EMT Lecture Notes TEC

8/15/2019 EMT Lecture Notes TEC

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ELECTROMAGNETIC THEORY [ EE2202 ]

1. INTRODUCTION

Sources and effects of electromagnetic fields – Vector fields – Different co-ordinatesystems- 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 - Dielectricstrength - Electric field in multiple dielectrics – Boundary conditions, Poisson’s andLaplace’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 andvector potential – Magnetic force – Torque – Inductance – Energy density – Magneticcircuits.

4. ELECTRODYNAMIC FIELDS

Faraday’s laws, induced emf – Transformer and motional EMF – Forces and Energy inquasi-stationary Electromagnetic Fields - Maxwell’s equations (differential and integralforms) – Displacement current – Relation between field theory and circuit theory.

5. ELECTROMAGNETIC WAVES

Generation – Electro Magnetic Wave equations – Wave parameters; velocity, intrinsicimpedance, 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 andpower.

TEXT BOOKS

1. Mathew N. O. SADIKU, ‘Elements of Electromagnetics’, Oxford University pressInc. First India edition, 2007.

2. Ashutosh Pramanik, ‘Electromagnetism – Theory and Applications’, Prentice-Hallof India Private Limited, New Delhi, 2006.

REFERENCE BOOKS

1. Joseph. A.Edminister, ‘Theory and Problems of Electromagnetics’, Secondedition, 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.

ELECTROMAGNETIC THEORY [EE2202]

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


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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 = a OA =

OA

OA

By the use of the unit vectors a x, a y, a z along the x, y, and z axes of a Cartesian co-ordinate system. A = A xax+A yay+A zaz

A = A =222

z y x A A A ++VECTOR ALGEBRA:

VECTORS MAY BE ADDED

A+B = ( ) z z y y x x a Aa Aa A ++ + ( ) z z y y x x a Ba Ba B ++ = ( ) ( ) ( ) z z z y y y x x x a B Aa B Aa B A +++++

SUBTRACTEDA -B = ( ) z z y y x x a Aa Aa A ++ - ( ) z z y y x x a Ba Ba B ++ = ( ) ( ) ( ) z z z y y y x x x a B Aa B Aa B A −+−+−

ASSOCIATIVE LAW:

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

DISTRIBUTIVE LAW:

K (A+B) = kA+kB

(K 1+K

2) =K

1A+K

2A

CUMULATIVE LAW:

A+B =B+A




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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 BA.B = A xBx+A yBy+A zBz

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

= A xBx (ax. a x) +A yBy (ay.ay) +A zBz (a z.az) +

Ax By (ax.ay) + …..+A zBy (a z.ay)

ax. a x = a y.ay = a z.az =1Cos θ in the dot product is unity. When angle is zero, θ=90 0. cos θ=0.All other dot productof the unit vectors are zero.

A. B =A xBx+A yBy+A zBz

CROSS PRODUCT LAW :

Cross product of two vectors is by definition

A x B = (AB Sin θ) an

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

Expanding the cross product in component from

Ax B = ( ) z z y y x x a Aa Aa A ++ x ( ) z z y y x x a Ba Ba B ++Which is conveniently expressed as a determinant?

ax ay azAx B = A x Ay A z

Bx By Bz

B X A = - A X B

B X A has the same magnitude but opposite direction.




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Ex: 2Given A=2a x+4a y-3a z and B =a x- ay, find A .B and A x B.

A .B = A xBx+A yBy+A zBz

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

= 2 - 4 + 0

=-2

ax ay az Ax B = A x Ay A z

Bx By Bz

= (A yBz-AzAy) a x-(A zBx-A xBz) a y+ (A xBy-AyBx) a z

= a x (0-3)-a y (0+3) +a z (-2-4)

= -3a x-3a y-6a z

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 of various 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 axis

direction.• In this system, if right hand is used then thumb indicates x axis, the forefinger indicates y

axis and middle finger indicates z axis, when three fingers are held mutually perpendicular to 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.




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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 a x, a y, a z are the base vectors of Cartesian coordinate system.

• Consider a point P (x 1, 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 = x 1ax+y1ay+z 1az• The magnitude of this vector in terms of three mutually perpendicular components given

by,

21

2

1

21 z y xrop ++=

aop =Unit vector along op=op

op

r

r

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’ .




