1 Engineering Electromagnetics Essentials Chapter 4 Basic concepts of static magnetic fields.

1

Engineering Electromagnetics

EssentialsChapter 4

Basic concepts of

static magnetic

fields

2

Objective

Recapitulation of basic concepts of static magnetic fields or magnetostatics

Topics dealt with

Coulomb’s and Gauss’s laws of static magnetic fields or magnetostaticsBiotSavart’s law Ampere’s circuital law Lorentz force experienced by a moving charge in a magnetic field Force experienced by a current-carrying conductor in a magnetic field

Background

Vector calculus expressions developed in Chapter 2Basic concepts of static fields developed in Chapter 3 as some of these concepts are extended to static magnetic fields in this chapter

3

Force between a pole of a magnet and a pole of another magnet

expressed by a law analogous to Coulomb’s law of electrostatics:

rmm a

r

qqF

2

21

4

Permeability of the medium

:, 21 mm qq Magnetic charges representing the poles:ra

Unit vector directed from the magnetic point charge qm1 to qm2

Distance between the magnetic point charge qm1 to qm2 :r

:

Relative permeability

0 r

Permeability of a free-space medium

70 104 H/m

(That the unit of permeability is H/m will be appreciated from the expression of inductance of a solenoid to be derived later, the unit of inductance being Henry or H)

4

HB

Magnetic flux density is related to magnetic field as electric flux density or electric displacement is related to electric field

rm ar

qB

24

Magnetic flux density due to a magnetic point charge qm at a distance r is analogous to electric field due to an electric point charge

BqF m

Force on magnetic point charge qm in a region of magnetic flux density analogous to force on electric point charge in a region of electric field

rm ar

qH

24

Magnetic field due to a magnetic point charge qm at a distance r is analogous to electric displacement due to an electric point charge

5

E

B

D

H

ED

HB

rar

qqF

2

21

4 r

mm ar

qqF

2

21

4

EqF

BqF m

rar

qE

24

r

m ar

qB

24

rar

qD

24

rm ar

qH

24

Electrostatic quantities/expressions

Magnetostatic quantities/expressions

Analogous electrostatic and magnetostatic quantities

6

S S

n QdSaDSdD

Gauss’s law and Poisson’s equation of magnetostatics

S S

n

QdSaESdE

S S

ndSaHSdH 0

Electrostatics Magnetostatics

S S

ndSaBSdB 0

analogous

analogous

Free magnetic charges do not exist

Free magnetic charges do not exist. A magnetic charge is always accompanied by an equal and opposite magnetic charge as in a permanent magnet. One cannot separate north and south poles of a magnet by breaking the magnet.

As many flux lines leave a volume enclosing a pole (magnetic charge) of a magnet as those enter the volume. Thus, magnetic flux lines always form closed loops.

In other words, magnetic flux lines are continuous

Gauss’s law

7

Electrostatics Magnetostatics

analogous

Free magnetic charges do not exist

0 H

Poisson’s equation

D

0 B

E

analogous

Poisson’s equation of magnetostatics is the manifestation of the absence of free magnetic charges (poles); it is another way of stating that magnetic flux lines are continuous.

Electric flux lines (as in an electric dipole) originate from a positive point charge and terminate on a negative point charge whereas magnetic flux lines originating from north pole of a magnet will return to the north pole via south pole.

8

BiotSavart’s Law

200

2

4)(

4

R

aldiHdBd

R

aldiHd

R

R

Element of magnetic field and element of magnetic flux density due to a direct current through an element of length of a filamentary conductor (say, a solid-cylindrical conductor of circular cross-section of radius approaching zero) is given by

( )idlCurrent element

i

O

P

Ra

OP=R

conductor y filamentarrough current th of

direction on the takesctor element veLength ld

sought isdensity flux

and field magnetic the where distance aat point the

element tocurrent thefrom directedr unit vecto theis

R

aR

9

Magnetic field due to a long thin wire carrying a direct currentWire is aligned along z axis carrying a direct current along z.

P is point where the magnetic field is sought.

O is the foot of perpendicular form P on the length of the wire.

O is considered as the origin z = 0.

OP = r

Wire extends from z = - to .

