EMT Lectures

8/12/2019 EMT Lectures

1/105

MODULE I

VECTOR ALGEBRA, VECTOR CALCULUS,COORDINATE SYSTEMS, VECTOR FIELDS

Compiled by: MKP for CEC S5 EC -July 2008

Syllabus Module I Vector analysis:

Vector algebra, Coordinate systems andtransformations-Cartesian, cylindrical and sphericalcoordinates. Constant coordinate surfaces. Vectorcalculus-Differential length, area and volume. Line,surface and volume integrals. Del operator. Gradient of ascalar, Divergence of a vector, Divergence theorem, Curlof a vector. Stocks theorem, Laplacian of a scalar.Classification of vector fields.


References Text Books:

1. Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford UniversityPress

2. Jordan and Balmain, Electromagnetic waves and radiating systems,Pearson Education PHI Ltd.

References:

1. Kraus Fleisch, Electromagnetics with applications, McGraw Hill2. William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill3. N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson

Education PHI Ltd.4. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical

Publishers.5. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill


Scalars and vectors A scalar is a quantity that has only magnitude.

TimeDistanceTemperatureSpeed

A vector is a quantity that has both magnitude anddirection.

ForceDisplacementVelocity


Unit vector A vector has both magnitude and direction.

A unit vector along is defined as a vector whose magnitude is

unity and whose direction is along vector . It is denoted by

Vector is completely specified in terms of its magnitude A anddirection

A

agnitude of A A A= =

A

A Aa

A A A

a A A

= =

A A A a=

Aa A


Vectors represented in rectangularcoordinate systems

Any vector in space can be uniquely expressed in terms of x, y andz coordinates using a rectangular coordinate system.

Y

X

Z

x A

y A

z A

xa ya

z a A

x x y y z z A A a A a A a= + +


2/105


Vectors represented in rectangularcoordinate systems

x x y y z z A A a A a A a= + +, , , , x y z A A Components of A in the direction of x y z

, ,

, , x y z a a a Unit vectors specifying the direction

of x y z axes


Position vector of a point in space A point P in Cartesian coordinate system may be expressed as itsx,y,z coordinates. The position vector of a point P is the directeddistance from the origin O to the point P. A point P (3,4,5) has the position vector

Y

X

Z

x A

y A

z A

xa ya

z a

p x x y y z z r OP A a A a A a= = + +P

3 4 5 p x y z r a a a= + +


Vector addition and subtraction

x x y y z z

x x y y z z

If A A a A a A a and

B B a B a B a

= + +

= + +

( ) ( ) ( ) x x x y y y z z z C A B A B a A B a A B a= + = + + + + +

( ) ( ) ( ) x x x y y y z z z D A B A B a A B a A B a= = + +


Distance vectorDistance vector is the displacement from one point to another.If two points A (A x,Ay,Az) and B (B x,By,Bz) are given, the distancevector from A to B is given by

( ) ( ) ( ) AB x x x y y y z z z r B A a B A a B A a= + +


Unit vector in the direction of given vector

Let be a vector in space given by

A unit vector in the direction of is given by

x x y y z z A A a A a A a= + + A

A

2 2 2

x x y y z z A

x y z

A a A a A aa

A A A

+ +=

+ +


Example 1 10 4 6 x y z If A a a a and = + 2 x y B a a find = +

( )

( ) 3

( ) 2

yi Component of A along a

ii Magnitude of A B

iii A unit vector along A B

+

: Answer ( 4) -i

( ) ( ) ( ) 3 30 12 18 2 x y z x yii A B a a a a a = + +

28 13 18 x y z a a a= +2 2 23 28 13 18 35.74 A B = =+ +


3/105


Example 1

( ) 2 14 2 6 x y z iii A B a a a+ = +

0.9113 0.1302 0.3906 x y z c a a a= +

2 22

A B A unit vector c along A B

A B

++ =+

2 2 2

14 2 6

14 2 6 x y z a a a +=

+ +


Vector multiplication-dot productScalar product or dot product: It is defined as the product ofmagnitudes of the two vectors and the cosine of the angle betweenthem.

Properties:

cos AB A B AB =

AB is the smaller angle between them

( ) :i Commutative Prope ty B B Ar =

( )ii When two vectors are perpendicular the angle =90 cos 90 0between them is =

cos90 0 A B AB = =

, If the dot product of two vectors are zero they are.erpendicular


Vector multiplication-dot product ( ) , , x y z iii Since a a a are mutually perpendicular

= 0 x y y z z xa a a a a a = =

( )iv When two vectors are parallel the angle between them 0 180is either or

cos0 A B AB AB or = =

cos180 A B AB AB = =

( ) .v The square of a vector is the square of its magnitude2cos0 A A AA A = =

2 2 A A=Compiled by: MKP for CEC S5 EC -July 2008

Vector multiplication-dot product

x x y y z z

x x y y z z

If A A a A a A a and

B B a B a B a

= + +

= + +

( ) ( ) . x x y y z z x x y y z z A B A a A a A a B a B a B a= + + + +

x x y y z z A B A B A B= + +

( )vi Scalar product is equal to the sum of products of their .corresponding components


Vector Product or cross productVector Product or cross product: Vector product of two vectorsand is denoted as and is defined as

Where is a unit vector perpendicular to and such thatforms a right handed system.Geometrically the cross product can be defined as a vector whosemagnitude is equal to the area of the parallelogram formed by

and whose direction is in the direction of advance of aright handed screw as is turned in to through the smallerangle.

