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Electromagnetic Field Theory

Dr. Sandeep Nagar

G. D. Goenka University

November 23, 2015

Dr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 1 / 349

Reference

The contents for the course has been prepared from the book:Principles of Electromagnetic6th edition Matthew N.O. Sadiku, S. V. KulkarniOxford University PressISBN - 9780199461851Students are advised to consult this or any other book for betterunderstanding of the subject.


Syllabus

Cartesian coordinates

cylindrical coordinates

Spherical coordinates

Vector calculus

Differential length, area and volume

Line, surface and volume integrals

Curl

Gradient

Divergence

Laplacian operators

Divergence theorem

Stokes theorem,

Greens Theorem


Basics of vector calculus

Revise the basics of vector calculus with topics

Definition of vector

Addition and subtraction of vectors

Dot and cross product


Coordinate system

Cartesian coordinates

Cylindrical coordinates



Differential length, area and volume

NEED:

While solving physical problems, especially in conditions whennon-uniformity is an issue, small pieces of length, surface and/orvolume can be defined where uniformity can be safely assumed.

Vectors usually direct towards a point in space rather than a bigpiece of line, surface or volume.


Differential Length, Surface, Area

Length is usually associated as displacement in physical terms.

Differential displacement, surface and volume in cartesian coordinatesare shown in Fig. 7

Figure: Differential elements in cartesian coordinates



~dl = dx ~ax + dy ~ay + dz ~az (1)

~dSx = dy .dz . ~ax (2)~dSy = dz .dx . ~ay (3)

~dSz = dx .dy . ~az (4)

dv = dx .dy .dz (5)

~dl and ~dS are vectors while dv is a scalarGeneral definition of surface element

~dS = dS . ~an

where:dS is area of surface element~an is unit vector normal to surface dS which directed away from thesurface (if surface is part of the volume, ~an points away from the). SeeFigure 7.



In figure 7, area of surface ABCD and PQRS will be same (same dS)value but opposite in direction.

The two areas will be:

~dSABCD = dy .dz . ~ax

and~dSPQRS = −dy .dz . ~ax

because ~dSABCD grown towards positive x-axis and ~dSPQRS growstowards negative x-axis


Differential Length, Surface, Area in cylindrical coordinates

Figure: Differential elements in cylindrical coordinates



~dl = d .ρ. ~aρ + ρ.dφ. ~aφ + dz . ~az (6)

~dSρ = ρ.dφ.dz . ~aρ (7)

~dSφ = dρ.dz . ~aφ (8)

~dSz = ρ.dφ.dρ. ~az (9)

dv = ρ.dρ.dφ.dz (10)


Differential Length, Surface, Area in spherical coordinates

Figure: Differential elements in spherical coordinates



~dl = dr .~ar + r .dθ.~aθ + r .sinθ.dφ. ~aφ (11)

~dSr = r2.sinθ.dθ.dφ.~ar (12)~dSθ = r .sinθ.dr .dφ.~aθ (13)

~dSφ = r .dr .dθ. ~aφ (14)

dv = r2.sinθ.dr .dθ.dφ (15)


Line, surface and volume integrals

By a line we mean a path along a curve in space (1D movement)

By a surface we mean a 2D surface in space

By a volume we mean a 3D volume in space

A line integral of vector field ~A around a curve L is given by∫L

~A.d~l =

∫ b

a|~A|cosθ.dl

where line L starts from a and ends at b and θ is the angle between ~A anddifferential length d~l


Line Integral

Circulation of ~A around L is shown by integration for a closed loopdepicted by: ∮

L

~A.d~l

Figure: Line integralDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 15 / 349

Surface Integral

Surface Integral

ψ =

∫S

~A. ~andS =

∫S

~A.d ~S (16)

Figure: Surface integral


Closed Surface Integral

A closed surface encloses a volume

A closed surface integral is shown by:

ψ =

∮S

~A.d ~S

Physicaly, it signifies net outward flux


Volume integral

Volume integral of a

V =

∫vρvdv

Its important to note that both ρv as well as dv are scalar quantifies.


Del/ Gradient Operator in cartesian coordinates

∇ symbol signifies the del / gradient operator

∇ =∂

dx~ax +

∂

dy~ay +

∂

dz~az

Gradient operator operates on a scalar and produces a vector.

~∇V Gradient of scalar V~∇2V Laplacian of scalar V~∇. ~A Divergence of vector ~A~∇× ~A Curl of vector ~A


Gradient operator in cylindrical coordinates

Since ρ =√

x2 + y2 and tanφ = yx hence:

∂

∂x= cosφ

∂

∂ρ− sinφ

ρ

∂

∂φ(17)

∂

∂y= sinφ

∂

∂ρ+

cosφ

ρ

∂

∂φ(18)

Which yields ~∇ operator in cylindrical coordinates as:

~∇ = ~aρ∂

∂ρ+ ~aφ

1

ρ

∂

∂φ+ ~az

∂

∂z(19)


Gradient Operator for speherical coordinates

Find out gradient operator for speherical coordinates when we know that:

ρ =√x2 + y2 + z2

and

tanφ =

√x2 + y2

z



∂

∂x= sinθcosφ

∂

∂r+

cosθcosφ

r

∂

∂θ− sinφ

ρ

∂

∂φ(20)



∂

∂y= sinθsinφ

∂

∂r+

cosθsinφ

r

∂

∂θ− cosφ

ρ

∂

∂φ(21)



∂

∂z= cosθ

∂

∂r− sinθ

r

∂

∂θ(22)



~∇ = ~ar∂

∂r+ ~aθ

1

r

∂

∂θ+ ~aφ

1

r .sinθ

∂

∂r(23)


Gradient

Gradient (~∇) is used to find the direction of rate of change of a scalarfield

Mathematical meaning:Consider two points P1 and P2in space where contours V1,V2 and V3 aredefined.

Figure: Gradient of a scalar fieldDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 26 / 349

Gradient

dV =∂V

∂xdx +

∂V

∂ydy +

∂V

∂zdz (24)

dV = (∂V

∂x~ax +

∂V

∂y~ay +

∂V

∂z~az).(dx ~axdy ~ay + dz ~az) (25)

If:

~G = (∂V

∂x~ax +

∂V

∂y~ay +

∂V

∂z~az) (26)

Then:

dV = ~G .d~l = Gcosθdl (27)

dV

dl= Gcosθ (28)


Gradient

G = ~∇V =dV

dl|max =

dV

dn

where θ = 900 i.e. dVdn is normal derivative

Direction of G = ~∇V is that of maximum rate of change of fieldmagnitude~∇V is defined in same cartesian coordinate system as the definition

of scalar field V


Gradient operator in different coordinate systems

Cartesian Coordinate system

~∇V =∂V

∂x~ax +

∂V

∂y~ay +

∂V

∂z~az (29)

Cylindrical Coordinate system

~∇V =∂V

∂ρ~aρ +

1

ρ

∂V

∂φ~aφ +

∂V

∂z~az (30)

Spherical Coordinate system

~∇V =∂V

∂r~ar +

1

r

∂V

∂θ~aθ +

1

rsinθ

∂V

∂ψ~aψ (31)


Formulas for ~∇ operator

~∇(V + U) = ~∇V + ~∇U

~∇(VU) = V ~∇U + U ~∇V

~∇(V

U) =

U ~∇V − V ~∇UU2

~∇n = nV n−1~∇V

Here V and U are scalars and n is an integer.


Meaning of gradient operation

The magnitude of ~∇V equals the maximum rate of change of V perunit distance~∇V points in the direction of the maximum rate of change in V .~∇V at any point is perpendicular to the constant V surface thatpasses through that point (See points P and Q in figure 31)

Figure: Gradient of a scalar field


Meaning of gradient operation

The projection of ~∇V in direction of a unit vector ~a (i.e. ~∇V .~a) iscalled the directional derivative of V along ~a.

It signifies the rate of change of V in the direction of ~a.

If ~A = ~∇V the V is called the scalar potential of ~A


Divergence

Divergence of a vector ~A is given by ~∇. ~A where:

~∇. ~A = limδ→0

∮~A.d ~S

δv(32)

here δv is the volume enclosed by the closed surface S in which P islocated.

In words:

The divergence of ~A at a given point P is the outward flux per unitvolume as the volume shrinks about P

It gives a measure about how much the vector field ~A diverges fromthis point.


Divergence

Figure: Divergence of a vector field at point P, (a) positive divergence (b)negative divergence (c) zero divergence


Divergence from a surface

Figure: Calculation of divergence (~∇. ~A) at a point P


Divergence in different coordinate systems

Cartesian Coordinates

~∇. ~A =∂Ax

∂x+∂Ay

∂y+∂Az

∂z

Cylindrical Coordinates

~∇. ~A =1

ρ

∂

∂ρ(ρAρ) +

1

ρ

∂Aφ∂φ

+∂Az

∂z

Spherical Coordinates

~∇. ~A =1

r2

∂

∂r(r2Ar ) +

1

rsinθ

∂

∂θ(Aθsinθ) +

1

rsinθ

∂Aφ∂φ


Properties of divergence of a vector field

Divergence of a vector field gives a scalar field as output.

Divergence of a scalar field makes no sense.

Divergence of sum of vector fields

~∇.(~A + ~B) = ~∇. ~A + ~∇. ~B

Divergence of a scalar product of a vector field

~∇.(V ~A) = V ~∇. ~A + ~A.~∇V


Divergence theorem/ Gauss theorem

Mathematical statement:∮S

~A.d ~S =

∫~∇. ~Adv

In words:Total outward flux of a vector field ~A through a closed surface S isthe same as volume integral of the divergence of ~A

Graphical

Figure: Guass Theorem


Proof of divergence theorem

Subdivide volume v into a large number of cells

suppose volume of kth cell is ∆vk and it is bounded by surface Skthen ∮

S

~A.d ~S =∑k

∮Sk

~A.d ~S =∑k

∮Sk~A.~S

∆vk∆vk

Since outward flux of one cell is inward flux to its neighbors, forinterior surface, the flux is canceled.⇒ Hence the Sum of surface integrals over surfaces shown by Sk ’s isequal to surface integral over the surface S .

Also, by definition :

~∇. ~A = limδ→0

∮~A.d ~S

δv


Proof of divergence theorem

Hence R.H.S = L.H.S ∮S

~A.d ~S =

∫v

~∇. ~Adv

Vector field ~A must be continuous in region defined by volume venclosed by surface S .


Curl

In Mathematical terms:curl ~A is:

A = ~∇× ~A = ( lim∆S→0

∮L~A.d~l

∆S)max ~an (33)

Here the surface(area) ∆S is bounded by the curve L and ~an is theunit vector normal to this surface(area).

In words:The curl of ~A is an axial (or rotational) vector whose magnitude is themaximum circulation of ~A per unit area as area tends to zero andwhose direction is the normal direction of the area when area isoriented so as to make the circulation maximum.


Curl of a vector field

Figure: Curl of ~A


Curl of ~A

Consider a differential area in yz plane (enclosed by circulation abcd )

A line integral for length L made up by going along the pathab → bc → cd → da

Mathematically:∮~A.d~l = (

∫ab

+

∫bc

+

∫cd

+

∫da

)~A.d~l


Curl of ~A

∮~A.d~l = (

∫ab

+

∫bc

+

∫cd

+

∫da

)~A.d~l

Using Taylor series expansion about the center point P(x0, y0, z0) andusing the fact that on side ab, d~l = dy ~ax and z = z0 − dz

2

∫ab

~A.d~l = dy [Ay (x0, y0, z0)− dz

2

∂Ay

∂z|P ] (34)

Similarly, on side bc, d~l = dz ~az and y = y0 − dy2

∫bc

~A.d~l = dz [Az(x0, y0, z0)− dy

2

∂Az

∂y|P ] (35)


Curl of ~A

Similarly, on side cd , d~l = dy ~ay and z = z0 − dz2

∫cd

~A.d~l = −dy [Ay (x0, y0, z0)− dz

2

∂Ay

∂z|P ] (36)

and also, on side da, d~l = dz ~az and y = y0 − dy2

∫da

~A.d~l = −dz [Az(x0, y0, z0)− dy

2

∂Az

∂y|P ] (37)


Curl of ~A

Substituting all the equations and making ∆S = dy .dz , we get:

lim∆→0

∮ ~A.d~l

∆S=∂Az

∂y− ∂Ay

∂z

Hence:

(curl ~Ax) =∂Az

∂y− ∂Ay

∂z

Similarly

(curl ~Ay ) =∂Ax

∂z− ∂Az

∂x

and

(curl ~Az) =∂Ay

∂x− ∂Ax

∂y


Physical meaning of curl of a vector field

Curl is the maximum value of the circulation of the field per unit area(circulation density)

Curl indicates the direction along which this maximum value occurs


Properties of curl

The curl of a vector field is another vector field.

