35675966 EMT Electromagnetic Theory MODULE I (1)

7/29/2019 35675966 EMT Electromagnetic Theory MODULE I (1)

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MODULE I

VECTOR ALGEBRA, VECTOR CALCULUS, COORDINATE SYSTEMS, VECTOR FIELDS


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Syllabus Module IVector analysis: Vector algebra, Coordinate systems and transformations-Cartesian, cylindrical and spherical coordinates. Constant coordinate surfaces. Vector calculus-Differential length, area and volume. Line, surface and volume integrals. Del operator. Gradient of a scalar, Divergence of a vector, Divergence theorem, Curl of a vector. Stocks theorem, Laplacian of a scalar. Classification of vector fields.

Compiled by: MKP for CEC S5 EC - July 2008


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References1. 2. Text Books: Mathew N.O. Sadiku, Elements of Electromagnetics, Oxford University Press Jordan and Balmain, Electromagnetic waves and radiating systems, Pearson Education PHI Ltd. References: Kraus Fleisch, Electromagnetics with applications, McGraw Hill William.H.Hayt, Engineering Electromagnetics, Tata McGraw Hill N.Narayana Rao, Elements of Engineering Electromagnetics, Pearson Education PHI Ltd. D.Ganesh Rao, Engineering Electromagnetics, Sanguine Technical Publishers. Joseph.A.Edminister, Electromagnetics, Schaum series-McGraw Hill

1. 2. 3. 4. 5.



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Scalars and vectorsA scalar is a quantity that has only magnitude.Time Distance Temperature Speed

A vector is a quantity that has both magnitude and direction.Force Displacement Velocity



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Unit vectorA vector

A has both magnitude and direction.

Magnitude of A = A = AA unit vector along A is defined as a vector whose magnitude is unity and whosedirection is along vector A . It is denoted by a A

aA

A A = = A A

A = AaAVector A is completely specified in terms of its magnitude A and direction a A



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Vectors represented in rectangular coordinate systemsAny vector in space can be uniquely expressed in terms of x, y and z coordinatesusing a rectangular coordinate system.Z

Az A = Ax a x + Ay a y + Az a z

az ax AxX

AAy ayY



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Vectors represented in rectangular coordinate systems A = Ax a x + Ay a y + Az a zAx , Ay , Az Components of A in the direction of x, y , z a x , a y , a z Unit vectors specifying the direction of x, y , z axes



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Position vector of a point in spaceA point P in Cartesian coordinate system may be expressed as its x,y,z coordinates. The position vector of a point P is the directed distance from the origin Oto the point P. A point P (3,4,5) has the position vector rp = 3a x + 4a y + 5azZ

AzP

rp = OP = Ax a x + Ay a y + Az a zAy

az ax AxX

Y

ay



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Vector addition and subtraction If A = Ax a x + Ay a y + Az a z and B = B x a x + B y a y + Bz a z

C = A + B = ( Ax + Bx ) a x + ( Ay + B y ) a y + ( Az + Bz ) a z D = A B Bx ) a x + ( Ay B y ) a y + ( Az Bz ) a z

Compiled by: MKP for CEC S5 EC

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

rAB = ( Bx Ax ) a x + ( B y Ay ) a y + ( Bz Az ) a z


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Unit vector in the direction of given vectorLet

A

be a vector in space given by

A = Ax a x + Ay a y + Az a z

A unit vector in the direction of

Ais given by

aA =

Ax a x + Ay a y + Az a z Ax 2 + Ay 2 + Az 2


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Example 1 If A = 10a x 4a y + 6a z and B = 2a x + a y find (i ) Component of A along a y (ii ) Magnitude of 3A B (iii ) A unit vector alongA + 2 B

Answer : (ii ) 3A B = ( 30a x 12a y + 18a z ) ( 2a x + a y ) = 28a x 13a3A B = 282 + 132 + 182 = 35.74Compiled by: MKP for CEC S5 EC

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(i )

4


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Example 1 (iii ) A + 2 B = 14a x 2a y + 6a z A unit vector c along A + 2 B = = 14a x 2a y + 6a z 142 + 2 2 + 62 A + 2B A +

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


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Vector multiplication

dot productScalar product or dot product: It is defined as the product of magnitudes of thetwo vectors and the cosine of the angle between them.