20

2

4)(

4

R

aldiHdBd

R

aldiHd

R

R

Elements of magnetic field and magnetic flux density are given by:

Pat considerdvector element current ldi

zaidzldi

z along considerdvector unit za

zald

alongcurrent of direction the takes

(Biot-Savart’s law)

10

P toP from directedvector unit

P and Pbetween distance 4 2

R

Rz

a

RR

aadziHd

Pat considerdvector element current

4 2

ldi

R

aldiHd R

zaidzldi

aaa Rz

sin

2/

aaa Rz

cos cos)2/sin(sin)sin(sin

reader thefromaway direction azimuthalin

paper theof plane the tonormalvector unit a

reader thefromaway direction azimuthalin vector unit a

11

24 R

aadziHd Rz

aaa Rz

cos

aR

dziHd

24

cos

tantanOPPO rz

drdz 2sec

aR

driHd

2

2

4

cossec

adR

riHd

sec

4 2

secrR

adr

iHd

cos

4

adr

iad

r

iHdH

cos

4cos

4

1

12

adr

iHd

cos

4

adr

iad

r

iHdH

cos

4cos

4

1

Integrating over the length of the wire we get

For the wire extending from z = - to , the integration limits become = -/2 and = /2 giving

2/

2/

cos4

cos4

ad

r

iad

r

iH

13

2/

2/

cos4

cos4

ad

r

iHad

r

iH

a

r

iad

r

iH

2/2/

2/

2/

][sin4

cos4

ar

ia

r

ia

r

iH

)11(

4)]2/sin()2/[sin(

4][sin

42/2/

a

r

ia

r

ia

r

iH

24

2)11(

4

(Magnetic field due to an infinitely long thin wire carrying a direct current)



14

Magnetic field due to a thin straight wire of finite length carrying a direct current

Point where to find magnetic field is not symmetric with respect to the wire of finite length

For the wire extending from z = -l2 to l1, the integration limits become = 1 and = 2 giving

a

r

iad

r

iad

r

iH

1

2

1

2

][sin4

cos4

cos4

Wire length AB = OA + OB = l1+ l2

a

r

ia

r

iH

)sin(sin

4][sin

4 211

2

22

2

22

22

1

11

sin

sin

rl

l

rl

l

a

lrlrr

iH

2

222

12 /1

1

/1

1

4

(Magnetic field at a point not symmetric with respect to the wire length)



15

Point where to find magnetic field is symmetric with respect to the wire length of finite length

(Magnetic field at a point symmetric with respect to the wire length)

Wire length AB = OA + OB = l1+ l2 = l/2+l/2 = l

a

lrlrr

iH

2222 )2//(1

1

)2//(1

1

4

a

lrlrr

iH

2

222

12 /1

1

/1

1

4

alrr

i

alrr

iH

2

22

)/2(1

1

2

)2//(1

2

4



16

Magnetic field at the centre of a a circular loop of wire carrying a direct current

ra

Current element

a

Ra

idl

Current loop

Z

Za

idl ia

24 R

aldiHd R


aR

aa

adlld

rR

2222 444

)()(

4 a

aidl

a

aaidl

a

aadli

R

aldiHd zrrR

dla

aiHdH z

24

adl 2

aa

aidl

a

aiHdH zz

2

44 22

zaa

iH

2

(magnetic field at the centre of a a circular loop of wire carrying a direct current)

17

Magnetic field at the centre of a regular polygonal loop of wire carrying a direct current

(Magnetic field at a point symmetric with respect to the wire length)

a

lrr

iH

2)/2(1

1

2

Recall the expression for magnetic field due to a straight wire of finite length

a

lar

iH

2)/2(1

1

2

ar

ll

a = perpendicular distance of each side of the polygon from its centre

l = length of each side of the polygon

(Expression for the magnetic field at the centre of the polygon loop of wire carrying a direct current due to a single side of the polygon)

18

a

lar

iH

2)/2(1

1

2

(Expression for the magnetic field at the centre of the polygon loop of wire carrying a direct current due to a single side of the polygon)

n

zalaa

iH

)

)/2(1

1(

2 2

(Expression for the magnetic field at the centre of the polygon loop of wire carrying a direct current due to n sides of the polygon obtained by summing up the contributions of sides)

zzznn

z aa

iaa

a

ial

a

ia

laa

iH

22

44)/2(

1

2 222

la

n 1for large number of sides of the polygon

Expression becomes identical with the expression for magnetic field at the centre of a circular loop of wire carrying a direct current, for large number of sides of the polygon when the latter tends to become a circle!