The curl of a scalar field V will make no sense.~∇× (~A + ~B) = (~∇× ~A) + (~∇× ~B)

Divergence of curl of a vector field vanishes i.e ~∇.( ~∇× A) = 0

Curl of gradient of a scalar field vanishes i.e ~∇× ~∇V = 0~∇× (V ~A) = V ~∇× ~A + ~∇× ~A

Figure: Curl of ~A around a point P: (a) curl at P points out of the page, (b) curlat P is zero.


Stoke’s theorem

∮L

~A.d~l =

∫S

(~∇× ~A.d ~S) (38)

Figure: Stoke’s theorem

In words:

Circulation of a vector field ~A around a closed path L is equal to thesurface integral of the curl of ~A over the open surface S bounded byL provided that ~A and ~∇× ~A are continous on S .


Proof of Stoke’s theorem

Proof is Stoke’s theorem is similar to Guass theorem.

Surface S is subdivided into a large number if cells.

Figure: Stoke’s theorem


Proof of Stoke’s theorem

∮L

~A.d~l =∑k

∮Lk

~A.d~l =∑k

∮lk~A.d~l

∆Sk∆Sk (39)

Interior paths will cancel all the circulations and only the path/line on theboundary will survive the calculations i.e.

∮L

~A.d~l =

∫S

(∆× ~A).d ~S (40)


Laplacian

Composite of gradient and divergence operators i.e. Laplacian of ascalar field V is the divergence of gradient of V .

∇2V = ~∇.~∇V


Laplacian

Cartesian coordinates:

∇2V =∂2V

∂x2+∂2V

∂y2+∂2V

∂z2(41)

Cylindrical coordinates

∇2V =1

ρ

∂

∂ρ(ρ∂V

∂ρ) +

1

ρ2

∂2V

∂ψ2+∂2V

∂z2(42)


∇2V =1

r2

∂

∂r(r2∂V

∂r) +

1

r2sinθ

∂

∂θ(sinθ

∂V

∂θ) +

1

r2sin2θ

∂2V

∂φ2(43)


Laplacian

A scalar field V is said to be Harmonic in a given region if itsLaplacian vanishes i.e. ~∇V = 0


Laplacian of vector field

~∇2 ~A = ~∇(~∇. ~A)− ~∇× ~∇× ~A (44)


Classification of vector fields

Figure: Vector fields

(a) ~∇. ~A = 0, ~∇× ~A = 0

(b) ~∇. ~A 6= 0, ~∇× ~A = 0

(c) ~∇. ~A = 0, ~∇× ~A 6= 0

(d) ~∇. ~A 6= 0, ~∇× ~A 6= 0


Types of vector fields

Solenoidal = Divergenceless ~∇. ~A = 0

No source no sinkExamples:Conduction current density in steady state

Irrotational ~∇× ~A = 0

Also called conservative field because as per Stoke’s theorem, lineintegral of ~A is independent of path.Example: Electrostatic field, gravitational field.


Unit 2: Electrostatics

Coulomb’s law, Electric field intensity, Electric field due to volume charge,line charge and sheet charge distribution, Electric flux density, Gausss Law,Applications of Gausss Law Symmetrical charge distribution, Differentialvolume element, Energy and potential in a moving point charge in anelectric field, potential gradient, Dipole, Energy density in electric fields,Current and current density, continuity of current, boundary conditions forconductors and dielectrics, method of images, Electrostatic boundary valueproblems: Poissions and Laplaces equations, Uniqueness theorem


Introduction to Electrostatics

Electrostatic fields are produced by static charge distribution.

Two fundamental laws governing electrostatic fields are:

Coulomb’s lawGuass’s law

Both laws are based on experimental studies

Both laws are inter-dependent


Coulomb’s law

Formulated in 1785 by French scientist Charles Augustin de Coulomb

Force F between two point charges Q1 and Q2 is:

Along the line joining them.Directly proportional to the product Q1Q2 of the charges.Inversely proportional to the square of distance (R) between them.

Equation for Coulomb’s law

F =kQ1Q2

R2

k is constant of proportionality

k =1

4πε0= 9× 109m/F

Where ε0 = 8.854× 10−12F/m (permittivity of free space)


Coulomb’s law

Substituting k we get:

F =1

4πε0

Q1Q2

R2

Figure: Charges separated at a distance


Coulomb’s law

In vector form:When charge Q1 and Q2 are placed at position vectors ~r1 and ~r2, then theforce is given by:

~F12 =1

4πε0

Q1Q2

R2~aR12

here~R12 = ~r2 − ~r1 alsoR = |~R12|Hence the unit vector ~aR12 =

~R12R

~F12 =1

4πε0

Q1Q2

R3~R12

~F12 = − ~F21


Coulomb’s law

Consider the sign of charge during calculations

Distance R must be large compared to linear dimensions of thecharges so that they can be considered point charges.

Charges must be static

What if we have many body problem instead of two body problem?


Coulomb’s law for many body problem

Resultant force ~F on a charge Q situated at position vector ~r due to Ncharges can be given by:

~F =Q

4πε0

N∑k=1

Qk(~r − ~rk)

|~r − ~rk |3


Electric field intensity

Electric field intensity (Electric field strength) is force per unit chargewhen placed in an electric field

~E = limQ→0

~F

Q

Write in terms of position vector and for N charges


Types of continous charge distribution

Figure: Charge distributions


Electric field due to continous charge distribution

Line charge distribution

dQ = ρLdl ⇒ Q =

∫LρLdl

Surface charge distribution

dQ = ρSdS ⇒ Q =

∫SρSdS

Volume charge distribution

dQ = ρvdv ⇒ Q =

∫vρvdv



Electric field intensity due to continous charge distribution can beconsidered in terms of summation of the fields make by numerouspoint charges which make up the above-said charge distribution.

Mathematically, this can be accomplished by replacing Q with ρldl ,ρSdS and ρvdv for line, surface and volume charge distribution andintegrating the same.



For a Line charge distribution

~E =

∫ρLdl

4πεR2~aR

For Surface charge distribution

~E =

∫ρSdS

4πεR2~aR

For a Volume charge distribution

~E =

∫ρvdv

4πεR2~aR

For ease of calculation of ρl , ρS and ρv : we shall consider that chargedistribution is uniform.



Electric field due to continuous line charge distribution

Electric field due to continuous surface charge distribution

Electric field due to continuous volume charge distribution



Figure: Evaluation of ~E due to a Line Charge distribution



Consider a line charge with uniform charge density ρL extending fromA toB along the z-axis

Differential charge element dQ associated with differential lineelement dl is given by:

dQ = ρLdl = ρLdz

Total charge thus becomes:

Q =

∫ zB

zA

ρLdz

For a Line charge distribution, electric field intensity is given by:

~E =

∫ρLdl

4πεR2~aR



It is important to define ~R in this problem.

We shall consider working in cartesian coordinate system

It is customary to represent the field point with coordinates (x , y , z)and source points with (x

′, y

′, z

′)

Thus the differential length dl exists only in z-direction:

dl = dz′

Thus ~R can be defined in terms of two position vectors:

~R = (x , y , z)− (0, 0, z′) = x ~ax + y ~ay + (z − z

′)~az

If we consider a horizontal length ρ along y -axis then

~R = (x , y , z)− (0, 0, z′) = ρ~aρ + (z − z

′)~az



Now given:~R = ρ~aρ + (z − z

′)~az

We can define magnitude of R as:

R = |~R|2 = x2 + y2 + (z − z′)2 = ρ2 + (z − z

′)2

This leads to definition of unit vector in direction of ~rho as:

~aRR2

=~R

|~R|3=

ρ~aρ + (z − z′)~az

[ρ2 + (z − z ′)2]3/2

This makes the expression for electric field as:

~E =ρL

4πε

∫ρ~aρ + (z − z

′)~az

[ρ2 + (z − z ′)2]3/2dz

′



For convenience, we can define the angles α, α1 and α2 as

R = [ρ2 + (z − z′)2]1/2 = ρ.secα

andz′

= OT − ρ.tanα

alsodz

′= −ρ.sec2α.dα

This results in ~E expressed as:

~E =−ρL4πε0

∫ α2

α1

ρ.sec2α[cosα~aρ + sinα~az ]

ρ2.sec2αdα



This ultimately leads to:

~E =−ρL

4πε0ρ

∫ α2

α1

[cosα~aρ + sinα~az ]dα

For a finite line charge:

~E =−ρL

4πε0ρ

∫ α2

α1

[−(sinα2 − sinα1)~aρ + (cosα2 − cosα1)~az ]dα

For infinite line charge: B(0, 0,∞) and A(0, 0,−∞) and α1 = π/2,α2 = −π/2. This makes z-component vanish, which results in:

~E =ρL

2πε0ρ~aρ


Electric field due to continuous Surface charge distribution

Figure: Evaluation of ~E due to a Surface Charge distribution



Consider an infinite sheet of charge in xy plane with uniform surfacecharge density ρS . i.e.

dQ = ρSdS

which gives total charge as:

Q =

∫ρSdS

Contribution to ~E field at point P(0, 0, h):

d ~E =dQ

4πε0R2~aR



Position vector can be defined as:

~R = ρ(−~aρ) + h ~az

Magnitude of position vector:

R = |~R| =√

(ρ2 + h2)

Unit vector ~aR can be defined as:

~aR =~R

R

differential charge will depend on differential surface dS as:

dQ = ρSρdφdρ



This enables us to write electric field as:

d ~E =ρSρdφdρ[−ρ~aρ + h ~az ]

4πε0[ρ2 + h2]3/2

Figure: Evaluation of ~E due to a Surface Charge distribution

For every element 1 there is an element 2 which contribute equallyand opposite.



Thus contributions due to Eρ cancel out and ~E has only componentfrom z direction

~E =

∫d ~Ez =

ρS4πε0

∫ 2π

φ=0

∫ ∞ρ=0

hρdρdφ

[ρ2 + h2]3/2~az

which leads to

~E =ρS2ε0

~az

When we consider xy plane, ~E has only z-component.

In general, E is directed normal to the infinite conducting sheet.



Electric field is independent of distance between sheet andobservation point!

Parallel plate capacitor has two sheets with equal and oppositecharges:

~E =ρS2ε0

~an +−ρS2ε0

(−~an) =ρSε0~an



Figure: Evaluation of ~E due to a Volume Charge distribution



For a uniform volume charge distribution ρv in differential volume dv ,total charge is given by:

dQ = ρvdv

Total charge will be:

Q =

∫ρvdv = ρv

∫dv = ρvvsphere = ρv

4πa3

3



Electric field d ~E at P(0, 0, z) due to differential charge volume:

d ~E =ρvdv

4πε0R2~aR

Where

~aR = cosα~az + sinα~aρ

Due to spherical symmetry of charge distribution, contributions of Ex

and Ey sum up to be zero leaving us with only Ez i.e:

Ez = ~E . ~az =

∫dE .cosα =

ρv4πε0

∫dv .cosα

R2



Ez is given by:

Ez =ρv

4πε0

∫dv .cosα

R2

Now dv , cosα and R needs to be calculated.

dv is given by:

dv = (r′)2(sinθ

′)dr

′dθ

′dφ

′

Using cosine rule

R2 = z2 + (r′)2 − 2Rr

′.cosθ

′(45)

(r′)2 = z2 + R2 − 2zR.cosα (46)



Calculating cosα:

cosα =z2 + R2 − (r

′)

2zR

Also the expression for cosθ′

can be written as:

cosθ′

=z2 + (r

′)− R2

2zR

Differentiating w.r.t θ′

keeping z and r′

to be fixed:

sinθ′dθ

′=

RdR

zr ′

Also we already derived

dv = (r′)2(sinθ

′)dr

′dθ

′dφ

′



Substituting all terms back to original expression, we obtain:

Ez =ρv

4πε0

∫ 2π

φ′=0dφ

′∫ a

r ′=0

∫ z+r′

R=z−r ′(r

′)2 rdR

zr ′dr

′ z2 + R2 − (r′)

2zR

1

R2

Final result for a point P(00, z) will be:

~E =Q

4πε0z2~az

Hence for any general point P(r , θ, φ), electric field can be written as:

~E =Q

4πε0r2~ar


Electric Flux density

Since we defined electric field as:

~E =Q1Q2

4πεR2~aR

Hence electric field at a particular point depends on dielectricproperties of material (related to ε value: ε0 is for free space).