A B A B

sin AB n A B A B a =

B

na

, n B and a

B A and B


Vector Product or cross product

A

B

A B

na AB sin

AB sin


4/105


Vector Product or cross productProperties:

( ) :i Anti commutative A B B A =

( ) ( ) ( )( ) :ii Distributive A B C A B A C + = +

( ) ( )( ) :iii Not Associative A B C A B C

( ) V .iv ector product of two parallel vectors is zero

sin0sin 0 AB n n A B A B a A B a = = =

( ) sin 0 0nv A A A A a = =



( ) 0 x x y y z z vi a a a a a a = = =

( ) 1.1.sin 90 x y z z vii a a a a and = = y z xa a a =

z x ya a a = y x z a a a =

z y xa a a =

x z ya a a =



xa

yaa

xa

ya a xa

yaa

xa

ya a


Vector Product or cross product

( )

x x y y z z

x x y y z z

viii If A A a A a A a and

B B a B a B a

= + +

= + +

x y z

x y z

x y z

a a a

A B A A A

B B B

=


Projection of a vector on another vectorScalar component of along is called projection of onand is given by

A B A B

cos B AB A A =

cos B AB A a =

Ba=

B

AB

cos AB A Ba


Projection of a vector on another vectorVector component of along is the scalar component multipliedby a unit vector along

B B

( ) B B B B B A A a A a a= =


5/105


Scalar triple productGiven three vectors the scalar triple product is definedas

and is represented as

Geometrically the scalar triple product is equal to the volume of aparallelepiped having as sidesProperties:

( ) ( ) ( ) A B C B C A C A B = =

, A B and C

B C

, A B and C

( )i A B C B C A C A B = =

( ) ( ) ( ). . A B C Bi e C A C A B = =


Scalar triple product(ii) A change in the cyclic order of vectors changes the sign of scalar

triple product.

A B C B A C =

( )

x x y y z z

x x y y z z

x x y y z z

iii If A A a A a A a and

B B a B a B a

C C a C a C a

= + +

= + +

= + +

x y z

x y z

x y z

A A

B C B B B

C C C

=


Vector triple product

, , For any three vectors A B C

( ) ( ) ( ) A B C B A C C A B =

bac-cab rule


Cylindrical Coordinate Systems Any point in space is considered to be at the intersection of threemutually perpendicular surfaces:

A circular cylinder ( =constant) A vertical plane ( =constant) A horizontal plane (z=constant)

Any point in space is represented by three coordinates P( , ,z) denotes the radius of an imaginary cylinder passing through P, or

the radial distance from z axis to the point P. denotes azimuthal angle, measured from x axis to a vertical

intersecting plane passing through P.z denotes distance from xy-plane to a horizontal intersecting planepassing through P. It is the same as in rectangular coordinatesystem.


Cylindrical Coordinate Systems

P( ,,z)

z=constant

=constant

=constant

z

X

Y

Z

: 0

0 2

Ranges

z

< >d we can make the following assumptions2

2 1 1 2andcosr r d r r r

20

cos

4Q d

Then V r

=

0 1 2

1 14

QV r r

=

cos wherer z r d d a d da a is a unit vector in r direction = =

0 1 0 24 4

Q QV

r r = 2 1

0 1 24Q r r

r r

=


48/105

Compiled by MKP for CEC S5 Batch July 2008

Electric dipoles: Potential and Electric field

20

4r Qd aThen V

r

=

20

4

r p aV r =

, By defining Qd p dipole moment =

The magnitude of dipole moment is equal to the product of charge

and distance and its direction is from Q to +Q The electric f ield due to the dipole with centre at the origin is given by

E V = 1 r

V V a a

r r

= +

There is no field variation along direction



20

cos

4Q d

Where V r

=1 r V V

E a ar r

= +

3 30 0

cos sin 2 4r

Qd Qd E a a

r r

= + ( )3

0

2cos sin4 r

pa a

r

= +

( )30

2cos sin4 r

p E a a

r

= +

20

4r p aV r

=When the dipole centre is at the origin,

'When the dipole centre is not at the origin, but at r ( )'

3'0

4

p r r V

r r

=



2 A point charge is a monopole and its field varies inversely as r and its potential varies inversely as r

3 The electric field due to a dipole varies inversely as r and its2 potential varies inversely as r

The electric field due to successive higher order multipoles4 5 6, , ,......varies inversely as r r r while their potential varies

3 4 5, , ,......inversely as r r r


Flux lines and equipotential surfaces An electric flux line is an imaginary line or path drawn in such a waythat its direction at any point is the direction of the electric field atthat point.

A surface on which the potential is the same throughout is calledequipotential surface.The intersection of an equipotential surface and a plane results in apath or line called equipotential line.No work is done in moving a charge from one point to another alongan equipotential line or surface.

This implies flux lines (direction of E) are always normal toequipotential surfaces.

0 E dl =


Electric dipoles and flux lines

+

Flux lines

Equipotential surface

Equipotential surfaces of a point charge


Electric dipoles and flux lines Z

Flux lines

Equipotential surface

0V >

0V RV

The electric field of spherical charge distribution is

0

0

3

r r

E or r a F R

=

( )12 E V W D E dV =

20

12 V

E dV =


Example 2

0

0

3 r

r E a

= 003r

E

=

2 22 0

209r

E

=

20

12 E V

W E dV = 2 2

020

0912 V

d r

V

= 0

22

0 0218 V

r dV

=

0

22

0 2 si8

n1 V

r r drd d

= 2 4

20

0 0 00

s18

in R

r r drd d

= = =

= 2 40 0 02

0

0

si18

n R

r d d r dr

= = =

= 2 5

0

00

418 5

Rr

=

2 50

0

245

R

=2 5

0

0

245 E

RW

=


Conduction and convection currentsThe current through a given area is the electric charge passingthrough the area per unit time. Its unit is Amperes.

One ampere current is produced if charge is transferred at the rateof one coulomb per second.The concept of current density is useful in defining the eventsoccurring at a point.If current I flows through a surface S, then current density is

dQ I

dt =

n I

J S

=

n And Assuming the current density is perpendic u I = J S lar

to the surface.


Conduction and convection currentsIf the current density is not normal to the surface

The total current flowing through the surface is

I = Jcos S = J S

S I = J dS

S

S


51/105


Convection currentsConvection current is produced when current flows through aninsulating medium such as liquid, rarefied gas, or a vacuum.It does not involve conductors and hence does not satisfy Ohmslaw.

A beam of electrons in a vacuum tube is an example of convectioncurrent.

l

S V

u

y


Convection currentsConsider a beam of electrons with volume charge density V flowingin the y direction with a velocity u y.

The current through the filament is

The y directed current density J y is given by

In general

y yu u a=

Q I t

= ( )V S l

t =

V l S t

= V ySu =

y V y I

J uS

= =

V J u =

V Convection current density J u =


Conduction currentsConduction current requires a conductor as a medium.

