TAGORE - files.teceee-in.webnode.infiles.teceee-in.webnode.in/200000080-99f329aec5/EMT-Lecture Notes-TEC.pdf · electromagnetic theory [ ee2202 ] (lecture notes) dr. k.srinivasan

ELECTROMAGNETIC THEORY [ EE2202 ]

(LECTURE NOTES)

Dr. K.SRINIVASAN& Mr. K.KARTHIKEAN

TAGORE ENGINEERING COLLEGE

Rathinamangalam, Chennai - 600127,

ELECTROMAGNETIC THEORY [EE2202]

1 Dr.K.Srinivasan &Mr. K.Karthikeyan /EEE/TEC - LECTURE NOTES

http://www.valliammai.co.in/

ELECTROMAGNETIC THEORY [ EE2202 ] 1. INTRODUCTION Sources and effects of electromagnetic fields – Vector fields – Different co-ordinate systems- vector calculus – Gradient, Divergence and Curl - Divergence theorem – Stoke’s theorem. 2. ELECTROSTATICS Coulomb’s Law – Electric field intensity – Field due to point and continuous charges – Gauss’s law and application – Electric potential – Electric field and equipotential plots – Electric field in free space, conductors, dielectric -Dielectric polarization - Dielectric strength - Electric field in multiple dielectrics – Boundary conditions, Poisson’s and Laplace’s equations – Capacitance- Energy density. 3. MAGNETOSTATICS Lorentz Law of force, magnetic field intensity – Biot–savart Law - Ampere’s Law – Magnetic field due to straight conductors, circular loop, infinite sheet of current – Magnetic flux density (B) – B in free space, conductor, magnetic materials – Magnetization – Magnetic field in multiple media – Boundary conditions – Scalar and vector potential – Magnetic force – Torque – Inductance – Energy density – Magnetic circuits. 4. ELECTRODYNAMIC FIELDS Faraday’s laws, induced emf – Transformer and motional EMF – Forces and Energy in quasi-stationary Electromagnetic Fields - Maxwell’s equations (differential and integral forms) – Displacement current – Relation between field theory and circuit theory. 5. ELECTROMAGNETIC WAVES Generation – Electro Magnetic Wave equations – Wave parameters; velocity, intrinsic impedance, propagation constant – Waves in free space, lossy and lossless dielectrics, conductors-skin depth, Poynting vector – Plane wave reflection and refraction – Transmission lines – Line equations – Input impedances – Standing wave ratio and power.

TEXT BOOKS 1. Mathew N. O. SADIKU, ‘Elements of Electromagnetics’, Oxford University press

Inc. First India edition, 2007. 2. Ashutosh Pramanik, ‘Electromagnetism – Theory and Applications’, Prentice-Hall

of India Private Limited, New Delhi, 2006.

REFERENCE BOOKS 1. Joseph. A.Edminister, ‘Theory and Problems of Electromagnetics’, Second

edition, Schaum Series, Tata McGraw Hill, 1993. 2. William .H.Hayt, ‘Engineering Electromagnetics’, Tata McGraw Hill edition, 2001. 3. Kraus and Fleish, ‘Electromagnetics with Applications’, McGraw Hill International Editions, Fifth Edition, 1999.



SUBJECT NAME: ELECTROMAGNETIC THEORYSUBJECT CODE: EE 2202BRANCH: EEESEMESTER: III

UNIT – I

CONTENTS:

1. SOURCES & EFFECTS OF ELECTROMAGNETIC FIELDS

2. VECTOR FIELD

3. CO-ORDINATE SYSTEMS

4. DIVERGENCE THEOREM

5. STOKES THEOREM



SOURCES OF ELECTROMAGNETIC FIELDS

NATURAL SOURCES OF ELECTROMAGNETIC FIELDS

• Electromagnetic fields are present everywhere in our environment but are invisible to thehuman eye.

• Electric fields are produced by the local build-up of electric charges in the atmosphereassociated with thunderstorms.

• The earth's magnetic field causes a compass needle to orient in a North-South direction andis used by birds and fish for navigation.

HUMAN-MADE SOURCES OF ELECTROMAGNETIC FIELDS

• Besides natural sources the electromagnetic spectrum also includes fields generated byhuman-made sources: X-rays are employed to diagnose a broken limb after a sport accident.

• The electricity that comes out of every power socket has associated low frequencyelectromagnetic fields

FIELD:

• Field is a function of that specifies a quantity everywhere in a region or a space.• If at each point of a region or space, there is a corresponding value of some physical

function then the region is called a field.

MAGNETIC FIELD:

• If the field produced is due to magnetic effects, it is called magnetic field.• The current carrying conductor produces a magnetic field.

ELECTRIC FIELD:

There are two types of electric charges• Positive and• Negative

Such charges produce a field around it, which is called an electric field. The moving chargesproduce a current.

ELECTROMAGNETIC FIELD:• Electromagnetic is a branch of physics (or) electrical engineering.• Electric and magnetic fields are related to each other. Such a field is called electromagnetic

field.• The field may be time varying or time independent.• Quantify the field, three-dimensional representation can be made easy by the use of vector

analysis.

ELECTROMAGNETIC QUANTITY:

The various quantities involved in the study of engineering electromagnetic can be classified as

• Scalars• Vectors



SCALAR:

A scalar is a quantity, which is wholly characterized by its magnitude.• Quantity whose value may be represented by a single real number• It may be positive or negative• The direction is not required in describing a scalar.

EX:♦ Mass♦ Density♦ Pressure♦ Volume♦ Volume resistivity♦ Speed♦ Temperature♦ Voltage

VECTOR:

A vector is a quantity, which is characterized by both magnitude and a direction.

• In electromagnetic vectors defined in two and three-dimensional spaces.• Vectors may be defined in n-dimensional space.

EX:♦ Force♦ Velocity♦ Acceleration♦ Displacement♦ Electric field intensity♦ Magnetic field intensity♦ Straight line from positive to negative of battery.

SCALAR FIELD:

The distribution of a scalar quantity with a definite in a space is called scalar field.

EX:♦ Temperature of atmosphere♦ Sound intensity in an auditorium♦ Light intensity in a room

VECTOR FIELD:

If a quantity, which is specified in a region to define a field, is a vector then the corresponding fieldis called a vector field.

• Gravitational force on a mass in a space• Velocity of particles in a moving fluid• Wind velocity of atmosphere• Voltage gradient in a cable



REPRESENTATION OF A VECTOR:

R A TERMINATING POINT

OSTARTING POINT

ü Two dimentional ,a vector can be represented by a straight line with an arrow in a plane.ü Length of the segment is called magnitude.ü Arrow indicates direction of a given co-ordinate system.ü OA = R

UNIT VECTOR:

The unit vector has only direction .its magnitude is always unity.

Unit vector = aOA =OAOA

By the use of the unit vectors ax, ay, az along the x, y, and z axes of a Cartesian co-ordinate system. A = Axax+Ayay+Azaz

A = A = 222zyx AAA ++

VECTOR ALGEBRA:

VECTORS MAY BE ADDED

A+B = ( )zzyyxx aAaAaA ++ + ( )zzyyxx aBaBaB ++

= ( ) ( ) ( ) zzzyyyxxx aBAaBAaBA +++++SUBTRACTED

A -B = ( )zzyyxx aAaAaA ++ - ( )zzyyxx aBaBaB ++

= ( ) ( ) ( ) zzzyyyxxx aBAaBAaBA −+−+−

ASSOCIATIVE LAW:

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

DISTRIBUTIVE LAW:

K (A+B) = kA+kB

(K1+K 2 ) =K1A+K2A

CUMULATIVE LAW:

A+B =B+A



DOT PRODUCT LAW:

Dot product of two vectors is, by definition

A.B = AB cos θ

Where, θ is the smaller angle between A and B

A.B = AxBx+AyBy+AzBz

Ex:1

The dot product obeys the distribute & scalar multiplication Laws

A. (B+C) = A.B +A.C

A. K B = K (A.B)

A. B = ( )zzyyxx aAaAaA ++ . ( )zzyyxx aBaBaB ++

= AxBx (ax. ax) +AyBy (ay.ay) +AzBz (az.az) +

Ax By (ax.ay) + …..+AzBy (az.ay)

ax. a x = ay.ay = az.az =1Cos θ in the dot product is unity. When angle is zero, θ=900. cos θ=0.All other dot productof the unit vectors are zero.

A. B =AxBx+AyBy+AzBz

CROSS PRODUCT LAW:

Cross product of two vectors is by definition

A x B = (AB Sinθ) an

Where θ is the smaller angle between A & B. an is unit vector.

Expanding the cross product in component from

Ax B = ( )zzyyxx aAaAaA ++ x ( )zzyyxx aBaBaB ++Which is conveniently expressed as a determinant?

ax ay azAx B = Ax Ay Az

Bx By Bz

B X A = - A X B

B X A has the same magnitude but opposite direction.



Ex: 2Given A=2ax+4ay-3az and B =ax- ay, find A .B and A x B.

A .B = AxBx+AyBy+AzBz

= 2(1) + 4(-1) + (-3) (0)

= 2 - 4 + 0

=-2

ax ay az Ax B = Ax Ay Az

Bx By Bz

= (AyBz-AzAy) ax-(AzBx-AxBz) ay+ (AxBy-AyBx) az

= ax (0-3)-ay (0+3) +az (-2-4)

= -3ax-3ay-6az

THE COORDINATE SYSTEMS:

• To describe a vector accurately and to express a vector in terms of its components, it isnecessary to have some reference directions. Such directions are represented in terms ofvarious coordinate systems.

• There are various coordinate systems available in mathematics, which are

v Cartesian or rectangular coordinate systemv Cylindrical coordinate systemv Spherical coordinate system

CARTESIAN COORDINATE SYSTEM:

• This is also called rectangular co-ordinate system.• This system has three coordinate axes represented as x, y, and z which are mutually at right

angles to each other’ s.• These three axes intersect at a common point called origin of the system. There are two

types of such system called

v Right handed systemv Left handed system

• The right handed system means if x axis is rotated towards y axis through a smaller angle,then this rotation causes the upward movement of right handed screw in the z axisdirection.

• In this system, if right hand is used then thumb indicates x axis, the forefinger indicates yaxis and middle finger indicates z axis, when three fingers are held mutually perpendicularto each other.

• In left handed system x and y axes are interchanged compared to right handed system. Thismeans the rotation of x axis into y axis through smaller angle causes the downwardmovements of right handed screw in the z axis direction.



NOTE:

• The right handed system is very commonly used.

• The base vectors are the unit vectors, which are strictly oriented along the directions of thecoordinate system. The base vectors are the unit vectors oriented in x, y, z-axis of thesystem. So ax, ay, az are the base vectors of Cartesian coordinate system.

• Consider a point P (x1, y1, z1) is the position vector by the distance from the origin,directed from origin to point P radius is called radius vector.

r op = x1ax+y1ay+z1az

• The magnitude of this vector in terms of three mutually perpendicular components given

by,

21

21

21 zyxrop ++=

aop =Unit vector along op=op

op

rr

DIFFERENTIAL ELEMENTS IN CARTESIAN COORDINATE SYSTEM

• Consider a point P( x, y ,z) in the rectangular co-ordinate system. Let us increase each co-ordinate by a differential amount. A new point P’ will be obtained having co ordinates (x+ dx, y + dy, z + dz).

Thus

dx= Differential Length in x direction.dy = Differential Length in y directiondz = Differential Length in z direction

• Differential vector Length also called elementary vector length can be represented as

dl= dx ax + dy ay + dz az

• This is the vector joining original P to new point P’ .



• Now point P is the intersection of three planes while point P’ is the intersection of newthree planes which are slightly displaced from original three planes.

• These six planes together define a differential volume, which is a rectangularparallelpiped. The diagonal of this parallel piped is the difference vector length.

• The distance of P’ from P is given by magnitude of the differential vector length,

222 dzdydxdl ++=

• Hence the differential volume of the rectangular parallelepiped is given by

dv= dx dy dz

NOTE:

dl is a vector

dv is a scalar

The differential surface element ds is represented as

andsds =

ds =Differential surface area of the element

=xds Differential vector surface area normal to x direction.

= dy dz ax

=yds Differential vector surface area normal to y direction.

= dx dz ay=zds Differential vector surface area normal to y direction.

= dx dy az



CYLINDRICAL CO-ORDINATE SYSTEM:

• The circular cylindrical co-ordinate system is the three –dimensional version of polar co-ordinate of analytic geometry.

• Three unit vectors in cylindrical co-ordinates are al , aφ and az .

• The unit vector al at a point P(l , φ , z ) is directed radically outward, normal to thecylindrical surface l = l 1 .

• It lies the plane φ = φ 1 and z =z1. The unit vector aφ is normal to plane φ = φ 1 pointsin the direction of increasing φ 1 lies in the plane z=z 1 and in tangent to the cylindricalsurface l =l 1.

• The unit vector az is the same as the unit vector az of the Cartesian co-ordinate system.

• The unit vectors are mutually perpendicular for each is normal to three mutuallyperpendicular surfaces.

al . aφ = az

• A differential volume element in cylindrical coordinates may be obtained by increasing l ,φ and z by differential increments dl , dφ , dz.

• The cylinders of radius l and (l +dl ), radial plane angles φ and (φ +dφ ), horizontalplanes at z and z + dz.

• The rectangular parallel piped having sides of length dl , l dφ and dz.



NOTE:d l and dz are dimensionally lengths, but dφ is not, l dφ is the length.

The surface areas of l d l dφ , dl dz and l dφ dz. And volume becomes l d l dφ dz.x = l cos φy = l sin φz = z

= 22 YX +

Φ= tan −

xy

Z = ZCartesian vector

A= Ax ax + Ay ay +Az az

Where each component is given as a function of x,y,z.

In Cartesian co-ordinates

A= A a + AΦ aΦ +Az az

Where each component is given as a function of ,Φ,Z.

A =A. a

AΦ =A . aΦ

Expanding these dot product,

A = ( Ax ax + Ay ay +Az az ) . a

= Ax ax . a + Ay ay . a

AΦ = ( Ax ax + Ay ay +Az az ) . aΦ

= Ax ax . aΦ+ Ay ay. aΦ

Az = ( Ax ax + Ay ay +Az az ) . az

= Az az . az

= Az

Since az . a and az . aΦ are zero.a aΦ az

ax . cosΦ -sinΦ 0ay. sinΦ cosΦ 0az . 0 0 1

The angle between ax and a as Φ, so ax . a = cosΦ. But the angle between ay and a is 90 -Φand ay . a = cos(90 -Φ) =sinΦ



SPHERICAL COORDINATESz

• ( u1,u2,u3 ) = ( R ,θ,ϕ ) A point P(R1, θ1, ϕ1 ) in spherical coordinates inspherical coordinates is specified as the intersection of the following 3surfaces :

• a spherical surface centered at the origin with a radius R = R1: a rightcircular cone with its apex at the origin, its axis coinciding with the +z axisand having a half angle θ = θ1; and a half plane containing the z axis andmaking an angle ϕ = ϕ1 with the xz plane.aR, aθ and aϕ are the base vectors.

• This system is also a right-handed system having the following relations

aR x aθ = aθaθ x aϕ = aR

aϕ x aR = aθ

• A vector in spherical coordinates is given by

A = aRAR + aθAθ + aϕAϕ

• For spherical coordinates the metric coefficients are h1 = 1, h2 = R and h3 =Rsinθ

• The differential length dl is given as

dl = aRdR + aθRdθ +aϕRsinθdϕ

• The differential surface area ds is given as

dsR = R² sinϕdϕdθdsθ = RsinθdϕdRdsϕ = Rdrdθ



• Differential Volume element in Spherical Coordinates

The differential volume dv is given as

dv = R² sinθdRdθdϕ

Conversion formulae

From spherical to Cartesian coordinates

x = Rsinθcosϕ

y = Rsinθsinϕ

z = Rcosθ

From Cartesian to spherical coordinates

R = ( x² + y² + z² )1/2

θ = tan –1 ( x² + y² )1/2z

ϕ = tan –1 (y/x)



DIVERGENCE THEOREM:

• Divergence is defined as the net outward flow of the flux per unit volume over a closed incrementalsurface.

• The volume integral of the divergence of a vector field over a closed surface S enclosing volume Vis equal to the volume integral of the divergence of integral taken through out the volume V.

AdvdsAvs .. ∇= ∫∫∫∫∫

VOLUME INTEGRAL:

• It is the integral over the region enclosing a volume V.

SURFACE INTEGRAL:

• Surface integral is the surface of the surface of the region .It is denoted by S.