For practical reasons, we would require a new vector which can beindependent of the material

~D = ε0~E

Electric flux given by the symbol Ψ can be defined as

Ψ =

∫~D.d ~S

SI unit of Ψ is Coulombs (C ) and that of ~D Electric flux density isC/m2


Electric Flux density

All formulas derived till now for ~E can also be written for ~D by simplymultiplying them with ε0


Gauss’s Law

In words:

Total electric flux Ψ through any closed surface is equal to the totalcharge enclosed by that surface.

Mathematically:

Ψ = Qenclosed

Now Psi can be written as:

Ψ =

∮dΨ =

∮S

~D.d ~S

And Q can be written as:

Q =

∫vρvdv


Gauss’s Law

This implies that

Q =

∮S

~D.d ~S =

∫vρvdv

By applying divergence theorem i.e.∮S

~D.d ~S =

∫v

~∇. ~Ddv

This leads to:

ρv = ~∇. ~D (47)

This is first of four maxwell’s equations

It states that volume charge density is same as divergence of electricflux density.


Gauss’s Law

It is important to note that both equations∮S

~D.d ~S =

∫v

~∇. ~Ddv

and

ρv = ~∇. ~D

state Guass law but in two forms i.e. Integral and Differential forms.

Guass law is alternate statement of coulomb’s law (Application ofdivergence theorem to Coulomb’s law results in Guass’s law)


Gauss’s Law

Practical use of Guass law is to calculate electric field due to aparticular charge distribution.

Procedure:

First find out: Does symmetry exists?If Symmetry exists then construct a mathematical closed surface(Guassian Surface) ~S such that ~D is normal or tangential to theGaussian Surface.When ~D is normal to surface, then ~D.d ~S = D.dS since angle between~D and d ~S is 0.When ~D is tangential to surface, then ~D.d ~S = 0 since angle between~D and d ~S is 900.

Lets apply this procedure to point charge, line charge distribution,surface charge distribution and volume charge distribution.


Gauss’s Law

For point charge

Suppose a point charge Q is placed at origin and we wish to find theelectric flux density at point P placed at distance r from origin

In this case, a sphere of radius P will form the Gaussian surface.

Figure: Gaussian surface about a point charge


Gauss’s Law

For point charge

Since ~D is normal to Gaussian surface

~D = Dr ~ar

Applying Guass law:

Q =

∮~D.d ~S = Dr

∮dS = Dr .4π.r

2

Which results in~D =

Q

4πr2~ar


Gauss’s Law

For infinite line charge

Figure: Gaussian Surface about a Infinite Line Charge


Gauss’s Law


Consider a infinite length line charge having uniform charge densitygiven by ρL lying along z-axis.

We aim to calculate the charge at point P situated at distance ρ.

Gaussian surface in this case is a cylindrical surface containing P.

Since ~D is constant and normal for the Gaussian surface:

~D = D ~aρ


Gauss’s Law


For an arbitrary length of line i.e. l we can write using Gauss law:

ρLl = Q =

∮~D.d ~S = Dρ

∮dS = Dρ2πρl

Hence ~D is given by:~D =

ρL2πρ

~aρ

Since ~D has no component along z-axis so top and bottom surfacewill not contribute in the integral term.


Gauss’s Law

For infinite sheet charge

Consider a sheet placed at z = 0 plane having uniform surface chargedensity of ρS (C .m−2)

As shown in figure:

Figure: Gaussian Surface about a Infinite Surface Charge


Gauss’s Law


We shall imagine Gaussian surface as a rectangle being cut by arectangular cross-section composed of charges from the sheet.

Since ~D is normal to the sheet ~D = Dz ~az .

Applying Guass’s law:

ρs

∫dS = Q =

∮~D.d ~S = DS [

∫top

dS +

∫bottom

dS ]

~D.d ~S = 0 for sides since ~D does not have any components along ~axand ~ay .


Gauss’s Law


If the area of top and bottom is A then one can write:

ρSA = Dz(A + A)

which implies:

~D =ρS2~az

which in turn implies that ~E is given by:

~E =~D

ε=ρS2ε~az


Gauss’s Law

For uniformly charged Sphere

Figure: Gaussian Surface about an uniformly charged sphere with two cases:r < a and r > a


Gauss’s Law



Gauss’s Law


Consider a uniformly charge sphere with radius a.

Two Gaussian surfaces can be considered with conditions

1 r < a2 r > a

~D is normal to surface of Gaussian surfaces and constant for a pointon them.

Enclosed charge for the case r < a is given by:

Qenc =

∫ρvdv = ρv

∫ 2π

φ=0

∫ 2π

θ=0

∫ r

r=0r2sin(θ)dr .dθ.dφ

which gives:

Qenc = ρv4

3πr2


Gauss’s law

Also,

Ψ =

∮~D.d ~S = Dr4πr2

Which gives

Dr4πr2 =4

3πa3ρv

Finally we get:

~D =a3

3r2ρv ~ar

for r ≥ a


Gauss’s law

~D =r

3ρv ~ar

for 0 < r ≤

~D =a3

3r2ρv ~ar

for r ≥ a


Gauss’s law

Figure: ~D variation with distance, with two cases: r < a and r > a


Electric potential

A quantity called potential (scalar) can be defined so that ~E can bedefined

Figure: Concept of electric potential derived using displacement of point charge Qin electric field ~EDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 109 / 349

Electric Potential

Work done (negative because work is done by external agent) in movingcharge Q from point A to point B:

W = −~F .d~l = −Q ~E .d~l

Hence total potential energy:

W = −Q∫ B

A

~E .d~l

A new quantity, potential energy per unit charge (unit = J/C = V ) canbe defined after which one can define potential difference VAB as

VAB =W

Q= −

∫ B

A

~E .d~l

Note: VAB is independent of path taken


Electric potential due to point charge

~E =Q

4πε0~ar

Now potential difference can be calculated as:

VAB = −∫ rB

rA

Q

4πε0~ar .dr ~ar

This results in:

VAB =Q

4πε0[

1

rA− 1

rB]

Which is same as:VAB = VA − VB

Where VA and VB are called absolute potentials at points A and Brespectively.VAB is potential at A w.r.t B


Electric potential

For point charges, its customary to take ∞ as the reference point,where potential is assumed to be zero (V∞ = 0).

If VA = 0 as rA → 0 then potential at any point rB → B will be givenby:

V =Q

4πε0r

In words: Potential at any point is the potential difference betweenthat point and a chosen point where potential is zero.

Potential at a distance r from the point charge is the work done perunit charge by an external agent in transferring a test charge frominfinity to that point

V = −∫ r

∞~E .d~l


Electric potential

If reference point is not considered to be infinity. then :

V =Q

4πε0r+ C

Where C is constant chosen as per reference point.


Electric potential

If the point charge Q is located at a position vector ~r ′ then:

V (~r) =Q

4πε0|~r − ~r ′ |Potential due to a point charge distribution:

V (~r) =1

4πε0

n∑k=1

Qk

|~r − ~r ′ |


Electric potential

Line charge distribution:

V (~r) =1

4πε0

∫L

ρL(~r ′)dl′

|~r − ~r ′ |

Surface charge distribution

V (~r) =1

4πε0

∫S

ρS(~r ′)dS′

|~r − ~r ′ |

Volume charge distribution

V (~r) =1

4πε0

∫v

ρv (~r ′)dv′

|~r − ~r ′ |


Relationship between ~E and V

Since potential is independent of path taken:

VBA = −VAB

This leads to the fact that

VAB + VBA = 0 =

∮~E .d~l

Physically, it will mean that net work done for moving a charge inclosed path will be zero.

Applying stokes theorem:∮~E .d~l =

∫(~∇× ~E ).d ~S = 0

~∇× ~E = 0

second Maxwell’s equation for electrostatics


Conservative nature of electric field

Figure: Conservative nature of electric field



By definition of V :

V = −∫~E .d~l

Differential potential:

dV = −~E .d~l = −Exdx − Eydy − Ezdz

and

dV =∂V

∂xdx +

∂V

∂ydy +

∂V

∂zdz



This leads to

Ex = −∂V∂x

and

Ey = −∂V∂y

and

Ez = −∂V∂z

Generalizing:

~E = −~∇VElectric field is gradient of a potential field.



Negative sign shows that direction of ~E is opposite to direction ofincrease of V~E directs from lower to higher value of V .


Electric dipole

An electric dipole is composed of two charges of equal magnitude andseparated by a small distance.

Figure: DipoleDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 121 / 349

Electric dipole

V =Q

4πε0[

1

r1− 1

r2]

results in

V =Q

4πε0[r2 − r1r1r2

]

If r d then r2 − r1 ' d .cos(θ) and r1r2 = r2.

V =Q

4πε0[dcos(θ)

r2]

Now:d .cos(θ) = ~d .~ar

where~d = d ~az


Dipole moment

Dipole moment can be defined as:

~p = Q ~d

Hence potential can be defined in terms of dipole moment:

V =~p.~ar

4πε0r2

Now we assumed that ~p is directed from −Q to +Q.

If dipole center is not at the origin but at ~r ′ becomes:

V (~r) =~p.(~r − ~r ′)

4πε0|~r − ~r ′ |3


Dipole

A point charge is a monopole and it’s electric field varies inversely asr2 while its potential field varies inversely as r~E due to a dipole varies inversely as r3 while its potential variesinversely as r2


Electric flux lines

Electric flux line is an imaginary path or line drawn in such a way thatits direction at any point is the direction of the electric field at thatpoint.

Lines which are tangential at Electric field density ~D at every point.

Equipotential surface is a surface on which the potential is sameeverywhere.

Equipotential line is line generated by intersection of equipotentialsurfaces.

On an equipotential surface, work done will be zero becauseVA − VB = 0 i.e ∫

~E .d ~S = 0

Flux lines normal to equipotential surfaces/lines.


Electric flux lines

Figure: Equipotential surface for (a) point chare (b) dipoleDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 126 / 349

Energy Density in electrostatic fields

Energy present in a collection of charges can be estimated by energytaken for making up that collectionSuppose we wish to position three point charges Q1, Q2, Q3 in aninitially empty space as shown in figure.

Figure: Assembly of charges



Since initially, the space is empty so to move Q1 to position P1, nowork is required to transfer Q1 i.i.

W1 = 0

Work done in transferring Q2 from infinity to P2 :

W2 = Q2V21

Work done in transferring Q3 from infinity to P3 :

W3 = Q3V31 + Q3V32

Total work done for the systems of charges:

WE = W1 + W2 + W3 = 0 + Q2V21 + Q3(V31 + V32)



If the charges were placed in reverse order (on same positions)

WE = W3 + W2 + W 1

i.e.WE = 0 + Q2V23 + Q1(V12 + V13)

Adding both equations:

WE =1

2(Q1V1 + Q2V2 + Q3V3)

For n point charges, the associated energy (Joules) will be:

WE =1

2

n∑k=1

QkVk



For line charge distribution:

WE =1

2

∫ρLVdl

For surface charge distribution:

WE =1

2

∫ρSVdS

For volume charge distribution:

WE =1

2

∫ρvVdv



Sinceρv = ~∇. ~D

Hence

WE =1

2

∫(~∇. ~D)Vdv

Which results into:

WE =1

2

∫v

(~∇.v ~D)dv − 1

2( ~D.~∇V )dv

Divergence theorem states that:∫v

(~∇.v ~D)dv =

∮S

(V ~D).d ~S



Hence:

WE =1

2

∮S

(V ~D).d ~S − 1

2( ~D.~∇V )dv

For point charges V ∝ 1r and ~D ∝ 1

r2

For dipoles V ∝ 1r2 and ~D ∝ 1

r3

In the term 12

∮S(V ~D).d ~S the term V ~D must at-least vary as 1

r3 and

d ~S at-least vary as 1r2 which means as as surface becomes larger,

integral term tends to zero.

Since ~E = ~∇V and ~D = ε0E2

WE =1

2( ~D.~∇V )dv =

1

2

∫v

( ~D. ~E )dv =1

2

∫vε0E

2dv



WE =1

2( ~D.~∇V )dv =

1

2

∫v

( ~D. ~E )dv =1

2

∫vε0E

2dv

Hence the energy density can be defined as

wE =dW

dv=

1

2~D. ~E =

1

2ε0E

2 =D2

2ε0

Which gives the total electrostatic energy as:

WE =

∫vwEdv


Electric field in material space

Until now we have entertained issue where charges were placed in emptyspaceNow we shall consider the charges in material space.