A conductor contains a large number of free electrons that provideconduction current due to an impressed electric field.When an electric field is applied the force on an electron withcharge e is

Since the electron is not in free space it will not be acceleratedunder the influence of the electric field.It suffers constant collisions with the atomic lattice and drifts fromone atom to another.If the electron with mass m is moving in an electric field with anaverage drift velocity the average change in momentum of the

free electron must match the applied force.

F eE =

E

E u


Conduction currents According to Newtons law,

This indicates that the drift velocity is directly proportional to theapplied field.If there are n electrons per unit volume the electronic charge densityis given byConvection current density is

mueE

=

eu E

m =

Average time interval

V ne = V u =

2ne E

m =

E =

J E =

between collisions

2

is the conduct n

i ityem v

=

POINT FORM OFOHMS LAW


Conductors A conductor contains free electrons which accounts for itsconductivity.In an isolated conductor, when an external electric field isapplied, the positive charges moves in the same direction as theapplied field.The negative charges moves in the opposite direction.These free charges accumulate on the surface of the conductor andform an induced surface charge .The induced charges set up an internal induced field whichcancels the externally applied field

e E

i E e E

Thus a perfect conductor cannot contain an electrostaticfield within it.


Conductors+++

+++++

_ _ _ _ _ _ _ _

e E

e E

e E

i E

i E

++++++++

_ _ _ _

_ _ _ _

0 E =

0V =

A perfect conductor cannot contain an electrostatic field within it.

e E

e E

e E


52/105


ConductorsInside a conductor Thus a conductor is an equipotential body.

Also, in a conductor, and as per the equation, theelectric field intensity

According to Gausss law,If the charge density

0 0 V=0 E V = =and which implies

J E =

0 E

0V =V S V

E dS dV = 0 E =

Inside a perfect conducter V ab ,E = 0, = 0 , V = 0


ConductorsWhen the two ends of the conductor are maintained at a potentialdifference V, the electric field is not zero inside the conductor.In this case there is no static equilibrium, since the applied voltageprevents the establishment of such equilibrium.

An electric field must exist inside the conductor to sustain the flow ofcurrent.The opposition to the flow is called resistance.

l E

V

I

- - - - -


ConductorsThe magnitude of the electric field is given bySince the conductor has a uniform cross section,By Ohms law

V E

l =

I J

A=

J E =

I V E

A l

= =

V l I A

=

l l R

A A

= = l Res istiv ity

=

l R

=


ConductorsResistance of a conductor having non-uniform cross section is

E dl V R

I E dS

= =


Example 1If calculate the current passingthrough

A hemispherical shell of radius 20 cm A spherical shell of radius 10 cm

33

1 (2cos sin ) A/mr J a ar = +

S I J dS =

2 si n r adS r d d =

33

1 (2cos sin ) A/mr J a ar = +

( ) ( )231 2cos sin sinr r S I a a r d d ar = + 1

2 cos sinS

I d d r

=


Example 1

31.4 A=

, 0 0.1 In the second case and r =

12 cos sin

S I d d

r =

2 /2

0 0

12 sin cos d d

r

= ==

/2

0

22 sin cos d

r

== /2

031.4 2sin cos d

==

0

22 sin cos I d

r

== 062.8 2sin cos d

== 0=


53/105


Continuity equation and relaxation time

dS J

dS

Bounding Surface S

inQ


Continuity equation and relaxation timeElectric charges can be neither created nor destroyed according tolaw of conservation of electric charges.Consider an arbitrary volume V bounded by surfaces S as shown infigure.

A net charge Q in exists within this region.If a net current I flows across this surface the charge in the volumemust decrease at a rate that equals the current.If a net current flows across the surface in to the volume, the chargein the volume must increase at the rate equal to the current.The current leaving the volume is the total outward flux of thecurrent density vector through the surface S.

----(1)inOUT S dQ

I J dS dt

= =



Substituting (2) and (3) in (1)

This equation is called continuity of current equation or continuityequation.

---(2)S V

Using divergence theorem, J dS = Jdv

----(3)in vvV V dQ d

= dv =dt dt t

= vV V

J dvt

= v J t

= v J t


Continuity equation and relaxation timeIt basically states that there can be no accumulation of charge atany point.For steady currents and henceThe total current leaving a volume is the same as t he total currententering it.Kirchhoffs current law follows from it.

=0vt

=0 J


Continuity equation and relaxation timeIf we introduce charge at some interior point of a given material withconstants and its effect can be obtained using the continuityequation.From Ohms lawFrom Gausss lawSubstituting in the continuity equation,

= J E

v D =

van d h en ce E

=

= v J t

= v E t

=v vt

=0v vt

+

=0v vt

+



This equation is a homogeneous linear ordinary differentialequation.Separating the variables,

Integrating both sides

=0v vt

+

/0= r

t T v v e

vv

t =

0ln lnv vt

= +

0ln v Where is a constant of integration

/0= r

t T v v e

r Where T

=


54/105


Continuity equation and relaxation timeWhen we introduce a volume charge density at an interior point in amaterial, it decays resulting in a charge movement from the interiorpoint at which it was introduced to the surface of the material.The time constant T r of this decay is called relaxation time orrearrangement time.

Relaxation time is the time it takes for a charge placed in the interiorof a material to drop to e -1 or 36.8 percent of its initial value.It is very short for good conductors and very long for gooddielectrics.For a good conductor the relaxation time is so short that most of thecharge will vanish from the interior point and appear at the surfacewithin a short time.For a good dielectric the relaxation time is very long that theintroduced charge remains at the same point.

191.53 10Cu s 51.2Quartz days


Boundary conditionsWhen the field exists in a medium consisting of two different media,the conditions the field must satisfy are called boundary conditions.For the electrostatic field t he following boundary conditions areimportant.

Dielectric dielectric interface.Conductor dielectric.Conductor free space.


Dielectric-dielectric Boundary conditionsConsider the boundary between two dielectrics with permittivities

1 21 0 2 0 r r = =and

a

b

cd

W

h

1

2

1t E

1n E 1 E

2t E

2 n E 2 E

1

2


Dielectric-dielectric Boundary conditionsThe fields in the two media can be expressed as

Apply the equation to the path abcda in the figure.

Tangential components of are equal at the boundary.undergoes no change on the boundary and it is continuous

across the boundary.