• The divergence of any vector A is given

)1(. −−−−−−∂

∂+

∂

∂+

∂∂

=∇z

Ay

Ax

AA zyx

dv = dx dy dz

• Taking volume integral on both sides

zA

yA

xA

Adv zyxvv ∂

∂+

∂

∂+

∂∂

=∇ ∫∫∫∫∫∫ . )

• Consider element in x direction

]dydzdxx

Axv

∂∂

= ∫∫

• But x1 and x2 be limits for x direction

AxAxAxdxx

Ax =−=∂

∂∫ 21

• Then

Axdydzdxdydzx

Axs∫∫∫∫∫ =

∂∂

∫∫= xs Axds -------------------------- (2)

• lllly the following terms

Aydsydxdydzy

Aysv ∫∫∫∫∫ =

∂∂

------------- (3)



)4(−−−−=∂

∂∫∫∫∫∫ Azdszdxdydz

zAz

sv

• From equation (1),put equa (2),(3),(4).

)dxdydzz

Ay

Ax

AAdv zyxvv ∂

∂+

∂

∂+

∂∂

=∇ ∫∫∫∫∫∫ .

= ( )AzdszAydsyAxdsxs ++∫∫

dsAAdv sv .. ∫∫∫∫∫ =∇

STOKES THEOREM:

• The surface integral of the curl of a vector field over an open surfaceis equal to the closed line integral of the vector of along the contourbounding the surface.

( ∇ x H).ds = H.dl s

Proof:

Consider an arbitrary surface this is broken up into incremental surfaces of area s∆ as shown in the fig. If H is any field vector, the by definition of the curl to one of these incremental surfaces.

)( NxHs

sdlH∆=

∆

∆∫ .

Where, N indicates normal to the surface and dl s∆ indicates that the closed path of an incrementalarea s∆ .



The curl of H normal to the surface can be written as

)( NaxHs

sdlH∆=

∆

∆∫ .

or

∫ ∆∇=∆ saxHsdlH N)(.

sxH ∆∇= ).(

Where aN is a unit vector normal to s∆ .

The closed integral for whole surface s is given by the surface integral of the normal component of curl H

∫ ∫∫∇=s

dsxHdlH ..

Problems:

1. Determine the volume of the sphere of radius R.

SOLUTION:

Difference volume in spherical coordinates is



2. Determine by integration the volume V of a region defined in a cylindricalCoordinate system as

SOLUTION: Differential volume in cylindrical coordinates is

3. Determine by integration the area S of a surface defined in a spherical coordinatesystem as

SOLUTION: Differential surface area in spherical coordinate S



1


UNIT – IICONTENTS:

1. COULOMB’ S LAW.

2. ELECTRIC FIELD INTENSITY.

3.FIELD DUE TO CHARGE DISTRIBUTION.

4. GAUSS’S LAW.

5. ELECTRIC POTENTIAL.

6. DIELECTRIC POLARIZATION.

7. BOUNDARY CONDITIONS.

8. POISSON’S AND LAPALACE EQUATIONS.

9. CAPACITANCE.

10. ENERGY DENSITY.

11. DIELECTRIC STRENGTH.



2

1. COULOMB’S LAW:

Ø According to coulomb’ s law force between any two point charges is

• Proportional to the magnitude of each charges;

• Inversely proportional to the square of the distance between them; and is

• Directed along the line joining the charge’ s

Ø The force of interaction also depends upon the medium in which the charges are situated.

Ø Consider the two point charges q1 and q2.

Ø Force of interaction between charges ,q1 and q2 may be formulated as follows:

)1(][ 1212

2112 −−−−= U

rqqkF

Where

v F12=force exerted by q1 on q2v U12=Unit vector directed from q1 to q2.v R12 =distance between the charges.v K=proportionality constant

MKS system of units:

v Force = Newton ( N )v Charge = Coulomb’ s( C )v Distance= Metres( m )

And04

1πξ

=k

Where

0 = 8.854 x 10-12 Farad/metre (or) 910361Χπ

0 = permittivity of free space (or) Dielectric constant

Equation (1) may be rewritten, on substitution for k as

122120

2112 ]

4[ U

rqqF

πε=

F12

q2

r12

U12

q1

Force between charges q1 and q2



3

1.1.LINEAR SUPERPOSITION:

• If a system consists of n point-charges, namely q1 ,q2 ……qn ,then the force on i th charge isgiven by the vector sum of all the individual forces given by coulomb’ s law. This is called linearsuperposition.

• The force Fi on the i th charge can be written by the mathematical expression given below.

jin

jj

ji

jin

jj jii U

r

qqFF ∑∑

≠=

≠= ==

11 2

011

4πεWhere

qj is any of the charges other than qi, and

rji ,the distance between the j th i th charges.

• For a homogeneous medium, the permittivity is

roεεε =Where

= absolute permittivity

r = relative permittivity

PROBLEM NO: 1

• Use the vector form of coulomb’ s law, consider a charge of 3 x 10-4 at P (1, 2, 3) and a

Charge of -10-4 C at Q (2, 0, 5) in a vacuum. To find force

TO FIND:

o Force

SOLUTION:

Q1=3x 10 -4

Q2 = -10-4

r12 = r 2 - r 1

= (2-1) ax+ (0-2)ay+(5-3)az

=ax – 2ay +2az

F2

r2

Q2a12

Q1

r12

r1 O



4

22212 2)2(1 +−+=r = 3

a12 =12

12

rr

a12 = ax-2ay+2az

3

F12 = Naaa

xx

x zyx )3

22(

910)36

1(4

)10(1039

44 +−−−

−−

ππ

F = -10 ax+20 ay-20az

PROBLEM NO: 2

• A 2 mC positive charge is located in vacuum at P1 (3,-2,-4), and a 5 C negative charge is atP2 (1, -4, 2) .(a) Find the vector force on the negative charge.(b) What is the magnitude of thecharge at P1 .

TO FIND:

• Vector force on the negative charge.• Magnitude of the charge at P1 .

SOLUTION:

Q1=2x 10 -6

Q2 = 5x10-6

P1 = (3,-2,-4)

P2 =(1,-4, 2)

r12 = r 2 - r 1

=-2ax – 2ay +6az

4412 =r

a12 =12

12

rr

a12 = -2ax-2ay+6az

44

r2

p2a12

p1

r12

r1 O



5

F12 = Naaa

xxxxx zyx )

44

622(

4410854.84)1052

12

12 +−−−

−

π

F = 6.16ax+6.16ay-18.48az

51004.2 −= xF

PROBLEM NO: 3

• A charge Q1 = -20 C is located at P (-6, 4, 6) and a charge Q2 = 50 C is located at R(5, 8, -2) in a Free space. Find the force exerted on Q2 by Q1 in vector form. The distances aregiven in metres.

P=-6ax+4ay+6az

R=5ax+8ay-2az

122120

212 4

aR

QQFπε

=

R12 =RPR =R-P

= ( ) ( ) ( )( )[ ]zyx aaa 624865 −−+−+−−

=11ax+4ay-8az

( )22212 8411 −++=R

=14.1774

a12 =

12

12

RR

=14.1774

8az-4ay11ax +

a12 =0.7758ax+0.2821ay-0.5642az

12212

66

2 )1774.14(10854.8410501020 a

xxxxxxF −

−−−=

π

= -0.0447(.7758ax+0.2821ay+0.5642az)

=-0.0346ax-0.01261ay+0.02522az

2222 )02522.0()01261.0()0346.0( −++=F

= 44.634 Nm.



6

PROBLEM NO: 4• Calculate E at M (3, -4,2) in free space caused by (a) a charge Q1=2 C at P1 (0,0,0)(b) a

charge Q2 =3 C at P2 (-1,2,3)(c) a charge Q1 =2 C at P1 (0,0,0) and a charge Q2 =3 C atP2 (-1, 2, 3).

Q1=2 C

Q2=3 C

P1= (0, 0, 0)

P2 = (-1, 2, 3)

( ) 385.5243

385.510854.84102

212

6

1zyx aaa

xxxxE

+−=

−

−

π

=345.33ax-460.44ay+230.22az

( ) 280.764

280.710854.84103

212

6

2zyx aaa

xxxE

−−=

−

−

π

E=E1+E2

=624.85ax -879.72ay+160.34 az

2.ELECTRIC FIELD INTENSITY:

• The electric field intensity (or electric field )at a point is defined as the force per unit chargeon a test charge being as small as possible in comparison with other charges forming thesystem.

t

t

QFE =

• Consider one charge fixed in position, say Q1, and move a second charge slowly around,exists everywhere a force on this second charge.

• This second charge is displaying the existence of a force field.

• This second charge is called as a test charge Qt.The force on it is given by coulomb’ s law,

to

tt a

rQQF 12

12

1

4πε=

• Writing the force as a force per unit charge gives,

( )14 12

12

1 →= to

t

t

t ar

QQQF

πε• Electric field intensity as the vector force measured by the unit Newton’s per coulomb the

force per unit charge.

( )2→=t

t

QFE

( )34 12

10

1 →= tt

ar

QEπε



7

( )44 2

0

→= RaR

QEπε

• r is the magnitude of the vector r , the directed line segment from the point at which the pointcharge Q is located to the point at which E is desired and aR is a unit vector in the r direction.

SPHERICAL CO-ORDINATE SYSTEM

• Q1 at the centre of a spherical co-ordinate system, the unit vector aR then becomes the radialvector ar and R is r.

• Hence

( )54 2

0

1 →= rar

QEπε

or

20

1

4 rQEr

πε=

CARTESIAN CO-ORDINATES

Q at the origin,

azayax zyxrR ++==

222

)(

zyx

zyxa azayax

r++

++=

( )6)(4 222222222222

0

→

+++

+++

++++= zyx a

zyxza

zyxya

zyxx

zyxQE

πε

Q located at the same point

azzayyaxxr '''' ++=zazyayxaxr ++=

R as r-r’ , and then

'

'

2'04 rr

rr

rr

QE−−

−=

πε

( )3'

0

'

4 rr

rrQ

−

−=

πε



8

PROBLEM NO: 5

• Find E at P (1,1,1) caused by four identical 3 n C point charges located at P1 ( 1,1,0 ) , P2

(-1,1,0),P3 (-1,-1,0), and P4 (1,-1,0).

r = ax+ay+ az

r1 = ax+ay

r2 = -ax+ay

r3 = -ax-ay

r4 =ax-ay

r – r1 = a

r – r2 =2ax+az

r – r3 =2ax+2ay+az

r- r 4 =2ay+az

mVx

xQo

−== −

−

96.26)10854.8(4

1034 12

9

ππε

E = 26.96( ) ( ) ( )

++

++

++

3335

2

9

22

5

21

zyzyxzxz aaaaaaaa

=az +2ax+2ay

E = 6.82 ax+6.82ay+32.8az

y

P3 (-1,-1,0)

P

O

.P2(-1,1,0)

P4 ( 1,-1,0)

Z

X P4 ( 1,-1,0)



9

PROBLEM NO: 6

v A charge Q1 = -20 C is located at P (-6, 4, 6) and a charge Q2 = 50 C is located at R (5, 8,-2) in a free space. Find the force exerted on Q2 by Q1 in vector form. The distances aregiven in metres.

P=-6ax+4ay+6az

R=5ax+8ay-2az

122120

212 4

aR

QQFπε

=

R12 =RPR =R-P

= ( ) ( ) ( )( )[ ]zyx aaa 624865 −−+−+−−

=11ax+4ay-8az

( )22212 8411 −++=R

=14.1774

a12=

12

12

RR

=14.1774

8az-4ay11ax +

a12 =0.7758ax+0.2821ay-0.5642az

12212

66

2 )1774.14(10854.8410501020 a

xxxxxxF −

−−−=

π = -0.0447(.7758ax+0.2821ay+0.5642az)

=-0.0346ax-0.01261ay+0.02522az

2222 )02522.0()01261.0()0346.0( −++=F

= 44.634 Nm.



10

PROBLEM NO: 7

Calculate E at M(3,-4,2)in free space caused by (a) a charge Q1=2 C at P1(0,0,0)(b) a charge Q2

=3 C at P2 (-1,2,3)(c)a charge Q1 =2 C at P1(0,0,0) and a charge Q2 =3 C at P2 (-1, 2, 3).

Q1=2 C

Q2=3 C

P1= (0, 0, 0)

P2 = (-1, 2, 3)

( ) 385.5243

385.510854.84102

212

6

1zyx aaa

xxxxE

+−=

−

−

π

=345.33ax-460.44ay+230.22az

( ) 280.764

280.710854.84103

212

6

2zyx aaa

xxxE

−−=

−

−

π

E=E1+E2

=624.85ax -879.72ay+160.34 az

3. TYPES OF CHARGE DISTRIBUTIONS:

• The forces and electric fields due to only point charges are considered.

• In addition to the point charges, there is possibility of continuous charge distributions along a line,on a surface or in a volume.

• Thus there are four types of charge distributions which are,

v Point charge

v Line charge

v Surface charge

v Volume charge

LINE CHARGE:

• It is possible that the charge may be spreaded all along a line, which may be finite or infinite.

• Such a charge uniformly distributed along a line is called a line charge.

• The charge density of a line charge is denoted as Lρ and defined as charge per unit length.



11

=Lρ Total charge in coulomb (C/m) Total length in meters

• Thus, Lρ Is measured in C/m. It is constant all along the length L of the line carrying the charge.

Method of finding Q from Lρ :

• In many cases Lρ is given to be the function of co-ordinates of the line.i.e, xL 3=ρ or24yL =ρ etc.

• In such a case it is necessary to find the total charge Q by considering differential length dl of theline.

• Then by integrating the charge dQ on dl, for the entire length, total charge Q is to be obtained.Such an integral is called line integral.

• Mathematically,dLdQ Lρ= =Charge on differential length dl.

lddQQL L

L∫ ∫== ρ

• If the line of length L is a closed path, then integral is called closed contour integral and denotedas

Q= ∫L

Ldlρ

• Sharp beams in a cathode ray tube or a charged circular loop of conductor are the examples of linecharge. The charge distributed may be positive or negative along a line.

PROBLEM NO: 8

• A charge is distributed on x-axis of Cartesian system having a line charge density of

./3 2 mCx µ Find the total charge over the length if 10m.Sol:

mCxL /3 2µρ = And L=10m along x-axis.

The differential length is dl=dx in x direction and corresponding charge isdxdldQ LL ρρ ==

Q=10

0

10

0

22

333∫ ∫

==

LL

xdxxdlρ

=1000 mCC 1=µSURFACE CHARGE:

• If the charge is distributed uniformly over a two-dimensional surface then it is called asurface charge or a sheet of charge.

• The two dimensional surface has area in square metres. Then the surface charge densityis denoted as Sρ and defined as the charge per unit surface area.

=Sρ Total charge in coulomb (C/m2) Total area in square meters



12

• The Sρ is constant over the surface carrying the charge.

Method of finding Q from Sρ :

• In case of surface charge distribution, it is necessary to find the total charge By consideringelementary surface area .dS

• The charge dQ on this differential area is given by .dSSρ• Then integrating this dQ over the given surface, the total charge Q is to be obtained.• Such an integral is called a surface integral and mathematically given by

Q= ∫ ∫=S S

S dSdQ ρ

• The plate of a charged parallel plate capacitor is an example of surface charge distribution.

• If the dimensions of the sheet of charge are very large compared to the distance at which theeffects of charge are to be considered then the distribution is called infinite sheet of charge.

VOLUME CHARGE

• If the charge distributed uniformly in a volume then it is called volume charge.• The volume charge density is denoted as Vρ and defined as the charge per unit volume.

Vρ = Total charge in coulomb (C/m3) Total volume in cubic meters

Method of finding Q from Vρ :

• In case of volume charge distribution, consider the differential; volume dV .• Then the charge dQ possessed by the differential volume is dVVρ .• Then the total charge within the finite volume is to be obtained by integrating the

dQ throughout that volume. Such an integral is called volume integral.• Mathematically it is given by

Q= ∫vol

V dVρ

• The charged cloud is an example of volume charge.

3.1 FIELD DUE TO A CONTINUOUS CHARGE DISTRIBUTION:

• A region of space filled with a number of charges separated by minute distances.

• This distribution of very small particles with a smooth continuous distribution described bya volume charge density.

EX:

Space between the control grid and the cathode in the electron –gun assembly of acathode ray tube with space charge.



13

• The volume charge density isv

ρ , having the unit of coulombs per cubic meter(C/m3).