Current and current density

Current through a given area is amount of charge passing throughthe area per unit time.

I =dQ

dt

Current density

~J =∆I

∆S

When current density vector is perpendicular to the surface areavector

∆I = Jn∆S


Current density

When ~J is not perpendicular to d ~S then

∆I = ~J.d ~S

Total current is given by:

I =

∫S

~J.d ~S

Three kinds of current densities:

1 convection current density (Does not involve conductors and does notfollow Ohm’s law)

2 conduction current density (Does involve conductors and follow Ohm’slaw)

3 displacement current density (Will be discussed later!)


Convection Current density

Consider a filament with flow of charge (charge density ρv ) at velocity~u

Figure: Current in filament


Convection Current density

Now:

∆I =∆Q

∆t= ρv∆S

∆l

∆t= ρv∆Suy

Current density in y direction:

Jy =∆I

∆S= ρvuy

In general~J = ρv ~u


Conduction current

When an electric field ~E is applied then a force ~F acts on charge, sayelectron, e which is given by:

~F = e ~E

Electron constantly collides with atoms arranged in the lattice.Suppose the average drift velocity is ~u and average time between twocollisions is τ .

As per Newton’s second law:

m~u

τ= −e ~E

Which implies:

~u = −eτ

m~E


Conduction current

For n electrons per unit volume:

ρv = −ne

Hence conduction current density can be given by:

~J = ρv ~u =ne2τ

m~E = σ ~E

This is expression for Ohm’s law in terms of conduction current.


Conductors in electric field

Electric behavior of conductors is given by expression for ~J

When an external electric field ~E is applied:

1 Positive charges move in the direction of the electric field ~F = +q ~E2 Negative charges move in the opposite direction of the electric field~F = −q ~E

These charges moves on the surface on opposite faces and makeinduced surface charge

This induced surface charge sets up an internal induced electric field~Ei

~Ei cancels Ee .



A conductor is equipotential body i.e. ~E = −~∇V = 0

Now consider Ohm’s law: ~J = σ ~E

To maintain a finite current density ~J for a perfect conductor(σ →∞), ~Ei must vanish.

~E → 0⇔ σ →∞

If ~E = 0, according to Gauss law: ρv = 0 which implies Vab = 0



When two ends of a conductor are maintained at a potentialdifference V , due to electromotive force, charges move and prevent aelectrostatic equilibrium.

An electric field must exist inside the conductor to sustain the flow ofcharges.~E 6= 0 inside the conductor.

The resistance provides the damping force to this movement ofcharges.



Suppose the conductor is of length l and cross-section area S .

Electric field

E =V

l

Also the charge density is given as:

J =I

S

andJ = σE



This gives:I

S= σE =

σV

l

Now we define resistance as:

R =V

I=

l

σS

We can define resistivity ρc = 1/σ which results in:

R =ρcS



For conductors of non-uniform cross-section:

R =V

I=

~E .d~l

ρ~E .d ~S

Joule’s law defines the power in case of electrostatics as electrostaticforces times velocity: ∫

ρvdv ~E .~u =

∫~E .ρv ~udv

which gives

P =

∫~E . ~Jdv



Power density:

wP =dP

dv= ~E . ~J = σ|~E |2

For a conductor of uniform cross-section dv = dSdl :

P =

∫LEdl

∫SJdS = VI = I 2R


Displacement current

When dielectrics are placed in an electric field, charges experienceforces but they are also bound strongly to nucleus.

Unlike conductors, where electrons can move within conductors,dielectric electrons can at-most displace themselves.

A dipole is formed where negative and positive charges are separatedby a distance.

This phenomenon is defined by polarization vector ~p:

~p = Q ~d


Atomic/Molecular Dipoles

Figure: Atomic/molecular dipoles



Since a dielectric have a number of atoms so combined effect of Ndipoles will be given by:

N∑k=1

Qk~dk

Intensity of polarization is given by dipole moment per unit volume:

~P =lim

∆v→0

∑Nk=1 Qk

~dk

∆v


Atomic/molecular dipoles

There are two kinds of dielectric materials:

1 Polar molecules are those which have permanent separations of charges

permanant dipole momentOn application of ~E , dipoles align to ~E

2 Non-polar molecules experience an induced dipole moment

non-permanant dipole momentOn application of ~E , dipole moment is created.


Dipoles moment

Figure: Dipole momentDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 152 / 349

Field due to Polarization

Consider a material with dipole moment per unit volume ( ~P )originating from dipole moments.

Potential due to this dipole moment:

dV =~P. ~aRdv

′

4πε0R2

Gradient in primed coordinates:

~∇′ =1

R=

~aRR2

This results in:

~P. ~aRR2

= ~P.~

∇′(1

R) = ~∇′ .

~P

R−

~∇′.~P

R


Field due to Polarization

substituting in expression for V :

V =

∫v ′

1

4πε0[ ~∇′ .

~P

R−

~∇′.~P

R]dv

′

Applying divergence to te first term:

V =

∫S ′

~P. ~a′n

4πε0RdS

′+

∫v ′

− ~∇′.~P

R]dv

′

Here ~a′n is outward unit normal to surface dS

′of the dielectric.

Two terms denotes potential due to surface and volume chargedistribution.ρps = ~P.~a and ρpv = −~∇.~P


Bound charges

ρps and ρpv are composed on bound charges which are different thanfree charges in a way such that, they are not free to move within adielectric; they are caused by displacement occurring due topolarization.

Total charges bound on surface:

Qb =

∮~P.d ~S =

∫ρpsdS

Total charge bound by this surface (within volume V ):

−Qb =

∫vρpvdv = −

∫v

~∇.~Pdv


Bound charges

Total charge = ∮ρPsdS +

∫vρPvdv = Qb − Qb = 0

This is logical because the dielectric was electrically neutral beforepolarization.


Dielectric with free charge

Lets consider the case where dielectric region contains free charge.

If ρv is free charge volume density and ρt is total volume chargedensity:

ρt = ρv + ρPv = ~∇.ε0~E

This implies:ρv = ~∇.ε0

~E − ρvThis results in:

ρv = ~∇.(ε0~E + ~P)


~D

We define a term~D = ε0

~E + ~P

which results in:ρv = ~∇. ~D

In words:Due to application of electric field ~E to dielectric material, fluxdensity is changed than it would have been in free space

Net effect of placing a dielectric inside an electric field ~E is to changethe electric field by the value ~P, to make it ~D

In this respect, free space can be defined a medium such that ~P = 0


Electric Susceptibility

We define:

~P = χeε0~E

Here χe is called the electric susceptibility because it determineshow much a material will polarize when an electric field will beapplied.


Dielectric constant

We know that~D = ε0

~E + ~P

and~P = χeε0

~E

hence~D = ε0(1 + χe)~E = ε0εr ~E

which can be written as~D = ε~E

whereε = ε0εr

andεr = 1 + χe =

ε

ε0


Dielectric constant

ε is permittivity of the dielectric

ε0 is permittivity of free space

εr is relative permittivity / dielectric constant

εr is a dimension-less quantity since

εr =ε

ε0

For free space εr = 1

For dielectric medium, εr > 1


Dielectric constant

So far we have discussed ideal dielectric material.

Practically, when ~E is sufficiently large, electrons are pulled out ofmolecules and bound charge becomes free charge. In this way, adielectric becomes a conductor. This phenomenon is called thedielectric breakdown.

Dielectric Strength is the maximum value of ~E which a dielectriccan tolerate without undergoing dielectric breakdown.


Types of dielectrics

Linear Dielectric: ~D varies linearly with ~E

Homogeneous: ε does not depend on coordinates.

Isotropic: ~D and ~E are in same direction.


Continuity equation

Since the total charge of the system must be conserved:

Iout = −dQin

dt=

∮~J.d ~S =

where Qin is the charge enclosed by the surface and Iout is currentcoming out of closed surface.

In words: Time rate of decrease of charge should be equal to netoutward flux

Using divergence theorem, one can write:∮~J.d ~S =

∫v

~∇. ~Jdv


Continuity equation

Also:

−dQin

dt= − d

dt

∫vρvdv = −

∫v

∂ρv∂t

dv

which implies: ∫v

~∇. ~Jdv = −∫v

∂ρv∂t

dv

This can be written as:

~∇. ~J = −∂ρv∂t

This is called continuity of current equation


Relaxation time

Ohm’s law:~J = σ ~E

Gauss’s law:

~∇. ~E =ρvε

Now:

~∇.σ ~E =σρvε

= −∂ρv∂t

which can be rearranged as a homogeneous linear differentialequation:

∂ρv∂t

+σ

ερv = 0


Relaxation time

Solving by separation of variables:

∂ρvρv

= −σε∂t

Upon integration:

ln(ρv ) = −σtε

+ ln(ρv0)

Hence the solution is:

ρv = ρv0 .e− t

Tr

whereTr =

ε

σ

and ρv0 is ρv at t = 0


Relaxation time

The solution of equation can be interpreted as follows:

When a charge is introduced at some interior point of a conductingmaterial, there is a decay of volume charge density as charge fromthat point moves towards the surface of the material.

Tr is called the relaxation/rearrangement time associated with thisprocess.

Technically, Tr is time it takes a charge placed in the interior of amaterial to drop to e−1 = 36.8 percent of initial value.

Its is small value for good conductors and large value for gooddielectric materials (insulators).


Boundary Conditions

So far we have considered homogeneous medium.

Inhomogeneous mediums are defined by boundaries.

Boundary conditions are conditions which a field must satisfy at theinterface/boundary of two dis-similar mediums.


Boundary Conditions

We start with Maxwell’s equations:∮~E .d~l = 0

and ∮~D.d ~S = Qenc

We also decompose the electric field in two orthogonal (perpendicularto each other) components:

~E = ~Et + ~En

Where Et represents the tangential component and ~En represents thenormal component.


Dielectric-Dielectric boundary condition

Figure: Dielectric-Dielectric boundary condition (a) defined in terms of ~E (b)

defined in terms of ~D



Consider two different dielectrics differentiated by their permittivity:

ε1 = ε0εr1

and

ε2 = ε0εr2

Electric field can be decomposed into orthogonal components:

~E1 = ~E1t + ~E1n

and

~E2 = ~E2t + ~E2n



Since ∮~E .d~l = 0

so

0 = E1t∆w − E1n∆h

2− E2n

∆h

2− E2t∆w + E2n

∆h

2+ E1n

∆h

2

Here Et = | ~Et | and En = | ~En|As ∆h→ 0, we get:

E1t = E2t

Tangential components of electric fields is same for both media i.e.tangential component of ~E is continuous across the boundary.



What about ~D ?~D1 = ε1

~E1 and ~D2 = ε2~E2

Now

E1t = E2t ⇒~D1t

ε1=

~D2t

ε2

~D does changes across the interface, hence it is discontinuous.



Consider the (b) figure, A Gaussian surface can be considered andapplying Gauss’s law:

~∇. ~E =ρvε

We can obtain(∆h→ 0):

∆Q = ρS∆S = D1n∆S − D2n∆S

This results inD1n − D2n = ρS

Where ρS is free charge density at interface.

If there is no free charge at interface (ρS = 0) then ~D is continuousat the interface.



When we consider ~En instead of ~Dn we find that:

ε1E1n = ε2nE2n

~En is discontinuous.

These equations are called continuity equations for ~E and must befollowed by ~E for inhomogeneous media.


Refraction of ~E across the interface

Figure: Refraction of ~E across the interface


Refraction of ~E across the interface

E1sinθ1 = Et1 = E2t = E2sinθ2

and

D1cosθ1 = Dn1 = D2n = D2cosθ2

which can also be written as

ε1E1cosθ1 = Dn1 = D2n = ε2D2cosθ2

Which results in:

tanθ1

tanθ2=ε1

ε2

This is the law of refraction at a boundary which is devoid of free charge.