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• 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 rectangular parallelpiped. 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 dz dydxdl ++=

• 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

= z ds Differential vector surface area normal to y direction.

= dx dy az




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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 a l , a φ and a z .

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

• It lies the plane φ = φ1 and z =z 1. The unit vector a φ is normal to plane φ = φ1 pointsin the direction of increasing φ1 lies in the plane z=z 1 and in tangent to the cylindricalsurface l = l 1.

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

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

a l . aφ = a z• A differential volume element in cylindrical coordinates may be obtained by increasing l ,φ and z by differential increments d l , d φ , dz.

• The cylinders of radius l and ( l +d l ), radial plane angles φ and ( φ +d φ), horizontal planes at z and z + dz.

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




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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 3

surfaces :• a spherical surface centered at the origin with a radius R = R1: a right

circular 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, a θ and aϕare the base vectors.

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

aR x a θ = aθaθ x aϕ= aR

aϕx aR = a θ

• A vector in spherical coordinates is given by

A = aRAR + a θAθ + aϕAϕ

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

• The differential length dl is given as

dl = aRdR + a θRdθ +aϕRsin θdϕ

• The differential surface area ds is given as

dsR = R² sin ϕdϕdθdsθ = Rsin θdϕdR dsϕ= Rdrd θ




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• Differential Volume element in Spherical Coordinates

The differential volume dv is given as

dv = R² sin θdRd θdϕ

Conversion formulae

From spherical to Cartesian coordinates

x = Rsin θcosϕ

y = Rsin θsinϕ

z = Rcos θ

From Cartesian to spherical coordinates

R = ( x² + y² + z² )1/2

θ = tan –1 ( x² + y² )1/2z

ϕ= tan –1 (y/x)




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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.

Advds Av s

.. ∇=∫∫∫ ∫∫ 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 A

y A

x A A z y x

dv = dx dy dz

• Taking volume integral on both sides

z A

y

A

x A

Adv z y xvv ∂∂+∂

∂+

∂∂=∇ ∫∫∫ ∫∫∫ . )

• Consider element in x direction

]dydz dx x

A xv ∂

∂=∫∫

• But x 1 and x 2 be limits for x direction

Ax Ax Axdx x

A x =−=∂∂∫ 21

• Then

Axdydz dxdydz x

Ax s∫∫ ∫∫∫ =∂

∂

∫∫ = x s Axds -------------------------- (2)• lll ly the following terms

Aydsydxdydz y

Ay sv ∫∫ ∫∫∫ =∂

∂------------- (3)




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)4(−−−−=∂

∂ ∫∫ ∫∫∫ Azdsz dxdydz z Az

sv

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

)dxdydz z

A y

A

x A

Adv z y xvv ∂∂+∂

∂+

∂∂=∇ ∫∫∫ ∫∫∫ .

= ( ) Azdsz Aydsy Axdsx s ++∫∫

ds A Adv 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 contour

bounding 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.

)( N xH s

sdl H ∆=∆

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




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

)( N a xH s sdl H

∆=∆∆∫ .

or

∫ ∆∇=∆ sa xH sdl H N )(.

s xH ∆∇= ).(

Where a N 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

ds xH dl H ..

Problems:

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

SOLUTION:

Difference volume in spherical coordinates is




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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




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1

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

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.




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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 q 1 and q 2.

Ø Force of interaction between charges ,q 1 and q 2 may be formulated as follows:

)1(][ 1212

2112 −−−−= U r

qqk F

Where

v F12=force exerted by q 1 on q 2v U12=Unit vector directed from q 1 to q 2.v R 12 =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)

910361

Χπ0 = permittivity of free space (or) Dielectric constant

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

122120

2112 ]

4[ U

r

qq F

πε=

F12q2

r 12

U12

q1

Force between charges q 1 and q 2




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3

1.1.LINEAR SUPERPOSITION:

• If a system consists of n point-charges, namely q 1 ,q2 ……q n ,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 linear superposition.

• The force F i on the ith

charge can be written by the mathematical expression given below.

ji

n

j j

ji

jin

j j jii U

r

qq F F ∑∑ ≠=≠= == 11 2

011

4πεWhere

q j is any of the charges other than q i, and

r ji ,the distance between the jth i th charges.