• The small amount of charge density Q∆ in a small volume v∆ is

vvQ ∆=∆ ρ

• vρ , Mathematically by using a limiting process on

vQ

vv ∆∆

=→∆ 0

limρ

• The total charge within some finite volume is obtained by integrating throughout that volume,

dvQv

v∫= ρ

• One integral sign is indicated, the differential dv signifies integration throughout a volume andhence a triple integration volume.

PROBLEM NO: 9

• Find the charge in the volume defined by mx 10 ≤≤ , 0 my 1≤≤ ,andmz 10 ≤≤ if P=30 x2y(µC/m3 ).What charge occurs for the limits01 ≤≤− y m?

SincedQ= ρv dv

Q ydxdydzx∫ ∫ ∫=1

0

1

0

1

0

230

= 5µC For charge in limits on y

PROBLEM NO: 10

• A point charge, Q =30nC, is located at the origin in Cartesian

co-ordinates. Find the electric flux density D at (1,3, -4) m.

RaR

QD 24π=

=

−+−

2643

)26(41030 9 azayaxx

π

= ( ) 211 /26

431018.9 mCazayaxx

−+−

D =91.8 pC/m2



14

PROBLEM NO: 11

• Three equal positive charges of 4 x 10-9 Coulombs each are located at three corners of asquare, side 20 cm. Determine the magnitude and direction of the electric field at thevacant corner point of the square.

E=E1+E2+E3

E1 = 20

1

4 rq

πε

1E =9X109( )2

9

2.104X

1E = 900 v/M = 3EE2 is along the diagonal

1E =9X109( )2

9

2.2104X

= 450V/m (along the diagonal)E1+E2 = vector directed along the diagonal, in the same direction as E2.

121 2EEE =+

=900 2

E=900 2 +450 mv /45172545 00 ∠=+∠

PROBLEM NO: 12

• Find the total charge inside a volume; having volume charge density as 10 Z2 e-0.1x sin πy C/m3. The volume is defined between -2 ,2≤≤ x 0 1≤≤ y and .43 ≤≤ z

ρv = 10 Z2 e-0.1x sin π y C/m3

Consider differential volume in Cartesian system as, dv=dxdydz

dvdQ vρ=∴ = 10 Z2 e-0.1x sin π y dx dy dz

` dvQvol

v∫= ρ

But now it becomes triple integration.

∫ ∫ ∫= = −=

=∴4

3

1

0

2

2z y x

Q 10 Z2 e-0.1x sin π y dx dy dz

∫ ∫= =

=∴4

3

1

0z y

Q 10 Z2 e-0.1x sin π y2

2

1.0

1.0−

−

−

xe dy dz



15

∫=

=∴4

3z

Q1

0

cos

ππy

2

2

2.02.0

1.01.0−

−

−

−−

xx ee dz

=10 0267.40coscos3

4

3

3

−

−−

ππ

πz

=10 0267.4113

344

3

33

−

−ππ

=316.162 C

4.ELECTRIC FLUX:

• Electric flux are the lines of force that are drawn to trace the direction in which a positivetest charge will experience a force due to the main charge.

ELECTRIC FLUX DENSITY:

• Electric flux density measured in coulombs per square meter, is given by D.Electric fluxis more descriptive.

• The direction of D at a point is the direction of flux lines at that point, and the magnitudeis given by the number of flux lines crossing a surface normal to the lines divided by thesurface area.

• at a radial distance r, where ,bra ≤≤

D= rar

Q24∏

• Inner sphere becomes smaller and smaller, still retaining a charge of Q, it becomes apoint charge in the limit, but the electric flux density at a point r meters from the pointcharge is given by

D= rar

Q24∏

• Q lines of flux are symmetrically directed outwards from the point and pass through animaginary spherical surface area of 4∏r2.

• Radial electric field intensity of a point charge in free space

rar

QE 204 ε∏

=

• In free space,D= E0ε (free space only)

• Volume charge distribution in free space

r=ar=b

r=a



16

E= ∫ ∏volR

V aR

dV2

04 ερ

D= ∫ ∏volr

V aR

dV24

ρ

4.1.GAUSS LAW:

Ø The electric flux passing through any closed surface equal to the total charge enclosed bythat surface.

∆s

θ θ D

s

ØLet us imagine a distribution of charge, shown as a cloud of point charges, surrounded bya closed surface of any shape.

ØThe closed surface may be the surface of some real material.

ØIf the total charge is Q1 then Q coulombs of electric flux will pass through the enclosingsurface.

ØAt every point on the surface the electric flux density vector D, will have some value Ds .

ØWhere subscript S merely reminds us that D must be evaluated at the surface, and Ds willin general vary in magnitude and direction from one point on the surface to another.

Ø∆s is incremental surface element is a vector quantity.

ØAt any point P consider an incremental element of surface ∆s and let Ds make an angle with ∆s.

ØThe flux crossing ∆s is then the product of the normal component of Ds and ∆s.

Ø∆Ψ= Flux crossing ∆s =Ds ,norm ∆s = Ds cos ∆s= Ds∆s

ØThe total flux passing through the closed surface is obtained by adding the differentialcontribution crossing each surface element ∆s.

P



17

Ψ = dsDd s∫ ∫=ψ

ØThe resultant integral is a closed surface integral, and since the surface element ds alwaysinvolves the differentials of two co-ordinates, such as dx dy, ρ d dρ,(or) r2sin ,theintegral is a double integral.

ØWe shall always place an S below the integral sign indicates a surface integral ∫s

.

ØDifferential ds is the signal for a surface integral.

ØA small circle on the integral sign itself to indicate that the integration is to be performedover a closed surface. Such surface is often called Gaussian surface.

∫=s

s dsD .ψ =Charge enclosed =Q

ØThe charge enclosed might be several point charges ,in which case

∑= nQQ or

In line charge

dLQ l∫= ρ or

a surface charge

∫=s

sdsQ ρ (Not necessarily closed surface)

ora volume charge distribution

dvQvol

v∫= ρ

dvdsDvol

vs

s ∫∫ = ρ.

ØSurface is equal to the charge enclosed

SPHERICAL CO-ORDINATE SYSTEM:

rar

QE 204πε

=

ED 0ε=

rar

QD 24π=



18

ØAt the surface of the sphere

rs ar

QD 24π=

ØThe differential element of area on a spherical surface is ,in spherical co-ordinates

φθθ ddrds sin2= = a2 sin d d

ds = a2 sin d d arØThe integrant is

sin4

. 22 a

aQdsDs π

= d d ar .ar

=π4

Q sin d d

ØLoading to the closed surface integral

∫∫=

=

=

=

πθ

θ

πφ

φ π0

2

0 4Q

sin d d

ØWhere the limits on the integrals have been choosen so that the integration is carriedover the entire surface of the sphere once.

ØIntegrating gives

= φθπ

ππ

dQ0

2

0

)(cos4∫

= φπ

π

dQ∫2

0 2

=QØA result showing that Q coulombs of electric flux are crossing the surface ,as we shouldsince the enclosed surface charge is Q coulombs.

5. POTENTIAL DIFFERENCE:

• The work done in moving a point charge Q from point B to A in the electricfield E is given by

∫−=A

B

dLEQW .

• If the charge Q is selected as unit test charge then from the above equation weget the work done in moving charge from B to A in the field E.

• This work done in moving unit charge from point B to A in the field E is calledpotential difference between the points B to A is denoted by V.



19

Potential ∫−==A

B

dLEVDifference .

• Thus work done per unit charge in moving unit charge from B to A in the field Eis called potential difference between the points B to A.

5.1. PONTENTIAL DUE TO POINT CHARGE:

• Consider a point charge, located at the origin of a spherical co-ordinate systemproducing E radially in all the directions.

• Assuming free space, the field E due to a point charge Q at a point having radialdistance r from origin is given by

rar

QE 204πε

=

• Consider a unit charge, which is placed at a point B, which is at a radial distanceof rB from the origin.

• It is moved against the direction of E from point B to point A.

• The point A is at a radial distance of rA from the origin.

• The difference length in spherical system is,

rddradL r += θaφ +r sinθdφaφ

• Hence the potential difference VAB between points A and B is given by,

∫−=A

BAB dLEV . But B=rB and A= rA

∫−=Ar

BrAB a

rQV ).(

4( 2

0πεrddrar + θaφ +r sinθdφaφ)

∫−=Ar

BrAB a

rQV )

4( 2

0πεdr

Q A

rA

B

rB



20

AAr

B

r

BAB

rQdrrQV ∫

−

−=−=−

−

144

1

0

2

0 πεπε

−−

−−=

−−=

BA

r

BAB rr

Qr

QVA 11

41

4 00 πεπε

Vrr

Qrr

QVBABA

AB

−=

+−−=

114

114 00 πεπε

• When rB > rA ,

AB rr11

< and VAB is positive.

• This indicates the work is done by external source in moving unit chargefrom B to A.

6. DIELECTRIC POLARIZATION:

• When the dipole results from the displacement of the bound charges, thedielectric is said to be polarized.

• Consider an atom of a dielectric; this consists of a nucleus with positivecharge and negative charges in the form of revolving electrons in the orbits.

• The negative charge is thus considered to be in the form of cloud ofelectrons.

• E applied is zero.

• The number of positive charges is same as negative charges and hence atomis electrically neutral.

• Due to symmetry, both positive and negative charges can be assumed to bepoint charges of equal amount, coinciding at the center.

• Hence there cannot exist an electric dipole.

• This is called unpolarized atom.

• When electric field E is applied, the symmetrical distribution of chargesgets disturbed.

• The positive charges experience a force F=QE while the negative chargesexperience a force F=-QE in the opposite direction.

• There is separation between the nucleus and center of the electron cloud as,such an atom is called polarized atom.

• An electron cloud has a center separated from the nucleus.

• This forms an electric dipole.

• The dipole gets aligned with the applied field.



21

• This process is called polarization of dielectrics.

• There are two types of dielectrics

v Non -polarv Polar

• In non-polar molecules, the dipole arrangement is totally absent of electricfield E.

• It results only when an externally field E is applied to it.

• In polar molecules, the permanent displacements between centers positiveand negative charges exist.

• Thus dipole arrangements exist without application of E.

• But such dipoles are randomly oriented.

• Under the application of E, the dopoles experience torque and they alignwith the direction of the applied field E.

• This called polarization of polar molecules.

Ex:v Non polar molecules:

• Hydrogen• Oxygen• Rare gases

v Polar molecules:

• Water• Sulphur dioxide• Hydrochloric acid

7. BOUNDARY CONDITIONS:

v When an electric field passes from one medium to other medium ,it is importantto study the conditions at the boundary between the two media.

v The conditions existing at the boundary of the two media when field passes fromone medium to other are called boundary conditions.

v Depending upon the nature of the media, there are two situations of theboundary condition,

• Boundary between conductors and free space.• Boundary between two dielectrics with different properties.

v The free space is nothing but a dielectric hence first case is nothing but theboundary between conductor and a dielectric.v For boundary conditions, the Maxwell’ s equations for electrostatics are required.

∫ = 0.dlE And ∫ = QdsD.



22

v Similarly the field intensity E is required to be decomposed into twocomponents namely tangential to the boundary (Etan) and normal to the boundary(ES).

NEEE += tan

7.1. BOUNDARY CONDITIONS BETWEEN CONDUCTOR AND FREE SPACE:

v Consider a boundary between conductor and free space; the conductor is idealhaving infinite conductivity.v Such conductors are copper, silver etc, having conductivity of the order of 106

S/m and can be treated ideal.• The field intensity inside a conductor is zero and the flux density insidea conductor is zero.• No charge can exist within a conductor. The charge appears on thesurface in the form of surface charge density.• The charge density within the conductor is zero.

v Thus E,D and ρv within the conductor are zero. While ρs is the surface chargedensity on the surface of the conductorv To determine the boundary conditions let us use the closed path and the gaussiansurface.

7.2 E AT THE BOUNDARY:v Let E be the electric field intensity, This E can be into two components:

• The components tangential to the surface ( Etan).• The components normal to the surface (EN).

v It is known that,

∫ = 0.dLEv The integral of E.dL carried over a closed contour is zero i.e. work done in

carrying unit positive charge along a closed path is zero.

v Thus the E at the boundary between conductor and free space is always in thedirection perpendicular to the boundary.

ED 0ε= for free space

0tan0tan == ED εv Thus the tangential component of electric flux density is zero at the boundary

between conductor and free space.

v Hence electric flux density D is also only in the normal direction at the boundarybetween the conductor and free space.



23

7.3. BOUNDARY CONDITIONS BETWEEN CONDUCTOR AND DIELECTRIC:v The free space is a dielectric with .0εε = Thus if the boundary is between

conductor and dielectric with 0εε = rε

E tan = D tan =0

DN =ρs

EN =r

ssεε

ρερ

0

=

8. POISSON’ S AND LAPLACE’S EQUATIONS:

Ø According to Gauss’ s law in point form, the divergence of electric flux density is equal tothe volume charge density.

vD ρ=∇

D=ε E

∇ (ε E)= ρ v

∇ . E=ερv

But E= - V∇

ερvV =−∇∇ )(

ερvV −=∇∇ )(

ερvV −=∇ 2

Ø This is the poisson equation.

CARTESIAN CO-ORDINATES SYSTEM:

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=∇∇zV

xyV

xxV

xV.

= 2

2

2

2

2

2

zV

yV

xV

∂∂

+∂∂

+∂∂



24

Ø This Poisson’ s equation for Cartesian co-ordinate system is written as

V2∇

∂∂

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=zV

xyV

xxV

x=-

ερv

Ø For cylindrical co-ordinate system, the Poisson’ s equation is

V2∇ = 2

2

2

2

2

11zVVV

r ∂∂

+

∂∂

+

∂∂

∂∂

φρρρ

ρ

SPHERICAL CO-ORDINATE SYSTEM:

Ø The Poisson’ s Equation Is

ερ

φθθθ

θθvV

rV

rrVr

rrV =

∂∂

+

∂∂

∂∂

+

∂∂

∂∂

=∇ 2

2

2222

22

sin1sin

sin11

Ø If the volume charge density ( )vρ is zero, then

02 =∇ V

Ø This is Laplace’ s equation. The operator 2∇ is called Laplacian operator.

9. CAPACITOR:

Ø A Capacitor is an electric device, which consists of 2 conductors separated by a dielectricmedium.

Ø Consider a capacitor composed of two conducting plates of area A separated by a dielectricmedium whose permittivity is ε.

Ø The space separation between the plates is d.

Ø If potential V is applied across the plates, the positive charge Q is deposited on one plate andthe negative charge –Q is deposited on other plate as shown. The net charge is zero.

Ø The capacitance of two conducting planes is defined as the ratio of magnitude of charge oneither of the conductor to the potential difference between conductors.

∈

d

V

C

+++++++

-------



25

Ø It is given by

C=VQ

Ø The unit 0f capacitance is coulombs/volt or farad.

ØAssume that there is a uniform charge density D over the plates and dielectric medium.

D= 2/ mCAQ

Ø It is also written in terms of electric field E.

D=εE

EAQ

ε=

Q=AεEØ But electric field is given by

E=V/d V/m

Ø Substituting the value of E in the above equation

Q=AεdV

dA

VQ ε

=

Ø Capacitance is given by

C=VQ

C=dAε

C=d

A rεε 0 farad

Ø Where 0εε = rε9.1.Capacitance of a parallel plate capacitor having two dielectric media

Ø Consider a parallel plate capacitor consists of 2 dielectrics as shown

Ø The relative permittivity of dielectric medium 1 and medium 2 are 21 rr andεεrespectively.

V1 V2

∈r1 ∈r2

d1 d-d1d



26

Ø If the potential across the capacitor is V, the potential difference across medium 1 andmedium 2 are V1 and V2 respectively.

V=V1+V2Ø Let E1 and E2 be the field intensities in the medium 1 and medium 2 respectively. Then,

V1=E1+d1

V2=E2 (d-d1)

V=V1+V2

=E1d1+E2 (d-d1)Ø The electric flux density D=Q/A will be the same in both the media. The electric flux

densities are given by

E1=011 εεε rr A

QD=

E2=021 εεε rr A

QD=

Ø The applied potential V=E1d1+E2 (d-d1)

V=

−+

210

11

rr

dddAQ

εεε

21

0

11

rr

dddA

VQ

εε

ε−

+=

21

0

11

rr

dddAC

εε

ε−

+=

Ø The capacitance

C=12

210

)1(1 rr

rr

dddA

εεεεε−+

Ø If medium 1 is air, 11 =rε and for medium 2 rr εε =2 .The capacitance of capacitor is

C=)1(1

0

dddA

r

r

−+εεε

9.2.Capacitance of a parallel plate capacitor having 3 dielectric media:

Ø Consider a parallel plate capacitor consists of 3 dielectrics as shown in figure

Ø Let 3,21 , rr εεε be the relative permittivity and thickness d1, d2, and d3 of the medium 1,medium 2 and medium 3 respectively.

