...[A/m 2J�

Volumetric current

Total current]...[A I

ˆ )r(

4

)(s

2 sdR

akrB Ro ′×′

= �′

��

πµ

For surface current distribution: kds’


u dq dtd

dqd dtdq

dI��

�

��

��

===

Bu dqBIdFd��

��

×=×=

BuqF��

×=

••Lorentz force:Lorentz force:

u�

B�

Is the force acting on charge q moving with velocity in the presence of magnetic field

Similar to the electrostatic force acting on a static electric charge, there is a magnetic force acting on a moving electric charge

( is the velocity)u�

BuqqEFFF me

��×+=+=

This force is known as Lorentz forceImportant notesThe electric force is usually in the direction of the electric field while, the magnetic force is perpendicular to the magnetic field

The electric force acts on a charged particle whether or not it is moving, while the magnetic force acts moving charged particle onlyThe electric force expends energy in displacing a charged particle, while the magnetic one does no work when the particle is displaced because it is perpendicular to the velocity


Example: Find the magnetic flux density due to a current carrying conductor at point P as shown in figure

B�

Solution:

RRRa α

II

PP

aa2α1α

xx dxdx

×××××××× B�

2

ˆI

4 Rad

Bd Ro ×= ��

�

πµ

12 a sin (1) )(I

4 R

dxo απ

µ=

But we have,xa /)180tan(tan −=−−= αα αcot ax −=��

(1) .......... d cosec 2 ααadx =Ra /)180sin(sin =−= αα (2) ....... cosec αaR =��

122

2

a cos

sin ) cosec (

4I α

ααα

πµ

deca

aBd o=�

�� 1a sin 4

I 2

1

�=α

α

ααπ

µd

aB o�

121 a ]c-[cos 4

I ααπ

µos

aB o=�

Special cases:


Example: Find the magnetic flux density at any point on the axis of a circular current loop of radius a flown through current I

B�

Solution:

II

aa

zz

xx

zz

yy

PP

θθθθθθθθ

RR

d ad α′=′�

Bd�

θsin)(2 dBdBz =

Bd�

θθθθθθθθ θθθθθθθθ

α′ dα′

Ra

2

ˆI

4 Rad

Bd Ro ×= ��

�

πµ

θsin2dBdB z =

θαπ

µsin

)1( (1) )(Ia

42 2

��

� ′=

Rdo

απ

µ ′++

= daa

o

2222z

a

)z(1

2

Ia

� ′+

=π

απ

µ0

2/322

2

)z(

1

2Ia

da

B oz

z2/322

2

a )z(2

Ia

aB o

+= µ�


3-Ampere’s circuital lawIt states that the line integral of the tangential component of H around

a closed path is equal to the net current enclosed by the path

enos

oc

ISdJdB µµ == ��

��

.. �� ∇=sc

SdBxdB��

��

).(. Jo

��µ=×∇ B

Differential formIntegral form

I

ld�

sd�

1

32

4

Notes about selection of the closed path:

-Its positive direction follows the RHR to follow the current path

-The positive direction of the closed area follows the RHR according to the path chosen

Loop1 encloses zero current

Loop2 encloses only part of the current

Loop3 encloses the total current

Loop4 encloses zero current


Example: Find the magnetic field due to an infinite thin wire carrying current I

B� II

r

ceno

c

IdB µ=� ��

.

IrB oµπ =2

φπµ

arI

B o ˆ2

=�

Solution:

B(r)B(r)

rr

Example: Find the magnetic field due to an infinite wire of radius a and carrying current I.

B�

II

aar

��

�= 222 r

aI

rB o ππ

µπ φπµ

aaIr

B o ˆ2 2=

�

Solution:

r < ar < a

IrB oµπ =2 φπµ

arI

B o ˆ2

=�

r > ar > a


Since the magnetic flux density are usually closed loopsB�

0 . =∇ B�

Differential formDifferential form��

Which means that:

� The magnetic field is solenoid field

0 . =∇�v

dvB�

Integral formIntegral form0. =�s

SdB��Using Gauss’s theorem

The total outward flux on any closed surface is equal to zero

Generally, the magnetic flux is given by:Generally, the magnetic flux is given by: �=s