• For a homogeneous medium, the permittivity is

r oεεε =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

r 12 = r 2 - r 1

= (2-1) a x+ (0-2)a y+(5-3)a z

=a x – 2a y +2a z

F2

r 2

Q2a12

Q1

r 12

r 1 O




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4

22212 2)2(1 +−+=r = 3

a12 =12

12

r r

a12 = a x-2a y+2a z

3

F12 = N aaa

x x

x z y x )3

22(

910)36

1(4

)10(1039

44 +−−−

−−

ππ

F = -10 a x+20 a y-20a z

PROBLEM NO: 2

• A 2 mC positive charge is located in vacuum at P 1 (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 P 1 .

TO FIND :

• Vector force on the negative charge.• Magnitude of the charge at P 1 .

SOLUTION:

Q1=2x 10-6

Q2 = 5x10-6

P1 = (3,-2,-4)

P2 =(1,-4, 2)

r 12 = r 2 - r 1

=-2a x – 2a y +6a z

4412 =r

a12 =12

12

r r

a12 = -2a x-2a y+6a z

44

r 2

p2a12

p1

r 12

r 1O




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5

F12 = N aaa

x x x x x z y x )

44

622(

4410854.84)1052

12

12 +−−−

−

π F = 6.16ax+6.16ay-18.48az

5

1004.2 −

= x F PROBLEM NO: 3

• A charge Q 1 = -20 C is located at P (-6, 4, 6) and a charge Q 2 = 50 C is located at R(5, 8, -2) in a Free space. Find the force exerted on Q 2 by Q 1 in vector form. The distances aregiven in metres.

P=-6a x+4a y+6a z

R=5a x+8a y-2a z

122120

212

4a

R

QQ F

πε=

R 12 =R PR =R-P

= ( ) ( ) ( )( )[ ] z y x aaa 624865 −−+−+−−

=11a x+4a y-8a z

( )22212 8411 −++= R

=14.1774

a12 =12

12

R R

=14.1774

8az-4ay11ax +

a12 =0.7758a x+0.2821a y-0.5642a z

12212

66

2 )1774.14(10854.8410501020

a x x x

x x x F −

−−−= π

= -0.0447(.7758a x+0.2821a y+0.5642a z)

=-0.0346a x-0.01261a y+0.02522a z

2222 )02522.0()01261.0()0346.0( −++= F

= 44.634 Nm.




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6

PROBLEM NO: 4• Calculate E at M (3, -4,2) in free space caused by (a) a charge Q 1=2 C at P 1 (0,0,0)(b) a

charge Q 2 =3 C at P 2 (-1,2,3)(c) a charge Q 1 =2 C at P 1 (0,0,0) and a charge Q 2 =3 C atP2 (-1, 2, 3).

Q1=2 C

Q2=3 C

P1= (0, 0, 0)

P2 = (-1, 2, 3)

( ) 385.5243

385.510854.84

102212

6

1 z y x aaa x

x x

x E

+−= −

−

π

=345.33a x-460.44a y+230.22a z

( ) 280.764

280.710854.84

103212

6

2 z y x aaa

x x

x E

−−= −

−

π

E=E 1+E2

=624.85a x -879.72a y+160.34 a z

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

Q F

E =

• 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 Q t.The force on it is given by coulomb’ s law,

t

o

t t a

r

QQ F 12

12

1

4πε=

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

( )14

1212

1 →= t o

t

t

t ar

QQQ F

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

force per unit charge.

( )2→=t

t

Q F

E

( )34

1210

1 →= t t

ar

Q E

πε




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7

( )44 20

→= Ra R

Q E

πε• r is the magnitude of the vector r , the directed line segment from the point at which the point

charge Q is located to the point at which E is desired and a R 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 a R then becomes the radialvector a r and R is r.