∈r2∈r1 ∈r3

d1 d3d2

d



27

Ø If E1, E2, E3 are the electric field intensity of the medium 1,2 and 3 respectively. Theapplied potential

V=V1+V2+V3

V=E1d1+E2d2+E3d3Ø Since the flux density D=Q/A is the same in three media, the electric field intensity

E1=1010 rr A

QDεεεε

=

E2=20 rA

Qεε

E3=30 rA

Qεε

Then

V= 321302010

dA

QdA

QdA

Qrrr εεεεεε

++

V=

++

3210

321

rrr

dddAQ

εεεε

321

0

321

rrr

dddA

VQ

εεε

ε

++=

Ø The capacitance is given by

321

0

321

rrr

dddA

C

εεε

ε

++=

9. 3. Capacitance Of An Isolated Sphere:

Ø Consider a sphere of radius r having charge Q coulombs as shown.

Ø The potential is the work done per unit charge in carrying a positive test charge frominfinity to the sphere. The absolute potential is given by

V= - ∫∞

r

Edr

= - 24 rdrY

peQ

Qr



28

V=r

Qε∏4

Ø The capacitance of isolated sphere isC=Q/V=4εΠr farads

Where

rεεε 0=Ø If the medium is air, then C becomes

C=4Π 0ε r farad.9. 4.Capacitance of concentric spheres:

Ø Consider 2 concentric spheres of inner radius of ‘ a’ and outer radius ‘b’ . Let rε be thepermittivity of dielectric medium between the inner and outer spheres as shown.

Ø If the charge Q is distributed uniformly over the outer surface of the inner sphere, therewill be equal and opposite charge induced on their inner surface of the inner sphere.

Ø The electric field intensity at any point in between inner sphere and outer sphere(a )br ≤≤ is given by

E= 24 rQε∏

(a )br ≤≤

Ø The potential difference between the spheres is

V= - ∫ ∏

a

b

drr

Q .4 2ε

= - ∫∏

a

b rdrQ

24 ε

=a

brQ

∏1

4 ε

=

−

∏ baQ 11

4 ε

=

−

∏ ababQ

ε4Ø The capacitance of 2 concentric spheres is

C=

−∏=

abab

VQ

ε4

C=4

−∏

abab

ε

9. 5.Capacitance of co-axial cables (cylinders):

Ø Consider a capacitance of inner radius ‘ a’ and outer radius ‘ b’ as shown.

Ø The relative permittivity of dielectric filled in between two co-axial cylinders is rε .



29

Ø The potential difference V is applied in between the two cylinders.

Ø The two cylinders are charged at the rate of lρ C/m. It is assumed that inner cylinder has

charge of mCl /ρ and outer cylinder has charge of - mCl /ρ . By applying Gauss’ s law, theelectric field at any distance r from the axis of cylinder is given by

rE l

ερ∏

=2

Ø The potential difference between two co-axial cables is

V= - ∫a

b

Edr

= - ∫∏

a

br

l

rdr

ερ

2

= -

∏ abl ln

2 ερ

Ø The capacitance of co-axial cable/unit length is

C= ( ) mFabV

l //ln

2 ερ ∏=

C= ( ) mFab

r //ln

2 0εε∏

9. 6.Capacitance of parallel conductors (Transmission lines):

ØConsider a two parallel conductor of radius ‘ a’ separated by a distance ‘d’ as shown.

ØIf the conductor A has charge of mCl /ρ along its length, this will induce charge of -

mCl /ρ on the conductor BØThe electric field intensity at any point P with a distance r from the conductor A isalgebraic sum of electric field intensity at P due to conductor A and conductor B.

E=)(22 rdr

ll

−∏+

∏ ερ

ερ

=

−+

∏ rdrl 11

2 ερ

Ø The potential difference between the conductors is given by

ba ρ



30

V= - ∫ Edr

= - drrdr

a

ad

l ∫−

−+

∏11

2 ερ

= -

−+

−∏ ada

adal lnln

2 ερ

= -

−∏ adal ln2

2 ερ

V=

−

∏ aadl ln

ερ

Ø The capacitance per unit length between two parallel conductors

C= mF

aadV

l /ln

−∏

=ερ

If d>>a

C= mF

ad

/ln

ε∏

Ø If the dielectric medium between two conductor is air (for transmission lines) 1=rε

C= mF

ad

/ln

0

∏ε

9.7.Serial and parallel combination of capacitors:

Ø Two capacitors are connected in series as shown fig

Ø If potential difference V is applied across the two capacitors, V1 and V2are the potential differences of capacitors C1 and C2 respectively.

V=V1+V2Ø Since the charge acquired by each capacitor is same, then

V=21 C

QCQ

+

Ø The equivalent capacitance C is given by

21 CQ

CQ

CQ

+=

21

111

CCC+=

Ø If N number of capacitors are connected in series ,the equivalent capacitance is given by

CnCCCC1.........

31

21

111

+++=



31

∑=

=N

n CnC 1

11

Ø The two capacitors of value C1 and C2 are connected in parallel as shown.

Ø The voltage source V is applied to this combination.

Ø The total charge acquired by the equivalent capacitor is the sum of the charges onindividual capacitors.

Q=Q1+Q2

=C1V+C2V

Ø The equivalent capacitance C is given by

CV=C1V+C2V

C=C1+C2

Ø If N number of capacitors is connected in parallel the equivalent capacitance is given by

C=C1+C2+C3+ ………….Cn

10.ELECTROSTATIC ENERGY:

Ø The capacitor stores the electrostatic energy equal to work done to buildup the charge.

Ø If a voltage source is connected across the capacitor, the capacitor charges.

Ø Potential is defined as the work done per unit charge.

V=dQdW

Ø The work done dW =V dQ

Ø But V=Q/C

dQCQdW =

Ø The capacitor is charged to the value of Q.The total work done is

W= ∫Q

dQCQ

0

=Q

QC

0

2

21

W =C

Q2

2



32

Ø But Q=CV

W= 2

21 CV or

W= QV21

joules

10.1.ENERGY DENSITY:

ØConsider a elementary cube of side d∆ parallel to the plates of a capacitor as shown.

+ - + -

+| - + -

+ -

ØThe capacitance of elemental capacitance is

dAC

∆=∆

ε

=( )

dd

∆∆ 2ε

= d∆εØ Energy stored in the elemental capacitor is

CW ∆=∆21 ( )2V∆

Ø But potential difference across the elementary cube is

dEV ∆=∆

Ø Where E is the electric field exist in the cube.

dC ∆=∆ ε

Ø The stored energy

21

=∆W ( )( )2.. dEd ∆∆ε

= ( )32

21 dE ∆ε

= V∆Ε 2

21

ε

QQ

V

C∆d



33

Ø Where ( )3dV ∆=∆ is elementary volume. The energy density is given by

ε21

=∆∆

VW

E2 ( Ε= εDQ )

EDVW .

21

=∆∆

Joules/m3

11. DIELECTRIC STRENGTH:

ØThe ideal dielectric is non-conducting but practically no dielectric can be ideal.

ØAs the electric field applied to dielectric increases sufficiently, due to the force exerted onthe molecules, the electrons in the dielectric become free.

Ø Under such large electric field, the dielectric becomes conducting due to presence oflarge number of free electrons.

ØThis condition of dielectric is called dielectric breakdown.

Ø All kinds of dielectrics such as solids, liquids and gases show the tendency ofbreakdown under large electric field.

ØThe breakdown depends on the nature of material, the time and magnitude of appliedelectric field and atmospheric conditions such as temperature, moisture, humidity etc.

NOTE:Ø The minimum value of the applied electric field at which the dielectric breaksdown is called dielectric strength of dielectric.

ØThe dielectric strength is measured in V/m or kV/cm.

ØIt can be stated as the maximum value of electric field under which a dielectric cansustain without breakdown.

ØOnce breakdown occurs, dielectric starts conducting and no longer behaves as dielectric.

ØHence all the dielectrics are assumed to be either ideal or are not in a breakdowncondition.



34

IMPORTANT QUESTIONSPART A

1. Define coulomb’ s law?2. Define Gauss’ s law?3. Define electric flux density at a point?4. Define Poisson’ s potential?5. Define dielectrics?6. What is the difference between dielectrics and insulators?7. Define dipole?8. E.dl =0 why and how?9. Define gradient and divergence?10. Name any two dielectric materials.11. Give the expression for capacitance of a parallel plate capacitor?12. Derive the poisons equations for electrostatics fields?13. State divergence theorem?14. Define electric field intensity?15. What is convention current?16. State the principle of superposition?17. Gauss’ s law can be applied only to ----------- surface.18. Find the force on a charge Q2=10µc at point (2, 0, 0) by a charge Q1=20µc at point (1, 0, 0) in free space. Dimensions are in meters.19. What are the significant physical differences between Poisson’ s and laplace‘ s equations?20. Under what conditions will the field intensity be solenoidal and irrotational?21. What is the use of Gauss’ s law?22. Express Laplace equation in cylindrical and in Cartesian co-ordinate system.23. Write Laplace and Poisson’ s equation.24. What are boundary conditions?25. What is a magnetic dipole?26. List the properties of electric field lines?27. Calculate the energy stored in a 10µF capacitor which has been charged to a voltage Of 400V.28. Define Surface and Volume charge densities.29. Fill up: sD.ds=∫v30. State Poisson’ s equation in Cartesian co-ordinates.31. Write down the equation of continuity in point form.32. Define electric potential?33. State the conditions at the interface between two dielectric surfaces.34. State the relation between magnetization magnetic flux density and field intensity.35. Write the expression for force between any two point charges using coulombs.36. Define the electric field at a point in space.37. State Laplace equations.38. Distinguish between conductor and dielectric.39. Define electric potential?40. Derive the energy stored in a capacitor.41. State the assumption made in coulomb’ s law?



35

PART -B

1. Derive an expression to determine electric field intensity at a point volume charge distribution andsurface charge distribution.2. State and derive the expression for Laplace and Poisson’ s equations.3. Derive the capacitance of a parallel plate capacitor having two dielectric media.4.Two small identical conducting spheres have charge of -1 nano coulombs and 2 nano coulombsrespectively .If the y are brought in contact and then separated by 4 cm what is the force between them?5. State and explain the boundary conditions of electric field between a dielectric and a conductor.6. Derive an expression for the energy stored in the capacitor.7. Determine the electric field at P(-0.2,0,-2.3) due to a point charge of +5 nC at Q(0.2,0.1,-2.5) in air. Alldimensions are in meters.8. Derive and expression for electric field intensity at a point P due to an electric dipole hence defineelectric dipole moment.9. State divergence theorem.10. Find the potential of a uniformly charged spherical shell of radius R at points inside and outside.11. Give the potential V=10sin cos /r2 find the electric flux density D at (2, /2,0).




UNIT –IIICONTENTS:

1.Lorentz Law of force

2. Magnetic field intensity

3. Biot–savart Law

4. Ampere’ s Law

5. Magnetic field due to straight conductors

6. Magnetic flux density (B)

7. Magnetic materials

8. Boundary conditions

9. Scalar and vector potential

10. Torque

11. Magnetic circuits.



1. FORCE ON A MOVING CHARGE:

• In electric field, force on a charged particle is

F=QE

• Force is in the same direction as the electric field intensity(positive charge)

• A charged particle in motion in a magnetic field forcemagnitude is proportional to the product of magnitudes of thecharge Q, its velocity V and the flux density B and to the sineof the angle between the vectors V and B.

• The direction of force is perpendicular to both V and B and isgiven by a unit vector in the direction of V x B.

• The force may therefore be expressed as

F=QV x B

• Force on a moving particle due to combined electric andmagnetic fields is obtained by superposition.

F=Q (E + V x B)

• This equation is known as Lorentz force equation.

1.1 FORCE ON A DIFFERENTIAL CURRENT ELEMENT:

• The force on a charged particle moving through a steadymagnetic field may be written as the differential; forceexerted on a differential element of charge.

dQdF =

• Convection current density in terms of the velocity of thevolume charge density

• Differential element of charge may also be expressed in termsof volume charge density.

dvdQ vρ=Thus,

dvVxBdF vρ=



JxBdvdF =

• JdV is the differential current element

IdLKdsJdv ==

• Lorentz force equation may be applied to surface currentdensity.

KxBdsdF =

• Differential current element

IdLxBdF =

• Integrating the above equations over a volume, surface openor closed

∫=vol

JxBdvF

∫=s

KxBdsF

• To a straight conductor in a uniform magnetic field

F= ∫ ∫−= BxdLIIdLxB

ILxBF =

• The magnitude of the force is given by the familiar equation

F=BILsinθ

• Where θ is the angle between the vectors representing thedirection of the current flow and direction of the magnetic fluxdensity.



2. MAGNETIC FIELD INTENSITY:

• The quantitative measure of strongness or weakness of themagnetic field is given by magnetic field intensity or magneticfield strength.

• The magnetic field intensity at any point in the magnetic fieldis defined as the force experienced by a unit north pole of oneWeber strength, when placed at that point.

• The magnetic flux lines are measured in webers (wb) whilemagnetic field intensity is measured in newtons/weber oramperes per metre (AT/m)

• It is denoted as H.

• It is a vector quantity.

• This is similar to the electric field intensity E in electrostatics.

3. BIOT-SAVART LAW:

• The law of biot –savart then states that at any point P themagnitude of the magnetic field intensity produced by thedifferential element is proportional to the product of thecurrent, the magnitude of the differential length, and the sineof the angle lying between the filament and a line connectingthe filament to the point P .the magnitude of the magneticfield intensity is inversely proportional to the square of thedistance from the differential element to the point P.

24 RIdLxadH R

π=

• The unit of the magnetic field intensity H are ampere’ s permeter (A/m)

dL1

aR12

R12 P

I1

Point 1

Point 2



• The law of Biot –savart is sometimes called Ampere’ s law forthe current element.

212

112 4

12

RxadLI

dH R

π=

• The magnetic flux density at any point P due to currentelement I dl is given by

ar

IdLdB 24πµ

=

• Where,

♦ µ=µ0µ r is permeability of the medium

♦ IdL is the current element

♦ r is the distance between the point P and currentelement

♦ a is the unit vector

• Its magnitude is

24sinr

IdLdBπ

θµ=

• The magnetic field intensity is given by

212

112 4

sinR

dLIdHπ

θ=

• This is referred to as Ampere’ s law for current element



4. AMPERE’ S CIRCUITAL LAW:

o The Ampere’ s circuital law states that,

• The line integral of magnetic field H around a closedpath is exactly equal to the direct current enclosed bythat path.

∫ = IdLH .

v In electrostatics, the gauss’ s law is useful to obtain the E incase of complex problems.

v Similarly in the magnetostatics, the complex problems canbe solved using a law called Ampere’ s law or Ampere’ s worklaw.

PROOF:

• Consider a long straight conductor carrying direct current Iplaced along Z-axis.

• Consider a closed circular path of radius r which enclosesthe straight conductor carrying direct current I.

• The point P is at a perpendicular distance r from theconductor.

• Consider dL at point P, which is in aφ direction, tangentialto circular path at point P.

x

Y

aφ

I

I

z



φφarddL =

• While H obtained at point P, from Biot-savart law due toinfinitely long conductor is,

φπa

rIH

2=

=dLH . φφ φπ

ardar

I .2

= φπ

rdr

I2

= φπ

dI2

• Integrating H.dL over the entire closed path,

φπ

π

φ

dIdLH∫ ∫=

=2

0 2.

[ ] πφπ

202

I=

=ππ

22I

=I=current carried by conductor.

• This proves that the integral H. dL along the closed pathgives the direct current enclosed by that closed path.



5. MAGNETIC FIELD DUE TO STRAIGHT CONDUCTOR:

• Consider a infinitely straight conductor carrying a current Iand also consider a current element IdL.

• Let P be any point at which magnetic field intensity is tobe measured at a distance ‘ r’ from the current element Idl.

• According to Biot savart’ s law, the magnetic flux density atany point P is given by

B= ∫ 20 sin

4 rdLI θ

πµ

• From ∆ ABC

θsin=ABAC

θsindLAC =• But arc

AC =rdθ

θθ rddL =sin

θd =r

dL θsin

B= ∫ rdI θ

πµ4

0

• Substitute the value of r in above equation

B= ∫π

θθπ

µ

0

0 .sin4

ddI

B= πθπ

µ0

0 )cos(4

−dI

B= 2.4

0

dI

πµ

• The magnetic flux density due to infinite conductor

B= 20 /2

mwbdI

πµ

• The magnetic field intensity due to infinite conductor

mad

IH /2π

=



6. MAGNETIC FLUX DENSITY:

• The total magnetic lines of force i.e. magnetic flux crossing aunit area in a plane at right angles to the direction of flux iscalled magnetic flux density.

