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FIELD THEORY
Sub Code : EC44 IA Marks : 25Hrs/Week : 04 Exam Hrs. : 03Total Hrs. : 52 Exam Marks : 100
1. Electric Fields 18 hoursa. Coulomb’s law and Electric field intensityb. Electric flux density, Gauss law and divergencec. Energy and potentiald. Conductors, dielectrics and capacitancee. Poisson’s and Laplace’s equations
2. Magnetic fields 14 hoursa. The steady magnetic fieldb. Magnetic forces, materials and inductance
3. Time varying fields and Maxwell’s equations 5 hours
4. Electromagnetic waves 15 hours
Text Books :
William H Hayt Jr and John A Buck, “Engineering Electromagnetics”, Tata McGraw-Hill,6th Edition, 2001
Reference books :
John Krauss and Daniel A Fleisch, “Electromagnetics with Application”, McGraw-Hill, 5th
Edition, 1999Guru and Hiziroglu, Electromagnetics Field theory fundamentals, Thomson Asia Pvt. Ltd IEdition, 2001Joseph Edminster, “Electromagnetics”, Schaum Outline Series, McGraw-HillEdward C Jordan and Keith G Balmain, “Electromagnetic Waves and Radiating Systems”, Prentice-Hall of India, II Edition, 1968, Reprint 2002.David K Cheng, “Field and Wave Electromagnetics”, Pearson Education Ais II Edition, 1989, IndianRepr-01
Introduction to Field Theory
The behavior of a physical device subjected to electric field can be studied either by Field approachor by Circuit approach. The Circuit approach uses discrete circuit parameters like RLCM, voltageand current sources. At higher frequencies (MHz or GHz) parameters would no longer be discrete.They may become non linear also depending on material property and strength of v and iassociated. This makes circuit approach to be difficult and may not give very accurate results.
Thus at high frequencies, Field approach is necessary to get a better understanding of performance ofthe device.
FIELD THEORY
The ‘Vector approach’ provides better insight into the various aspects of Electromagneticphenomenon. Vector analysis is therefore an essential tool for the study of Field Theory.
The ‘Vector Analysis’ comprises of ‘Vector Algebra’ and ‘Vector Calculus’.
Any physical quantity may be ‘Scalar quantity’ or ‘Vector quantity’. A ‘Scalar quantity’ is specifiedby magnitude only while for a ‘Vector quantity’ requires both magnitude and direction to bespecified.
Examples :
Scalar quantity : Mass, Time, Charge, Density, Potential, Energy etc.,Represented by alphabets – A, B, q, t etc
Vector quantity : Electric field, force, velocity, acceleration, weight etc., represented by alphabetswith arrow on top.
etc.,B,E,B,A
Vector algebra : If C,B,A
are vectors and m, n are scalars then
(1) Addition
laweAssociativC)BA()CB(A
laweCommutativABBA
(2) Subtraction)B(-AB-A
(3) Multiplication by a scalar
lawveDistributiBmAm)BA(m
lawveDistributiAnAmAn)(m
laweAssociativ)A(mn)A(nm
laweCommutativmAAm
A ‘vector’ is represented graphically by a directed line segment.
A ‘Unit vector’ is a vector of unit magnitude and directed along ‘that vector’.
Aa is a Unit vector along the direction of A
.
Thus, the graphical representation of A
and Aa are
AaAorA /AaAlso
actor Unit veAVector
A
AA
A
Product of two or more vectors :
(1) Dot Product ( . )
πθ0,B}θCOSA{ORθCOSB(AB.A
B
B
θCosA
A
θCosB
A
A . B = B . A (A Scalar quantity)
(2) CROSS PRODUCT (X)
C = A x B = nθSINBA
C xAB xA)CB( xA
A xB-B xA
vectorsofsystemhandedrightaformCBAsuch thatdirected
BandAofplanelar toperpendicurunit vectoisnand
)θ0(BandAbetweenangleis'θ'where
Ex.,
π
CO-ORDINATE SYSTEMS :
For an explicit representation of a vector quantity, a ‘co-ordinate system’ is essential.