AB is the smaller angle between themProperties:

A B = AB cos AB

(i ) Commutative Property: A B = B A

(ii ) When two vectors are perpendicular the angle between them is =90 cos90 = 0A B = AB cos90 = 0

If the dot product of two vectors are zero, they are perpendicular.Compiled by: MKP for CEC S5 EC

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dot product (iii ) Since a x , a y , a z are mutually perpendicular ax a y = a = 0(iv ) When two vectors are parallel the angle between them is either 0 or 180A B = AB cos0 = AB or A B = AB cos180 = AB

( v ) The s

uare of a vector is the s

uare of its magnitude.A A = AA cos0 = A2 A2 = A2Compiled by: MKP for CEC S5 EC

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dot product( vi ) Scalar product is e

ual to the sum of products of their corresponding components.

If A = Ax a x + Ay a y + Az a z and B = B x a x + B y a y + Bz a z x + Ay a y + Az a z ) ( Bx a x + B y a y + Bz a z )

= Ax Bx + Ay B y + Az Bz


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Vector Product or cross productVector Product or cross product: Vector product of two vectors A and B is denoted as A

B and is defined as

A

B = A B sin AB an Where an is a unit vector perpendicular to A and B such that A, B and an forms aright handed system. Geometrically the cross product can be defined as a vectorwhose magnitude is e

ual to the area of the parallelogram formed by A and B andwhose direction is in the direction of advance of a right handed screw as A isturned in to B through the smaller angle.


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Vector Product or cross productA

B

AB sin

B

an

A

AB sin


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Vector Product or cross productProperties:

(i ) Anti commutative : A

B = B

A(ii ) Distributive : A

B + C = A

B + A

C (iii ) Not Associative : A

B

C A

v ) Vector product of two parallel vectors is zero. A

B = A B sin AB an = A B sin 0an = 0 ( v ) A

A = A A sin 0an = 0Compiled by: MKP for CEC S5 EC

July 2008

(

() (

) (

) (

)

)


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( vi ) a x

a x = a y

a y = a z

a z = 0 ( vii ) a x

a y = 1.1.sin 90 a z = a z and

a y

az = ax az

ax = a y a y

a x = a z az

a y = a x a x

a z = a yCompiled by: MKP for CEC S5 EC

July 2008


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ay

ax

az

a y

ax az

az

az

ay ax a x

ay


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Vector Product or cross product ( viii ) If A = Ax a x + Ay a y + Az a z and B = B x a x + B y a y + Bz a z

ax A

B = Ax Bx

ay Ay By

az Az Bz


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Projection of a vector on another vectorScalar component of A along B is called projection of A on B and is given by

AB = A cos AB

= A aB cos AB = A aBA AB

aB

A cos AB

BCompiled by: MKP for CEC S5 EC

July 2008


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Projection of a vector on another vectorVector component of A along B is the scalar component multiplied by a unit vector along B

AB = AB aB = A aB aB

(

)


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Scalar triple productGiven three vectors A, B and C the scalar triple product is defined as

A B

C = B C

A = C A

Band is represented as A B C

(

)

(

)

(

)

Geometrically the scalar triple product is e

ual to the volume of a parallelepiped having A, B and C as sides Properties:

(i )

A B C = B C A = C A B i.e. A B

C = B C

A = C A

B

(

)

(

)

(

)


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

A B C = B A C

(iii ) If A = Ax a x + Ay a y + Az a z and B = B x a x + B y a y + Bz a z a x + C y a y + Cz azAx A B C = Bx Cx Ay By Cy Az Bz CzCompiled by: MKP for CEC S5 EC

July 2008


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Vector triple productFor any three vectors A, B, CA

B

C = B AC C A Bbac

cab rule

(

)

(

)(

)


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Cylindrical Coordinate SystemsAny point in space is considered to be at the intersection of three mutually perpendicular surfaces:A circular cylinder (=constant) A ve

tical plane (=constant) A ho

izontal plane (z=constant)