• Hence

( )54 20

1 →= r ar

Q E

πε or

20

1

4 r

Q E r πε

=

CARTESIAN CO-ORDINATES

Q at the origin,

az ayax z y xr R ++==

222

)(

z y x

z y xa az ayaxr

++++=

( )6)(4 2222222222220

→

++

+

++

+

++++

= z y x a z y x

z a

z y x

ya

z y x

x z y x

Q E

πε

Q located at the same point

az z ay yax xr '''' ++= zaz yay xaxr ++=

R as r-r’, and then

'

'

2'04 r r

r r

r r

Q E

−−

−=

πε

( )3

'0

'

4 r r

r r Q

−

−=πε




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8

PROBLEM NO: 5

• Find E at P (1,1,1) caused by four identical 3 n C point charges located at P 1 ( 1,1,0 ) , P 2

(-1,1,0),P 3 (-1,-1,0), and P 4 (1,-1,0).

r = a x+ay+ a z

r1 = a x+ay

r2 = -a x+a y

r3 = -a x-ay

r4 =a x-ay

r – r 1 = a

r – r 2 =2a x+a z

r – r 3 =2a x+2a y+az

r- r 4 =2a y+az

mV x

xQ

o

−== −−

96.26)10854.8(4

1034 12

9

ππε

E = 26.96( ) ( ) ( )

++++++ 3335

2

9

22

5

21

z y z y x z x z aaaaaaaa

=a z +2a x+2a y

E = 6.82 a x+6.82a y+32.8a z

y

P 3 (-1,-1,0)

P

O

.P 2(-1,1,0)

P 4 ( 1,-1,0)

Z

X P 4 ( 1,-1,0)




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9

PROBLEM NO: 6

v A charge Q 1 = -20 C is located at P (-6, 4, 6) and a charge Q 2 = 50 C is located at R (5, 8,-2) in a free space. Find the force exerted on Q 2 by Q 1 in vector form. The distances aregiven in metres.

P=-6a x+4a y+6a z

R=5a x+8a y-2a z

122120

212

4a

R

QQ F

πε=

R 12 =R PR =R-P

= ( ) ( ) ( )( )[ ] z y x aaa 624865 −−+−+−−

=11a x+4a y-8a z

( )22212 8411 −++= R

=14.1774

a12=12

12

R R

=14.1774

8az-4ay11ax +

a12 =0.7758a x+0.2821a y-0.5642a z

12212

66

2 )1774.14(10854.8410501020

a x x x

x x x F −

−−−= π = -0.0447(.7758a x+0.2821a y+0.5642a z)

=-0.0346a x-0.01261a y+0.02522a z

2222 )02522.0()01261.0()0346.0( −++= F

= 44.634 Nm.




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10

PROBLEM NO: 7

Calculate E at M(3,-4,2)in free space caused by (a) a charge Q 1=2 C at P 1(0,0,0)(b) a charge Q 2

=3 C at P 2 (-1,2,3)(c)a charge Q 1 =2 C at P 1(0,0,0) and a charge Q 2 =3 C at P 2 (-1, 2, 3).

Q1=2 C

Q2=3 C

P1= (0, 0, 0)

P2 = (-1, 2, 3)

( ) 385.5243

385.510854.84

102212

6

1 z y x aaa x

x x

x E

+−= −

−

π

=345.33a x-460.44a y+230.22a z

( ) 280.764

280.710854.84

103212

6

2 z y x aaa

x x

x E

−−= −−

π

E=E 1+E2

=624.85a x -879.72a y+160.34 a z

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.




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= 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, x L 3=ρ or 24 y L =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.

l d dQQ L L

L∫ ∫ == ρ• If the line of length L is a closed path, then integral is called closed contour integral and denoted

as Q= ∫

L Ldl ρ

• 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 mC x µ Find the total charge over the length if 10m.Sol :

mC x L /32µρ = And L=10m along x-axis.

The differential length is dl=dx in x direction and corresponding charge isdxdl dQ L L ρρ ==

Q=

10

0

10

0

22

33

3∫ ∫ == L

L x

dx xdl ρ

=1000 mC C 1=µSURFACE CHARGE:

•If the charge is distributed uniformly over a two-dimensional surface then it is called a

surface charge or a sheet of charge.

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

=S ρ Total charge in coulomb (C/m 2) Total area in square meters




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• The volume charge density isvρ , having the unit of coulombs per cubic meter(C/m 3).

• 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 ∆∆=

→∆ 0limρ

• 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 and

hence a triple integration volume.