19

Magnetic field due to a circular loop of wire carrying a direct current at a point lying on the axis of the loop off from its centre

24 R

aldiHd R


32 44 R

Raidl

R

aldiHd R

SPR

aRR

aidlldi

R

P is the point on the z-axis, at a distance d from the centre O of the circular current loop, where the magnetic field is to be found

dOP

diameter.on S and S points opposite two

at considered are elementscurrent Two

Element of magnetic field at P due to current element at S:

3(SP)4

SP

aidlHd

20

Element of magnetic field at P due to current element at S:

3(SP)4

SP

aidlHd

2/322 )(4

)(

da

adaaadliHd zr

zrzr adaaaa

daR

OPSO OPSO SP

)(SP 2/122

rz

zr

aaa

aaa

2/322 )(4

)(

da

adaadliHd rz

S)at element current todue

Pat field magnetic of (element

loopcurrent circular theof radius a

21

Sat element current todue

Pat field magnetic ofElement

Sat element current todue

Pat field magnetic ofElement

2/322 )(4

)(

da

adaadliHd rz

diameter)on S and Sat directions oppositein (with ra

While summing up the contribution to the magnetic field at P due to all diametrically opposite current elements, the second term containing the unit vector will cancel out since unit vectors at diametrically opposite points though equal in magnitude are in opposite directions.

dl

da

aai

da

adlaiHdH zz

2/3222/322 )(4)(4

adl 2

ada

aaiH z

2

)(4 2/322

2

zada

aiH

2/322

2

)(2 (magnetic field due to a circular loop of wire carrying a direct

current at a point lying on the axis of the loop off from its centre)

22

zaa

iH

2

(magnetic field due to a circular loop of wire carrying a direct current at a point lying on the axis of the loop off from its centre)

zzz ada

ia

da

aiaa

da

aiH

2/3222/322

2

2/322 )(2)(22

)(4

loopcurrent circular of area 2 a

Special case: d = 0

zaa

iH

32

zada

aiH

2/322

2

)(2

zm aa

pH

32

moment dipole magnetic ipm

(magnetic field at the centre of a a circular loop of wire carrying a direct current as a special case d = 0)

23

Magnetic dipole moment

moment) dipole (magnetic ipm

zm aa

pH

32

)sincos2(4

13

0

aap

rE r

(expression for electric field due to a short dipole at a distance r on the axis of the dipole in terms of the dipole moment p)

rar

pE

3

02

moment) dipole (electric qlp

(expression for magnetic field at the centre of a a circular current loop of wire)

(expression for electric field due to a short dipole at a distance r at an angle from the axis of the electric dipole in terms of the electric dipole moment p (already derived)

ED

0

)sincos2(4

13

aap

rD r

)sincos2(4

13

aap

rH rm

toanalogous is HD

(expression for magnetic field due to a short dipole at a distance r at an angle from the axis of the magnetic dipole in terms of the electric dipole moment pm)

24

)sincos2(4

13

aap

rH rm

(expression for magnetic field due to a short dipole at a distance r at an angle from the axis of the magnetic dipole in terms of the electric dipole moment pm)

HB

0

)sincos2(1

4 30

aar

pB rm

Electric dipole Magnetic dipole

)sincos2(4

13

0

aap

rE r

)sincos2(

1

4 30

aar

pB rm

04/1 4/0

0 0/1

)sincos2(4

13

aap

rD r

)sincos2(

4

13

aap

rH rm

qlp )( 2aiipm

Analogous electric and magnetic dipole quantities

25

Ampere’s circuital law in integral form

Line integral of magnetic field around a closed path of any arbitrary shape on a plane perpendicular to a current is equal to the current enclosed by the path.

consideredpath closed on the Ppoint aat field magneticH

dl

Closed pathi

P

.

H

l

ildH

Ampere’s circuital law

current theofdirection in the screw theof

motion axialan cause that willscrew a ofrotation of sense theisdirection its and

is of magnitude thesuch that point at thector element velength

i

dlldld

26

Let us take up the problem of an infinitely long current-carrying wire

Length element

dl dla

Closed circular path

Zi

rP

aHH

Ampere’s circuital law simplifies the problems that enjoy geometrical symmetry in magnetostatics as does Gauss’s law of electrostatics.