1 1 1t n E E E = +

2 2 2t n E E E = +

0l E dl =

1 1 2 2 2 1 02 2 2 2t n n t n nabcdah h h h

E dl E w E E E w E E

= + + =

1 2 0t t E w E w =1 2t t E E =

1 2t t E E = E

t E


Dielectric-dielectric Boundary conditions

The tangential component of under goes some change across theboundary.

So is said to be discontinuous across the boundary.The boundary conditions for the normal components are obtained byapplying Gausss law on a small pill box shaped volume as in thefigure.

1 2t t E E = 1 21 2

t t D D

= 2 1 1 2t t D D =

D

D

S D dS Q =

1 2n n S D S D S Q S = =

1 2n n S D D = 0h Assuming



h

1

2

1t D

1n D1 D

2t D

2n D2 D

1

2

S


55/105



If no free charge exists at t he boundary,

Normal components of are equal at the boundary.undergoes no change on the boundary and it is continuous

across the boundary.

The normal component of under goes some change across theboundary.So is said to be discontinuous across the boundary.

1 2n n S D D =0S =

1 2n n D D=

1 2n n S D D =

1 2n n D D= D

n D

1 2n n D D= 1 1 2 2n n E E =1 2

2 1

n

n

E E

=

D

n D



1 1t D

1n D

1 D

2t D

2n D2 D

1t E 1n E

1 E

2t E

2n E 2 E

2

1

2

1

2



1 2t t E E =

1 1 2 2sin sin - - (1) E E =1 2n n D D=

1 1 2 2cos cos D D =

1 1 1 2 2 2cos cos - - - (2) E E =

1 1 2 2

1 1 1 2 2 2

(2)co

sin sis cos

n) /(1

E E

E E

=

2 1 1 2tan tan =

1 1 0 1 1

2 2 0 2 2

tantan

r r

r r

= = = 1 12 2

tantan

r

r

=

This is called law of refractionCompiled by MKP for CEC S5 Batch July 2008

Conductor-Dielectric boundary conditions

a

b

cd

W

h

0 r Dielectric =

t E n E

E

0Conductor E =


Conductor-Dielectric boundary conditions

0 r Dielectric = D

n D t D

0Conductor E =

h

S


Conductor-Dielectric boundary conditionsThe interface between a perfect conductor and a dielectric shown infigure.

Apply the equation to the path abcda in the figure.

The boundary conditions for the normal components are obtained byapplying Gausss law on a small pill box shaped volume as in thefigure.

0l E dl =

0 0 0 02 2 2 2t n nabcdah h h h

E dl E w E w E

= + + =

0t E w =

0t E =

S D dS Q =

0n D S S Q = 0h Assuming

n S Q

DS

= = 0n r n S D E = =

0t D = 0 0t r t D E = = 00t r t D E = =


56/105


Conductor-Dielectric boundary conditionsNo electric field exists inside a perfect conductor.External electric field, if any, is normal to the conductor surface asgiven by n S D =


Conductor-Free space boundary conditionsThis is a special case of conductor-dielectric conditions.It can be obtained by replacing in the equation

As in the earlier case

0n r n S D E = =1r =

0n n S D E = =

0 0t r t D E = =0 0t t D E = =

0n n S D E = =


Conductor-Free space boundary conditions

0 r Free space = D

n Dt D

0Conductor E =

n E t E

E


Poissons and Laplaces equationGausss law in point form is given by

Substituting in the above equation,

We know that

This equation is known as Poissons equation.

V D =

D E =

V E =

E V =

( ) V V =

V V

=

2. ., V i e V

=

2 V V

=


Poissons and Laplaces equation

x y z x y z V V V

V a a a a a a x y z x y z

= + + + +

V V V x x y y z z

= + +

2 2 2

2 2 2

V V V x y z

= + +

2 2 22

2 2 2V V V V V

x y z

= + + =

2 2 2

2 2 2V V V V

x y z

+ + =


Poissons and Laplaces equationIn the case of a charge free region, Poissons equation reduces toLaplaces equation as given below.

In cylindrical and spherical coordinate systems

2

0V =

2 2 2

2 2 2 0V V V

x y z

+ + =

2 22

2 2 2

1 10

V V V V

z

= + + =

22 2

2 2 2 2

1 1 1sin 0

sin sinV V V

V r r r r r r

= + + =


57/105


Poissons and Laplaces equationLaplaces equation is very useful in finding out the potential V of aset of conductors maintained at different potentials as in the case ofcapacitor plates.Electric field can be obtained once we obtain potential V from E

E V =


General Procedure for solving Poissons and Laplacesequations to find capacitance

Solve Laplaces ( if v = 0 ) or Poissons equation ) if ( v 0 ) usingDirect integration if V is a function of one variableSeparation of variables if v is a function of more than one variable.The solution at this point is not unique but expressed in terms of unknownintegration constants.

Apply boundary conditions to determine a unique solution for V.

After obtaining V, find using and using

Find the charge induced on a conductor using where s=D n and Dn is the component of normal to the conductor.

If required find the capacitance between the conductors using C=Q/V

E E V = D D E =

D sQ dS =


Uniqueness Theorem

Proof: ( by contradiction )Proof: ( by contradiction )

Assume that there are two solutions V 1 and V 2 of Laplacesequations both of which satisfies the given boundary conditions.

V1 and V 2 must reduce to the same potential along the boundary

If a solution to Laplaces Equation can be found that satisfies theboundary conditions, then the solution is unique.

21 0V =

22 0V =

( )2 2 1 0That is V V =( )2 1 d Letting V V V = 2 0d V =

( )2 1 0d V V V = =


Uniqueness TheoremFrom divergence theoremThe above equation is true for any vector, so let be the vector

V S dV A dS =

d d A V V =

A

( ) ( ) ------(1)d d d V S d V V V V dV dS =

Using the vector identity ( ) A A A = +

( ) ( ) ( )d d d d d d V V V V V V = +

2 0d Putting V =

( )22

d d d V V V = +

( ) ( )2

----(2)d d d V V V =


Uniqueness TheoremPutting eq (2) in eq (1)

( )2

d d d V S dV dS V V V =

0d Putting V = ( )2

0d V

V dV = 2

0d V

V dV =

0d V = ( )2 1 0V V =

2 1 .Which implies V V is a constant every where

2 1 0 At the boundary we have seen that V V =

2 1 1 2 0 .So V V or V V everywhere = =

1 2 .So V and V cannot be different solutions of the same problem .Uniqueness theorem is proved


Example 1Using Laplaces theorem obtain the potential distribution betweentwo spherical conductors separated by a single dielectric. The innerspherical conductor of radius a is at a potential V 0 and the outerconductor of radius b is at potential zero. Also evaluate the electricfield.