PROBLEM NO: 9

• Find the charge in the volume defined by m x 10 ≤≤ , 0 m y 1≤≤ ,andm z 10 ≤≤ if P=30 x 2y(µC/m 3 ).What charge occurs for the limits01 ≤≤− y m?

SincedQ= ρv dv

Q ydxdydz x∫ ∫ ∫ =1

0

1

0

1

0

230

= 5 µC 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.

Ra

R

Q D 2

4π=

=

−+−

26

43)26(4

1030 9 az ayax xπ

= ( ) 211 /26

431018.9 mC

az ayax x

−+−

D =91.8 pC/m 2




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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=E 1+E2+E 3

E1 = 20

1

4 r q

πε

1 E =9X109

( )29

2.

104 X

1 E = 900 v/M = 3 E E2 is along the diagonal

1 E =9X109

( )29

2.2

104 X

= 450V/m (along the diagonal)E1+E 2 = vector directed along the diagonal, in the same direction as E 2.

121 2 E E E =+ =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 Z 2 e-0.1x sin πy C/m 3. The volume is defined between -2 ,2≤≤ x 0 1≤≤ y and .43 ≤≤ z

ρv = 10 Z 2 e-0.1x sin π y C/m 3Consider differential volume in Cartesian system as,

dv=dxdydzdvdQ vρ=∴

= 10 Z 2 e-0.1x sin π y dx dy dz

` dvQvol

v∫ = ρ

But now it becomes triple integration.

∫ ∫ ∫ = = −=

=∴4

3

1

0

2

2 z y x

Q 10 Z 2 e-0.1x sin π y dx dy dz

∫ ∫ = =

=∴4

3

1

0 z y

Q 10 Z 2 e-0.1x sin π y2

2

1.0

1.0 −

−

− xe

dy dz




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∫ =

=∴4

3 z

Q1

0

cosππ y

2

2

2.02.0

1.01.0 −

−

−−− x x ee

dz

=10 0267.40coscos

3

4

3

3 −−−

ππ

π z

=10 0267.411

334

4

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 ,br a ≤≤D= r a

r Q

24 ∏• Inner sphere becomes smaller and smaller, still retaining a charge of Q, it becomes a

point charge in the limit, but the electric flux density at a point r meters from the pointcharge is given by

D= r ar

Q24 ∏

• Q lines of flux are symmetrically directed outwards from the point and pass through animaginary spherical surface area of 4 ∏r 2.

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

r ar

Q E 2

04 ε∏=

• In free space,D= E 0ε (free space only)

• Volume charge distribution in free space

r=ar=b

r=a




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E= ∫ ∏vol RV a

RdV

204 ε

ρ

D= ∫ ∏vol r V a

RdV

24ρ

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

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

Ø The closed surface may be the surface of some real material.

Ø If the total charge is Q 1 then Q coulombs of electric flux will pass through the enclosingsurface.

Ø At every point on the surface the electric flux density vector D, will have some value D s .

Ø Where subscript S merely reminds us that D must be evaluated at the surface, and D s 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.

Ø At any point P consider an incremental element of surface ∆s and let D s make an anglewith ∆s.

Ø The flux crossing ∆s is then the product of the normal component of D s and ∆s.

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

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

P




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Ψ = ds Dd s∫ ∫ =ψ Ø The 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) r 2sin ,theintegral is a double integral.

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

.

Ø Differential ds is the signal for a surface integral.

Ø A 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 ds D .ψ =Charge enclosed =Q

Ø The 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)

or a volume charge distribution

dvQvol

v∫ = ρdvds D

vol v

s s ∫ ∫ = ρ.

Ø Surface is equal to the charge enclosed

SPHERICAL CO-ORDINATE SYSTEM:

r ar

Q E 2

04πε=

E D 0ε=

r ar Q

D 24π=




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Ø At the surface of the sphere

r s ar Q

D 24π=

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

φθθ d d r ds sin2= = a 2 sin d d

ds = a 2 sin d d a r Ø The integrant is

sin4

. 22 aaQ

ds D s π= d d a r .a r

=π4

Q sin d d

Ø Loading to the closed surface integral

∫ ∫ ==

=

=

πθ

θ

πφ

φ π02

0 4Q sin d d

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

Ø Integrating gives

= φθπππ

d Q

0

2

0

)(cos4∫

= φππ

d Q∫

2

0 2

=QØ A 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

dL E QW .