Conductor-dielectric boundary condition

Figure: Conductor-dielectric boundary condition


Conductor-dielectric boundary condition

We follow the same procedure and use the fact that ~E = 0 inside theconductor i.e for normal component E1n = 0

0 = 0∆w − 0∆h

2− E2n

∆h

2− E2t∆w + E2n

∆h

2+ 0

∆h

2

As ∆h→ 0 then

Et = 0

Similarly for the Gaussian surface, one can obtain:

∆Q = Dn.∆S − 0.∆S

which gives

Dn =∆Q

∆S= ρS


Conductor-Free Space boundary condition

Figure: Conductor - Free Space boundary condition


Free space is simply a special dielectric with εr = 1 so same boundaryconditions such as previous case exist i.e:

Dt = ε0Et = 0

and

Dn = ε0En = ρSDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 182 / 349

Electrostatic boundary value problem

Until now we have been dealing with situations where either chargedistribution and/or potential is known and we wish to find the ~E inentire region using Gauss law or the fact that ~E = −~∇.

These situations rarely exist for real life measurements.

Now we shall deal with issues where electrostatic conditions of chargeand/or potentials are defined at the interface and we wish to find the~E or V for entire region.

Such problems are called boundary value problems.


Poisson’s and Laplace’s equation

We know from Gauss’s law that

~∇. ~D = ~∇.ε ~E = ρv

since ~E = −~∇V so the equation can be rewritten as:

∇2V = −ρvε

In a free-charge free space, Poisson’s equation become Laplace’sequation i.e.

∇2V = 0


Uniqueness theorem

For particular boundary conditions, solutions of Poisson’s / Laplace’sequation is unique, regardless of method used. This is called theUniqueness Theorem.

Proof relies on the method of contradiction where we assume thatthere exist two solutions and then we show them to be same.


Proof of uniqueness theorem

Assuming that V1 and V2 are both solution to Laplace’s equation, then

∇2V1 = 0

and∇2V2 = 0

and lets assume that boundary condition is

V1 = V2

If we consider the difference in potential:

Vd = V2 − V2

this implies:

∇2Vd = 0

andVd = 0



We now consider divergence theorem:∫v

~∇. ~A.dv =

∮S

~A.d ~S

Consider:~A = Vd

~∇Vd

Now we consider a vector identity

~∇. ~A = ~∇.(Vd~∇Vd) = Vd∇2Vd + ~∇Vd .~∇Vd = ~∇Vd .~∇Vd

Now: ∫v

~∇Vd .~∇Vd =

∮SVd~∇Vd .d ~S

RHS must vanish.



∫v|~∇Vd |dv = 0

Since integration is always positive, hence:

~∇Vd = 0

This means that Vd is constant everywhere ⇒ Vd = V1 = V2


General Procedure for solving Laplace’s / Poisson’sequation

1 Solve Laplace’s / Poisson’s equation by1 Direct Integration when V is function of one variable2 Separation of variables when V is function of more than one variable.

Solution at this point is not unique by expressed in terms of unknownintegration constants.

2 Apply boundary conditions to determine the unique solution of V bygiving values to the unknown integration constants.

3 By applying ~∇V = ~E and D = ε~E calculate ~E and ~D4 Find Q using the fact

Q =

∫ρSdS

whereρS = Dn

where Dn is the normal component of ~D


Magnetostatics

SyllabusBiot-Savarts Law, Amperes circuit law, application of amperes law,magnetic flux density, magnetic scalar and vector potential, Magneticforces on a moving charge, magnetic materials, magnetic boundaryconditions.


Magnetostatistics

Magneto-statics is the study of static magnetic fields.


Biot-Savart law

dH ∝ I .dl .sinα

R2

The constant of proportionality is k whose value is 14π in SI units, which

makes Biot-savart law as:

dH =I .dl .sinα

4π.R2

Which can be written in vector form as:

d ~H =Id~l × ~aR

4πR2=

Id~l × ~R

4πR3

Figure: Direction of ~dHDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 191 / 349

Biot-Savart law

Figure: Biot-Savart Law


Biot-Savart law

Magnetic field intensity dH at a point P due to a differential currentcarrying element I .dl is proportional to I .dl and sine of angle betweencurrent carrying element and point P (i.e. sinα) and inverselyproportional to the square of the distance between P and element.


Magnetic field due to distributed current elements

For line current:

~H =

∫L

I ~dl × ~aR4πR2

For surface current:

~H =

∫S

~KdS × ~aR4πR2

For volume current:

~H =

∫v

~Jdv × ~aR4πR2

Here ~K and ~J are surface and volume current density

I ~dl ≡ ~KdS ≡ ~Jdv


Magnetic field due to distributed current elements

Figure: Current distributions: (a) Line current, (b) surface current, (c) volumecurrent


Magnetic field due to straigh filamentary conductor

Figure: Magnetic field due to straight filamentary conductor



due to d~l , d ~H is given by:

d ~H =Id~l × ~R

4πR3

Now d~l = dz ~az and ~R = ρ~aρ − z ~az which leads to

d~l × ~R = ρdz ~aφ

Hence the total magnetic field intensity is given by:

~H =

∫Iρdz

4π[ρ2 + z2]3/2~aφ



Now:

z = ρcotα

anddz = −ρcosec2αdα

This gives the total magnetic field intensity as:

~H =I

4πρ(cosα2 − cosα1) ~aφ



Semi-finite conductor α1 = 900, α2 = 00 makes

~H =I

4πρ~aφ

infinite conductor: α1 = 1800, α2 = 00

~H =I

2πρ~aφ


Ampere’s circuit law

In words: Line integral of tangential component of ~H around a closedpath is same as net current enclosed by the path.∮

~H.d~l = Ienc

It is similar to Guass law

When current distribution is symmetrical, its is easy to find ~H.

Ampere law is a special case of Biot-Savart law.


Biot Savart law from Ampere Law

By applying stokes law, we get:

Ienc =

∮L

~H.d~l =

∫S

(~∇× ~H).d ~S =

∫S

~J.d ~S

This yields:

~∇× ~H = ~J

This is third maxwell’s equation.This essentially Ampere’s law in differential form.Since ~J 6= 0, hence magneto-static field is non-conservative field.


Infinite current length

Consider a long filamentary current I along z -axis.To analyze ~H at P, we imagine a closed path (Amperian path: pathon which Ampere law is followed) crossing P.For infinite current filament

~H =I

2πρ~aφ

Figure: ~H due to filamentary current


Infinite current length

If ρ is constant then ~H is constant i.e. magnetic field is same allaround the Amperian path.

This is analogous to Gaussian surface in electrostatics.

For closed path:

I =

∫Hφ ~aφ.ρdφ ~aφ = Hφ

∫ρdφ = Hφ.2πρ

which yields

~H =I

2πρ~aφ

As expected.


Infinite sheet of current

Figure: ~H due to infinite sheet of current



Consider a sheet in z = 0 plane having uniform current density~K = Ky ~ay .

As per Ampere law: ∮~H.d~l = Ienc = Kyb

For calculating ~H, we consider the infinite sheet to be composed of asmall filaments.

d ~H above or below the sheet due to a pair of filamentary (which isgiven by ~H = I

2πρ ~aφ)

As evident from the figure, resultant ~H has only x- componentbecause current travels in y -direction and current element is definedin z-direction.



~H on one side is sheet will be opposite in sign to the other side.

~H = H0 ~ax : z > 0~H = −H0 ~ax : z < 0

Also evaluating the line integral:∮~H.d~l = (

∫ 2

1+

∫ 3

2+

∫ 4

3+

∫ 1

4) ~H.d~l

which gives

= 0(−a) + (−H0)(−b) + 0(a) + H0(b)

Resulting finally as: ∮~H.d~l = 2H0b = kyb


title

i.e.

H0 =1

2Ky

This gives:

~H =1

2Ky ~ax : z > 0

~H = −1

2Ky ~ax : z < 0

In general:

~H =1

2~K × ~an

Where ~an is a unit normal vector directed from current sheet to the pointof interest.


Magnetic flux density

Just like ~D = ε0~E one can define

~B = µ0~H

Where µ0 is permeability of free space whose value is:

µ0 = 4π × 10−7

Units: Henry/mMagnetic flux through a surface:

Ψ =

∫S

~B.d ~S

Unit = WeberMagnetic flux density is found in W /m2 or Tesla.


Magnetic flux lines

Figure: Magnetic Flux Lines


Magnetic flux lines

Magnetic flux lines is the path to which ~B is tangential at every point inmagnetic field.It is the line along which magnetic needle of a compass will orient itself.Direction of ~B is taken as the north pointing direction.Magnetic field lines are closed and do not cross each other.


Magnetic flux lines

Figure: (a) Electric field lines need not be closed ⇒ electric monopoles exist (b)Magnetic field lines must be closed ⇒ magnetic monopoles do not exist


Gauss Law for magnetostatics

Since magnetic field lines are closed, hence∮~B.d ~S = 0

.

Although magneto-static field is not conserved, magnetic flux isconserved quantity.

Now ∮~B.d ~S =

∫v

~∇. ~Bdv = 0

which implies:

~∇. ~B = 0

This is Fourth Maxwell’s equation.


Maxwell’s equations

Differential form Integral form Remarks~∇. ~D = ρv

∮S~D.d ~S =

∫v ρvdv Gauss Law

~∇. ~B = 0∮S~B.d ~S = 0 No magnetic monopole

~∇× ~E = 0∮L~E .d~l = 0 Conservative ~E

~∇× ~H = ~J∮L~E .d~l =

∫S~J.d ~S Ampere Law


Magnetic Scalar and Vector Potentials

Just as electro static potential V is defined to be ~E = −~∇V we canwither define a scalar magnetic potential Vm or even a vectorpotential ~A.

When ~J = 0 then~H == ~∇Vm

Since ~J = ~∇× ~H we can write

~∇2Vm = 0

Since ~∇. ~B = 0 so we can define a vector potential ~A such that

~B = ~∇× ~A

Also:

Ψ =

∮L

~A.d~l


Force due to magnetic field

Force due to magnetic field can be considered for three cases:

1 Force on moving charged particle

2 Force on current element in an external ~B field

3 Force between two current elements


Magnetic force on moving charged particle

Electrostatic force on charge Q due to electric field ~E is given by:

~Fe = Q ~E

If charge Q moves with velocity ~u in magnetic field ~B the magneticforce is given by:

~Fm = Q~u × ~B

When both electric and magnetic fields are present:

~F = ~Fe + ~Fm = Q(~E + ~u × ~B)

~F is called Lorentz force.


Force on current element

We know that convection current is given by:

~J = ρv ~u

Now we also know:

Id~l = ~KdS = ~Jdv

So current element can be written as:

Id~l = ρv ~udv = dQ~u =dQ

dtd~l

Hence magnetic force:

d ~Fm = Id~l × ~B

Total magnetic force:~Fm =

∮L Id

~l × ~B


Force on current element

For line element:~Fm =

∮LId~l × ~B

For surface:

~Fm =

∮S

~kds × ~B

For volume:

~Fm =

∮v

~Jdv × ~B


Force between two current elements

Figure: Force two current element


Force between two current elements

Force d(d ~F1) on element I1d~l1 due to the field d ~B2:

d(d ~F1) = I1d~l1 × d ~B2

Biot-Savart law states that:

d ~B2 =µ0I2d~l2 × ~aR21

4πR221

This results in the force product to be:

d(d ~F1) =µ0I1d~l1 × (I2d~l2 × ~aR21)

4πR221

Total force:

~F1 =µ0I1I2

4π

∮L1

∮L2

d~l1 × (d~l2 × ~aR21)

R221


Magnetic Torque and Moment

If a current carrying loop is placed parallel to magnetic field, itexperiences a force that tends to rotate it.

Mathematically, this is given by torque.

Torque (Mechanical moment of force) is given by:

~T = ~r × ~F

Units: N.m


Magnetic Torque and Moment

Figure: Rectangular planar current carrying loop in uniform planar magnetic field


Magnetic Torque

Consider a rectangular loop of length l and width w placed in uniformmagnetic field ~B.

Since d~l ||~B along side 12 and 34 hence the force is exerted on thesesides is zero.

The total force:

~F = I

∫ 3

2d~l × ~B + I

∫ 1

4d~l × ~B

substituting values of length:

~F = I

∫ l

0dz~az × ~B + I

∫ 0

ldz~az × ~B


Magnetic Torque

If we assume ~F0 = IBl (Because ~B is uniform) then

~F = ~F0 − ~F0 = 0

Equal and opposite force acting on exactly opposite points producesno couple but when they act on different points on the loop, a coupleis created.

If normal to the plane of loop makes α with ~B:

| ~T | = |~F0|wsinα = BIlwsinα = BISsinα

where S = lw is cross-sectional area of loop.


Magnetic Torque

We define magnetic dipole moment:

~m = IS~an

Unit: A/m2

Its direction is normal to the loop.