Different systems used :
Sl.No. System Co-ordinate variables Unit vectors1. Rectangular x, y, z ax , ay , az2. Cylindrical ρ, , z aρ , a , az3. Spherical r, , ar , a , a
These are ‘ORTHOGONAL‘ i.e., unit vectors in such system of co-ordinates are mutuallyperpendicular in the right circular way.
r,z,zyxi.e.,
RECTANGULAR CO-ORDINATE SYSTEM :
Zx=0 plane
azp
y=0 Yplane ay
ax z=0 planeX
yxz
xzy
zyx
xzzyyx
aa xa
aa xa
aa xa
0a.aa.aa.a
az is in direction of ‘advance’ of a right circular screw as it is turned from ax to ay
Co-ordinate variable ‘x’ is intersection of planes OYX and OXZ i.e, z = 0 & y = 0
Location of point P :
If the point P is at a distance of r from O, then
If the components of r along X, Y, Z are x, y, z then
arazaya xr rzyx
Equation of Vector AB :
ZandYX,alongBofcomponentsareB&B,Band
ZandYX,alongAofcomponentsareA&A,Awhere
A-BABorBABA
thenaBaBaBBOBand
aAaAaAAOAIf
zys
zys
zzyyxx
zzyyxx
Dot and Cross Products :
get wegrouping,andby term termproducts'Cross'Taking
)aBaBa(B x)aAaAa(AB xA
CABABA)aBaBa(B.)aAaAa(AB.A
zzyyxxzzyyxx
zzyyxxzzyyxxzzyyxx
zyx
zyx
zyx
BBBAAAaaa
B xA
)AB.(ABABABlengthVector
where
AB
ABa
ABalongVectorUnit
CandB,Asidesofoidparallelopaof volume therepresents)C xB(.A(ii)
parallelareBandA0θ0θSin then0B xA
larperpendicuareBandA90θi.e.,0θCos then0B.A(i)
vectors,zerononareCandB,AIf
CCCBBBAAA
)C xB(.A
AB
0
zyx
zyx
zyx
B
B
AB
0 A
A
Differential length, surface and volume elements in rectangular co-ordinate systems
zyx
zyx
adzadyadxrd
dzzrdyrdx
xrrd
azaya xr
y
Differential length 1-----]dzdydx[rd 1/2222
Differential surface element, sd
1. zadxdy:z tor2. zadxdy:z tor ------ 23. zadxdy:z tor
Differential Volume element
dv = dx dy dz ------ 3
zdx p’
p dzdy
r
rdr
0 y
xOther Co-ordinate systems :-
Depending on the geometry of problem it is easier if we use the appropriate co-ordinate system thanto use the Cartesian co-ordinate system always. For problems having cylindrical symmetrycylindrical co-ordinate system is to be used while for applications having spherical symmetryspherical co-ordinate system is preferred.
Cylindrical Co-ordiante systems :-z
P(ρ, , z) x = ρ Cos y = ρ Sin
az r ρ z = z
ap r y
zzy / x tanφ
yxρ1-
22
ρ
x
0
1zrhar
ρrh;aρaraCosρaSinρ-r
1ρrh;aha
ρraSinaCos
ρr
1------dzzrdrρd
ρrrd
azaSinρaCosρr
azaya xr
zz
yx
ρρρyx
zyx
zyx
z
Thus unit vectors in (ρ, , z) systems can be expressed in (x,y,z) system as
2222
z
zzz
y
xyxρ
(dz))dρ(ρdrdand
2------adzadρaρdrd,Furtherorthogonalareaanda,a;aa
aCosaSinaaCosaSin-a
aSinaCosaaSinaCosa
yx
Differential areas :
zz
adz)ρ(dads3-------a.)dρ((dz)ads
a.)dρ()ρ(dads
Differential volume :
4-----dzdρdρdor(dz))dρ()ρ(dd
Spherical Co-ordinate Systems :-Z X = r Sin Cos
Y = r Sin Sin z p Z = r Cos
R r
0 y Y
x r Sin
X
dddrSinr vd
ddrrSdddrSinrSd
ddSinrSd
adSinradradrRd
dRdRdrrRRd
aCosaSin-R/Ra
aSinaSinCosaCosCosR/Ra
aCosaSinSinaCosSinrR/
rRa
aCosraSinSinraCosSinrR
2
2
2r
r
yx
zyx
zyxr
zyx
General Orthogonal Curvilinear Co-ordinates :-z u1 a3 u3
a1u2
a2y
x
Co-ordinate Variables : (u1 , u2, u3) ;Hereu1 is Intersection of surfaces u2 = C & u3 = Cu2 is Intersection of surfaces u1 = C & u3 = Cu3 is Intersection of surfaces u1 = C & u2 = C
33
22
11
321
333222111
33
22
11
321zyx
133221
321321
uRh,
uRh,
uRh
;factorsscaleareh,h,hwhereaduhaduhaduh
duuRdu
uRdu
uRRdthen
u&u,uoffunctionsarezy, x,&azaya xRIf
0a.a&0a.a,0a.aifOrthogonalisSystemu&u,u tol tangentiaorsubnit vectarea,a,a
Co-ordinate Variables, unit Vectors and Scale factors in different systems
Systems Co-ordinate Variables Unit Vector Scale factors
General u1 u2 u3 a1 a2 a3 h1 h2 h3
Rectangular x y z ax ay az 1 1 1
Cylindrical ρ z a ρ a az 1 ρ 1
Spherical r ar a a 1 r r sin
Transformation equations (x,y,z interms of cylindrical and spherical co-ordinate systemvariables)
Cylindrical : x = ρ Cos , y = ρ Sin , z = z ; ρ 0, 0 2 - < z <
Sphericalx = r Sin Cos , y = r Sin Cos , z = r Sinr 0 , 0 , 0 2
)u,u,(uAAand)u,u,(uAA)u,u,(uAA wherefieldVectoraisaAaAaAA&
fieldScalara)u,u,u(VVwhere
AhAhAhuuu
ahahah
hhh1A x
)Ah(hu
)Ah(hu
)Ah(huhhh
1A.