Any point in space is

ep

esented by th

ee coo

dinates P(,,z) denotes the

adius of an imagina

y cylinde

passing th

ough P, o

the

adial distance f

om z axis to the point P. denotes azimuthal angle, measu

ed f

om x axisto a ve

tical inte

secting plane passing th

ough P. z denotes distance f

om xy-plane to a ho

izontal inte

secting plane passing th

ough P. It is the same as in

ectangula

coo

dinate system.Compiled by: MKP fo

CEC S5 EC - July 2008


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Cylind

ical Coo

dinate SystemsZ z=constant =constant z P(,,z) Y

Ranges : 0 < 0 < 2 =constant

X

< z < Com

iled by: MKP for CEC S5 EC

July 2008


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Cylindrical Coordinate SystemsP(3,45,8) Z =3

z=8 P(3,45,8) Y =45

XCompiled by: MKP fo



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Cylind

ical Coo

dinate SystemsZ

a z az

a z a a

a

Y

XCompiled by: MKP fo



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Cylind

ical Coo

dinate SystemsA vecto

in cylind

ical coo

dinate system may be specified using th

ee mutuallype

pendicula

unit vecto

s a , a , a z fo

m a

ight handed system because an RH sc

ew when

otated f

om a to a moves towa

ds a z These unit vecto

s specify di

eons along , and z axes. Using these unit vecto

s any vecto

A may be exp

essed as

a , a , a z

A = A a + A a + Az a zThe magnitude of the vecto

is given by

A =A 2 + A 2 + A z 2Compiled by: MKP fo



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Cylind

ical Coo

dinate Systems

a a = a a = a z a z = 1 a a = a a z = a z Compiled by: MKP fo



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Relationship between cylind

ical and

ectangula

coo

dinate systemsz Z

= x2 + y2 = tan 1 z=z y x

P(,,z) o

(x,y,z)

z y x

x = cos y = sin

z=zY

x = cos

y = sin X

a z a aCompiled by: MKP fo



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ical and

ectangula

coo

dinate systemsY

a a

ax a

a cos

a

a sin X

a x = a cos a sin Com


July 2008


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Relationshi

between cylindrical and rectangular coordinate systemsY

a a

ay

a sin a a cos

X

a y = a cos + a sin Compiled by: MKP fo



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ical and

ectangula

coo

dinate systemsY

ay

a

ax

X

a = a x cos + a y sin Compiled by: MKP fo



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ical and

ectangula

coo

dinate systemsY

a a x

ay

a

ax

X

a = a y cos a x sin Com


July 2008


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Relationshi

between cylindrical and rectangular coordinate systems a x = a cos a sin

a y = a cos + a sin equations (2) az = az a = a x cos + a y sin a = a y cos a x sin az = az

equations (3) Compiled by: MKP fo



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T

ansfo

mation of vecto

s between cylind

ical and

ectangula

coo

dinate systemsSubstituting the equations (2) in the gene

al equation fo

a vecto

in

ectangula

coo

dinates,

A = Ax ( a cos a sin ) + Ay ( a cos + a sin ) + Az a z A Ax sin + Ay cos ) a + Az a zA = Ax cos + Ay sin


A = Ax sin + Ay cos

Az = Az equations (4) Compiled by: MKP fo



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T

ansfo

mation of vecto

s between cylind

ical and

ectangula

coo

dinate systemsSubstituting the equations (3) in the gene

al equation fo

a vecto

in cylind

ical coo

dinates,

A = A ( a x cos + a y sin ) + A ( a x sin + a y cos ) + Az a z x + ( A sin + A cos ) a y + Az a z

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

Ax = A cos A sin

Ay = A sin + A cos Az = Az

equations (5) Compiled by: MKP fo



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T

ansfo

mation of vecto

s between cylind

ical and

ectangula

coo

dinate systemsT

ansfo

mation of a vecto

exp

essed in

ectangula

coo

dinates (Ax.Ay,Az) to cylind

ical coo

dinates (A.A,Az) can be achieved using equations (4). The set of equations (4) can be exp

essed in mat

ix fo

m as

A cos sin 0 Ax A = sin cos 0 A y 0 1 Az


July 2008


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Transformation of vectors between cylindrical and rectangular coordinate systemsTransformation of a vector expressed in cylindrical coordinates (A.A,Az) to

ectangula

coo

dinates (Ax.Ay,Az) can be achieved using equations (5). The set of equations (5) can be exp

essed in mat

ix fo

m as

Ax cos sin 0 A A = sin cos 0 A y 0 1 Az


July 2008


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Example

2Convert the following points expressed in cylindrical coordinates to rectangularcoordinate system. (i) P(2,5/6,3) x = cos (ii) Q(4,4/3,