A long straight wire along z carrying current i and a circular closed path of radius r considered on a plane perpendicular to Z axis passing through the point P where the magnetic field is to be found

l

ildH

(Ampere’s circuital law)

adlld

aHH

l

iadlaH

l

iHdl

The problem enjoys cylindrical symmetry and the magnetic field magnitude remains constant at all points on the closed path of radius r.

l

idlH l

rdl 2

irH 2r

iH

2

ar

iaHH

2

27

a

r

iH

2

Expression for magnetic field due to a long straight current carrying wire has been obtained so simply by Ampere's circuital law doing away with the evaluation of integrals required in the application Biot-Savart’s law to find the expression.

Let us find the magnetic field both inside and outside a long solenoid carrying a direct current and also its inductance

Length of the solenoid is considered to be very large compared to the radius of the and distance between two consecutive turns of the solenoid. Therefore, we can take the magnetic field due to the solenoid to be independent of z.

Pa b

cd

d

a b

c

P z

Current directed away from reader

Current directed towards reader

28

Biot-Savart’s law can be used to appreciate that the direction of the magnetic field due to currents, both out of and into the plane of the paper, will be azimuthal, as in the problem of a long straight current carrying wire.

At a point outside the solenoid, at a radial distance much larger than the solenoid radius, the magnetic field contributed by the anticlockwise flux lines due to the currents out of the plane of the paper will balance the magnetic field contributed by the clockwise flux lines due to currents into the plane of the paper, thereby making the magnetic field outside the solenoid nil.

We can also appreciate this with the help of Ampere’s circuital law. For the application of Ampere’s circuital law, take a rectangular path abcd in the cross-sectional plane passing through the point P outside the solenoid where the magnetic field is sought. We purposely take the side cd of the rectangle to be at a different distance from its side ab

l

ildH

(Ampere’s circuital law applied to path abcd)

daab bc cdabcd

0ldHldHldHldHldH

Current i enclosed by the path abcd is nil since the number of turns corresponding to the current going out of the plane of the paper is equal to the number of turns corresponding to the current going into the plane of the paper.

Magnetic field outside the solenoid

Pa b

cd

d

a b

c

P z



29

cdab

ldHldH 0

daab bc cdabcd

0ldHldHldHldHldH

Magnetic fields transverse to the solenoid axis due to the current turns cancel out and therefore the integrands of the second and fourth integrals each becomes equal to zero.

0)()( outsideoutside cd

zz

ab

zz adlaHadlaH

directed oppositelybut z alongeach arely which respective cd and ab sides at the

fields magnetic of magnitudes theare and outsideoutside HH

0outsideoutside cdab

dlHdlH

0outsideoutside lHlH

say ,cdab l

outsideoutside HH

nil. toequaleach and put tohave weTherefore,

axis. solenoid thefrom distancesdifferent at are cd and ab since

fromdifferent be willpresent iffact that thesContradict

outsideoutside

outside

outside

HH

H

H

0outsideoutside HH

(magnetic field outside the solenoid is nil)

Pa b

cd

d

a b

c

P z



30

Magnetic field inside the solenoid

adba cb dcdcba

nlildHldHldHldHldH

reader. towardsdirected

paper ofout turns theof currents theencloses and

Ppoint he through tpasses dcbapath r Rectangula

l

ldH law) circuital s(Ampere' enclosedcurrent

solenoid rough thecurrent th

solenoid theoflength unit per turnsofnumber

cbda

i

n

l

Magnetic field outside the solenoid being zero, the integrand of the third term becomes zero. Magnetic fields transverse to the solenoid axis due to the current turns canceling out, the integrands of the second and fourth terms each become nil.

ad dcba

0ldHldHldH

ba

nlildH

ba

nliadlaH zz

ba

nlidlH

Pa b

cd

d

a b

c

P z



ba

ldl

nliHl niH

zaHH

zaniH

Expression for magnetic field inside the solenoid

z

z

adlld

aHH

31

Inductance of a solenoid

zaniH

(magnetic field inside the solenoid)

zaniHB

00 (magnetic flux density inside the solenoid found thus uniform over the cross section of the solenoid)

dSaBS

nB

turnsinglet, (magnetic flux linked with a single turn of the solenoid)

nidSnidSaaniSS

zzB 000turnsinglet,

(magnetic flux linked with a single turn of the solenoid)