SdB��

.ψ

•Basic equations of magnetostatic field in vacuum

Differential form Integral form

0B . =∇�

Jo

��µ=×∇ B �� =

so

c

SdJ��

��

..dB µ

0 . � =s

SdB��

�The magnetic field lines have neither source (start point) or sink (end point)�It is not possible to have an isolated magnetic poles (or magnetic charges)


5-Magnetization in material mediaOn atomic scale we have three sources which generates magnetic dipoles and its moment:

1- electron orbiting 2- electron spin 3- nuclear spin

[Amp/m] lim0

��

�

�

��

�

�

∆=

�∆

→∆ V

mM v

i

V

�

�

The magnetization vector within a magnetic material equals the magnetic dipole moment per unit volume

The most effective ones are the electron orbiting and electron spin which results in an angular momentum

It gives rise to a current I that encircles a surface dS which is called orbital magnetic moment

In atoms vector sum of the individual orbital magnetic moments gives thetotal orbital magnetic moment

The net effect of these magnetic moments of an atom is a magnetic current loopwhose magnetic dipole moment equals:im

�

]....[ 2AmsIdmi

�� =


Diamagnetic material

Before application of ext. field oB�

The dipoles are randomly oriented and thus the vector sum of dipole moment of each atom equals zero

After application of ext. field oB�

The ext. field redistribute the electrons orbits and creates induced mag. moment opposite to . oB

�

oB�

oinside BB��

<

(Ex.: (Gold, water)

Since dipoles are randomly oriented , hence the vector sum of dipole moment of each atom might equal zero or not (intrinsic dipole moment) depending on the material

Application of an external field will disturb the electrons in orbits and redistribute them which results in induced magnetic dipoles which aligned in opposite direction to the external one but the intrinsic dipoles (if any) will aligned in the direction of the external


Before application of ext. field oB�

After application of ext. field oB�

Non zero magnetic dipole moment of each atom (intrinsic magnetic moment)

� The magnetic moment of each atom will be aligned with .

� Induced magnetic moment will be created opposite to .

oB�

oB�

(Ex.: Aluminum alloys)

oB�

oinside BB��

>

Either diamagnetic or paramagnetic materials can be considered as free-space (µµµµr=1.00005 or 0.9992).

� The ferromagnetic material are such material which exhibit large paramagnetic effect

� Ex.: Iron, Nickel & Cobalt (µµµµr=105 : 109 )

Paramagnetic material

Ferromagnetic material


[Amp/m] lim0

��

�

�

��

�

�

∆=

�∆

→∆ V

mM v

i

V

�

�

......... �∆ v

im�

the total dipole moment of an atom due to

orbital motion counted over the volume ∆∆∆∆V.

oB�

VMaterial

∆∆∆∆∆∆∆∆VV

0r�′′′′

S

Given

M�

)r(M)r(J m′′′′××××∇∇∇∇ ′′′′====′′′′ ��

n)r(M)r(J sm ××××′′′′====′′′′ ��[A/m]

[A/m2]

From we can get the equivalent volume and surface magnetic currents and their resultant magnetic fields:

M�


)( mo JJB��

+=×∇ µ

After magnetization of material, the resultant total magnetic field is due to two types of currents and (neglecting Jsm for unbounded medium), so

mJ J��

( ) MJB o

��×∇+=×∇ µ/

JMB

o

��

=

��

�−×∇

µ

H�

MBH o

��−= µ/ )( MHB o

��+= µ

��

�� &&

JH��

=×∇ AmpereAmpere’’s circuital law (Diff. form)s circuital law (Diff. form)

AmpereAmpere’’s circuital law (Integ. form)s circuital law (Integ. form)

Then,

�� =×∇ss

SdJSdH��

.).( Using Stock’s theorem

ensc

ISdJdH == ��

��

..