• It is denoted as B and is a vector quantity.

• It is measured in Weber per square metre (wb/m2).

• Which is also called Tesla(T).

• This is similar to the electric flux density D in electrostatics.

B= H0µ free space only Wb/m2 (or) tesla (or) Gauss.

6.1 MAGNETIC FLUX:

• The magnetic flux through a surface area is the normalcomponent of magnetic field times µ over the area.

∫∫= Hdsm µφ

• Where µ is the permeability of the medium.(H/m)

7. MAGNETIC MATERIALS:• All material shows some magnetic effects. In many

substances the effects are so weak that the materials areoften considered to be non magnetic.

• A vacuum is the truly nonmagnetic medium.

• Material can be classified according to their magneticbehavior into

v Diamagnetic

v Paramagnetic

v Ferromagnetic

v Antiferromagnetic

v Ferromagnetic

v Super paramagnetic



DIAMAGNETIC:

• In diamagnetic materials magnetic effects are weak.

• Atoms in which the small magnetic fields produced by themotion of the electrons in their orbit and those producedby the electron spin combine to produce a net field ofzero.

• The fields produced by the electron motion itself in theabsence of any external magnetic field.

• This material as one in which the permanent magneticmoment m0 of each atom is zero. Such a material istermed diamagnetic.

ANTIFERROMAGNETIC:

• In anti-ferromagnetic materials the magnetic moments ofadjacent atoms align in opposite directions so that the netmagnetic moment of a specimen is nil even in thepresence of applied field.

FERRIMAGNETIC:

• In ferromagnetic substance the magnetic moments ofadjacent atoms are also aligned opposite, but themoments are not equal, so there is a net magneticmoment.

• It is less than in ferromagnetic materials.

• The ferrites have a low electrical conductivity, whichmakes them useful in the cores of ac inductors andtransformers.

• Since induced currents are less and ohmic losses arereduced.



8. BOUNDARY CONDITIONS:

• A boundary between two isotropic homogeneous linearmaterials with permeability 11µ and µ 2.

• The boundary condition on the normal components isdetermined by allowing the surface to cut a smallcylindrical gaussian surface.

• Applying gauss’ s law for the magnetic field.

0. =∫s

dsB

• We find that

021 =∆−∆ SBSB NN

21 NN BB =

212

1NN HH =

µµ

• The normal component of B is continuous, but the normal

component of H is discontinuous by the ratio2

1

µµ

.

• The relationship between the normal components of M,isfixed once the relationship between the normalcomponents of H is known .

• For linear magnetic materials, the result is written simplyas

122

121

2

122 N

m

mNmN MHM

µχµχ

µµ

χ ==

• Next, Ampere’ s circuital law

∫ = IdLH .

• Is applied about a small closed path in a plane normal tothe boundary surface.

• Taking trip around the path, we find that



LKLHLH tt ∆=∆−∆ 21

• Boundary may carry a surface current K whose componentnormal to the plane of the closed path is K.Thus

KHH tt =− 21

• The direction are specified more exactly by using the crossproduct to identify the tangential components,

KxaHH N =− 1221 )(

• Where aN12 is the normal at the boundary directed fromregion 1 to region 2.

• An equivalent formulation in term of the vector tangentialcomponents may be more convenient for H:

KHH tt =− 21 x 12Na

• For tangential B, we have

KBB tt =−2

2

1

1

µµ

• The boundary condition on the tangential component ofthe magnetization for linear material is therefore

KMM mtm

mmt 21

1

22 χ

χχ

−=

• The last three boundary conditions on the tangentialcomponents are much simpler, the current density is zero.

• This is a free current density, and it must be zero ifneither material is a conductor.

9. SCALAR AND VECTOR POTENTIAL:

9.1. SCALAR MAGNETIC POTENTIAL:

Ø Ampere’ s law stated that the line integral of the field Haround a closed path is equal to the current enclosed.



∫ = IdLH .

Ø If no current is enclosed i.e.

J=0

∫ = 0Hdl

Ø Magnetic field H can be expressed as negative gradientof a scalar function.

∫−= dlHVm .

Ø This scalar potential also satisfies lapalace equation

Ø In free space

but

H

B

0.

0.

0 =∆

=∆

µ

mVH −∇=

0).(0 =−∇∇ Vmµ

020 =∇− Vmµ

02 =∇ Vm



VECTOR MAGNETIC POTENTIAL:

Ø Scalar magnetic potential exists if there is no currentenclosed i.e. ∫ = 0.dLH .

Ø If current is enclosed, the potential which depends uponcurrent element (vector quantity) is no more scalars butit is vector quantity.

Ø Since the divergence of a vector is a scalar, vectorpotential is expressed in curl.

0. =∇ B

xAB ∇=

Ø Where A is magnetic vector potential. Take curl on bothsides

xAxxB ∇∇=∇

Ø By the identity

AAxAx 2).( ∇−∇∇=∇∇

But JxB µ=∇

JAA µ=∇−∇∇ 2).(

For the steady dc

0).( =∇ A

Then 2∇− A Jµ=

)(222 zJzyJyxJxAzzAyyAxx ++−=∇+∇+∇ µØ Equating

JxAx µ−=∇2

JyAy µ−=∇2

JzAz µ−=∇2



Ø They are in the form of Poisson’ s equation. from theabove equation ,magnetic vector potential can be writtenas

dvrJxAx

v

)(4 ∫=πµ

dvrJyAy

v

)(4 ∫=πµ

dvrJzAz

v

)(4 ∫=πµ

Ø The general magnetic vector potential can be expressedas

dvrJA ∫∫∫=

πµ

4



10. TORQUE:

Ø When a current loop is placed parallel to a magneticfield, forces act on the loop that tends to rotate it.

Ø The tangential force multiplied by the radial distance atwhich it acts is called torque or mechanical moment onthe loop.

Ø Consider the rectangular loop of ‘ l’ and breadth ‘ B’carrying a current i in a uniform magnetic field of fluxdensity B.

Ø The force on the loop

F=Bil.Ø If the loop plane is parallel to the magnetic field ,the

total torque on the loop

T=2 x torque on each side

= 2 xforce x distance

=2(Bil).b/2

=Bilb

=BIA

Ø Torque is given by

T=BIA

l

B

Axis of rotation

θ

F F t

θFtF

Axis of rotation

b

θ



Ø If loop plane makes an angle θ with respect to fluxintensity B, the tangential component of the force is

θcosFFts =

Ø The total torque on the loop T=BIA cos θ.

Ø The magnetic moment of loop is IA

m =IA

T=mBcosθ

Ø In vector form

T=m x B

M=T/B

Ø The magnetic moment is defined as the maximum torqueon loop per unit magnetic induction.

11. MAGNETIC CIRCUITS:

• In magnetic circuits are analogous to the electric circuits.

• The common examples of the magnetic circuits aretransformers, toroids motors, generators, relays andmagnetic recording devices.

• An electric circuit forms a circuit (i.e. closed path) throughwhich current flow.

• Similarly in magnetic circuits, magnetic lines of flux arecontinuous and can form closed paths.

• A single magnetic line of flux or all parallel magnetic linesof flux may be considered as magnetic circuits.

• Electromotive force in an electric circuit, in magneticcircuit called magnetiomotive force (m.m.f.)

• The magnetomotive force (m.m.f.) is defined as,

em =NI= ∫ dLH .



• The SI unit of m.m.f. is ampere(A)

• In electric circuit, resistance is defined as the ratio ofvoltage to current given by

R=V/I

• In magnetic circuit, reluctance as the ratio of themagnetomotive force to the total flux.



IMPORTANT QUESTION

PART A

1. State Biot –Savart Law?

2. What is magnetic vector potential?

3. Name any two dielectric materials.

4. Name any two applications of Ampere’ s Law?

5. Define magnetic flux density?

6. What is the difference between scalar and vector magnetic potential?

7. Compare the usefulness of Ampere‘ s Circuital Law and Biot-Savart Law indetermining B of a current carrying circuit.

8. A conductor 1.5 m long carries a current of 50 A at right angles to a magnetic field ofdensity 1.2T. Calculate the force on the conductor.

9. What is the expression for the torque experienced by a current carrying loop, placed in a magnetic field.

10. Define magnetic flux?

11. Define Amperes law?

12. Obtain H due to infinitely long, straight filament of current I.

PART B

1. State and explain ampere’ s Circuital Law for least two specific cases.

2. State and explain Biot savar’ s law.

3. Using the biot savart’ s law in H, find the magnetic field intensity at a point on the axisof a circular loop of radius ‘ a’ carrying a current I. The point is at a distance (on the Zaxis) from the centre of the loop.

4. A steady direct current I amps flows in a wire bent in the form of a square of side a.Assuming that the Z axis passing through the centre of the square is normal to theplane of the square ,Find the magnetic field intensity H at any point on the axis.

5. Obtain the flux density produced by an infinitely long straight wire carrying a current I,at any point distant ‘ a’ normal to the wire.

6. In a cable the solid inner conductor of radius ‘ a’ carries I amps.



SUBJECT NAME: ELECTROMAGNETIC THEORY

SUBJECT CODE: EE 2202

BRANCH: EEE

SEMESTER: III

UNIT –IV

ELECTRODYNAMIC FIELDSCONTENTS:

1. FARADAY’ S LAWS

2. INDUCED EMF - TRANSFORMER EMF AND MOTIONALEMF

3. MAXWELL’ S EQUATIONS (DIFFERENTIAL ANDINTEGRAL FORMS)

4. DISPLACEMENT CURRENT

5. RELATION BETWEEN FIELD THEORY AND CIRCUITTHEORY.



1. FARADAY’ S LAWS:

• According to faraday’ s law, the induced e.m.f. is equal to the time rate of change of magnetic flux linking with the closed circuit.

• Faraday’ s law can be stated as

e =dtdN φ

− volt

• Where

N=number of turns in the circuit

e =induced e.m.f.

• Let us assume single turn circuit i.e. N=1, then faraday’ s law can be stated as,

e= voltsd

dφ

• The minus sign in equations (1) and (2) indicates that thedirection of the induced e.m.f. is such that to produce acurrent which will produce a magnetic field which willoppose the original field.

LENZ LAW:

• Thus according to Lenz law, the induced e.m.f. acts toproduce an opposing flux.



2. INDUCED E.M.F:

TRANSFORMER E.M.F AND MOTIONAL E.M.F:

• Let us consider faraday’ s law. The induced e.m.f.is a scalarquantity measured in volts. Thus the induced e.m.f. is givenby,

)1(. −−−−= ∫ dLEe

• The induced e.m.f in equation (i) indicates a voltage about aclosed path such that if any part of the path is changed thee.m.f. will also change.

• The magnetic flux passing through a specified area is givenby

)1(. −−−= ∫s

dSBφ

• Where

B=magnetic flux density

• Using above result ,equation (2) can be rewritten as

)2(. −−−−−= ∫s

dSBdtde

• From equation (1) and (2) ,we get,

== ∫ dLEe . )3(. −−−−− ∫s

dSBdtd

• There are Two Conditions for induced e.m.f.



FIRST CONDITION:

v The closed circuit in which e.m.f. is induced is stationary andthe magnetic flux is sinusoidally varying with time.

v Form equation (3), the magnetic flux density is the quantityvarying with time.

v Partial derivative to define relationship as B may be changingwith the co-ordinates as well as time.

v Hence we can write,

=∫ dLE. )4(. −−−−∂∂

− ∫s

dStB

v This is similar to transformer action and e.m.f. is calledtransformer e.m.f.

v Using Stoke’ s theorem, a line integral can be converted to thesurface integral as

)5(.).( −−−−∂∂

−=∇ ∫∫ dStBdSxE

SS

v Assuming that both the surface integral taken over identicalsurfaces.

dStBdsxE

S

.).( ∫ ∂∂

−=∇

v Hence finally,

)6(−−−∂∂

−=∇tBxE

v Equation (6) represents one of the Maxwell’ s equations.

v If B is not varying with time, then equations (4) and (6) givethe results in the electrostatics.

∫ = 0.dLE0=∇xE



SECOND CONDITION:

v Magnetic field is stationary, constant not varying with time whilethe closed circuit is revolved to get the relative motion betweenthem.

v This action is similar to generated action; hence the inducede.m.f. is called motional or generated e.m.f.

v Consider that a charge Q is moved in a magnetic field B at avelocity v.

v Then the force on a charge is given by,

)7(−−−= QvxBF

v But the motional electric field intensity is defined as the force perunit charge.

v It is given by

)8(−−−== vxBQEEm

v Thus the induced e.m.f is given by

)9().(. −−−−=∫ ∫ dLvxBdLE m

v Equation (9) represents total e.m.f. induced when a conductor ismoved in a uniform constant magnetic field.

v If in case ,the magnetic flux density is varying with time, thenthe induced e.m.f. is the combination of transformer e.m.f. andgenerator e.m.f. is given by,

∫∫ ∫ +∂∂

−= dLvxBdStBdLE ).(..



3. MAXWELL’ S EQUATIONS:

v Maxwell’ s equations are nothing but a set of four expressionsderived from

Ø Ampere’ s circuit law

Ø Faraday’ s law

Ø Gauss’ s law

v This is for electric field.

v In magnetic field it is derived from

Ø Gauss’ s law

v These four expressions can be written in following forms

Ø Point form or differential form

Ø Integral form

MAXWELL’ S EQUATION DERIVED FROM AMPERE’ S CIRCUIT LAW:

v According to Ampere’ s circuit law, the line integral of magneticaround a closed path is equal to the current enclosed by thepath.

v Replacing current by the surface integral of conduction currentdensity J over an area bounded by the path of integration of FL,we get more general relation as,

v Above expression can be to conduction current density madefurther general by adding displacement current as follows,



v Equation (I-a) is Maxwell’ s equation derived from Ampere’ scircuit equation is in integral form in which line integral of H iscarried over the closed bounding the surface S over which theintegration is carried out on R.H.S. In theory, closed path iscalled Mesh.

v Hence the equation considered above is also Mesh equation orMesh relation.

v Applying Stoke’ s theorem to L.H.S. of the equation (I-a), weget,

v Assuming the surface considered for both the integrations issame, we can write,

v Above equation is the Point form or differential form ofMaxwell’ s equation derived from Ampere’ s circuit law.

MAXWELL’ S EQUATION DERIVED FROM FARADAY’ S LAW:

INTEGRAL FORM:

v Now consider Faraday’ s law which relates e.m.f. induced in acircuit to the time rate of decrease of total magnetic flux linkingthe circuit. In general we can write,

v This is Maxwell’ s equation derived from Faraday’ s lawexpressed in integral form.



MAXWELL’ S EQUATION DERIVED FROM FARADAY’ S LAW:

POINT FORM OR DIFFERENTIAL FORM:

v Using Stokes theorem, converting line integral of equation (2)to the surface integral,

v Assuming that the integration is carried out over the samesurface on both the sides, we get,

v This is Maxwell’ s equation derived from Faraday’ s lawexpressed in point form or differential form.

MAXWELL’ S EQUATION FOR ELECTRIC FIELDS DERIVED FROMGAUSS’ S LAW:

INTEGRAL FORM

v According to Gauss’ s law, the total flux out of the closed surfaceis equal to the net charge within the surface.

v This can be written in integral form as,

v If we replace R.H.S. of above equation by the volume integral ofvolume charge density Pv through the volume enclosed by thesurface S considered for integration at L.H.S. of equation (3), weget more general form of equation given by

v This equation is called Maxwell’ s equation for electric fieldsderived from Gauss’ s law, expressed in integral form andapplied to a finite volume.



MAXWELL’ S EQUATION FOR ELECTRIC FIELDS DERIVED FROMGAUSS’ S LAW:

POINT FORM OR DIFFERENTIAL FORM:

v Using divergence theorem, we can write

v Assuming same volume for integration on both the sides,

v This is Maxwell’ s equation for electric fields derived fromGauss’ s law which is expressed in point form or differentialform.

MAXWELL’ S MAGNETIC FIELD EQUATION INTEGRAL FORM:

DERIVED FOR GAUSS’ S LAW:

v For magnetic fields, the surface integral of over a closedsurface S is always zero, due to non existence of monopole inthe magnetic fields.

v This is Maxwell’ s magnetic field equation expressed in integralform. This is derived for Gauss’ s law applied to the magneticfields.