au v
h1a
u v
h1a
u v
h1V
3213332122
32111332211
321
332211
321
332211
321
3213
2312
1321321
333
222
111
Vector Transformation from Rectangular to Spherical :
z
y
x
zyx
zyx
rzryrxr
zyxr
rr
RrrRS
zzyyxxR
AAA
a.aa.aa.aa.aa.aa.aa.aa.aa.a
AAA
asA,A,A torelatedareA,A,AwhereaAaAaA
a)a.A(a)aA(a)aA(A:Spherical
aAaAaAA:rRectangula
R
Field Theory
A ‘field’ is a region where any object experiences a force. The study of performance in the presenceof Electric field )E(
, Magnetic field () is the essence of EM Theory.
P1 : Obtain the equation for the line between the points P(1,2,3) and Q (2,-2,1)
zyx a2-a4-aPQ P2 : Obtain unit vector from the origin to G (2, -2, 1)
Problems on Vector Analysis
Examples :-1. Obtain the vector equation for the line PQ between the points P (1,2,3)m and Q (2, -2, 1) m
ZPQ P (1,2,3)
Q(2,-2,-1)0
Y
X
)a2-a4-a(
a3)-(-1a2)-(-2a1)-(2
a)z-(za)y-(ya) x-(xPQ vectorThe
zyx
zyx
zpqypqxpq
2. Obtain unit vector from origin to G (2,-2,-1)
G
G
0
)a0.333-a0.667-a(0.667a
3(-1)(-2)2G
GGa,orunit vectThe
)a-a2-a(2
a0)-(za)0-(ya)0-(xG vectorThe
zyxg
222
g
zyx
zgygxg
3. Given
zyx
zyx
a5a2-a4-B
aa3-a2A
B xA(2)andB.A(1)find
Solution :)a5a2-a(-4.)aa3-a(2B.A(1) zyxzyx
= - 8 + 6 + 5 = 3Since ax . ax = ay . ay = az . az = 0 and ax ay = ay az = az ax = 0
(2)524132aaa
B xAzyx
= (-13 ax -14 ay - 16 az)
4. Find the distance between A( 2, /6, 0) and B = ( 1, /2, 2)Soln : The points are given in Cylindrical Co-ordinate (ρ,, z). To find the distance between twopoints, the co-ordinates are to be in Cartesian (rectangular). The corresponding rectangular co-ordinates are (ρ Cos, ρ Sin, z)
2.6421.73AB)(
a2a1.73-
a0)-(2a1)-(1a1.73-
a)A-(Ba)A-(Ba)A-(BAB
a2aa2a2
Sina6
CosB&
aa1.73a6
Sin2a6
Cos2A
22
zx
zyx
zzzyyyxxx
zyzyx
yxyx
5. Find the distance between A( 1, /4, 0) and B = ( 1, 3/4, )Soln : The specified co-ordinates (r, , ) are spherical. Writing in rectangular, they are (r Sin
Cos , r Sin Sin , r Cos ).Therefore, A & B in rectangular co-ordinates,
1.7320.5)0.5(2
)AB.AB(AB
a(-0.707)a0.707)(-a1.414-
a)A-(Ba)A-(Ba)A-(BAB
)a0.707a0.707(
)a4
3CosaSin4
3SinaCos4
3Sin(B
)a0.707a0.707(
)a4
Cos1a0Sin4
Sin1a0Cos4
Sin(1A
1/2
1/2
zyx
zzzyyyxxx
yx
zyx
yx
zyx
6. Find a unit vector along AB in Problem 5 above.
AB
ABaAB = [ - 1.414 ax + (-0.707) ay + (-0.707) az]1.732
1
= )a408.0a0.408-a0.816-( zyx
7. Transform ordinates.-ColCylindricainFinto)a6a8-a(10F zyx
Soln :
a)a.F(a)a.F(a)a.F(F zzppCyl
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