• 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 called potential difference between the points B to A is denoted by V.




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Potential ∫ −== A

B

dL E V Difference .

• 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 system producing 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

r ar Q

E 204πε

=

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

• The difference length in spherical system is,

rd dradL r += θaφ +r sin θdφaφ• Hence the potential difference V AB between points A and B is given by,

∫ −= A

B AB dL E V . But B=r B and A= r A

∫ −= Ar

Br AB ar

QV ).(

4( 2

0πε rd dra r + θaφ +r sin θdφaφ)

∫ −= Ar

Br AB ar

QV )

4( 2

0πεdr

Q A

r A

B

r B




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A Ar

B

r

B AB

r Qdr r

QV ∫ −−=−=

−−

144

1

0

2

0 πεπε−−−−=−−=

B A

r

B AB r r

Qr

QV

A 114

14 00 πεπε

V r r

Qr r

QV

B A B A AB −=+−−=

114

114 00 πεπε

• When r B > r A , A B r r 11 < and V AB 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 of electrons.

• 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 be point 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.




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• This process is called polarization of dielectrics.

• There are two types of dielectrics

v Non -polar v 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 the boundary 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 the boundary between conductor and a dielectric.v For boundary conditions, the Maxwell’s equations for electrostatics are required.

∫ =0.dl E And ∫ = Qds D.




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

N E E E += 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 10 6

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 conductor v 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 ( E tan).• The components normal to the surface (E N).

v It is known that,

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

E D 0ε= for free space0tan0tan == E D ε

v 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 boundary

between the conductor and free space.




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7.3. BOUNDARY CONDITIONS BETWEEN CONDUCTOR AND DIELECTRIC:

v The free space is a dielectric with .0εε = Thus if the boundary is betweenconductor and dielectric with 0εε = r ε

E tan = D tan =0

D N =ρs

E N =r

s sεε

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.

v D 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:

∂

∂

∂

∂+

∂

∂

∂

∂+

∂

∂

∂

∂=∇∇ z

V

x y

V

x x

V

xV .

=2

2

2

2

2

2

z V

yV

xV

∂∂+∂

∂+∂∂




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Ø It is given by

C=V Q

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

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

D= 2/ mC AQ

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

D=εE

E AQ ε=

Q=A εEØ But electric field is given by

E=V/d V/m

Ø Substituting the value of E in the above equation

Q=A εd V

d A

V Q ε=

Ø Capacitance is given by

C=V Q

C=d

Aε

C=d

A r εε0 farad

Ø Where 0εε = r ε9.1.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 r r and εεrespectively.

V V

∈ ∈r2

d d-

d




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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 r r A

Q D =

E2=021 εεε r r A

Q D =

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

V=

−+

210

11

r r

d d d A

Qεεε

21

0

11

r r

d d d A

V Q

εε

ε−+

=

21

0

11

r r

d d d A

C

εε

ε−+

=

Ø The capacitance

C=12

210

)1(1 r r r r

d d d A

εεεεε−+

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

C=)1(1

0

d d d A

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 Let3,21

, r r 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




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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 r r A

Q Dεεεε =

E2=20 r A

Qεε

E3=30 r A

Qεε

Then

V= 321302010

d A

Qd

A

Qd

A

Q

r r r εεεεεε ++

V=

++

3210

321

r r r

d d d A

Qεεεε

321

0

321

r r r

d d d A

V Q

εεε

ε++

=

Ø The capacitance is given by

321

0

321

r r r

d d d A

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 r dr

Y peQ

Qr




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V=r

Qε∏4

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

Where

r εεε 0=Ø If the medium is air, then C becomesC=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 r ε be the permittivity 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 byE= 24 r

Q

ε∏(a )br ≤≤

Ø The potential difference between the spheres is

V= - ∫ ∏a

b

dr r

Q.

4 2ε

= - ∫ ∏a

b r dr Q

24 ε

=

a

br

Q

∏1

4 ε =

−∏ ba

Q 114 ε

=

−

∏ ababQ

ε4Ø The capacitance of 2 concentric spheres is

C=

−∏= abab

V Q ε4

C=4

−∏

ab

abε

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 r ε .