Hence one can write the magnetic torque as:

~T = ~m × ~B


A Magnetic Dipole

A bar magnet or a small filamentary loop = magnetic dipole.Magnetic vector potential is defined as:

~A =µ0I

4π

∮d~l

r

Figure: Magnetic dipoleDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 226 / 349

A Magnetic Dipole

For r >> a , ~A has only φ dependence:

~A =µ0Iπa

2sinθ

4πr2~aφ

In terms of magnetic moment:

~A =µ0 ~m × ~ar

4πr2

Where ~m = Iπa2~az and ~az × ~ar = sinθ~aφ


A magnetic dipole

Now magnetic flux density is:

~B = ~∇× ~A

Which gives:

~B =µ0m

4πr2(2cosθ~ar − sinθ~aθ)

Figure: ~B due to current loop and bar magnetDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 228 / 349

A magnetic dipole

Assume an isolated magnetic charge signified by its pole strength anddepicted by Qm.

Suppose that the length of magnetic bar is l.

Dipole moment can be written as Qml

Torque can be defined as:

~T = ~m × ~B = Qm~l× ~B

Here~l points from South to North.

Force acting on magnetic charge:

~F = Qm~B


A magnetic dipole

A current carrying loop and bar magnet are equivalent magnetically,so

T = QmlB = ISB

which yields:

Qml = IS


Magnetization in materials

Similar to electric polarization, we can talk about magnetic polarization asper the nature of magnetic field.Atomic dipoles are made up by the unpaired valence electrons. Together,they make up the dipole moment of a magnetic material. Under theinfluence of an external magnetic field, this dipole moment experiencemagneto-static force, which characterizes the nature of magneticinteractions.


Magnetization

If there are N atoms in ∆v volume where kth atom has magneticmoment:

~M = lim∆v→0

∑Nk−1 ~mk

∆v

For differential volume d ~m = ~Mdv′.

Vector magnetic potential ~A due to d ~m is:

d ~A =µ0~M × ~aR

4πR2dv

′=µ0~M × ~R

4πR3dv

′

Now:

~R

R3= ~∇′ 1

R


Magnetization

So:

~A =µ0

4π

∫~M × ~∇′ 1

Rdv

′

Now:

~M × ~∇′ 1

R=

1

R~∇′ × ~M − ~∇′ ×

~M

R

Thus:

~A =µ0

4π

∫v ′

~∇′ × ~M

R− µ0

4π

∫v ′~∇′ ×

~M

R


Magnetization

As per vector identity:∫v ′~∇′ × ~F = −

∮ ′

S

~F × d ~S

Applying the vector identity to second integral:

~A =µ0

4π

∫v ′

~∇′ × ~M

Rdv

′+µ0

4π

∮S ′

~M × ~anR

dS′

We can write ~∇× ~M = ~J and ~M × ~an = ~K we can write:

~A =µ0

4π

∫v ′

~JbRdv

′+µ0

4π

∮S ′

~kbRdS

′


Magnetization

Hence we can define:

Bound volume current density:

~Jb = ~∇× ~M

Units: A/m2

Bound surface current density:

~Kb = ~M × ~anUnits: A/m~an is unit vector normal to the surface.


Magnetization

In free space ~M = 0:

~∇× ~H = ~Jf

which is same as

~∇×~B

µ0= ~Jf

In a material ~M 6= 0:

~∇×~B

µ0= ~Jf + Jb = ~J = ~∇× ~H + ~∇× ~M

which yields:

~B = µ0( ~H + ~M)


Linear magnetization

For linear materials: ~M depends linearly on ~H i.e.

~M = χm~H

where χm is dimensionless quantity called magnetic susceptibility

We can write:~B = µ0(1 + χm) ~H = µ ~H

or

~B = µ0µr ~H

whereµr = 1 + χm =

µ

µ0

Where µ is called the permeability (Henry.m)


Classification of magnetic materials

Figure: Classification of magnetic materials


Magnetic Boundary Conditions

Just as in electrostatics, we have to find boundary conditions for twodis-similar magnetic medias.

Figure: Magnetic boundary conditions


Magnetic Boundary Conditions

We shall use two laws of magneto-statics:∮~B.d ~S = 0

and ∮~H.d~l = I

We shall obtain:

~B1n = ~B2n

or

µ1~H1 = µ2

~H2

This essential means that normal component of ~B is continuouswhereas normal component of ~H isn’t.


Magnetic boundary conditions

Similarly, for closed path

K .∆w = H1t .∆w +H1n.∆h

2+H2n.

∆h

2−H2t .∆w−H2n.

∆h

2−H1n.

∆h

2

As ∆→ 0 we finally get

H1t − H2t = K

i.e.

B1t

µ1− B2t

µ2= K


Magnetic Boundary Condition

In general:

( ~H1 − ~H2)× ~an12 = ~K

Where ~an12 is the unit vector normal to interface and is directed frommedium 1 to 2.

For boundaries which are free of current i.e. ~K = 0

~H1t = ~H2t

or

B1t

µ1=

B2t

µ2

which essential signifies that tangential component of ~H is continuous.


Magnetic refraction

If the fields make and angle θ to the normal of interface:

B1cosθ1 = B1n = B2n = B2cosθ2

Also

B1

µ1sinθ1 = H1t = H2t =

B2

µ2sinθ2

Diving both we get:

tanθ1

tanθ2=µ1

µ2


Inductors and Inductances

A closed conducting loop carrying current I produces a magnetic field~B

This causes a flux Ψ to pass through such that:

Ψ =

∮~B.d ~S

This flux passes through each turn. Suppose there are N turns, thenone can define λ (flux linkage) as:

λ = NΨ

The the material surrounding the coiled conductor is linear then Ψ isproportional to current I producing it:

λ ∝ I ⇒ λ = LI


Inductors and Inductances

L is called Inductance of the circuit.

The circuit or part of circuit which have inductance is called aninductor.

Mathematically inductance can be defined as:

L =λ

I=

NΨ

I

Unit: Henry

Like capacitance, inductance gives an estimate about total magneticenergy stored in a material.

Wm =1

2LI 2


Inductor and Inductances

When two such circuits interact, we consider the term mutualinductance.

Figure: Mutual inductance of two current carrying coils



Four fluxes are involved here:

Ψ12 =

∫S1

~B2.d ~S

Hence mutual inductance M12 and M21 can be defined as:

M12 =λ12

I2=

N1Ψ12

I2and

M21 =λ21

I1=

N2Ψ21

I1

If the medium surrounding the coils is linear (absence of aferromagnetic material):

M12 = M21



Similarly self-inductances L1 and L2 can be defined as:

L1 =λ1

I1=

N1Ψ1

I1

and

L2 =λ2

I2=

N2Ψ2

I2

Total energy:

Wm =1

2L1I

21 +

1

2L2I

22 ±M12M21I1I2

Positive sign for last term is taken when currents in two circuits flowsuch that magnetic fields strengthen each other.


Magnetic Energy

Figure: Magnetic Energy


Magnetic Energy

Consider a differential volume dv .

Let the volume be covered with conducting sheets on top and bottomsurface with current ∆I

Inductance for this volume:

∆L =∆Ψ

∆I=µH∆x∆y

∆I

Also using ∆I = H∆y we can write:

∆Wm =1

2∆L∆I 2 =

1

2µH2∆x∆y∆z =

1

2µH2dv

Magneto-static energy density:

wm = lim∆→0

∆w

∆v=

1

2µH2 =

1

2B.H =

1

2µB2


Magnetostatic energy

Total magneto-static energy:

Wm =1

2

∫~B. ~H


Waves

Until now, we have considered time-invariant ~E (x , y , z) and ~H(x , y , z)

Now we shall consider dynamic/time-varying ~E (x , y , z , t) and~H(x , y , z , t)


Faraday law

Michael Faraday discovered in 1831 that static magnetic field do notproduce a current flow but time-varying magnetic fields produce aninduced voltage (electromotive force-emf ) in a closed circuit, which inturn causes the flow of current.

Faraday’s law:

Vemf = −dλ

dt= −N dΨ

dt

where N is number of turns and Ψ is magnetic flux through each turn.

Negative sign signifies that induced voltage acts to oppose the fluxproducing it. This is Lenz law which essentially signifies that directionof current flow in the circuit is such that induced magnetic fieldproduced by induced current will oppose the original magnetic field.


Faraday Law

Figure: Circuit showing emf-producing electric field ~Ef and electrostatic field ~Ee

~E = ~Ef + ~Ee

Where Electrostatic field: ~Ee = −~∇V and ~Ef = 0 outside the battery andhave opposite direction to ~Ee inside the battery.


Faraday Law

We know that:~E = ~Ef + ~Ee

For closed circuit, we would integrate of obtain:∮L

~E .d~l =

∮L

~Ef .d~l + 0 = −∫ N

P

~Ef .d~l

Through the battery.∮~E0.d~l = 0 because electrostatic field is conservative in nature.

~Ef is non-conservative.


Relationship between electric and magnetic fields

For a circuit of single turn (N = 1):

Vemf = −dΨ

dt=

∮L

~E .d~l = − d

dt

∫S

~B.d ~S

because

Ψ =

∫~B.d ~S

and Stoke’s theorem.

Here S is surface area of a circuit bounded by the closed path L.


Creating the variation of flux with time

A stationary loop in time varying magnetic field.

A time varying loop area in static magnetic field.

Time varying loop area in time varying magnetic field.


stationary loop in time varying magnetic field

Figure: stationary loop in time varying magnetic field


stationary loop in time varying magnetic field

Now:

Vemf =

∮L

~E .d~l = −∫S

∂ ~B

∂t.dS

EMF produced by time varying current (producing time varying ~B) ina stationary loop is called transformer emf.

Applying stokes law to the middle term:∫S

(~∇× ~E ).d ~S = −∫S

∂ ~B

∂t.dS

This implies:

(~∇× ~E ) = −∂~B

∂tThis is Maxwell’s equation for time varying magnetic field:


Moving loop in static ~B field

Figure: Moving loop in static ~B field


Moving loop in static ~B field

Magneto-static force on charge moving with uniform velocity ~u is:

~Fm = Q~u × ~B

We define the motional emf as:

~Em =~FmQ

= ~u × ~B

Hence:

Vemf =

∮L

~Em.d~l =

∮L(~u × ~B).d~l

Applying stokes theorem and equating integrands:

~∇× ~E = ~∇× (~u × ~B)


Moving loop in time varying ~B field

Combining transformer and motional emf:

Vemf =

∮L

~E .d~l = −∫S

∂ ~B

∂t.dS +

∫L(~u × ~B).d~l

which implies:

~∇× ~E = −∂~B

∂t+ ~∇× (~u × ~B)



For static EM field:~∇× ~H = ~J

Since divergence of a curl is zero, hence:

~∇.(~∇× ~H) = 0 = ~∇. ~J

But current continuity equation requires:

~∇. ~J = −∂ρv∂t6= 0

This equation of static fields needs modifications for time-varyingfields



We shall add a current density term:

~∇× ~H = ~J + ~Jd

Hence

~∇.(~∇× ~H) = 0 = ~∇. ~J + ~∇. ~JdSince current continuity equation demands:

~∇. ~J = −∂ρv∂t6= 0

Hence:

~∇. ~Jd = −~∇. ~J =∂ρv∂t

=∂

∂t(~∇. ~D) = ~∇.∂

~D

∂t



This implies that displacement current density ( ~Jd) can be definedapart from conduction current density ( ~J = σ ~E ) ,as:

~Jd =∂ ~D

∂t

which in turn implies that new maxwell’s equation for transient EMfields will be:

~∇× ~H = ~J +∂ ~D

∂t



Displacement current can be defined as:

Id =

∫~Jd .d ~S =

∫∂ ~D

∂t.d ~S


Time varying potential

For static EM waves, electric scalar potential:

V =

∫v

ρvdv

4πεR

and magnetic vector potential:

~A =

∫v

µJdv

4πεR

Now we would like to examine the potential under time varyingconditions



~A was defined on the fact that ~∇. ~B = 0 which gives

~∇× ~A = ~B

This relation holds good for time varying magnetic fields too.

Now:

~∇× ~E = −d ~B

dt= −∂(~∇× ~A)

∂twhich implies that:

~∇× (~E +∂(~A)

∂t) = 0

The solution to which comes out to be:

~E +∂(~A)

∂t= −~∇V



So if ~A and V are known then ~B and ~E can be found.