1)

y = sin

z=z(i ) x = cos = 2cos 5 = 1.732 6 5 y = sin = 2sin =1 6P(2, (ii ) P(4, 5 ,3) P( 1.732,1,3) 6 4 , 1) P( 2, 3.464, 1) 3Com


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Exam

le

1Convert the following

oint ex

ressed in rectangular coordinate system to cylindrical coordinates and sketch the location of the

oint. P(x=3,y=4,z=5) Z = x2 +y2 = x 2 + y 2 = 32 + 42 = 5 = tan 1 y 4 = tan 1 = 53.1 x 3=5

= tan 1 z=z

y x

P(3, 4,5) P(5, 53.1 ,5)

z=5 P(5,53.1,5) Y X =53.1Com


July 2008


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Exam

le

3Convert the vector

= 4a x 2a y 4a z located at A(2,3,5) in to cylindrical cdinates.

A cos sin 0 Ax A = sin cos 0 A y 0 1 Az

cos sin 0 4 = sin cos 0 2 0 1 4 0

A = 4cos 2sin At the

oint P(2,3,5)A = 4cos56.3 2sin 56.3

A = 4sin 2cos Az = 4 = tan 1 y 3 = tan 1 = 56.3 x 2

A = 4sin 56.3 2cos56.3

Az = 4

= ( 4cos56.3 2sin 56.3) a + ( 4sin 56.3 2cos56.3) a 4a z

= 0.556a 4.44a 4a zCompiled by: MKP

o

CEC S5 EC

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Example

4Exp

ess the vecto

coo

dinates.

xy 2 za x + x 2 yza y + xyz 2a z

in cylind

ical

x = cos

y = sin

A cos

sin 0 Ax A = sin

cos 0 A

y 0 1 Az 2 A cos sin 0 xy z A = sin cos 0 x 2 yz 0


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Example

4 (Contd)A = xy 2 z cos + x 2 yz sin A = xy 2 z sin + x 2 yz cos

Az = xyz 2Put x = cos y = sin

A = 3 z cos2 sin 2 + 3 z sin 2 cos2 = 2 3 z cos2 sin 2 A = 3 z 3 sin Az = 2 z 2 sin cos Compiled by: MKP

o

CEC S5 EC

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Sphe

ical coo

dinate systemZ Z

Y

Y

=constant X

=constant X

Compiled by: MKP

o

CEC S5 EC

July 2008


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Sphe

ical coo

dinate systemZ

P(

, , )

Y

Ranges : 0

< 0 X

0 < < 2

Com


July 2008


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S

herical coordinate systemZ

ar

ar a a

a aY

r

Ranges : 0 r < 0 X

0 < < 2Com


July 2008


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S

herical coordinate systemThree unit vectors of the s

herical coordinate system are shown in the figure. Unit vector ar lies along the radially outward direction to the s

herical surface. It lies on the cone =constant and the lane =constant The unit vector a is normato the conical surface and lies in =constant

lane and is tangential to the s

herical surface. Unit vector a is the same as in cylindrical coordinate system. Itis normal to =constant

lane and is tangential to both the cone and the s

here. The unit vectors are mutually

er

endicular and forms a right handed set. An RH screw when rotated from ar to a will move it towards a direction.