DERIVED FROM GAUSS LAW APPLIED TO THE MAGNETICFIELDS:

DIFFERENTIAL FORM OR POINT FORM:

v Using divergence theorem, the surface integral can beconverted to volume integral as

v But being a finite volume, dv o

v This is differential form or point form of Maxwell’ s equationderived from Gauss law applied to the magnetic fields.

v summarizes Maxwell’ s equations



4. DISPLACEMENT CURRENT:

ü In static electromagnetic fields ,according to Ampere’ s circuitallaw

)1(−−−=∇ JxHü Taking divergence on both sides,

JxH .).( ∇=∇∇

ü According to vector identity, ‘ divergence of the curl of any vectorfield is zero. Hence we ca write,

JxH .).( ∇=∇∇ =0---- (2)

ü Equation of continuity is given by,

)3(. −−−∂

∂−=∇

tJ ρν

ü From equation (3) it is clear when 0=∂

∂t

ρν,and then only

equation (2) becomes true.

ü Thus equation (2) and (3) are not compatible for time varyingfields.

ü Equation (1) by adding one unkown term say N.

ü Then equation (1) becomes,

)4(−−−+=∇ NJxH

ü Again taking divergence on both the sides

0..).( =∇+∇=∇∇ NJxH

ü Ast

J∂

∂−=∇

ρϑ. , to get correct condition.

tvN

∂∂

=∇ρ.

ü But according to Gauss’ s law



Dv .∇=ρü Thus replacing vρ by D.∇

).(. Dt

N ∇∂∂

=∇

tD

∂∂

∇= .

ü Comparing two sides of the equation,

)5(−−−−∂∂

=tDN

ü Ampere’ s circuital law in point form as,

)6(−−−−∂∂

+=∇tDJxH c

ü In equation (6) is conduction current density denoted by Jc.



5. RELATIONSHIP BETWEEN FIELD THEORY AND CIRCUIT THEORY:

ü The comparison of circuit theory and field theory on basis ofthe analysis of transmission line is as given below.

FIELD THEORY:

v The analysis of the transmission line from the field point ofview is carried out starting withMaxwell equations based onFaraday’ s law and Ampere circuital law with conductioncurrent J=0.

v Maxwell equation based on Ampere circuital law is

=∇xHtE

∂∂

ε

§ For a wave traveling in X direction,

)1(−−∂

∂−=

∂∂

tE

xH yz ε

§ Similarly Maxwell ‘ s equation based onFaradays law is

tHxE∂

∂−=∇ µ

§ For a wave traveling in X-direction,

)2(−−−∂

∂−=

∂

∂

tH

xE zy µ

§ Differentiating equation (i) with respect to tand (ii) with respect tox,

§ Wave equation as given by

2

2

2

2 1xE

tE yy

∂

∂=

∂

∂

µεv The velocity of the wave or signal along a lossless line given by

µε1

=v

• For air or vacuum the velocity is given by



csmxv == /103 8

v The intrinsic impedance of the medium is given by,

• 0ηεµ

η ==

For air or vacuum , 1== rr εµ

Ω=∴ 3770ηv The line constants are given as follows

1)(

tan−+=

=

mjj

tnconspropagatio

ϖεσϖµ

γ

mNpjj

tnconsattenuatio

/)(Re

tan

ϖεσϖµ

α

+=

=

mradjj

tonspha

/)(Im

tansec

ϖεσϖµ

β

+=

=

v The line voltage between two lines is given by

∫=2

1

.line

line

dLEv

The line current is given by

∫=inearoundonel

dLHI .

• In field theory, current density J is used.

• Mathematically the Ohms law is given by

∫ ∫= dLHZdLE ..



CIRCUIT THEORY:

v The analysis of the transmission line from the circuitpoint of view is carried out by considering it as a fourterminals network. Out of these four terminals two are atsending while other two are at receiving end.

v For a lossless line ,the differential voltage and current perdistance is given by

)(idtdIL

dxdv

−−−−= and

)(iidtdVC

dtdI

−−−−=

v Differentiating first equation with respect to distance anddifferentiating second equation with respect to time t,wecan get the transmission line wave equation as givenby,

2

22

2

2 1dx

VdLCdt

Vd=

v The velocity of the wave or signal along a lossless line isgiven by,

LCv 1

=

v The characteristics impedance of the line is given by,

Ω++

==CjGLjR

IVZ Line ω

ϖ

• For a lossless line,

Zline= Ω=CL

line



v The secondary line constants are given as follows,

1)((

tan−+=++=

=

mjCjGLjR

tnconspropagatio

βαωω

γ

mNpCjGLjR

tnconsattenuatio

/))((Re

tan

ϖϖ

α

++=

=

mradCjGLjR

tonspha

/))((Im

tansec

ϖϖ

β

++=

=

v The line voltage is given by

• The line current is given by

I= )()( 21 xtee

YZ

Vxtee

YZ

V jxjx βωβω αα +++



IMPORTANT QUESTIONS

PART A

1. Compare field theory with circuit theory?

2. State Lenz’ s law?

3. What is Displacement current?

4. Define Displacement current and Eddy current?

5. What is the relation between field theory and circuit theory?

6. Give two application’ s of Lenz’ s Law?

7. What is the static maxwell’ s equation each term?

8. Give Lorentz force equation explains each term.

9. Write down the differential form of Maxwell’ equation?

10. What is the significance of Displacement current density?

11. Explain the significance of displacement current and eddy current?

12. State two applications of Eddy current?

13. Write down the Maxwell equation corresponding to faraday’ s law, in point form.

14. Define curl of a vector.

15. State faraday’ s law of electromagnetic induction.

16. State 4 similarities between electric and magnetic circuits.

17. Give Ampere’ s law of Maxwell equation in point form and integral form.

18. What is convection current?



PART B

1. Determine the electromotive force using faraday’ s law using relevant expressions.

2. Derive Maxwell’ s equation from Ampere’ s law and Faraday’ s law .Express theequation in phasor from for time harmonic fields.

3. Derive the expression for magnetic vector potential.

4. Derive the expression for displacement current.

5. Explain in detail about Lenz’ s law and their significance.

6. Compare circuit theory with field theory?



DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING

ASSIGNMENT DETAILS

COURSE CODE: EE1201 COURSE NAME: ELECTROMAGNETIC THEORYACADEMIC YEAR: 2005-2006 SEMESTER: ODD/EVEN

YEAR/SEMESTER: II / IIIASSIGNMENT-IV

PART- A

1. Compare field theory with circuit theory?

2. State Lenz’ s law?

3. What is Displacement current?

4. Define Displacement current and Eddy current?

5. Give two application’ s of Lenz’ s Law?

6. What is the static maxwell’ s equation each term?

7. Give Lorentz force equation explains each term.

8. Write down the differential form of Maxwell’ equation?

9. What is the significance of Displacement current density?

10. Give Ampere’ s law of Maxwell equation in point form and integralform.

PART- B1. Determine the electromotive force using faraday’ s law using

relevant expressions.

2. Derive Maxwell’ s equation from Ampere’ s law and Faraday’ s law.Express the equation in phasor from for time harmonic fields.

3. Derive the expression for displacement current.

4. Explain in detail about Lenz’ s law and their significance.

5. Compare circuit theory with field theory?



SUBJECT NAME: ELECTROMAGNETIC THEORY

SUBJECT CODE: EE 2202

BRANCH: EEE

SEMESTER: III

UNIT –V

ELECTROMAGNETIC WAVESCONTENTS:

1. Generation of electromagnetic waves

2. Electro Magnetic Wave equations

3. Electromagnetic wave in perfect dielectric

4. Electromagnetic wave in lossy dielectric

5. Electromagnetic wave in good conductor

6. Skin depth

7. Poynting vector

8. Plane wave reflection and refraction.



1. GENERATION OF ELECTROMAGNETIC WAVES:

v Electromagnetic waves are generated by moving electrons.

v An electron generates an electric field, which we can visualize as

lines radiating from the electron.

v If the electron moves, say it vibrates back and forth, then this

motion will be transferred to the field lines and they will become

wavy.

v In turn, the moving electron generates a magnetic field that will

also become wavy from the motion of the electron.

v These combined electrical and magnetic waves reinforce one

another.

v This kind of wave is called an electromagnetic wave and light is

such a wave.

v Since all matter contains electrons and all these electrons are in

motion, as are the atomic nuclei they spin around, all matter

generates electromagnetic waves.



v Since all electromagnetic waves travel at the same speed (c) the

frequency of the waves is determined by the frequency of the

vibrating electrons that generate them.

v Hot substances have more energy and their component atoms

vibrate more rapidly than those of cold bodies.

v Thus the peak energy radiated by hot bodies has a higher

frequency, shorter wavelength, than that of cooler bodies.

v Wein’ s Law expresses the relationship of the peak frequency of a

black body to its absolute temperature.

v Wein's Law

lmax = a/T

a = 2989 if l is measured in microns

2. ELECTRO MAGNETIC WAVE EQUATIONS:

IN LOSSELESS MEDIUM:

v From Maxwell’ s equation, we have

v performing curl operations on both sides



v we know that ∇.B = 0

v from Maxwell’ s equation, we have


v we know that ∇.E = 0 ( in free space)



IN CONDUCTIVE MEDIUM:



v We know that ∇.B = 0


v Performing curl operations on both sides



v We know that ∇.E = 0 ( in free space)

v The wave equations for sinusoidal time variations is given as:

In losseless medium:

In conducting medium:

(1) ∇² E + (ω² µ0 - jωµσ)E = 0(2) ∇² H + (ω² µ0 - jωµσ)H = 0

v The above equations can be written as



3. ELECTROMAGNETIC WAVE IN PERFECT DIELECTRIC:

v If a medium, through which the uniform plane wave is propagating,

is perfect dielectric (which is also called lossless dielectric), then the

conductivity is zero i.e. σ = 0.

v Let the permittivity permeability of the medium be c = o Er and

respectively.

v The propagation constant y is given by,

v From equation (I) it is clear that, propagation constant is purely

imaginary.

v It indicates in a perfect dielectric medium, attenuation constant α is

zero.

v Let us select value of which gives propagation of wave in positive

z-direction.



v Similarly an intrinsic impedance for a perfect dielectric medium is

given by,

v Thus intrinsic impedance η is real resistive.

v That means phase angle of intrinsic is zero.

v But the phase angle of intrinsic impedance is zero means phase

difference between E and H is zero.

v In other words, for a perfect dielectric, both the fields E and H , are

in phase.

v As in perfect dielectric, σ= 0, attenuation constant (α) is also zero.

v As wave propagates, only the phase ( ) changes. Thus no

attenuation i.e.α = 0 means no loss.

v KEY POINT:

So perfect dielectric medium is also called loss less dielectric.

v The velocity of propagation in the perfect dielectric is given by,

v If λ is the wavelength of one cycle of the propagating wave then

velocity is given by



4. ELECTROMAGNETIC WAVE IN LOSSY DIELECTRIC:

v Practical1y all the dielectric materials exhibit some conductivity. So

we can not directly a assuming it zero.

v Obviously as compared to the results obtained for perfect medium,

the results for lossy dielectric will be different.

v The propagation constant y in lossy dielectric is given by

v Rearranging the terms

v From equation (2), it is clear that the propagation constant for lossy

dielectric medium is different than that for lossless dielectric

medium, due to the presence of radical factor.

v When σ becomes zero as in case with perfect dielectric, the radical

factor becomes unity and we can obtain the propagation constant γ

for perfect dielectric.

v It is also clear from equation (2) that the attenuation constant σ is

not zero.

v By substituting the values ofω,σ , µ and, the attenuation constant

(σ) and phase constant 3 may be calculated.

v The presence of a indicates certain loss of signal in the medium,

hence such medium is called lossy dielectric.

v When a wave propagates in a lossy dielectric, amplitude of the

signal decays exponentially due to the factor



v For forward as well as backward waves, the amplitude decays

exponentially.

v As σ is not zero, the intrinsic impedance becomes a complex

quantity. It is given by,

v Being a complex quantity η is represented in polar form as shown in

above equation.

v This angle θn indicates phase difference between the electric and

magnetic fields.

v Thus in lossy dielectric, the electric and magnetic fields are not in

time phase.

v The intrinsic impedance can be expressed as

v The angle θn is given by

v This angle depends on the properties of the lossy dielectric medium

as well frequency of a signal. For low frequency signal, co becomes

very small. Then



5. ELECTROMAGNETIC WAVE IN GOOD CONDUCTOR:

v A practical or good conductor is the material which has very high

conductivity.

v In general, the conductivity is of the order of i0 Dim in the good

conductors like copper, aluminum etc.

v For good conductors,

v The propagation constant γ is given by,

v

v Thus for good conductor,



v For good conductor, α and are equal and both are directly

proportional to the square root of frequency (f) and conductivity

(σ).

v The intrinsic impedance of a good conductor is given by

v The angle of intrinsic impedance is 45° . As we have already studied

that for perfect dielectric i.e. zero conductivity, the intrinsic

impedance angle is zero and for the good conductor angle is 45° .

v Moreover the intrinsic impedance has only a positive angle. This

clearly indicates that the field H may lag the field E by at the most

45° .



v Consider only the component of the electric field E x travelling in

positive z-direction.

v When it travels in good conductor, the conductivity is very high and

attenuation constant α is also high.

v Thus we can write such a component in phasor form as

v When such a wave propagates in good conductor.

6. SKIN EFFECT:

v Consider a conductor made up of a large number of fine strands of

wires and carrying the same current.

v A strand at the center is linked by all the internal flux in the

conductor, whereas a strand on the surface is not linked by the

internal flux.

v The inductance and reactance of the strand at the center is greater

than that of the strand at the surface.

v If the current is permitted to vary with time and all strands still

carry the same current, then the voltage drop along the center

strand will be greater than that along an outside strand.

v This effect is, however, a direct violation of Kirchhoff’ s law.

v Therefore the currents carried by the strands cannot be equal in

magnitude, since the impedances are unequal.

v The interior strand thus carries less current than the outer so as to

produce equal impedance drops along the strands.

v This phenomenon is known as Skin effect.

v The depth of penetration or skin depth is defined as that depthwhich the wave has been attenuated to



7. POYNTING VECTOR:

v Poynting Theorem: As electromagnetic waves propagate through

space from their source to distant receiving points, there is a

transfer of energy from the source to the receivers.

v There exists a simple and direct relation between the rate of this

energy transfer and the amplitudes of electric and magnetic field

strengths of the electromagnetic wave.

v This relation can be obtained from Maxwell’ s equations

v Multiplying equation (1) by E we get,



8. REFLECTION BY PERFECT CONDUCTOR:

NORMAL INCIDENCE:

v When an electromagnetic wave traveling in one medium

impinges upon a second medium having a different dielectric

constant, permeability, or conductivity, the wave in general

will be partially transmitted and partially reflected.

v In case of a plane wave in air incident normally upon the

surface of a perfect conductor, the wave is entirely reflected.

v For fields that vary with time neither E nor H can exist within

a perfect conductor so that none of the energy of the incident

wave can be transmitted.

v Since there can be no loss within a perfect conductor, non of

the energy is absorbed.

v As a result the amplitude of E and H in the reflected wave are

the same as in the incident wave, and the only difference is in

the direction of power flow.

v The expression for the electric field of the incident wave is(1)

And the surface of the perfect conductor is taken to be the

x=0 plane as shown in figure, the expression for the reflected

wave will be



v Er is determined from the boundary conditions.

v According to the boundary conditions the tangential component

of E must be continuous across the boundary and E is zero

within the conductor.

v The tangential component of E just outside the conductor must

also be zero.

v For this the sum of the electric field strengths in the initial and

reflected waves add to give zero resultant field strength in the

plane x=0.

v Therefore Er = - Ei ( 3 )

v The amplitude of the reflected electric field strength is equal to

that of the initial electric field strength, but its phase has been

reversed on reflection.

v The resultant electric field strength at any point a distance -x

from the x=0 plane will be the sum of the field strengths of the

incident and reflected waves at that point and will be given by



v If Ei is chosen to be real,

v Equation (6) shows that the incident and reflected waves

combine to produce a standing wave, which does not

progress.

v The magnitude of the electric field varies sinusoidally with

distance from the reflecting plane.

v It is zero at the surface and at multiples of half wavelength

from the surface.

v It has a maximum value of twice the electric field strength of

the incident wave at distances from the surface that are odd

multiples of a quarter wavelength.

v Since the boundary conditions require that the electric field

strength be reversed in phase on reflection in order to

produce zero resultant fields at the surface, the magnitude

field strength must be reflected without reversal of phase.

v If both magnetic and electric field strengths were reversed,

there would be no reversal of direction of energy propagation.

v Therefore the phase of the reflected magnetic field strength

Hr is the same as that of the incident magnetic Ht at the

surface of the reflection x = 0.



v The expression for the resultant field will be

v The resultant magnetic field strength H also has a standing wave

distribution. It has maximum value at the surface of the

conductor and at multiples of a half wavelength from the

surface, whereas the zero points occur at the odd multiples of a

quarter wavelength from the surface.

v From the boundary conditions for H it follows that there must be

a surface current Js A/m, such that Js = Ht (at x = 0).

v Since Et and Ht were in time phase in the incident plane wave , a

comparison of equation (6) and (11) shows that Et and Ht are 90

degrees out of time phase because of the factor in equation(6).

v That Et and Ht are 90 degrees apart in the time phase can be

seen more clearly by rewriting (6) and (11).

v Replacing – j by its equivalent e-M_.___ and combining this with the

.