))(( turnsinglet, nlBB (magnetic flux linked with nl turns of length l of the solenoid, n being the number of turns per unit length)

))(())(( 0turnsinglet, nlninlBB

lni

nlni

iL B 2

00 ))((

(Weber or Wb)

(Wb/A or henry or H)

solenoid) theof area sectional-(cross S

dS

(expression for inductance of a solenoid) That the practical unit of permeability is H/m is clear from this expression for inductance.

32

Inductance per unit length of a coaxial cable

B

Element of strip of infinitesimal area

dr

I

lI

a

r

b

ldr

r

IHB

2

)( 00 ldrBd B

r

drIld B

2

0

a

bl

Iabl

I

rlI

r

drl

Id b

a

b

a

BB

ln2

)ln(ln2

][ln22

00

00

a

bl

Iab

lI

IB ln

2

ln2 Inductance 0

0

a

bln

2lengthunit per Inductance 0

Apply Ampere’s circuital law to a closed circular path on a cross-sectional plane perpendicular to the axis of the cable of radius r between r = a (radius of the inner conductor) and r = b (radius of the outer conductor) carrying current I. You can get the following expression for magnetic flux density in the region between the conductors:

(element of magnetic flux density linked with a rectangular element of strip of length l, width dr and area ldr in the region between the conductors on a radial cross-sectional plane)

Magnetic flux linked with the rectangular area on a radial plane between the conductors can be obtained by integration as follows:

(expression for inductance and inductance per unit length of a coaxial cable)

33

Ampere’s circuital law in differential form

E

D

AB

2a

3a

1a

1 1h du

2 2h du

P

3dSArea element

Sense of integral rotation

In electrostatics, we obtained Poisson’s equation in point form or differential form from Gauss’s law in integral form. Similarly, in magnetostatics, we can obtain Ampere’s circuital law in differential or point form from its integral form.

l

ildH

(Ampere’s circuital law in integral form)

Apply Ampere’s circuital law to a closed element of area )( 22113 duhduhdS 3a

which is taken normal to the unit vector in generalised curvilinear system of coordinates, and which encloses the point P where to obtain Ampere’s circuital law in point or differential form.

33. dSaJi

l

dSaJldH 33.

current density regarded approximately as constant over element of area dS3

34

l

dSaJldH 33.

E

D

AB

2a

3a

1a

1 1h du

2 2h du

P

3dSArea element


33

.aJdS

ldHl

332211 aJaJaJJ

333322113

).( JaaJaJaJdS

ldHl

333 0

JdS

ldH

S

Ltl

In the limit S3 0, corresponding to the area element shrinking to the point P, the current density may be taken as exactly constant and the approximation can be removed from the sign of equality.

33)( JH

(by the definition of the component of the curl of a vector)

35

33)( JH

E

D

AB

2a

3a

1a

1 1h du

2 2h du

P

3dSArea element


3a

(taking the area element perpendicular to )

We have obtained

11)( JH


Similarly, we can obtain

2a

1a

22)( JH


Similarly, we can also obtain

332211332211 )()()( aJaJaJaHaHaH

Combining the components we then obtain the differential form of Ampere’s circuital law as follows

JH

(Ampere’s circuital law in differential form)

36

Lorentz force on a moving point charge in a magnetic field

Show that an electron when it is shot with a dc velocity perpendicular to a uniform dc magnetic field of flux density of magnitude B executes a circular motion with an angular frequency called the electron cyclotron frequency.

BvqF

r

mvvBe

2

BBr

r

v

rT

222

B

BTc

222

Brm

Brev

A point charge moving with a velocity in a region of magnetic field experiences a force called Lorentz force given by

placed is chargepoint in which density flux magnetic

chargepoint of velocity

chargepoint ofamount

B

v

q

of magnitude theis sin FqvBF

BvF

product cross theofdirection theis ofDirection

Cyclotron frequency

You can balance Lorentz force with the centrifugal force to obtain

electron ofmotion circular of radius

motioncircular in electron of velocity

electron of ratio mass-to-charge of magnitude

mass electronic

charge electronic of magnitude

r

v

m

e

Time period T of electronic circular motion is given by

(angular electron cyclotron frequency)

(Lorentz force)

37

Relation between the current density, volume charge density and beam velocity in a beam of charge flow Consider a current due to the flow of charge particles, for instance electrons,

taken as point charges all with a constant velocity v. This current is called the conduction current in a conducting medium or convection current in a fee-space medium.