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

Ø The two cylinders are charged at the rate of l ρ C/m. It is assumed that inner cylinder hascharge of mC l /ρ and outer cylinder has charge of - mC l /ρ . By applying Gauss’ s law, theelectric field at any distance r from the axis of cylinder is given by

r E l

ερ∏= 2Ø The potential difference between two co-axial cables is

V= - ∫ a

b

Edr

= - ∫ ∏a

br

l

r dr

ερ

2

= -

∏ abl ln

2 ερ

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

C=( )

m F abV

l //ln

2 ερ ∏=

C=( )

m F ab

r //ln

2 0εε∏

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

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

Ø If the conductor A has charge of mC l /ρ along its length, this will induce charge of -mC l /ρ on the conductor B

Ø The 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 r d r

l l

−∏+∏ ερ

ερ

=

−+∏ r d r l 11

2 ερ

Ø The potential difference between the conductors is given by

b

a ρ




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V= - ∫ Edr = - dr

r d r

a

ad

l ∫ −

−+∏11

2 ερ

= -

−+−∏ ad a

ad al

lnln2 ερ

= -

−∏ ad al ln2

2 ερ

V=

−

∏ aad l ln

ερ

Ø The capacitance per unit length between two parallel conductors

C= m F

aad V

l /ln

−∏= ερ

If d>>a C= m F

ad

/ln

ε∏

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

C= m F

ad

/ln

0

∏ε

9.7.Serial and parallel combination of capacitors:

Ø Two capacitors are connected in series as shown figØ If potential difference V is applied across the two capacitors, V1 and V2

are 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

QC Q +

Ø The equivalent capacitance C is given by

21 C

Q

C

Q

C

Q

+=

21

111

C C C +=

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

CnC C C C 1

.........3

12

11

11 +++=




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Ø But Q=CV

W= 221

CV or

W= QV 21

joules

10.1.ENERGY DENSITY:

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

+ - + -

+| - + -

+ -

Ø The capacitance of elemental capacitance is

d A

C ∆=∆

ε

=( )

d d

∆∆ 2ε

= d ∆εØ Energy stored in the elemental capacitor is

C W ∆=∆21 ( )2V ∆

Ø But potential difference across the elementary cube is

d E V ∆=∆Ø Where E is the electric field exist in the cube.

d C ∆=∆ εØ The stored energy

21=∆W ( )( )2.. d E d ∆∆ε

= ( )3221

d E ∆ε

= V ∆Ε 221 ε

QQ

V

C

∆d




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

ε21=∆

∆V W

E2 ( Ε = ε DQ )

E DV

W .

2

1=∆∆

Joules/m3

11. DIELECTRIC STRENGTH:

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

Ø As 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 of large number of free electrons.

Ø This condition of dielectric is called dielectric breakdown.

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

Ø The 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.

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

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

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

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




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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 Q 2=10µc at point (2, 0, 0) by a charge Q 1=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 10 µF capacitor which has been charged to a voltage Of 400V.28. Define Surface and Volume charge densities.

29. Fill up : s D.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?




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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 /r 2 find the electric flux density D at (2, /2,0).




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

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.




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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=Q V 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=




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JxBdvdF =

• JdV is the differential current element

IdL Kds Jdv ==

• 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

JxBdv F

∫ = s

KxBds F

• To a straight conductor in a uniform magnetic field

F= ∫ ∫ −= BxdL I IdLxB

ILxB F =• The magnitude of the force is given by the familiar equation

F=BILsin θ

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




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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 magnetic

field intensity is inversely proportional to the square of thedistance from the differential element to the point P.

24 R IdLxa

dH Rπ

=

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

dL

aR 12

R 12 P

I1

Point 1

Point 2




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

2

12

112

4

12

R

xadL I dH R

π

=

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

ar

IdLdB 24π

µ=

• 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

IdLdB

πθµ=

• The magnetic field intensity is given by

212

112

4

sin

R

dL I dH

πθ=

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




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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.

∫ = I dL H .

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 can

be 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 a φ direction, tangentialto circular path at point P.

x

Y

aφ

I

I

z




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φφard dL =

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

φπ a

r I

H 2

=

=dL H . φφ φπ ard a

r I

.2

= φπ

rd r

I 2

= φπ

d I

2

• Integrating H.dL over the entire closed path,

φπ

π

φd

I dL H ∫ ∫

=

=2

0 2.