Integrating both sides with respect to the surface S;

These formulas for Ampere’s law are used for any material with permeability µµµµµµµµ

The source of is the conduction steady current

6- Relation between HB��

&


HM��

∝ HconstM��

.)(= HM m

��χ=

χχχχm …… is a dimensionless quantity called magnetic susceptibility

HB��

µ=

rmo

µχµµ =+= 1 is the relative permeability

)( MHB o

��+= µ

)( HHB mo

��χµ += )1( mo H χµ +=

�

Hro

�µµ= H

� µ=

Hr

� )1( −= µ

HM r

�� )1( −= µ

HB

Mo

��

�−=

µH

H

o

��

−=µ

µ

•Basic equations of magnetostatic field in material media

Differential form Integral form

0B . =∇�

J��

=×∇ H �� =sc

SdJ��

��

..dH

0 . � =s

SdB��


Example: Find the magnetic field intensity due to N-turns toroidal coil wounded over ferromagnetic material with µµµµ and carrying current I

H�

)0(2 =rH π

Solution:

r < a

ensc

ISdJdH == ��

��

..

0=H�

φπa

rNI

H ˆ2

=�)(2 NIrH =π

a < r < b

)(2 NINIrH −=π

r > b

0=H�

HB��

µ=&

0=B�

φπµ a

rNI

B ˆ2

=�

0=B�

&

&

&


7-Boundary conditionsNormal field:Normal field: Tangential field:Tangential field:

0. =�s

SdB��

0)]( ˆ. ˆ.[ 120=∆+∆−∆

→∆hfSnBSnBLim

h

��

0 (as ∆∆∆∆h�0)

0)B-B( . n 12 ====��

or

S∆2n

1n

h∆µµµµ1 11

22µµµµ2

1B�

2B�

n

1n2n BB ====

Applying the equation,

ensc

ISdJdH == ��

��

..

��

��

∆=∆+∆−∆→∆

)]( . .[ 120 sh

JhfHHLim ττ

Applying the equation,

0 (as ∆∆∆∆h�0)

sJn��

=× )H-H( 12 or sJ=− ττ 12 HH

B2n = B1n �� µµµµ2 H2n = µµµµ1 H1n

µµµµ 2 H2 cosθθθθ2 = µµµµ 1 H1 cosθθθθ1 … … … … .(1)

Also, we have H2ττττ = H1ττττ ��

H2 sinθθθθ2 = H1 sinθθθθ1 … … … … .(2 )

Eqn. (2) ÷÷÷÷ Eqn. (1) gives,

2

1

2

1

tantan

µµ

θθ = (refraction law)(refraction law)

h∆

11

22

1H�

2H�

�∆

cµµµµ1

µµµµ2

××××××××××××××××

sJ�

1H�

2H�

θθθθθθθθ11

θθθθθθθθ22

µµµµ1

µµµµ2××××

××××××××

0Js ====�

××××


Example: Discontinuity in at current sheetH�

sJ�

×××××××××××××××× ×××××××××××××××× ×××××××××××××××× ×××××××××××××××× ××××××××××××××××

x

y

hg

H�Solution:

ensc

ISdJdH == ��

��

..

g . g . . sJHgH =+

xs a

2J

H ±±±±====�

… . Effect of the sheet without any incident . H�

Js x

yJs/2

- Js/2

H=Js/2��

Effect of the sheet on the incident : H�

τ1H

sJsJ=− ττ 12 HH

sJ+= ττ 12 HHsJ�

Med. 1 (µµµµo)

Med. 2 (µµµµo)

1H�

τ2H�


8- Magnetic energy

Electrostatic FieldElectrostatic Field Magnetostatic FieldMagnetostatic Field

• Electric Energy density (wElectric Energy density (wee):): • Magnetic Energy density (wMagnetic Energy density (wmm):):

�=v

e wW dv e

E . 21 ��

Ewe ε=

[Joule/m3]

2|| 21

E�

ε=D . 21 ��

E=

ε

2||

21 D

�

=

HHwm

�� .

21 µ=

[Joule/m3]

2|| 21

H�

µ=B . 21 ��

H=

µ

2||

21 B

�

=

•• Energy density (WEnergy density (Wee):): •• Energy density (WEnergy density (Wmm):):

�=v

E dV E . 21 ��

ε

2

21

CV= [Joule]

�=v

m wW dv m �=v

dV H . H 21 ��

µ

2

21

LI= [Joule]

•• Capacitance (C):Capacitance (C):

2e

VW2

C =

(Energy definition)

2

2IW

L m=[Farad] [Henery]

•• Inductance (L):Inductance (L): (Energy definition)

magnetostatic field in free space - guc - …eee.guc.edu.eg/courses/communications/comm402... ·...

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