0V

ab

24 sbQb

=

24 saQ

b

=


58/105


Example 1

22 2

2 2 2 2

1 1 1sin 0

sin sinV V V

V r r r r r r

= + + =

Laplaces equation in spherical coordinates is

Since V is a function of r only2 2

2

10

V V r

r r r

= =

2

1 0Since we get

r 2 0V r

r r

=

Integrating with respect to r 2 1V

r K r

=

21,

V r That K is

r

=


Example 1 gain integrating with respect to r 1 2

K V K

r = +

1 2 Constants K and K are found out by applying boundary.conditions

0( ) ,

( ) , 0

i r a V V

ii r b V

= =

= =

10 2

K V K a

= +

120

K K

b= +

1eqations

1Solving eqations

1 0 ( )ab

K V b a

=

02 ( )

V a K

b a

=

0 0

( (

) )V ab V a

V b b

T r

ha a

en =


Example 1

0 0

( ) ( )V ab V a

V b a r b a

= E V =

1 1 sinr

V V V E a a a

r r r

= + +

Since V is a function of r only r V

E ar

=

0 0

( ) ( ) r

V ab V ab a r r b

E aa

=

02 ( ) r

V ab E a

b a r =

02 (

V/m) r

V ab E a

b a r =


Example 2Find the potential at any point between the plates of a parallel platecapacitor and electric field.

++++++++

+ 0 y = y d =

z

y

0V


Example 2Laplaces equation in rectangular coordinates is

Since V is a function of y only

2

2 0V

y

=

2 2 2

2 2 2 0V V V

x y z

+ + =

Integrating with respect to y 1V

K y

=

Again integrating with respect to y 1 2V K y K = +

1 2 Constants K and K are found out by applying boundary .conditions

( ) , 0ii y d V = =0( ) y 0,i V V = =


Example 21 2 Substituting in V K y K = + 2 0 K V = 01

V K

d =

00

V V y V

d = +

E V = V a y

=

00 y

V y V a y d = +

0 V ad

=

0 V E ad

=0

yV

E ad

=


59/105


Example 3Find the potential distribution between the conductors, and thecapacitance per unit length.

ab

0V

1m


Example 3Laplaces equation in cylindrical coordinates is

2 22

2 2 2

1 10

V V V V

z

= + + =

Since V is a function of only 1

0V

=

10Since we get

0V

=

Integra ting with respect to 1V

K

=

Aga in integrating with respect to 1 2lnV K K = +


Example 3

1 2 Constants K and K are found out by applying boundary .conditions

0( ) ,

( ) , 0

i a V V

ii b V

= =

= =

(1)eqations

( ) 1Solving eqations

0 1 2lnV K a K = +

1 20 ln K b K = +

01 ln( / )

V K

b a= 0

2 lnln( / )V

K bb a

=


Example 3

0 0ln lnln( / ) ln( / )

V V V

bb a b a

+=0

ln( / )ln( / )

bV

a b =

0ln( / )

Voltsln( / )

bV V

a b =

E V = V a = 0

ln( / ) ln( / )

bV a

a b

= 0

ln( / )V

ab a

=

0 V/mln( / )

V E a

b a =

' , pplying gauss s law for the inner conduc tor

S D dS Q =


Example 3

0

2C=

ln( / )

QV b a

=2

C= F/mln /

( )b a

0 ln( / )S

V a d dza Q

b a

=

0

ln( / )S

V d dz Q

b a

=

0

ln( / ) S V

d dz Qb a

= 1 20

0 0ln( / )V

dz d Qb a

=

02ln( / )

V Q

b a =


60/105

MODULE III

Magnetostatics

Compiled by MKP for CEC S5 batch August 2008

Syllabus Magnostatics and Maxwells equations:

Magnostatic fields - Biot Savart law - Amperes circuital law - Applications of Amperes circuital law - Magnetic flux density -Magnetic scalar and vector potentials. Magnetic forces, materials

and devices - Forces due to magnetic fields - Magnetic torque andmoment - Magnetic dipole - Magnetization in materials -Classification of magnetic materials - Magnetic boundary conditions- Inductors and inductances - Magnetic energy - Magnetic circuits -Faradays law - displacement current. Time harmonic fields -Maxwells equations for static fields and time varying fields - wordstatement.

Compiled by: MKP for CEC S5 - July 2008

References Text Books:

1. Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford UniversityPress

2. Jordan and Balmain, Electromagnetic waves and radiating systems,Pearson Education PHI Ltd.

References:

1. Kraus Fleisch, Electromagnetics with applications, McGraw Hill2. William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill3. N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson

Education PHI Ltd.4. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical

Publishers.5. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill6. K.D. Prasad , Electromagnetic fields and waves, Sathya Prakashan

Compiled by: MKP for CEC S5 - July 2008

Concept of Current elementDirect currents flows only in closed loops.We can find out the contributions to the magnetic field due todifferential lengths of such current carrying conductors.

A current element is the current I flowing through a differential vectorlength of a filamentary conductor.

A filamentary conductor is the limiting case of a cylindrical conductorof circular cross section as the radius approaches zero.

dL

IdL

I


Biot-Savarts Law Biot-Savarts law states that the magnetic field intensity dH produced at a point P by a differential current element Idl isproportional to the product Idl and the sine of the angle between the element and the line joining P to the element

and is inversely proportional to the square of the distance Rbetween P and the element and its direction can beobtained by right handed screw rule.