[ ] πφπ

202

I =

= ππ22 I

=I=current carried by conductor.

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




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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 sin4 r dL I θ

πµ

• From ∆ABC

θsin= AB AC

θsindL AC =• But arc

AC =rd θ

θθ rd dL =sin

θd =r

dL θsin

B= ∫ r

d I θ

π

µ

4

0

• Substitute the value of r in above equation

B= ∫ π

θθπ

µ0

0 .sin4

d d I

B= πθπ

µ0

0 )cos(4

−d I

B= 2.4

0

d I

πµ

• The magnetic flux density due to infinite conductorB= 20 /

2mwb

d I

πµ

• The magnetic field intensity due to infinite conductor

mad

I H /

2π=




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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/m 2).

• Which is also called Tesla(T).

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

B= H 0µ free space only Wb/m 2 (or) tesla (or) Gauss.

6.1 MAGNETIC FLUX:

• The magnetic flux through a surface area is the normalcomponent of magnetic field times µover 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




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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 of zero.

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

• This material as one in which the permanent magneticmoment m 0 of each atom is zero. Such a material istermed diamagnetic.

ANTIFERROMAGNETIC:

• In anti-ferromagnetic materials the magnetic moments of adjacent 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 of adjacent 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.




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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

ds B

• We find that

021 =∆−∆ S BS B N N

21 N N B B =

212

1 N N H H =µ

µ

• The normal component of B is continuous, but the normalcomponent of H is discontinuous by the ratio

2

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

m N m N M H M µχ

µχµµχ ==

• Next, Ampere’ s circuital law

∫ = I dL H .

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

• Taking trip around the path, we find that




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L K L H L H t t ∆=∆−∆ 21

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

K H H t t =− 21

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

K xa H H N =− 1221 )(

• Where a N12 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:

K H H t t =− 21 x 12 N a

• For tangential B, we have

K B B t t =−

2

2

1

1

µµ

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

K M M mt m

mmt 21

1

22 χχ

χ −=

• 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 if neither 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.




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∫ = I dL H .

Ø If no current is enclosed i.e.

J=0

∫ =0 Hdl

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

∫ −= dl H V m .

Ø This scalar potential also satisfies lapalace equation

Ø In free space

but

H

B

0.

0.

0 =∆

=∆

µ

mV H −∇=

0).(0 =−∇∇ Vmµ

020 =∇− Vmµ

02 =∇ Vm




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VECTOR MAGNETIC POTENTIAL:

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

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

xA B ∇=

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

xA x xB ∇∇=∇

Ø By the identity

A A xA x 2).( ∇−∇∇=∇∇

But J xB µ=∇

J A A µ=∇−∇∇ 2).(

For the steady dc

0).( =∇ A

Then 2∇− A J µ=

)(222 zJz yJy xJx Az z Ay y Ax x ++−=∇+∇+∇ µ

Ø Equating

Jx Ax µ−=∇2

Jy Ay µ−=∇2

Jz Az µ−=∇2




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

dvr Jx

Axv

)(4 ∫ = πµ

dvr

Jy Ay

v

)(4 ∫ = πµ

dvr

Jz Az

v

)(4 ∫ = πµ

Ø The general magnetic vector potential can be expressedas

dvr J

A ∫∫∫ = πµ

4




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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

F F t

θFtF

Axis of rotation

b




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

θcos F F ts =

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

Ø The magnetic moment of loop is IA

m =IA

T=mBcos θ

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,

e m =NI= ∫ dL H .




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• The SI unit of m.m.f. is ampere(A)

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

R=V/I

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




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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 of density 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 the plane 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.




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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 .




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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 =dt d

N 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

d φ

• 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.




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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,

=∫ dL E . )4(. −−−−∂∂−∫

s

dS t

B

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

transformer e.m.f.

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

)5(.).( −−−−∂∂−=∇ ∫ ∫ dS t B

dS xE S S

v Assuming that both the surface integral taken over identicalsurfaces.

dS t

Bds xE S

.).( ∫ ∂∂−=∇v Hence finally,

)6(−−−∂∂−=∇

t B

xE

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

v If B is not varying with time, then equations (4) and (6

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