Consider two identical cross-sectional planes each of cross-sectional area separated by a distance numerically equal to v being the distance a charged particle covers during a second. Consequently, all the electrons within the volume v will flow through a cross-sectional plane of area in a second. The number of such electrons is equal to the number of electrons per unit volume n multiplied by the volume v, that is n v. Multiplying this number n v by the charge e carried by each electron we get the current i = n v e through the cross-sectional area and dividing i by we get current density J:

envi

density)(current /)( vnevenvJ

density) charge (volume ne

(relation between current density, volume charge density and beam velocity in a beam of charge flow)

38

BvendlFd

ndl

:electrons ofnumber on force Lorentz ofElement BvqF

(Lorentz force)

Force on a current carrying conductor placed in a magnetic field

electrons ofnumber of

eachby carried charge electronic

of magnitude theis

ndl

e

eq

BldiFd

Consider a current element placed in a region of a steady magnetic field (a wire of infinitesimal length dl and cross-sectional area through which a direct current i passes. Let us find the element of Lorentz force on the moving charge particles constituting the current element. In the length element dl, the number of charge particles is (n)(dl) = n dl, where n is the number of charge particles per unit volume of the current element and dl is its volume. Magnetic field is assumed over the length element

va

avv

v

v

ofdirection in ther unit vecto theis

BavendlFd v

BadlenvFd v

))((

envenvi

adlld v

(element of force on a current element carrying a direct current in a magnetic field)

B

dlLength element

Current element idl

BldiFdF

Magnetic field even if it is non-uniform may be regarded as constant over the element of length dl. However, we have to integrate the element of force over the entire length of the conductor to find the force on it due to the magnetic field:

39

BldiFd

(element of force on a current element carrying a direct current in a magnetic field)

Appreciate that like currents attract and unlike currents repel.

212221 BldiFd

Consider two long straight parallel conducting wires carrying direct currents i1 and i2 each parallel to z axis. You can express the force on a current element on the wire carrying current i2 due to the magnetic field of the current i1.as

z

x

adlld

ad

iB

22

1021 law) circuital sAmpere' using e(obtainabl

2

)2

()( 102221 xz a

d

iadliFd

y

y

xz

ad

dliiFd

ad

idliFd

aad

idliFd

2

))(2

)((

))(2

)((

221021

102221

102221

Negative sign in the above force expression indicates that the force experienced by the wire carrying current i2 is in the negative y direction, that is, towards the wire carrying current i1. In other words, the force between the wires is that of attraction which means that ‘like currents attract’. If the direction of current i2 is reversed to make the two currents unlike, the direction of the element of length vector on the wire carrying i2 will also reverse giving zadlld

22

which in turn gives

yad

dliiFd

22210

21 which indicates that the force is in the positive y direction or the force is that of attraction. This, in other words means that ‘unlike currents attract’.

40

Magnetic Vector Potential due to a Steady Current

Finding magnetic field from magnetic vector potential is an alternative approach to Biot-Savarts’ law. We will see that magnetic field can be found from magnetic vector potential as electric field can be found from electric potential.

200

2

4)(

4

R

aldiHdBd

R

aldiHd

R

R

Recall Biot-Savarts’ law:

RRaR /

( )idlCurrent element

i

O

P

Ra

OP=R

RRR

3

11

3

0

4 R

RldiBd

R

ldiBd1

40

R

ldiG

GGG

/1

identity)(vector )(

ldi

Rldi

RBd

11

40

41

However, you may note that the second term of the right-hand side of the above expression becomes zero since in that term the curl operation is taken with respect to the field-point variables (x,y,z) while the current element term on which this operation is taken is represented by the source point variables (x/,y/,z/, say):

ldi

Rldi

RBd

11

40

0 ld

AdR

ldiBd

4

0

where the element of vector potential die to a current element turns out to be

R

ldiAd

4

0

AAdBdB

R

ldiAdA

4

0

We can find the magnetic vector potential by integration:

42

Find magnetic field due to a long current carrying straight wire using vector potential approach

Z

X

Y

θ

ra

a

a

O

N

M

r

R

P (r,θ,Ф)

dz'

z

z

Consider current element at the point M on z axis at a distance z/ from the origin O. The problem enjoys cylindrical symmetry. Element of magnetic vector potential due to the current element at P (r,,z) is

R

azdiAd z

4

0

2/122

2/122

])[(

])NP()NM[(PM

rzz

R

2/1220

])[(4 rzz

azdiAd z

azdR

riB

3

0

4

2/122

0

])[(4 rzz

azdiAdB z

arzz

zd

rrzz

azdB z

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

Expansion of curl in the right hand side in cylindrical system of coordinates and remembering azimuthal symmetry of the problem / = 0

2/122 ])[(PM rzzR

43

2/

2/

02/

2/

233

0 cos4

secsec4

adr

iadr

r

riB

z

z

azdR

riB

3

0

4

sec

sec

tan2

rR

drzd

rzz

ar

iB

20

azdR

riB

3

0

4Integration limits taken as z/ = - and z/ = for a long wire

Z

X

Y

θ

ra

a

a

O

N

M

r

R

P (r,θ,Ф)

dz'

z

z

ar

ia

r

i

ar

i

ar

iB

24

)]1(1[4

)]2/sin()2/[sin(4

][sin4

00

0

2/2/

0

a

r

iH

2

0/ BH

(expression for ,agnetic field due to a long current carrying straight wire which is one and the same as that obtained by Biot-Savarts law)

44

Z

X

Y

AB

Ф

θ

O

Ф'Ф'

P(r,θ,Ф)r

lId

lId

You can find magnetic field due to a circular current loop at a point off the axis using the magnetic vector potential approach however now treating the problem in spherical polar system of coordinates unlike cylindrical system of coordinates used to treat the problem of a long straight current carryong wire.

)sincos2(4

13

aap

rH rm

)sincos2(4 3

20

aar

aiB r

Hope you will be motivated to deduce the following expression using the vector potential approach in terms of the magnetic dipole moment pm = ia2, the other symbols having their usual significance (see the text of the book for the details of deduction):

(expressions which become identical with the expressions predicted earlier from the corresponding expressions derived for electric field quantitiesobtained by drawing an analogy between the electric and magnetic dipoles)

45

Summarising Notes Basic concepts of static magnetic fields or magnetostatics have been developed.

Magnetostatic quantities analogous to the corresponding electrostatic quantities have identified.

Coulomb’s and Gauss’s laws as well as Poisson’s and Laplace’s equations of magnetostatics have been developed on the same line of electrostatics.

Absence of free magnetic charges (poles) makes it possible to write Gauss’ law and Poisson’s equation of magnetostatics analogous to those of electrostatics.

Gauss’s law of magnetostatics and Poisson’s equation of magnetostatics have been appreciated as the manifestation of the absence of free magnetic charges (poles) or that of the continuity of magnetic flux lines.

Biot-Savart’s law predicts magnetic field due to a current element as does Coulomb’s law predict electric field due to a point charge.

Use of Biot-Savart’s law has been demonstrated in illustrative examples, for instance, in the problem of finding the magnetic fields due to a long filamentary steady current, a filamentary current of finite length, a circular loop of current and a polygonal loop of current.

Ampere’s law has made the problem of finding the magnetic field due to a steady current simpler in problems that enjoy geometrical symmetry, as has done Gauss’s law in electrostatic problems.

46

Readers are encouraged to go through Chapter 4 of the book for more topics and more worked-out examples and review questions.

Ampere’s circuital law in integral form has been extended to derive the law in differential form.

Force experienced by a moving point charge placed in a magnetic field is given by Lorentz force equation.

Concept of Lorentz force equation has been extended to find the force on a current carrying conductor placed in a magnetic field.

Concept of magnetic vector potential due to a steady current has been developed and its application to finding magnetic field as an alterative to Biot-Savarts’s law discussed.

1 Engineering Electromagnetics Essentials Chapter 4 Basic concepts of static magnetic fields.

Documents

Transcript of 1 Engineering Electromagnetics Essentials Chapter 4 Basic concepts of static magnetic fields.