2

sin Idl dH

R

2

sin Idl dH k

R =

1 / 4k is a proportionality constant whose value is

2

sin4

Idl dH

R

=


Biot-Savarts Law

2

sin4

Idl dH

R

=

This equation can be modified by incorporating the direction

of the magnetic field intensity

2

4

R I dl adH R

=34

I dl R R

=

and R R

R R a R

= =

Idl Current element Ra Unit vector directed from current element to P

R Distance from current element to point P

dH Magnetic field intensity at P


61/105


Biot-Savarts Law

R

P

I

dl Ra

dH

2

4

R I dl adH R

=

2

sin4

Idl dH

R

=

2

sin

4 n Idl

dH a R

=


Current elements

I

I dl KdS JdV

J

K

Line current Surface current Volume current


Concept of surface current density When current flows through a sheet of vanishingly small thicknesswe cannot measure current density in amperes per square meteras it becomes infinite.In this case surface current density is measured in amperes permeter width and is designated as

If the surface current density is uniform, the total current in any widthb is I=Kb where the width b is measured perpendicular to thedirection of current flow.

J

K

b

I

K


Concept of surface current density For a non-uniform surface current density, we have to integrate Kover the path of interest.

Where dN is a differential element of the path across which thecurrent is flowing.Let S be the cross sectional area of the wire. Then

Let dv=dSdh , where dh is the t hickness of the surface current. Then

I KdN =

IdL J SdL Jdv= =

( ) Jdv JdSdh Jdh dS KdS = = =

IdL Jdv KdS


Biot-Savarts Law In terms of distributed current sources, the Biot-Savarts law may beexpressed in the following ways:

2

4 R

Lor line current

I dl a H

R

=

2

4

R

S for sur

KdS a face cu e t

Rrr n H

=

2

4 R

V for

Jdv avo H

Rlume current

=


Magnetic field of a linear conductor

A

R

P

dl

I

z

y x

2

B

0

1

z

0,0,z

cot z =

a

z za

z R a za =

2 2 R z = +


62/105



dl Small current element in the conductor AB P Point where the magnetic field is required.

Perpendicular distance between the conductor and the point P

1 2, Angles subtended by the lower and upper ends of AB.

Let

By Biot - Savart's law the contribution dH at P due to an element

(0,0,z)dl at is

34 I dl R

dH R

=

z z But andl = dza R z d a a =



( ) z z dl R = dza z a a

= dza

3 4 I dz

dH a R

=

3 4 I dz

H a R

= cot But z = 2cosecdz = d

2 2 3/2 4 ( ) I dz

a z

=+

( )2

1

2 2

3/22 2 2

4 I cosec d

H a+ cot

= ( )

2

1

2 2

3/22 2

4 I cosec d

a1+cot

=



2

1

2 2

3 3

4 I cosec d

acosec

= 2

1

2 2

3/22 2

4 cos

I cosec d H a

ec

=

2

1

sin4 I

a d

=

( )2 1 cos cos4 I

a =

( )2 1 cos cos4 I

H a = l a a a =

[ ] 21

cos4 I

a

=


Magnetic field of a linear conductorWhen the conductor is semi-infinite , so that point a is now at O(0,0,0)while B is at

When the conductor is of infinite length , point A is at

1 2(0,0, ) 90 , 0 = = and

( )cos 0 cos 904 I

H a =

4 I

a =

4 I

H a =

1 2 (0,0, ) 180 , 0 B is at = = while and

(0,0, )

( )cos0 cos1804 I

H a =

2 I

a =

2 I

H a = l a a a =

l a a a =


Magnetic field at the centre of a circular current loop

90 = I dl

Ra X

Y

Z

O

P

2

sin4

n Idl

dH R

a

=

I a



The magnetic field intensity at O is given bywhere is the field intensity at O due to any current elementThe direction of at any point P on the circular wire is given bythe tangent at P in t he direction of current flow.The unit vector at P directed towards O is along the radius PO sothat = 90 o.

Total field intensity at the centre of the circular wire is obtained byintegrating dH around the circular path.

H dH =

dH Idl

2

4

R I dl adH R

=

2

sin4

n Idl

dH R

a

=

2

sin904

n Idl

aa

=

24 z dl

a I

a =

2 4 z I

H d al a

=

Idl


63/105



2 4 z I

H d al a

=

2 4 z I

H d al a

= 2 2 4 z I

aaa

=

2 z

I

aa=

2 z I

H a

a=


Magnetic field at a line through the centre of a circularcurrent loop

P

X

Y

Z

O

I

Rh

a

'dl dl

'dH

' ydH ydH

dH

' z dH z dH

dH

R

2 2 sin

a

a h =

+

2 2 R a h= +

Ra



Let P be a point at a distance h from the centre of a circular currentloop.Consider two diametrically opposite elements of the loop dl and dl .The field intensity at P distant R from the current element is given by

Since and are perpendicular

2

4

R I dl adH R

=2

sin4

ndl

a I

R

=

dl Ra

24n

Idl dH

Ra

=

n Ra A unit vector perpendicular to the plane containing dl and a

24 Idl

dH R =Compiled by MKP for CEC S5 batch August 2008


The field is oriented at an angle to the plane of the loop.The diametrically opposite element dl will also produce a field ofmagnitude equal to dH .Its component parallel to the plane of the loop gets cancelled.The components along the z axis gets added up.

24 Idl

dH R

=

2 sin4 z Idl

dH R

=

2 2 2

2 2 sin

a But and R a h

a h = = +

+

( )2 2 2 24 z Idl a

dH a h a h

=+ + ( )

3/22 24

Iadl

a h =

+



The resultant field intensity at the point P is given by integrating theZ components of the field contributions of all the current elements.

( )3/22 24 z

IadH dl

a h =

+

( )3/22 24

z Ia

H dH dl a h

= =+

( )3/22 24

Iadl

a h =

+

( )3/22 2

24

Iaa

a h

=

+ ( )2

3/22 22

Ia

a h=

+

( )2

3/22 2

2 z

Ia H a

a h=

+ ( )2

3/22 2

2 z

Ia H a

a h=

+


Example 1Find the magnetic field at (0,0,5) due to side OA and BO of thetriangular loop carrying a current of 10A and lying in the xy plane.