IMPORTANT QUESTIONS

PART A

1. State poynting theorem.

2. What is transmission –LINE parameters?

3. What is loss tangent?

4. Define wave?

5. What are the conditions to be satisfied for plane waves?

6. What is the significance of intrinsic impedance?

7. Write the expression for plane electromagnetic waves propagating

in a dielectric media

8. Find the velocity of a wave in a lossless medium having a relative

permeability of 5 and relative permeability of 2.

9. Define the desired boundary conditions for the conductor free space

boundary in electrostatics.

10. What are the boundary conditions of electromagnetic wave at the

interface between two losses less dielectric media?

11. What are boundary conditions?

12. Define skin depth.

13. What do you mean by the uniform plane wave?

14. What is meaning of wave?

15. Define propagation constant.

16. Define attenuation constant and phase constant.

17. What is perfect dielectric medium?

18. What is lossy dielectric?

19. Explain skin effect.

20. Define depth of penetration.

21. What is poynting vector?



PART B

1. Derive poynting‘ s theorem.

2. Derive the wave equation starting from Maxwell’ s equation forfree space.

3. Derive the expressions describing propagation of uniform planewave in good conductor?

4. What is skin effect? What is skin depth? What is its relation withattenuation constant, conductivity and frequency?

5. Explain propagation of uniform plane in perfect dielectric .what islossless dielectric?



Electromagnetic TheorySemester - III (EEE)

Time : Three Hoursl [Maximum Marks : 100

Answer ALL questions

PART - A (10x2= 20 Marks)

Q.l Giae an example for diuerging field nnd for curling field.

Q.2 Stnte diaergence theorem and mention the significance of the theorem.

Q.3 using Gauss's law, deriae the capacitance of a coaxial cable.

Q.4 Deriae Poisson's equation the electrostntic field.

Q.5 A conductor located nt x = 0.5 m, A = 0 and 0<z<2.0m carries a current of L0 Ain the n, directiott. Along the length of the conductor B = 2.5 arT. Find the torqueabout the x axis.

Q.6 Draw the mngnetic field pattern inside and outside the circular conductor withwrifurm current density.

Q.7 Find the total flux induced in the loop of constant width'l '. The tength'x'isincreased uniformly with the time by moaing the sliding conductor at a unifurmaelocity 'u'. The flux density is snme and normal to the plane of the loop as shown.

Fig. 1

Q.8 Find the nuerage power loss/uolume for a dielectric haaing e, = 2 and tan6 = 0.000t5,if E = 1.0 kV/m at 500 MHz.

Q.9 Cnluilate the skin depth and u)aae uelocity at 2 MHz in Aluminium withconductiaity 40 MS/m and p, = 1.

Q.10 Giaen E=En,sin(at -Bf) a, in free spnce, sketch E and H at t = 0.PART - B (5x15 = 80 Marks)

Q.11 i) Explain when nnd how an electromagnetic walre is generated, I2+al

ii) Deriue the electromagnetic waae equations in free space and mention the types ofsolutions [8+21

(P - 102)



Electrgmagnetic Theorv p - 103 Nov./Dec. - 2003

Q'12 al i) Deriae the electrostntic botmdnry conclitions at the interface of two dielectric media. .

b) i) Deriua tlrc cxpressions for scalar ,rr?r\r, due to a point charge nnd a ring chnrge,12+Gl

iil A totnl chnrge of 100 nC is unifurmly distributecl arottnd a circulnr ring of 1.0 mradius. Firtd tlrc potentinl nt a point on the nxis 5.0 nt nbozte the plane iy in, ring.Compare zuith the result zuhere nll charges nre at the origin in the form'of a poiriclmrge.

tglQ'13 a) i) Explnin mngnetizntiott in mngnetic mnterials nnd explain how the ffict of

mngnetizntiott is tnken into account in the caluilntion of B/H. t10Iii) Find H in n magnetic material

7) When p = 0.000018 H/m nnd M - 120 A/m.2) When B - 300 pT and magnetic susceptibility = 29.

ORb) Derizte the mngnetic force betzueen tzao parallel condtrctors carrying equal ytTents in the

i) Sanrc direction 1gl

ii) Opposite direction.

field inside nnd outside the conclttctor ?

b') i) Deternine the nmplitude of the reflected nnd trnnsmittetl Eof two media zuitlt the foilowing properties. Medium 7 : e,Medi,m 2 ; free spoce. Assunte normal incidence and th.emediwn L nt the interfnce is 1..5 ntV/m.

ii) Dcriue nll the formulae used.

l10Ithe potential nnd electric

t5l

t3l

t3l

t8l

snd H at the interface= 8 . 5 , p r = 7 , o - 0amplitude of E in tlrc

t8l

t8Iotrtr

Q'14 al From the fundnmentnl lsws deriae the Maxwell's equntions nnd the need for tIrcMnxzuell's contributiott

.to electromagnetic theory. State the eqrtations in bothdifferential nnd integrnl form. tlfl

ORb) Ex\tlnin the relationship between the fielcl tlrcory snd circuit theory using n simple

RLC series ciruit. Also explnin the lintitations o| tlrt circuit theory. 112+41Q'15 al A plane trnaelling 70nae hns a peak electric fietd intensity E as 6 kV/*. If the

medium is lossless zuith Er- = 3 and p, - 1, find the aelocity of the EM u)(n)e, peakPOyNT/NG ucctor, impedance of the medittm nnd the ptit, ialue of the magneticfield H. Deriae all the formulae used.

' r tg+gl

OR



Electromagnetic TheorySemest€r - III (EEE)

Time : Three Hoursl [Maximum Marks : 100

Answer AtL questions

PART - A 00x2= 20 Marks)

Q.l How the unit aectors are dfined in cylindrical co-ordinate systems?

e.2 What is the physical significance of the term "diaergence of a aector field"?

Q.3 A unifurm line charge with P, =SpClmlies along the x-axis. FindE at (3,2, 1) m.

Q.4 Define dielectric strength. What is the dielectric strength of co-axial cable ?

Q.5 Calculate inductance of a ring shaped coil haaing a mean diameter of 20 cm woundon a wooden core of 2 cm- diameter. The winding is unifurmly distributed andcontains 200 turns,

Q.6 Distinguish magnetic scalar potential and magnetic aector potential

Q.T Write down the erpression for the e.mf. induced in the mouing loop in static B field.

Q.8 Compare field theory with circuit tluory.

Q.9 What do you mean by "Dqth of penetration".

Q.10 For a lossy dielectric material haaing p, -- l, tr - 48, a - 20 S/m calculate thepropagation constant at n frequency of 16 GHz.

PART B - (5x16 = 80 Marks)

e.tt i) Determine the constant c such that the oector F = (x+ay)i * (V +bz)i + (x+cz_)kwill be solenoidal. I4l

ii) GiaenA -ZrcosQf, + rI, in cytinilrical co-ordinate. For the contour shown aertfuStoke's theorem.

Fig. 1

Q.12 a) i) A circular disc of radius 'a' m is charged uniformly with a charge density of

ps C / *2 . Find the electric field at a point 'h' m from the disc along its axis. I8l

ii) A circular disc of L0 cm radius is chnrged unifurmty with a total charge of 10-u9:

Find the electric intensity at a point 30 cm away fro* the disc along the axis. I8l

(P - 104)



Electromagnetic Theory p - 105 Mav/June - 2006

OR

bl i) Deriae the expression for energv density in electrostntic .fields.iil A cnpncitor consists of srynred two metnl plates each L00 cm side plnced parallel

nnd 2 mm npart. The spnce betwcen the plntcs is filled zuith a dielectric haaing arelntiae permittiaity of 3.5. A potentinl drop of 500 V is maintained between lhe

l8l

plntes. Cnlculate

1) The cnpncitnncc

3) The electric flux densitrl

Q.13 a) i) Find the field intcnsity at ans sltown.

2) The chnrge of capncitor

4) The potentinl grndient

point P due to a strniglrt conductor cnrryhtg current

t8II

l8l

I__.>

Fig.2ii) Find E nt the centre of nn eEilateral triangtrlar loop of side 4 m carrying current

of 5 A. IBI

ORD i) Derizte the expression for co-efficient of coupling in terrns of muttral and self

inductnrrces.

ii) An iron ring with a cross-sectionnl area of 3 cmz and a nrcnn circumference of75 cm is wowtd zuitlt 250 turns zuire carrying n arrent of 0.3 A. The relatiaepernteability of the ring is 7500. Cnlculnte tlrc flux established in the ring. t8l

Q.1a a) With necessary axplanation, deriae the Maxwell's equation in dffirential and integralforms. 116l

OR

of electromagnetic induction I8l

current? Write down the expression for the

b) i) Write slnrt notes on Fnradny's laws

ii) Whnt tlo yott mean by displacementtotnl current density.

Q.15 a) i) what is the physical significance of the Poynting aector?

ii) Stnte nnd explain the Poynting's theorem.

OR

bl A plnne rDoag propagoting tlrouglt n medium with er = 8, Fr - 2 hasE = 0.5, sin(lbt -Bz) a* V/m. Determine

i) 9 ii) The loss tangent iii) Wnrte impedance ia) Waae aelocity a) E field. tlGI

DTIB

l8l

l4l

Ir2l

t.<--L----+



Electromagnetic TheorySemester - III (EEE)

Time : Three Hoursl [Maximum Marks : 100Answer ALL questions

PART - A (10x2= 20 Marks)

Q.1 State the diaergence theorem nnd giae an example for diaergence field.

Q.2 what is the electric field around a long transmission line?

Q.3 Calculate the totnl charge enclosed by ufue of 2 m sides, centered nt the origin andwith the edges pnrnllel to the axes ulrcn the electric flux densita oaer the cttbe is[ = [1oxu ls]u.(Cl*z).

Q.4 Define dielectric strength of n materinl. Mantion the san'te for nir.

Q.5 PIot the unrintion of H inside nnd otttsitlc n circulnr condttctor uitlr tutifortn currentdensity.

Q.6 Plnne ! = 0 carries a unifurm current of 30 a, mA/m. Cnlctilnte the mngnetic fieldintensity nt (7, 70, - 2) m in rectnngtilar co-ordinate sustem.

Q.7 Deriae an expression fo, 'loss tangent' in an instrlating material and mention the

practical sigrtificnnce o.f the same.

Q.8 A medium hns constant conductiaity of 0.1 mho/rn, pr -- l, El = 30. when theseparameters do not change zuith tlrc frequency, check whether the medium behaaes likea conductor or a dielectric at S0 kHz and 10 GHz.

Q.9 In free space E (2, t) = 100sin(rr-gr)d*(Vl4 Find tlrc totnl porL,er pnssirtgtlrough a square area of side 25 mm, in the z _ 0 plnne.

Q.10 Giuen E (2, t) =700sin(at-Pz)ao(Vlm)infree space, sketch E and H at t - 0.P A R T B - ( 5 x 1 5 = 8 0 M a r k s )

Q.11 a) i) What are the major sources of electromagnetic fields (any fiae) ?

iil What are the positiae and negatiue fficts EM fietds on liaing things ?

iii) what are the E and H field linits for puhlic exposures ?

iu) ciae any one example to reduce thc e,ffect af EM .fierd.

ORb) i) Explain the electric field distribution inside and ottside a condttctor. I4l

ii) Explain the principle of electroststic shielding. l4liii) Draw the equipotential lines nnd E lines inside and arottnd a metal sphere. t8l

(P - 106)

IsltsII2lt4l



Electromagnetic Theory P -107 Nov./Dec. - 2006

Q.12 a) Deriztc tha elactric fialtl lncl ytsf clttisi distribution nnd the cnpncitnnce peroJ n conxinl cnblc.

OR

b\ Explnitt itt detill tlrc ltehsrtiour o.f n dielcctric medium in electric field.

Q.13 a) Cnluilsta B nntl H due to n long solenoid.

unit lengtlt[6+5+5]

t15I

[8+8]

OR

bl i) Deriue .for force nnd tortTue hr a mttgttctic fiald using rttotor os an example.

ii) Fitrcl tlrc torque nbout tka V nxis for the tuto cottductors of lengtlt l, cnrryingurrcnt ht opposite directions, scparnted by n fixed distnnce us, in the uniformntngrrctic field irt x dircctiotr. [8+81

Q.14 a) Dcriaa tlrc Mnxrucll's equntiorrs in ltoth differentinl and integrnl forms. 1L5l

OR

bl Explnin tlrc rlifferarft nu'tlrods of t'.m.f. induction zuith neccssary goaerning equntionsnnd utitlt sttitnhle exontples. t16l

Q.15 a) i) Dcriue tlrc electronmgnetic roaaa cquntion (in frcquency domain) and thepropngntiort constmt nnd intrinsic impednnce.

ii) Explnin thc propngntion of'EM wnaes inside a conductor.

OR

b) i) Dariua tha trnnsnission orrd reJlaction coafficients nt the interfnce of tuo media fornornml incidence. t8I

iil A frce spnce-silaar interfncc Ims E (ittcident) = 700 V/m on the free space side. The

freryrcncy is 1-S MHz snd the siluer constnnts are er = V, = 1, o - 61..7 MSlm.Deterntine E (reflected), E (trnnsmitted) at the interfnce. t8l

treD

t8l

l8l



ELECTROMAGNETIC THEORY(EC1253)

1.State stokes theorem. The line integral of a vector around a closed path is equal to the surface integral of the normal component of its curl over any surface bounded by the path H.dl = (ÑxH)ds 2.State coulombs law. Coulombs law states that the force between any two point charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. It is directed along the line joining the two charges. F=Q1Q2 / 4πεr2 ar 3.State Gauss law for electric fields The total electric flux passing through any closed surface is equal to the total charge enclosed by that surface. 4.Define electric flux. The lines of electric force is electric flux. 5.Define electric flux density. Electric flux density is defined as electric flux per unit area. 6.Define electric field intensity. Electric field intensity is defined as the electric force per unit positive charge. E =F/ Q =Q/4πεr2 V/m



7.Name few applications of Gauss law in electrostatics. Gauss law is applied to find the electric field intensity from a closed surface.e.g)Electric field can be determined for shell, two concentric shell or cylinders etc. 8.What is a point charge? Point charge is one whose maximum dimension is very small in comparison with any other length. 9.Define linear charge density. It is the charge per unit length. 10.Write poisson’s and laplace ’s equations. Poisson ‘s eqn: Ñ2V= - ρv / ε Laplace’ s eqn: Ñ2V= 0 11.State the condition for the vector F to be solenoidal. Ñ·F =0 12. .State the condition for the vector F to be irrotational. ÑxF =0 13.Define potential difference. Potential difference is defined as the work done in moving a unit positive charge from one point to another point in an electric field.