Com


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S

herical coordinate systemA vector

A in s herical coordinate system may be ex ressed as A = Ar ar + A a + A a a

, a , a a

e unit vecto

s along

, , di

ections

Magnitude o

the vecto

is given by

A =

A

2 + A 2 + A 2

The unit vecto

s

a

a

= a a = a a = 1 a

a = a a = a a

= 0

a

, a , a

a

e mutually o

thogonal. Thus

a

a = a a

a = a

a

a

= aCompiled by: MKP

o

CEC S5 EC

July 2008


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, , in te

ms o

x, y , z

= x2 + y2 + z2 = tan 1x2 + y2 z

T

ans

o

mation o

va

iablesZ

=

sin

x, y , z in te

ms o

, , x =

sin cos y =

sin sin z =

cos

= tan

1

y x

P(x,y,z) o

(,,z) o

(

,,) z

z =

cos y Y

x

x = cos

y = sin XCompiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

va

iables

= x2 + y2 + z2

x =

sin cos y =

sin sin

= tan 1

x2 + y2 z

z =

cos

= tan

1

y x

Compiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

vecto

s90

a x = a

sin cos + a cos cos a sin

Z =

sin

a y = a

sin sin + a cos sin + a cos a z = a

cos a sin

z

a

a

a

a

z =

cos

a

a

y

Y

az ay

xy = sin

x = cos

ax

X

Compiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

vecto

s The unit vecto

s a x , a y , a z a

e to be exp

essed in te

ms o

unit vectosphe

ical coo

dinates a

, a , a a x consists o

the p

ojections o

a

, a he x axis. In o

de

to

ind this p

ojection,

i

st

ind the p

ojection on the xyplane and then on to the

e

ui

ed unit vecto

s. a x = sin cos a

+ cos c a a y = sin sin a

+ cos sin a + cos a equation (1)

a z = cos a

sin a

Compiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

vecto

s90

a

= a x cos sin + a y sin sin + a z cos a = a x cos cos

Z =

sin

a = a x sin + a y cos

a

a

z

a

90

xy = sin

z =

cos

y

Y

az ax ay

x = cos

XCompiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

vecto

s90

a

= a x cos sin + a y sin sin + a z cos

Z =

sin

a = a x cos cos + a y sin cos a z sin a = a x sin + a y c

z

a

a

a

a

ax

z =

cos

a

y

Y

az ay

xy = sin

x = cos

ax

XCompiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

vecto

s The unit vecto

s a

, a , a a

e to be exp

essed in te

ms o

unit vecto

s gula

coo

dinates a x , a y , a z a

consists o

the p

ojections o

a x , a y ,a z on the

axis. In o

de

to

ind this p

ojection,

i

st

ind the p

ojection on the =constant plane and then on to the

e

ui

ed unit vecto

s. a

= sin cos+ sin sin a y + cos a z a = cos cos a x + cos sin a y sin a z equation (2)

a = sin a x + cos a y

Compiled by: MKP

o

CEC S5 EC

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T

ans

o

mation o

vecto

sSubstituting e

(1) in the gene

al e

uation o

a vecto

in

ectangula

coo

dinates,


A = Ax ( sin cos a

+ cos cos a sin a )

+ Ay ( sin sin a

+ cos sin a + cos a )

+ Az ( cos a

sin a ) A = ( Ax sin cos + Ay sin sin + Az cos ) a

+ ( Ax cos cos + Ay cos aCompiled by: MKP

o

CEC S5 EC

July 2008

+ ( Ax sin + Ay cos ) a


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T

ans

o

mation o

vecto

sA

= Ax sin cos + Ay sin sin + Az cos A = Ax cos cos + Ay cos sin n + Ay cos This can be

ep

esented in mat

ix

o

m as

Ar sin cos sin sin cos Ax A = cos cos cos sin sin A Compiled by: MKP for CEC S5 EC

July 2008


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Transformation of vectorsSubstituting e

(2) in the general e

uation of a vector in spherical coordinates,

A = Ar ar + A a + A a A = A

( sin cos a x + sin sin a y + cos cos a y ) + A ( cos cos a x + cos sin a y sin a z )

A = ( A

sin cos + A cos cos sin A ) a x

+ ( A

sin sin + A cos sin + A cos ) a y

+ ( A

cos A sin ) a z

Compiled by: MKP

o

CEC S5 EC

July 2008


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T

ans

o

mation o

vecto

sAx = A

sin cos + A cos cos sin A Ay = A

sin sin + A cos sin + in This can be

ep

esented in mat

ix

o

m as

Ax sin cos cos cos sin Ar A = sin sin cos sin cos A Compiled by: MKP for CEC S5 EC

July 2008


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Example 1 Given the point P(

2,6,3)and vector A = ya x + ( x + z )a y express P and A in Cartesian, cylindrical and spherical coordinates.