Y

Z

2

1

10 AO

A B

1

X


64/105


Example 1

X

Z

O

A

2

( )2 1 cos cos4 I

H a =

5

1 2

a

l a a a =

l xa a=

5 = 1cos cos90 0 = =

22

cos29

=

10 2 4 5 29

H a =

10 I =

0.059.1 a =

x ya a a a = =

0.059.1 y H a=

To find H due to OA


Example 1

Y

Z

2

5

10 AO

A

B

1

2

5 =27

( )2 1 cos cos4 I

H a =

l a a a =

5 =

1cos cos105.7 0.272 = =

2cos cos90 0 = =

10 I =

( )1

1 22

5180 sin

5 2

= +

2 90 =

1 2105.7 , 90 = =

To find H due to BO

1


Example 1( )2 1 cos cos4

I H a

=

( )10 0 0.2724 5

H a = +

0.272 0.0433

2a a

= =

l a a a =

cos45 cos45l x ya a a= 0.707 0.707 x ya a=

( ) 0.707 0.707 x y z a a a a =

l z a a=

0.707 0.707 y xa a=

( ) 0.0433 0.707 0.707 y x H a a= 0.0306 0.0306 x y H a a= +


Example 2 A circular loop located on x2+y 2= 9 carries a direct current of 10 Aalong . Determine magnetic field at (0,0,4) and (0,0,-4)a

( )2

3/22 2

2 z

Ia H a

a h=

+

3a = 4h = 10 I =

( )2

3/22 2

10 3

2 3 4 z H a

=+

90 250 z

a= 0.36 z a=

(0,0,4)at

3a = 4h = 10 I = ( 0, 0, 4)at

( )( )

2

3/222

10 3

2 3 4

z H a=

+

90

250 z a= 0.36 z a=

(0,0,4)

(0, 0, 4)

3a =

H

H

10 I = 4

Z


Amperes Current law Amperes circuital law states that the line integral of the tangentialcomponent of around a closed path is the same as the net currentenclosed by the path .

Line integral of the magnetic field around a closed path is calledmagneto motive force.So amperes law can be stated as : The magneto motive force around aclosed path is equal to the current enclosed by that path .

encl H dl I =

H

H

enc S ut I J dS =

S l J dS H dl =

S l J dS H dl =


Amperes Current law

encl H dl I =

I


65/105


Amperes Current law Applying Stokes theorem to the LHS of the above equation

This is the Amperes law in differential form or point form and is thethird of the Maxwells equations.

( )S S J dS H dS =

H J =

H J =


Magnetic field of an infinite line current

Z

Y

X

P

I

O

Amperian Path

dl


Magnetic field of an infinite line currentLet an infinitely long filamentary current element be placed along thez axis.To determine H at a point P we select a closed amperian path thatpasses through P.Since this path encloses the whole current,

enc l I H dl =

l

I H a dla = l

H dl = 2 H =

2 I

H =

2 I

H a =

2

I H a

=


Magnetic field of a coaxial cable

a

b

t

1 L

2 L

3 L

4 L

I I +

Amper ian path sZ


Magnetic field of a coaxial cableConsider an infinitely long transmission line consisting of twoconcentric cylinders having their axes along the Z a xis.The inner conductor has radius a and carries current I while theouter conductor has inner radius b and thickness t and carriesreturn current I .The magnetic field is to be evaluated for the four regionsseparately.

0 a

a b

b b t

b t

+ +



1 0 ' Fo r the regio n a apply Am pere s law to pa th L

2enc

I I J dS d d

a

= =

Current is assumed to be uniformly distributed over thesec .cross tio n

2 z I

aa

= z and dS d d a =

1enc L

H dl I J dS = =

2

I d d

a

=


66/105


Magnetic field of a coaxial cable2

2 2 2enc I

I a

=

2

2

I a =

1

2

2 L H dl

I

a

=

1

2

2

L H a dla

I a =

1

2

2 L H d

al

I

=2

22 H I

a

=

2

2

2

I

a H

=

22 I

a H

=

22 I

H a

= 0 a



2 ' Fo r the regio n a b ap ply Am pere s law to pa th L

2enc L

H dl I J dS = =

2

L

H a dl I a =2 Since whole current is enclosed by the path L

1 L H dl I =

2 H I =

2 H

I

=

2 I

H = a b



3 ' Fo r the regio n b b t ap ply Am pe re s law to pa th L +

3 L H dl I J dS = +

3

L

H a dla I J dS = +

3 L H dl I J dS = +

2 H I J dS = +

The J in this case is the current density of the outer

conductor Compiled by MKP for CEC S5 batch August 2008


( )2 2 z

I J a

b t b =

+

2 H I J dS = +

( )2 22 z z S

I I a d d a

b t b H

+

=

( )2 2 S I

I d d b t b

= +

( )

2

2 2

0 b

I I d d

b t b

=

= +



( )

2

2 22 2

2b H

I I

b t b

+

=

( )2 22 2 22 I b I

b bt t b =

+ +

( )2 222

I b I bt t

= +

( )2 221 2

b I

t bt

= + ( )

2 2

21

2 2

b I H

t bt

= +

b b t +


Magnetic field of a coaxial cable4 ' Fo r the regio n b t ap ply Am pe re s law to pa th L +

4

0 L

H dl I I = =

2 0 H =0 H =

0 H =

b t +

2 02 I

a H a

=

2 I

H a b =

( )2 22

12 2

b I

H t bt

b b t

=

++

0 t H b +=


67/105



H

0

12 a

12 b

a b b t +

H Vs IMPORTANT:

Outside the co-axial cablemagnetic field is zero. This is

the basic principle behindelectro magnetic shielding


Magnetic field by a toroid A toroid can be considered as a solenoid bent into a circle asshown. We can apply Amperes law along the circular path insidethe toroid.

encl L H dl I =

encl I NI =

N is the number of loops in the toroid,and I is the current in each loop

2 NI

B a

=

encl L Hdl I =

encl L

H dl I =2 encl H I =

2 H NI =

2

NI

H =

2 NI

H a =


Magnetic field by a solenoidWhen the coils of the solenoid are closely spaced, each turncan be regarded as a circular loop, and the net magnetic fieldis the vector sum of the magnetic field for each loop.This produces a magnetic field that is approximately constantinside the solenoid, and nearly zero outside the solenoid.


Magnetic field by a solenoidThe ideal solenoid is approached when the coils are very closetogether and the length of the solenoid is much greater than itsradius. Then we can approximate the magnetic field as constantinside and zero outside the solenoid.