14.Define potential. Potential at any point is defined as the work done in moving a unit positive charge from infinity to that point in an electric field. V=Q / 4πεr 15.Give the relation between electric field intensity and electric flux density. D=εE C/m2

16.Give the relationship between potential gradiant and electric field. E= - ÑV 17.What is the physical significance of div D ? Ñ·D=-ρv The divergence of a vector flux density is electric flux per unit volume leaving a small volume. This is equal to the volume charge density. 18. Define current density. Current density is defined as the current per unit area. J= I/A Amp/m2

19.Write the point form of continuity equation and explain its significance. Ñ·J= - ρv / t 20.Write the expression for energy density in electrostatic field. W=1 / 2 εE2

21.Write the boundary conditions at the interface between two perfect dielectrics. i)The tangential component of electric field is continuous i.e)Et1=Et2



ii)The normal component of electric flux density is continuous I.e)Dn1=Dn2 22.Write down the expression for capacitance between two parallel plates. C=εA / d 23.What is meant by displacement current? Displacement current is nothing but the current flowing through capacitor. J= D / t 24.State point form of ohms law. Point form of ohms law states that the field strength within a conductor is proportional to the current density. J=σE 25 Define surface charge density. It is the charge per surface area. 26.State amperes circuital law. Magnetic field intensity around a closed path is equal to the current enclosed by the path. H·dl=I 27.State Biot –Savarts law. It states that the magnetic flux density at any point due to current element is proportional to the current element and sine of the angle between the elemental length and inversely proportional to the square of the distance between them dB=µ 0Idl sinθ / 4πr2 28.Define magnetic vector potential. It is defined as that quantity whose curl gives the magnetic flux density. B=Ñ x A =µ / 4π J/r dv web/m2



29.Write down the expression for magnetic field at the centre of the circular coil. H = I/2a. 30.Give the relation between magnetic flux density and magnetic field intensity. B =µ H 31.Write down the magnetic boundary conditions.

i) The normal components of flux density B is continuous across the boundary. ii) The tangential component of field intensity is continuous across the boundary.

32.Give the force on a current element.

dF = BIdlsinθ

33..Define magnetic moment. Magnetic moment is defined as the maximum torque per magnetic induction of flux density. m=IA

34.State Gauss law for magnetic field.

The total magnetic flux passing through any closed surface is equal to zero. B.ds =0

35.Define a wave.

If a physical phenomenon that occurs at one place at a given time is reproduced at other places at later times , the time delay being proportional to the space separation from the first location then the group of phenomena constitutes a wave.



36. Mention the properties of uniform plane wave.

i) At every point in space ,the electric field E and magnetic field H are perpendicular to each other. ii)The fields vary harmonically with time and at the same frequency everywhere in space.

37.Write down the wave equation for E and H in free space. Ñ2H– µ 0ε0 2H / t 2 =0.

38.Define intrinsic impedance or characteristic impedance. It is the ratio of electric field to magnetic field.or It is the ratio of square root of permeability to permittivity of medium.

39.Give the characteristic impedance of free space.

377ohms

40.Define propagation constant.

Propagation constant is a complex number γ =α +jβ where α is attenuation constant β is phase constant γ = jωµ (σ +jωε)

41.Define skin depth

It is defined as that depth in which the wave has been attenuated to 1/e or approximately 37% of its original value. ∆ = 1/α = 2 / jωσ

42.Define Poynting vector. The pointing vector is defined as rate of flow of energy of a wave as it propagates. P =E X H

43. State Poyntings Theorem.



The net power flowing out of a given volume is equal to the time rate of decrease of the the energy stored within the volume- conduction losses. 44.Give significant physical difference between poisons and laplaces equations. When the region contains charges poisons equation is used and when there is no charges laplaces equation is applied. 45.Give the difficulties in FDM. FDM is difficult to apply for problems involving irregular boundaries and non homogenious material properties. 46.Explain the steps in finite element method. i) Discretisation of the solution region into elements. ii) Generation of equations for fields at each element iii) Assembly of all elements iv) Solution of the resulting system

47.State Maxwells fourth equation. The net magnetic flux emerging through any closed surface is zero. 48. State Maxwells Third equation The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume. 49.State the principle of superposition of fields. The total electric field at a point is the algebraic sum of the individual electric field at that point. 50.Define ohms law at a point Ohms law at appoint states that the field strength within a conductor is proportional to current density.



51.Define self inductance. Self inductance is defined as the rate of total magnetic flux linkage to the current through the coil. 52.Define pointing vector. The vector product of electric field intensity and magnetic field intensity at a point is a measure of the rate of energy flow per unit area at that point. 53.Give the formula to find potential at a point which is surrounded by four orthogonal points in FDM. V0= ¼(V1+V2+V3+V4) 54.Give the formula to find potential at a point which is surrounded by six orthogonal points inFDM. V0= ¼(V1+V2+V3+V4 +V5+V6) 55.State Lenz law. Lenz’s law states that the induced emf in a circuit produces a current which opposes the change in magnetic flux producing it. 56.What is the effect of permittivity on the force between two charges? Increase in permittivity of the medium tends to decrease the force between two charges and decrease in permittivity of the medium tends to increase the force between two charges. 57.State electric displacement. The electric flux or electric displacement through a closed surface is equal to the charge enclosed by the surface.



58.What is displacement flux density? The electric displacement per unit area is known as electric displacement density or electric flux density. 59.What is the significance of displacement current? The concept of displacement current was introduced to justify the production of magnetic field in empty space. It signifies that a changing electric field induces a magnetic field .In empty space the conduction current is zero and the magnetic fields are entirely due to displacement current. 60.Distinguish between conduction and displacement currents. The current through a resistive element is termed as conduction current whereas the current through a capacitive element is termed as displacement current. 61.Define magnetic field strength. The magnetic field strength (H) is a vector having the same direction as magnetic flux density. H=B/µ 62.Give the formula to find the force between two parallel current carrying conductors. F=µI I1 / 2πR 63.Give the expression for torque experienced by a current carrying loop situated in a magnetic field. T = IABsinθ 64What is torque on a solenoid? T = NIABsinθ 65.Explain the conservative property of electric field. The work done in moving a point charge around a closed path in a electric field is zero.



Such a field is said to be conservative. / E.dl = 0 66.Write he expression for field intensity due to a toroid carrying a filamentary current I H=NI / 2пR 67.What are equipotential surfaces? An equipotential surface is a surface in which the potential energy at every point is of the same vale. 68.Define loss tangent. Loss tangent is the ratio of the magnitude of conduction current density to displacement cuurrent density of the medium. Tan δ = σ / ωε 69.Defie reflection and transmission coefficients. Reflection coefficient is defined as the ratio of the magnitude of the reflected field to that of the incident field. 70. Define transmission coefficients. Transmission coefficient is defined as the ratio of the magnitude of the transmitted field to that of incident field. 71.What will happen when the wave is incident obliquely over dielectric –dielectric boundary? When a plane wave is incident obliquely on the surface of a perfect dielectric part of the energy is transmitted and part of it is reflected .But in this case the transmitted wave will be refracted, that is the direction of propagation is altered.



72.What is the expression for energy stored in a magnetic field? W = ½ LI2 73.What is energy density in magnetic field? W = ½ µH2 74.Distinguish between solenoid and toroid. Solenoid is a cylindrically shaped coil consisting of a large number of closely spaced turns of insulated wire wound usually on a non magnetic frame. If a long slender solenoid is bent into the form of a ring and there by closed on itself it becomes a toroid. 75.Describe what are the sources of electric field and magnetic field? Stationary charges produce electric field that are constant in time, hence the term electrostatics. Moving charges produce magnetic fields hence the term magnetostatics. 76.What are the significant physical differences between Poisson ‘s and laplace ‘s equations. Poisson ‘s and laplace ‘s equations are useful for determining the electrostatic potential V in regions whose boundaries are known. When the region of interest contains charges poissons equation can be used to find the potential. When the region is free from charge laplace equation is used to find the potential. 77.State Divergence Theorem. The integral of the divergence of a vector over a volume v is equal to the surface integral o f the normal component of the vector over the surface bounded by the volume.



78.Give the expression for electric field intensity due to a single shell of charge E = Q / 4πεr2 79.Give the expression for potential between two spherical shells V= 1/ 4π (Q1/a – Q2/b) 80.Define electric dipole. Electric dipole is nothing but two equal and opposite point charges separated by a finite distance. 81.What is electrostatic force? The force between any two particles due to existing charges is known as electrostatic force, repulsive for like and attractive for unlike. 82.Define divergence. The divergence of a vector F at any point is defined as the limit of its surface integral per unit volume as the volume enclosed by the surface around the point shrinks to zero. 83.How is electric energy stored in a capacitor? In a capacitor, the work done in charging a capacitor is stored in the form of electric energy. 84.What are dielectrics? Dielectrics are materials that may not conduct electricity through it but on applying electric field induced charges are produced on its faces .The valence electron in atoms of a dielectric are tightly bound to their nucleus. 85.What is a capacitor? A capacitor is an electrical device composed of two conductors which are separated through a dielectric medium and which can store equal and opposite charges ,independent of whether other conductors in the system are charged or not.



86.Define dielectric strength. The dielectric strength of a dielectric is defined as the maximum value of electric field that can b applied to the dielectric without its electric breakdown. 87.What meaning would you give to the capacitance of a single conductor? A single conductor also possess capacitance. It is a capacitor whose one plate is at infinity. 88.Why water has much greater dielectric constant than mica.? Water has a much greater dielectric constant than mica .because water ha a permanent dipole moment, while mica does not have. 89.What is lorentz force? Lorentz force is the force experienced by the test charge .It is maximum if the direction of movement of charge is perpendicular to the orientation of field lines. 90.Define magnetic moment. Magnetic moment is defined as the maximum torque on the loop per unit magnetic induction. 91.Define inductance. The inductance of a conductor is defined as the ratio of the linking magnetic flux to the current producing the flux. L = NФ / I 92.What is main cause of eddy current? The main cause of eddy current is that it produces ohmic power loss and causes local heating.



93.How can the eddy current losses be eliminated? The eddy current losses can be eliminated by providing laminations. It can be proved that the total eddy current power loss decreases as the number of laminations increases. 94.What is the fundamental difference between static electric and magnetic fild lines? There is a fundamental difference between static electric and magnetic field lines .The tubes of electric flux originate and terminates on charges, whereas magnetic flux tubes are continuous. 95.What are uniform plane waves? Electromagnetic waves which consist of electric and magnetic fields that are perpendicular to each other and to the direction of propagation and are uniform in plane perpendicular to the direction of propagation are known as uniform plane waves. 96.Write short notes on imperfect dielectrics. A material is classified as an imperfect dielectrics for σ <<ωε, that is conduction current density is small in magnitude compared to the displacement current density. 97.What is the significant feature of wave propagation in an imperfect dielectric ? The only significant feature of wave propagation in an imperfect dielectric compared to that in a perfect dielectric is the attenuation undergone by the wave. 98.What is the major drawback of finite difference method? The major drawback of finite difference method is its inability to handle curved boundaries accurately. 99.What is method of images? The replacement of the actual problem with boundaries by an enlarged region or with image charges but no boundaries is called the method of images.



100.When is method of images used? Method of images is used in solving problems of one or more point charges in the presence of boundary surfaces.



Part-B 1.Find the electric field intensity of a straight uniformly charged wire of length ‘L’m and having a linear charge density of +λ C/m at any point at a distance of ‘h’ m. Hence deduce the expression for infinitely long conductor. Hints: Field due to charge element is given by: dE = λdi/ 4πξr2

Ex=λ [cos α1+cosα2] /4πεh

Ey=λ [sin α1-sinα2] /4πεh

For infinitely long conductor

E = λl / 4πεh

2.Derive the boundary relations for electric fields. Hints: i)The tangential component of the electric field is continuous at the surface .Et1 = Et2 ii)The normal component of the electric flux density is continuous if there is no surface charge density. Dn1 = Dn2 3.Find the electric field intensity produced by a point charge distribution at P(1,1,1)caused by four identical 3nC point charges located at P1(1,1,0) P2(-1,1,0) P3(-1,-1,0) and P4(1,-1,0). Hints: Find the field intensity at P by using the formula Ep = 1/4επ[( Q1/r1p

2 u1p ) +(q2/r2p2 u2p) +(q3/r3p

2 u3p)+(q4/r4p2)u4p)]



4.A circular disc of radius ‘a’ m is charged with a charge density of σC/m2 .Find the electric field intensity at a point ‘h’m from the disc along its axis. Hints: Find the field due to the tangential and normal components Total field is given by E =ρs /2ε [1-cos α] 5. Four positive charges of 10–9 C each are situated in the XY plane at points (0,0) (0,1) (1,0) and (1,1).Find the electric field intensity and potential at (1/2 ,1/2). Hints: Find the field intensity at point using the formula E = Q / 4πεr2 ur Find the potential at point using the formula V = Q / 4πεr Find the field intensity at the point due to all four charges by using the superposition principle. 6. Given a electric field E = (-6y/x2) x + 6/x y + 5 z .Find the potential difference VAB given A(-7,2,1) and B( 4,1,2) Hint: Find the potential using the formula v=-/E.dl and substitute the points



7.Derive an expression for potential difference between two points in an electric field. Hint: The potential difference between two points r1 and r2 is V = V1 –V2 V = Q / 4πεr1 _ Q / 4πεr 2 8.Find the magnetic flux density at a point Z on the axis of a circular loop of radius ‘a’ that carries a direct current I. Hints: The magnetic flux density at a point due to the current element is given by dB = µIdl / 4π r2

B = µIa2 / 2(a2 + z2)3/2 9.Determine the force per meter length between two long parallel wires A and B separated by 5cm in air and carrying currents of 40A in the same direction. Hints: Calculate the force per metre length using the formula F/L = µI1I2 / 2πd In the same direction force is attractive. 10.Derive an expression for magnetic vector potential. Hint:



magnetic vector potential is A = µ / 4π ///J / r dv 11.Derive the magnetic boundary relations. i)The tangential component of the magnetic field is continuous across the boundary .Ht1 = Ht2 ii)The normal component of the magnetic flux density is continuous across the boundary Dn1 = Dn2 12.Find the magnetic field intensity at a distance ‘h’m above an infinite straight wire carrying a steady current I. Hints: The magnetic flux density is calculated starting from Biot savarts law. The magnetic flux density at any point due to aninfinite long conductor is given by B = µI / 2πd 13.Two conducting concentric spherical shells with radii a and b are at potentials V0 and 0 respectively. Determine the capacitance of the capacitor. Hint: Derive the capacitance between concentric spheres using the formula C = Q /V = 4πε [ ab /(b-a) ]



14State and derive an expression for Poyntings theorem. Hints: The net power flowing out of a given volume v is equal to the time rate of decrease of the energy stored within the volume minus the conduction losses. 15.Find the forces /length between two long straight parallel conductors carrying a current of 10A in the same direction. A distance of 0.2m separates the conductors. Also find the force/length when the conductors carry currents in opposite directions. Hints: Calculate the force per metre length using the formula F/L = µI1I2 / 2πd In opposite direction force is repulsive 16 Derive an expression for torque acting on a loop. Hints :When a current loop is placed parallel to a magnetic field forces act on the loop that tends to rotate the tangential force times the radial distance at which it acts is calledtorque or mechanicl moment of the loop. T = m X B 17.Derive an expression for energy and energy density in a electric field. Energy =CV2/2 Energy density = εE2/2



18. .Derive an expression for energy and energy density in a magnetic field. Energy =LI2/2 Energy density = µH2/2 19.Derive all the maxwells equations. Hints: i)Maxwells equation from electric Gauss law. ii) Maxwells equation from magnetic Gauss law. iii)Maxwells equation from Amperes law. iv) Maxwells equation from Faradays law. 20.Derive an expression for displacement, conduction current densities. Also obtain an expression for continuity current relations Hints: Displacement current density Jd = εδE/δt Conduction current density Jcond = σE 21.Derive the general Electromagnetic wave equation. Hint: Starting from the maxwells equation from Faradays law and Amperes law derive the Equation ˘ 2 E - µ σ(δ E/ δt )-µε (δ2 E/δt2 ) 22.Briefly explain reflection by a perfect dielectric when a wave is incident normally on a perfect dielectric and derive expression for reflection coefficient. Hints: When a plane electromagnetic wave is incident on the surface of aperfect dielectric part of the energy is transmitted and part of it is reflected.



Er / Ei = ( 2 – 1) /( 2 + 1) 23. Briefly explain reflection by a perfect dielectric when a wave is incident normally on a perfect conductor. Hints :When the plane wave is incident normally upon the surface of a perfect conductor the wave is entirely reflected. Since there can be no loss within a perfect conductor none of the energy is absorbed. E (x,t) = 2Ei sinβx sinω t 24. Derive the relation between field theory and circuit theory for an RLC series circuit. Hints : Starting from field theory erquation for a series RLC circuit derive the circuit equation V= IR + L dI/dt +(1 /C) / Idt 25.State and explain Faradays and Lenzs law of induction and derive maxwells equation. Hints: The total emf induced in a circuit is equal to the time rate of decrease of the total magnetic flux linking the circuit. ˘ X E = -δB/ δt



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TAGORE - files.teceee-in.webnode.infiles.teceee-in.webnode.in/200000080-99f329aec5/EMT-Lecture Notes-TEC.pdf · electromagnetic theory [ ee2202 ] (lecture notes) dr. k.srinivasan