Now ~∇. ~D = ρv so:

~∇. ~E =ρvε

= −∇2V − ∂

∂t(~∇. ~A)

this implies

∇2V − ∂

∂t(~∇. ~A) =

ρvε



We know that:

~B = ~∇× ~A

Taking curl on both sides and using the fact:

~∇× ~∇× ~A = µ ~J + εµ∂

∂t( ~∇V − ∂ ~A

∂t) = µ ~J − εµ~∇∂V

∂t− εµ∂

2 ~A

∂t2

According to a vector identity:

~∇× ~∇× ~A = ~∇(~∇. ~A)−∇2 ~A

Thus:

∇2 ~A− ~∇(~∇. ~A) = −µ ~J + εµ~∇∂V∂t

+ εµ∂2 ~A

∂t2



To define a vector field uniquyely, its curl and divergence must bedefined. Now curl os ~A is already defined as

~∇× ~A = ~B

We shall choose to define the divergence of ~A as:

~∇. ~A = −µε∂V∂t

This will be termed as Lorentz condition for potential


Retarded potentials

By imposing Lorentz conditions, we get two wave equations:

∇2V − µε∂2V

∂t2= −ρv

ε

and

∇2 ~A− µε∂2 ~A

∂t2= −µ ~J

Potentials V and ~A are termed as retarded electric potential andretarded magnetic potential

Spped of propagation of wave is:

u =1√µε

In free space, u = c i.e speed of light


EM wave propagation

We shall now solve maxwell’s equations for following environments:

Free spaceσ = 0, ε = ε0, µ = µ0

Lossless dielectrics

σ = 0, ε = εr ε0, µ = µrµ0|σ << ωε

Lossy dielectricsσ 6= 0, ε = εr ε0, µ = µrµ0

Good conductors

σ ' ∞, ε = εr ε0, µ = µrµ0|σ >> ωε

where ω is angular frequency of the EM wave.


Wave theory: Basics

A general scalar wave equation in 1− D (z-direction) can be writtenas:

∂2E

∂t2− u2∂

2E

∂z2= 0

where u is wave-velocity.

Two solutions are:E+ = f (z − ut)

andE− = g(z + ut)

General solution can be written as linear superposition of solutions as:

E = f (z − ut) + g(z + ut)


Soltutions to wave equation

The functions f and g can have the forms like:

z ± ut

orsink(z ± ut)

orcosk(z ± ut)

ore jk(z±ut)

where k is a constant


Harmonic solutions

If we particularly assume the harmonic solutions (e jωt), the waveequation simplifies to:

d2Es

dz2+ β2Es = 0

where β = ω/u and Es is the phasor form of E .

solutions:E+ = Ae j(ωt−βz)

and

E− = Be j(ωt+βz)

and

E = Ae j(ωt−βz) + Be j(ωt+βz)


Understanding harmonic solutions

Consider the real part of E gives:

E = Asin(ωt − βz)

Properties:

A is amplitude of wave and has same units as E .(ωt − βz) is the phase (in radians) of wave.ω is angular frequency (radians/second)β is phase constant or wave number (radians/meter)


Harmonic solutions

Figure: Sine wave along with its characteristics propertiesDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 278 / 349

Wave properties

We can define wavelength (meters):

λ = uT

Frequency:

f =1

T

and

ω = 2πf =2π

λ

and

β =ω

u

group velocity:u = f λ


Wave propagation in lossy dielectrics

Most general case of wave propagation is that in lossy dielectric.Other cases can be derived from the same.

A lossy dielectric is a medium in which an EM wave loses power asit propagates due to poor conduction (σ 6= 0).

Consider such a lossy dielectric which is linear, homogeneous,isotropic and free of charge ρv = 0.



Maxwell’s equations:

~∇. ~Es = 0 (48)

~∇. ~Hs = 0 (49)

~∇× ~Es = −jωµ ~Hs (50)

~∇× ~Hs = (σ + jωε)~Es (51)



taking curl on both sides of third equation:

~∇× ~∇× ~Es = −jωµ~∇× ~Hs

Using vector identity:

~∇× ~∇× ~A = ~∇(~∇. ~A)−∇2 ~A

Since ~∇. ~E = 0 so

−~∇2 ~Es = −jωµ(σ + jωε)~Es

If we assume:γ2 = jωµ(σ + jωε)

where γ is termed as propagation constant.



Finally we get:~∇2 ~Es − γ2 ~Es = 0

By similar logic we can obtain:

~∇2 ~Hs − γ2 ~Hs = 0

These equations are called Helmholtz equations


Helmholtz equations

Since γ is a complex quantity:

γ = α + jβ

Hence- Re γ2 = β2 − α2 = ω2µε and|γ2| = β2 + α2 = ωµ

√σ2 + ω2ε2

Solving equations for α and β we get:

α = ω

√µε

2[

√1 + [

σ

ωε]− 1]

and

β = ω

√µε

2[

√1 + [

σ

ωε] + 1]


Electric field propagation

If we assume that wave propagates along ~az direction and ~Es has onlyx-component i.e.

~Exs(z)~ax

Substituting in~∇2 ~Es − γ2 ~Es = 0

we get:(~∇2 − γ2)~Exs = 0

which results in

[d2

dz2− γ2]Exs(z) = 0

because differential terms for x and y results in zero.



Hence finally we obtain

[d2

dz2− γ2]Exs(z) = 0

Its solution can be given as:

Exs(z) = E0e−γz + E

′0eγz

where E0 and E′0 are constants.

Field must be finite at infinity. This requires E′0 = 0

Inserting a time factor e jωt

~E (z , t) = Re[E0e−αze j(ωt−βz)~ax ]



This ultimately yields:

~E (z , t) = E0e−αzcos(ωt − βz)~ax

Figure: Propagation of |~E | in a lossy mediaDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 287 / 349

Magnetic field propagation

SImilar steps can be undertaken for propagation of magnetic field:

~H(z , t) = H0e−αzcos(ωt − βz) ~ay

where

H0 =E0

η

η is a complex quantity known as intrinsic impedance:

η =

√jωµ

σ + jωε= |η|e jθn


Magnetic field propagation

Also

|η| =

√µε

[1 + ( σωε)2]1/4

andtan2θn =

σ

ωε

where 0 ≤ θn ≤ 450

Using above equations, magnetic field can be written in terms ofcomponents of electric field:

~H(z , t) =E0

|η|e−αzcos(ωt − βz − θn)~ay


EM wave propagation

Now:~E (z , t) = E0e

−αzcos(ωt − βz)~ax

and~H(z , t) =

E0

|η|e−αzcos(ωt − βz − θn)~ay

α is called the attenuation constant for a medium (dB/m).

β is measure of phase shift per length i.e phase constant or wavenumber.

θn is phase difference between ~E an ~H.


Loss tangent

Loss tangent is derived from the ratio of ~Js and ~Jds :

| ~Js || ~Jds |

=|σ ~Es ||jωε~Es |

=σ

ωε= tanθ

tanθ is called loss tangent as it gives an idea about how lossy is amedium.

If loss tangent is small = good dielectric

If loss tangent is large = good conductor


Plane waves in lossless dielectrics

For lossless dielectrics:

σ = 0

ε = ε0εr

µ = µ0µr

which results in:

α = 0, β =√µε, u =

ω

β=

1√µε, λ =

2π

β

~E and ~H are in-phase.


Plane waves in free space

For free space

σ = 0

ε = ε0

µ = µ0

which results in:

α = 0, β = ω[√µ0ε0] =

ω

c, u =

1

[√µ0ε0]

= c , λ =2π

β

Also θn = 0, η = η0 where η0 is called intrinsic impedance of freespace

η0 =1

[√

1µ0ε0

]= 120π = 377Ω



Electric field:~E = E0cos(ωt − βz)~ax

Magnetic field:

~H = H0cos(ωt − βz)~ay =E0

η0cos(ωt − βz)~ay

Direction of ~E , ~H and propagation direction:

~ak × ~aE = ~aH

and~ak × ~aH = −~aE

and~aE × ~aH = ~ak



Figure: (a)Plot of ~E and ~H as function of z at t = 0, (b) Plot of ~E and ~H as seenfront front of propagation direction.


Plane wave in perfect conductor

For free space

σ ' ∞ε = ε0

µ = µ0µr

which results in:

α = β =

√ωµσ

2=

√πf µσ

u =ω

β=

√2ω

µσ

λ =2π

β

η =

√ωµ

σ]450


Plane wave in perfect conductor

~E leads ~H by 450

Electric field:~E = E0e

−αzcos(ωt − βz)~ax

Magnetic field:

~H =E0

[√

ωµσ ]

cos(ωt − βz − 450)~ay

Skin depth:As shown in above equations, ~E , ~H attenuates by a factor e−αz .The distance δ by which it attenuates by e−1 = 37% is called skindepth:

E0e−αδ = E0e

−1 ⇒ δ =1

α


Skin depth

Skin depth gives a measure to which an EM wave can penetrate amedium.

Figure: Skin depth


Poynting vector

Maxwell’s equation:

~∇× ~E = −µ∂~H

∂t

and

~∇× ~H = σ ~E + ε∂ ~E

∂t= ~J +

∂ ~D

∂t

Taking a dot product on both sides for second equation:

~E .(~∇× ~H) = ~E . ~J + ~E .∂ ~D

∂t

Now we know

~∇.(~E × ~H) = −~E .(~∇× ~H) + ~H.(~∇× ~E )


Poynting vector

~H.(~∇× ~E )− ~E .(~∇× ~H) = ~E . ~J + ~E .∂ ~D

∂t

Since

~∇× ~E = −µ∂~H

∂tso

− ~H.∂~B

∂t− ~E .(~∇× ~H) = ~E . ~J + ~E .

∂ ~D

∂t

which can be written as:

−~E .(~∇× ~H) = ~E . ~J + ε~E .∂ ~E

∂t+ µ ~H.

∂ ~H

∂t


Poynting vector

Two time derivatives can be rearranged as:

ε~E .∂ ~E

∂t=

∂

∂t(

1

2~D. ~E )

and

µ ~H.∂ ~H

∂t=

∂

∂t(

1

2~B. ~H)

which ultimately gives

−~E .(~∇× ~H) = ~E . ~J +∂

∂t(

1

2~D. ~E ) +

∂

∂t(

1

2~B. ~H)


Poynting vector

Integrating over the volume

−∫v

~E .(~∇× ~H) =

∫v

~E . ~J +

∫v

∂

∂t(

1

2~D. ~E ) +

∫v

∂

∂t(

1

2~B. ~H)

Applying divergence theorem to right hand side:∮s(~E × ~H).d ~S =

∫v

~E . ~J +

∫v

∂

∂t(

1

2~D. ~E ) +

∫v

∂

∂t(

1

2~B. ~H)


Poynting vector

Finally: ∮S

(~E × ~H).d ~S =

∫vσE 2 +

∂

∂t

∫v

[ε∂E 2

2+ µ

∂H2

2]

LHS is total power leaving the volume

First term on RHS is rate of decrease in energy stored in electric andmagnetic fields

Ohmic power dissipated

The equation is known as Poynting theorem and ~E × ~H is called thePoynting vector


Reflection of a plane wave at normal incidence

Consider the case as below:

Figure: Reflection of a plane wave at normal incidence



Incident wave: ~Ei , ~Hi is traveling in +~az in medium 1 where wesuppress the time factor e jωt :

~Eis(z) = Eioe−γ1z~ax

and~His(z) = Hioe

−γ1z~ay =Ei0

η1e−γ1z~ay

Reflected wave: ~Er , ~Hr is traveling in −~az in medium 1 where wesuppress the time factor e jωt :

~Ers(z) = Eroe−γ1z~ax

and~Hrs(z) = Hroe

−γ1z(−~ay ) = −Ero

η1e−γ1z~ay



Transmitted wave: ~Et , ~Ht is traveling in +~az in medium 2 where wesuppress the time factor e jωt :

~Ets(z) = Etoe−γ2z~ax

and~Hts(z) = Htoe

−γ2z(~ay ) = −Eto

η2e−γ2z~ay



Medium 1 has incident and reflected wave whereas medium 2 hastransmitted wave.

~E1 = ~Ei + ~Er

and~H1 = ~Hi + ~Hr

Medium 2 only contain reflected wave:

~E2 = ~Et

and~H2 = ~Ht



At z = 0 boundary conditions dictate that tangential components of~E , ~H are continuous.