H

Apply Amperes law for the closed path 1234

N turns in alength of L

xa


Magnetic field by a solenoid

12 23 34 41 L

H d l H d l H d l H d l H d l = + + +

encl L H d l I =

0 0 H L H L= + + =

=C u r r e n t e n c l o se d b y t h e p a t h i s N I

H L N I =

N I H

L=

N I B

L =

x N I

B a L

=

times number of turns per unit length N

B I I L

= =

x N I

H a L

=


Magnetic field by a long cylindrical conductor

Find the magneticfield inside andoutside an infinitelylong cylindrical

conductor havingradius R andcarrying a current I

z a


68/105



A long straight wire of radius R carries a steady current I that isuniformly distributed through the cross-section of the wire.In region where > R choose a circle of radius centered on thewire as a path of integration. Along this path, H is again constant inmagnitude and is always parallel to the path.

encl L H dl I =

encl L H dl I =

( )2 encl H I =2 H I =

2 I

H

=

2

I H a

=

2

I a

=

2

I H a

=

2

I a

=



In region where < R choose a circle of radius centered on thewire as a path of integration. Along this path, H is again constant inmagnitude and is always parallel to the path.

encl L H dl I =

encl L H dl I =

( )2 encl H I =

2

encl I H

=

encl S I J dS =

2

2

I R =

2 z I

J a R

=

2

encl z z S

I I a dSa

R =

z dS dSa=

2 S

I dS

R =

22

I R

=



2

2 2 I

H R

=

2 2 I

H R

=

2 2 I

B R

=

2 2 I

B a R

=

2 2 I

a R

=

2 2 I

H a R

=

2

2 I

H a R

=



H

R

2 for

2 I

a R R

= >1 so that we can approximate thevalue of and as follows

( ) j j = +

1 j

j

= +

1Since or

j =

Compiled by MKP for CEC S5 batch September 2008

Wave propagation in good conductors

j =

1 2 1

2

j

+ = ( )

2

1 2 j = +

( )12

j = + j = +

Equating the real and imaginary parts

2

= =


Wave propagation in good conductorsThe wave equation in this case is of the form

When we consider the forward traveling wave only

A high frequency uniform plane wave suffers attenuation as itpasses through a lossy medium.Its amplitude gets multiplied by the factor, e - x where is theattenuation constant.

1 2 or z z z z

y m m E E e e E e e = +

1 2( , ) cos( ) cos( ) z z

y m m E z t E e t z E e t z = + +

1( , ) cos( ) z

y m E z t E e t z =



ze

z

1m E

y E

y


Wave propagation in good conductorsSkin depth is defined as the distance the wave must travel to haveits amplitude reduced by a factor of e-1.The exponential multiplying factor is unity at z=0 and decreases to1/e when z=1/ So skin depth is

Thus we see that the skin depth decreases with an increase infrequency.The intrinsic impedance of a medium may be expressed in terms of skin depth.

1 =Skin depth

2

=

1=

f



= j

j

+

fo r a co nd uc to j

r

45

= 1But, =2

2

245

2 =

22

245

=

245

= ( )1 1 j

= + ( )1 1 j

= +


104/105


Wave propagation in good conductorsThe phenomenon by which field intensity in a conductor rapidlydecreases with increase in frequency is called skin effect.

At high frequencies the fields and the associated currents areconfined to a very thin layer of the conductor surface.

1 =Skin depth


Poynting theoremThe vector product at any point is a measure of the rateof energy flow per unit area at that point ; the direction of energyflow is perpendicular to E and H in the direction of the vector

P E H =

E H

2/P E H W atts m=


Poynting theorem- Proof Maxwells first curl equation states that

This equation can be rewritten as

Pre dotting the equation (2) with we get,

We have the vector identity

D H J

t

= +

D E

J H H t t

= =

E i

( ) (1) D E E E J E H E E H t t

= =

i

i i i i

( ) ( ) ( ) (2) E H H E E H =

( ) ( ) ( ) E H H E E H =


Poynting theorem - Proof Putting eq (2) in (1)

( ) ( ) E E J H E E H E t

=

i i

( ) B E E J H E H E t t

=

i i

( ) H E H E H E t t

=

i

( ) (3) H E H E E H t t

=

i

2 21 1 (4)2 2

H E But H H and E E

t t t t

= =


Poynting theorem - Proof Substituting equation (4) in equation (3), we get

Integrating both sides of equation (5) over a volume V, we get

( )2 21 12 2 E J H E E H t t =

i

( )2 2

2 2 H E E H

t t =

( )2 2

(5)2 2

H E E H

t = +

( ) ( )2 2

2 2v v H E

E J d d E H d t

= +

i


Poynting theorem - Proof We apply divergence theorem to the last term on the LHS, convertthe volume integral into a surface integral and get

( ) ( )2 2

2 2v S H E

E J d d E H dS t

= +

i

( ) ( )2 2

2 2S v H E

E H dS E J d d t

= + +

i

( ) ( ) 2 2S v B H E D

E H dS E J d d t

= + +

i


105/105


Poynting theorem

Term 1 Term 2 Term 3

( ) ( )2 2

2 2S v E E H E d

t d H dS J

+ +

= i

Ingoing power fluxover the surface S

Total dissipated powerwithin the volume V atany instant due to ohmiclosses

Rate of decrease dueto total electromagneticenergy stored withinthe volume V

Magnetic energystored within thevolume V

Electrostatic energystored within thevolume V


Poynting theorem - Interpretation

2

2 E

d

Stored electrical energy

Stored magnetic energy

Ohmic losses

Power out

Power in

( )S

E H dS

( )v E J d

i

2

2 H

d



The first term represents the rate of flow of energyinward through the surface of the volume or ingoing power flux overthe surface S

The second term represents total dissipated powerwithin the volume V at any instant.For a conductor of cross sectional area A carrying a current I andhaving a voltage drop E per unit length, the power loss is EI wattsper unit length.The power dissipated per unit volume is

Total power dissipated in a volume is

( )S E H dS

( )v E J d

i

( )v E J d

i

. E I

E J watts per unit volume A

=



The first part of the third term is the stored electric energyper unit volume ( electrostatic energy density ) of the electric field.

The second part of the third term is the stored magneticenergy stored per unit volume ( magnetic energy density ) of themagnetic fieldThe volume integral of the sum represents total electromagneticenergy stored within the volume V

212

E

212

H

Documents

EMT Lectures