Solution :At point P( x =

2, y = 6, z = 3) = x 2 + y 2 = 4 + 36 = 6.32 6 y = tan 1 = tan 1 = 108.43 2 x

z= z=3

r = x 2 + y 2 + z 2 = 4 + 36 + 9 = 7 = tan

1

x2 + y2 40 1 = tan = 64.62 z 3Com


July 2008


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Exam

le 1 (Contd)P ( 2,6,3) = P (6.32,108.43,3) = P (7,64.62 ,108.43 ) A cos A = sin Az 0 sin cos 0 0 Ax 0 A cos A = sin Az 0

sin cos 0

0 y 0 x + z 1 0

But x = cos , y = sin . Substituting , A cos A = sin Az 0 sin cos 0 0 sin

A = { cos sin + ( cos + z ) sin }a

+ { sin 2 + ( cos + z ) cos } aCompiled by: MKP

o

CEC S5 EC

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Example 1 (Contd)At P, =6.32, =108.43 , z=3cos = 0.316 Substituting , sin = 0.9487

A = 0.9487a 6.008aIn the sphe

ical system, Ar sin cos sin sin cos Ax A = cos cos cos sin sin Compiled by: MKP for CEC S5 EC

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Example 1 (Contd) Ar sin cos sin sin cos y A = cos cos cos sin sin x +

But x = r sin cos , y =

sin sin , z =

cos . Substituting ,

sin sin Ar sin cos sin sin cos A = cos cos cos sin

A = {r sin 2 cos sin +

(sin cos + cos )sin sin } a

+{

sin cos sin cos +

(sin cos + cos )cos sin}a +{

sin sin2 +

(sin cos + cos )cos} a

Compiled by: MKP

o

CEC S5 EC

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Example 1 (Contd)At P,

=7, =108.43 , =64.62cos = 0.316cos = 0.4286

sin = 0.9487sin = 0.903

Substituting ,

A = 0.8571a

0.4066a 6.008a

Compiled by: MKP

o

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Example 2Exp

ess the vecto

10 B = a

+

c os a + a

in

Ca

tesian coo

dinates.

ind B ( 3,4,0)

Solution : Ax sin cos cos cos sin Ar A = sin sin cos sin cos A

Bx sin cos cos cos sin

10 / r B = sin sin cos sin cos

Compiled by: MKP for CEC S5 EC July 2008


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Example 2 (Contd) Bx sin cos cos cos sin 10 / r B = sin sin cos sin cos

10 Bx = sin cos +

sin2 cos sin

10 By = sin sin +

cos2 sin + cos

10 Bz = cos

cos sin

= x + y + z = tan2 2 2

1

x2 + y2 z

= tan 1

y x

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July 2008


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Example 2 (Contd)sin =

=

x2 + y2 x2 + y2 + z2

cos =

z =

z x2 + y2 + z2

sin =

y

=

y x2 + y2cos =

x

=

x x2 + y2

10 x 2 + y 2 x2 + y2 + z2 x z2 x y Bx = 2 + 2 x + y2 + z2 x2 + y2 x + y2 + z2 x2

+ y2 x2 + y2 10x xz 2 y = 2 + x + y2 + z2 ( x 2 + y 2 + z 2 )( x 2 + y 2 ) x 2+ y 210 x 2 + y 2 x2 + y2 + z2 y z2 y y By = 2 + 2 x + y2 + z2 x2 + y2 x + y2 + z2 x2+ y2 x2 + y2 10 y yz 2 x = 2 + + x + y2 + z2 x2 + y2 x 2 + y 2 + z 2 )( x 2 + y2 ) (Compiled by: MKP

o

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July 2008


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Example 2 (Contd)z x2 + y2 10 z Bz = 2 2 2 x +y +z x2 + y2 + z2

and B = Bx a x + B y a y + Bz a zAt ( 3,4,0) x = 3, y = 4, z = 030 4 B x = + 0 = 2 25 5 40 3 By = +0 = 1 25 5By = 0