At z = 0:

~E1tan = ~E2tan ⇒ ~Ei (0) + ~Er (0) = ~Et(0)⇒ Eio + Ero = Eto

and

~H1tan = ~H2tan ⇒ ~Hi (0) + ~Hr (0) = ~Ht(0)⇒ 1

η1(Eio + Ero) =

1

η2Eto

This yields:

Ero =η2 − η1

η2 + η1Eio ,Eto =

2η2

η2 + η1Eio


Reflection and Transmission coefficients

Reflection coefficient:

Γ =Ero

Eio=η2 − η1

η2 + η1

Transmission coefficient:

τ =Eto

Eio=

2η2

η2 + η1

Relationship:τ = 1 + Γ

Both Γ and τ are dimensionless

o ≤ |Γ| ≤ 1


Special case

Suppose medium 1 is perfect dielectric (σ1 = 0) and medium 2 isperfect conductor (σ2 ' ∞).

In this case: η2 = 0: Γ = −1 and τ = 0 meaning that wave is totallyreflected.

This is expected because electric field inside conductor should vanishmeaning there is no transmitted wave.

When totally reflected wave mixes with incident wave to make astanding wave consisting of two traveling waves ~Ei , ~Er making astationary pattern because both travel with equal amplitude but inopposite directions.


Standing waves

Standing wave in medium 1:

~E1s = ~Eis + ~Ers = (Eioe−γ1z + Eroe

−γ1z)

But

Γ =Ero

Eio= −1, σ1 = 0, α1 = 0, γ1 = jβ1

Hence:

~E1s = −Eio(e jβ1z − e−jβ1z)~ax = −2jEiosinβiz~ax

which means that

~E1 = Re(~E1sejωt) = 2Eiosinβ1zsinωt~ax


Standing waves

We obtained:~E1 = 2Eiosinβ1zsinωt~ax

By similar procedure we can obtain:

~H1 =2Eio

η1cosβ1zcosωt~ay

Figure: Standing waveDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 312 / 349

Standing wave ratio

Standing wave ratio s can be defined as:

s =|~E1|max

|~E1|min

=| ~H1|max

| ~H1|min

=1 + Γ

1− Γ

Also:

|Γ| =s − 1

s + 1

Since |Γ| ≤ 1 hence 1 ≤ s ≤ ∞SWR is a dimensionless quantity

SWR can also be expressed in dB as:s(dB) = 20log10s


Transmission Lines Introduction

Until now we discussed infinite space for the medium

Now we shall discuss guiding mediums of finite dimensions and studypropagation of EMW.

These guiding mediums are called transmission lines

They are essentially made up of two or more parallel conductors usedto connect source and load

Source is the EM power generator (oscillator, hydroelectric generator,transmitter)

Load is EM reception station (antenna, oscilloscope etc.)


Transmission line parameters

1 R = resistance per unit length

2 G = conductance per unit length

3 C = capacitance per unit length

4 L = inductance per unit length



Transmission line parameters are uniformly distributed over thetransmission line.

Figure: Distribution of line parameters over a transmission line



For each line, conductors are characterized by σc , µc , εc = ε0 andhomogeneous dielectric separating the conductors is characterized byσ, µ, ε.

G 6= 1/R because R shows the AC resistance per unit length ofconductor and G is the conductance per unit length due to dielectricmedium separating the conductors.

L is the external inductance per unit length since effects of internalinductance is negligible at high frequency.

For all transmission lines:LC = µε

and

G

C=σ

ε


Coaxial cable

Figure: Coaxial line connecting a generator and load (b) ~E , ~H fields in coaxial line


Coaxial cable

When switch S is closed:

Inner conductor is made positive w.r.t outer one~E will point radially outward from inner conductor to outer conductingsheath.Ampere law states that ~H field encircles the current carrying conductor.Poynting vector ~E × ~H points along the transmission line.


Transmission line equations

Transmission line equations are also called as telegrapher’sequations since telegraph was first direct application of transmissionline based EMW communication system.

Figure: L-type equivalent circuit model of a differential length ∆z for atwo-conductor transmission line



The model is called the L-type equivalent circuit.

Here we assume that the wave propagates along the +z-direction(from generator to load).

By applying Kirchoff’s voltage law to outer loop:

V (z , t) = R∆zI (z , t) + L∆z∂I (z , t)

∂t+ V (z + z∆z , t)

which can also be simply written as

V (z + z∆z , t)− V (z , t)

∆z= RI (z , t) + L

∂I (z , t)

∂t



Taking limits:

lim∆z→0

(V (z + z∆z , t)− V (z , t)

∆z) = RI (z , t) + L

∂I (z , t)

∂t

Hence

−∂V (z , t)

∂t= RI (z , t) + L

∂I (z , t)

∂t


Transmission line equation

Similarly, applying Kirchoff’s current law to the main node of thecircuit:

I (z , t) = I (z+∆z , t)+∆I = I (z+∆z , t)+G∆zV (z+∆z , t)+C∆z∂V (z + ∆z , t)

∂t

We shall ultimately get:

−∂I (z , t)

∂t= GV (z , t) + C

∂V (z , t)

∂t



We obtained two differential equations:

−∂V (z , t)

∂t= RI (z , t) + L

∂I (z , t)

∂t

and

−∂I (z , t)

∂t= GV (z , t) + C

∂V (z , t)

∂t


Tranmission line equations

If we assume harmonic time dependence:

V (z , t) = Re[Vsejωt ]

andI (z , t) = Re[Ise

jωt ]

Hence our differential equations simplifies as:

−dVs

dz= (R + jωL)Is

and

−dIsdz

= (G + jωC )Vs

These are coupled differential equations where Vs and Is are coupled.



Taking the second derivative:

d2Vs

dz2= (R + jωL)(G + jωC )Vs

which yields:d2Vs

dz2− γ2Vs = 0

Similarly:d2Isdz2− γ2Is = 0

Hereγ = α + jβ =

√(R + jωL)(G + jωC )



These are similar to wave equations we have already studied

γ is propagation constant

α is attenuation constant

β is phase constant

Wavelength and wavenumber can be written as:

λ =2π

β

andu =

ω

β= f λ



Solutions:Vs = V+

o e−γz + V−o e−γz

andIs = I+

o e−γz + I−o e−γz

Here V+o ,V

−o , I

+o , I

−o are wave amplitudes

The + and − sign depicts the wave traveling in +z and −z directions.

We also obtain:

V (z , t) = Re[Vs(z)e jωt ] = V+o e−αzcos(ωt−βz)+V−o e−αzcos(ωt−βz)


Transmission Line parameters

Characteristic impedance

Zo =V+o

I+o

= −V−oI−o

=R + jωL

γ=

γ

G + jωC

which can be simplified to:

Zo =

√R + jωL

G + jωC= Ro + jXo


Lossless line

R = G = 0

Conductor is perfect σc ' ∞ and dielectric is lossless σ = 0.

α = 0

γ = jβ = jω√LC

u =ω

β=

1√LC

= f λ

Xo = 0

Zo = Ro =

√L

C


Distortionless line

A signal is usually composed of a number of frequencies having thierspecific amplitudes.

If attenuation is frequency dependent then signal is distorted.

When α is frequency independent and phase constant β linearlydepends on frequency, transmission line is termed distortion-less.

Such a line has:

R

L=

G

C


Distortionless line

Propagation constant:

γ =

√RG (1 +

jωL

R)(1 +

jωC

G)

which results in

γ =√RG (1 +

jωL

R) = α + jβ

whereα =√RG , β = ω

√LC

Note that here α is independent of ω or f

β linearly depends on ω or f .


Distortionless line

Zo =

√R(1 + jωL/R)

G (1 + jωC/G )=

√R

G=

√L

C= Ro + jXo

which gives

Ro =

√R

G=

√L

C

andXo = 0

and

u =ω

β=

1√LC

= f λ


Input Impedance

consider a transmission line of length ` characterized by γ,Zo

connected to a load ZL.

Figure: (a) Input impedance of a line terminated by a load (b) equivalent circuitDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 334 / 349

Input impedance

Let the transmission line has generator at z = 0 and load at z = `.

Vs = V+o e−γz + V−o e−γz

and

Is =V+o

Zoe−γz − V−o

Zoe−γz

These wave equations explain the behavior of voltage and current.To find V+

o ,V−o , terminal conditions must be given say

Vo = V (z = 0), Io = I (z = 0)

Substituting them we get

V+o =

1

2(Vo + Zo Io)

and

V−o =1

2(Vo − Zo Io)


Input impedance

If input impedance is Zin, then the input voltage Vo and Io :

Vo =Zin

Zin + ZgVg , Io =

Vg

Vin + Vg

On the other hand, if we are given conditions at load end:

VL = V (z = `), IL = I (z = `)

we get:

V+o =

1

2(VL + Zo IL)eγ`

and

V−o =1

2(VL − Zo IL)e−γ`


Input impedance

When input impedance Zin = Vs(z)Is(z) is given at any point on the

transmission line:

Zin =Vs(z)

Is(z)=

Zo(V+o ) + V−o

V+o )− V−o

Utilizing the fact that:

coshγ` =eγ` + e−γ`

2, sinhγ` =

eγ` − e−γ`

2, tanhγ` =

eγ` − e−γ`

eγ` + e−γ`

This yields:

Zin = Zo [ZL + Zotanhγ`

Zo + ZLtanhγ`]

This expression stands true for Lossy transmission line.


Input impedance

For lossless line γ = jβ which leads to tanhjβ` = jtanβ`

Hence the expression for input impedance:

Zin = Zo [ZL + jZotanβ`

Zo + jZLtanβ`]

The equation essentially means that input impedance variesperiodically along the length of line

β` is called electrical length of the line (expressed indegree/radians).


Standing wave ratio

We can define Voltage reflection coefficient at load:

ΓL =V−o eγ`

V+o e−γ`

=ZL − Zo

ZL + Zo

Standing wave ratio:

s =Vmax

Vmin=

Imax

Imin=

1 + |ΓL|1− |ΓL|

Imax = Vmax/Zo and Imin = Vmin/Zo

Relationship of input impedance and standing wave ratio

Zin|max =Vmax

Imin= sZo ,Zin|min =

Vmin

Imax=

Zo

s


Average power

Average power:

Pav =1

2Re[Vs(`)I ∗s (`)] =

|V+0 |2

2Zo(1− |Γ|2)

Factor of 1/2 comes into play since we are dealing with peak values,instead of r.m.s values.

First term refers to input power and second term refers to reflectedpower.

Maximum power is delivered to load when Γ = 0 i.e. due toimpedance matching, there isn’t any reflected wave.


Smith Chart

Smith chart is a graphical means of calculating transmission lineparameters.

Graphical indication of impedance of a transmission line as one movesalong the line.


Procedure to make a smith chart

1 Construct a unit circle (radius = |Γ| ≤ 1)

2 We shall use the relations:

Γ =ZL − Zo

ZL + Zo

and

Γ∠θL = Γi + jΓr

where Γi and Γr are real and imaginary parts of reflection coefficientΓ.

3 Calculating normalized impedance

zL =ZL

Zo= r + jx



Γ can be written as:

Γ = Γr + jΓi =zL − 1

zL + 1

which results in:

r =1− Γ2

r − Γ2i

(1− Γr )2 + Γ2i

and

x =2Γi

(1− Γr )2 + Γ2i



Rearranging the terms leads to the expression:

[Γr −r

1 + r]2 + Γ2

i = [1

1 + r]2

and

[Γr − 1]2 + [Γi −1

x]2 = [

1

x]2

These equations are similar to

(x − h)2 + (y − k)2 = a2

i.e. a circle of radius a and centered at (h, k)Hence first equations is resistance circle withcenter at:

(Γr , Γi ) = (r

r + 1, 0)

and radius at1

1 + rDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 344 / 349

r-circles

Figure: Radii and centers of r - circles


x-circles

Similar to r - circles, one can define x - circles:center at:

(Γr , Γi ) = (1,1

x)

and radius at1

x

While r is always positive, x can be both positive (for inductiveimpedance) and negative (for capacitive impedance).


x-circles

Figure: Radii and centers of x - circles


Smith charts

Superimposing r-circles and x-circles present Smith charts.

Figure: Smith ChartsDr. Sandeep Nagar (G. D. Goenka University) Electromagnetic Field Theory November 23, 2015 348 / 349

Reading Smith Charts

When r = 2 on r-circle and x = 1 on x-circle, input impedance is

z = 2 + j

s-circles i.e. constant standing wave ratio circles can also be drawnare centered at origin with s varying from 1 to ∞Value of standing wave ratio is found using the cross-point of s-circleand Γr axis.


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