1 engineering electromagnetics essentials chapter 4 basic concepts of static magnetic fields
TRANSCRIPT
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)
reader thefromaway direction azimuthalin
paper theof plane the tonormalvector unit a
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)
reader thefromaway direction azimuthalin
paper theof plane the tonormalvector unit a
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
reader thefromaway direction azimuthalin
paper theof plane the tonormalvector unit a
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
(Biot-Savart’s law)
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
(Biot-Savart’s law)
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
Current directed away from reader
Current directed towards reader
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
Current directed away from reader
Current directed towards reader
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
Current directed away from reader
Current directed towards reader
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
Sense of integral rotation
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
Sense of integral rotation
3a
(taking the area element perpendicular to )
We have obtained
11)( JH
(taking the area element perpendicular to )
Similarly, we can obtain
2a
1a
22)( JH
(taking the area element perpendicular to )
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.