Substituting , B = 2a x + a y

B = 2 a x + a y

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July 2008


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Constant coo

dinate su

acesI

we keep one o

the coo

dinate va

iables constant and allow the othe

two to va

y, constant coo

dinate su

aces a

e gene

ated in

ectangula

, cylind

ical andsphe

ical coo

dinate systems. In the Ca

tesian system, i

we keep x constant andallow y and z to va

y, an in

inite plane x=constant is gene

ated. Thus we can have in

inite planesX=constant Y=constant Z=constant

These su

aces a

e pe

pendicula

to x, y and z axes

espectively. Inte

section o

two planes is a line. x=constant, y=constant is the line RPQ pa

allel to z axis. Inte

section o

th

ee planes is a point. x=constant, y=constant, z=constant is the point P(x,y,z). Any point P may be de

ined as the inte

section o

th

ee o

thogonal planes.Compiled by: MKP

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July 2008


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Constant coo

dinate su

acesZ

x=constant

Y

z=constantX

y=constantCompiled by: MKP

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July 2008


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Constant coo

dinate su

acesZ

x=constant

Q P

Y R

z=constantX

y=constantCompiled by: MKP

o

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July 2008


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Constant coo

dinate su

acesO

thogonal su

aces in cylind

ical coo

dinate system can be gene

ated as=constant =constant z=constant

=constant is a ci

cula

cylinde

, =constant is a semi in

inite plane with its edgealong z axis, z=constant is an in

inite plane as in the

ectangula

system. Theinte

section o

two su

aces z=constant, =constant is the ci

cle QPR o

adius The inte

section o

su

aces z=constant, =constant is a semi in

inite line. The inte

section o

th

ee su

aces p

oduces a point. =constant, =constant, z=constant isthe point P(,,z)Compiled by: MKP

o

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July 2008


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Constant coo

dinate su

acesZ

=constant z=constant

p

Y

=constantX

Compiled by: MKP

o

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July 2008


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Constant coo

dinate su

acesZ

=constant

z=constantp

Y

X

=constantCompiled by: MKP

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CEC S5 EC

July 2008


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Constant coo

dinate su

acesO

thogonal su

aces in sphe

ical coo

dinate system can be gene

ated as

=constant =constant =constant

=constant is a sphe

e with its cent

e at the o

igin, =constant is a ci

cula

cone with z axis as its axis and o

igin at the ve

tex, =constant is a semi in

inite plane as in the cylind

ical system. The inte

section o

two su

aces

=constant, =constant is a semi ci

cle passing th

ough Q an P The inte

section o

th

ee su

aces p

oduces a point.

=constant, =constant, =constant is the point P(

,, ) Anpoint P may be de

ined as the inte

section o

these o

thogonal planesCompiled by: MKP

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July 2008


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Constant coo

dinate su

acesZ=constant

=constant

p

=constant

Compiled by: MKP

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July 2008


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Di

e

ential elements in

ectangula

coo

dinate systemsz ZAP

dy QS

B

dzR

D

C

dx

z

az axx

ayy

Y

igu

e(1)

XCompiled by: MKP

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July 2008


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Di

e

ential length, a

ea and volume in Ca

tesian coo

dinatesDi

e

ential displacement is given by

dl = dxa x + dya y + dza zDi

e

ential no

mal a

ea is given by

dS = dydza x = dxdza y Di

e

ential volume is given by

= dzdya z

dv = dxdydzdl and dSa

e vecto

s whe

e as dv is a scala

.

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July 2008


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Di

e

ential length, a

ea and volume in Ca

tesian coo

dinatesI

we move

om P to Q, dl = dya y I

we move

om Q to S, dl = dya + dza y z I

we move

om D to Q, dl = dxa + dya + dza x y z In gene

al the di

e

ential su

acea

ea is de

ined as dS = dsa n whe

e dS is the a

ea o

the su

ace element and an is a unit vecto

no

mal to the su

ace dS. The di

e

ent su

aces in

igu

e(1)is desc

ibed as

ABCD dS = dydza x PQRS dS = dydza x BCRQ dS = dydza y

ADSP dS = dydza y ABQP dS = dxdya z DCRS dS = dxdya zCompiled by: MKP

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July 2008


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Di

e

ential no

mal a

eas in

ectangula

coo

dinate systemsz Z

dy

dx dz dz

aydy

az

ax

dx

Y

XCompiled by: MKP

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July 2008


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Di

e

ential elements in cylind

ical coo

dinate systems

Y

XCompiled by: MKP

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July 2008


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Di

e

ential elements in cylind

ical coo

dinate systemsZ

A dS

D P

Bdz

R

Q d

CY

azX

a aCompiled by: MKP

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July 2008


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Di

e

ential no

mal a

eas in cylind

ical coo

dinate systemsDi

e


dl = d a + d a + dza zDi

e

ential no

mal a

ea is given by

dS = d dza = d dza Di

e


= dd az

dv = d d dz

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July 2008


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Di

e

ential no

mal a

eas in cylind

ical coo

dinate systemsz Z

ddz

dz

a

azd

a

d

dY

XCompiled by: MKP

o

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July 2008


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Di

e

ential elements in sphe

ical coo

dinate systemsZ

sin d

d

d

d

d

Y

XCompiled by: MKP

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July 2008


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Di

e

ential elements in sphe

ical coo

dinate systemsZ

d

d

d

sin d

d

Y

XCompiled by: MKP

o

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July 2008


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Di

e

ential no

mal a

eas in sphe

ical coo

dinate systemsZ

sin d

d

a

d

sin d

ad

a

dY

XCompiled by: MKP

o

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July 2008


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Di

e

ential no

mal a

eas in sphe

ical coo

dinate systemsDi

e


dl = d

a

+

d a +

sin d a Di

e

ential no

mal a

ea is given by

dS =

2 sin d d a

=

sin d

d a =

d

d a Di

e


dv =

2 sin d

d d

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July 2008


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Example 1

o

the object shown below calculate: (i) The distance BC (ii) The distance CD (iii) The su

ace a

ea ABCD (iv) The su

ace a

ea ABO (v) The su

ace a

ea AO

D D(5,0,10) (vi) The volume ABDC O

(0,0,10)

C(0,5,10)

B(0,5,0) O(0,0,0) A(5,0,0)

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Example 1Solution:The object has cylind

ical symmet

y hence it is convenient to solve the p

oblemusing cylind

ical coo

dinates. o

this

i

st we have to conve

t all the pointsto cylind

ical coo

dinates.

A(5,0,0) A(5,0 ,0) B (0,5,0) B 5, ,0 2 C (0,5,10) C 5, ,10 D ( 5,0 ,10 )

(0,0,10)

C 5, ,10 2

D (5,0,10) D ( 5,0 ,10 )

O(0,0,0)

B 5, ,0 2

A(5,0 ,0)

Com


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Exam

le 1(Contd.)(i ) Along BC , dl = dz; hence

(ii ) Along CD, dl = d , = 5. Hence CD = /20

BC = dl = dz = 100

10

d = 5 0 = 2.5 /2 10

/2

(iii )

or ABCD, dS = d dz; = 5. HenceA

ea ABCD = dS = A

ea ABO = dS = =0

z =0

d dz = 5

/2

0

d dz = 250 5

10

(iv ) or ABO, dS = d d ; z = 0. Hence

/2 =0 =0

5

d d =

/2

0

d d = 6.250

Com iled by: MKP for CEC S5 EC

July 2008


99/197

Exam

le 1(Contd.)( v )

or AO

D, dS = d dz; = 0. Hence A

ea AO

D = dS = 5

=0 z =0

10

d dz = 50

( vi )

o

volume ABDC

O, dv = d dzd . Hence Volume ABDC

O = dv = 5 0 5

=0 =0 /2

/2 10z =0

dzd d

= d 0

d dz = 62.5z =0

10

Com


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Exam

le 2The object given below is

art of a s

herical shell described as

3 r 5

60 90 45 60Calculate(i) The distance DH (ii) The distance

G (iii) The su

ace a

ea AEHD (iv) The su

ace a

ea ABDC (v) The volume o

the object

EAB

HG

DC

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Example 2Solution :(i ) Along DH dl =

sin d ;

= 3, = 90DH =

dl = 12

60 = 45

3 sin 90 d = 3 sin 90

60 45

d

= 3

= 0.785

(ii ) Along G

dl =

d ;

= 5, = 60 , 60 to90 90 G

=

dl = = 60

5d = 5

6

= 2 .6 1 7


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Exam

le 2 (Contd)(iii ) On AEHD dS = r 2 sin d d ;

= 3 60 to 90 , 45 to 60

Area AEHD = dS = = 960

60

= 45 = 6090

90

9 sin d d

= 45

d

= 60

sin d 90

= 9

12

[ cos ]60

= 1.178


July 2008


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Exam

le 2 (Contd)(iv ) On ABCD dS = rd dr ; = 45 60 to 90 , r 3 to 5

Area ABCD = dS = 90 60

5

r =35

5

90 = 60

d d

= d

d

3

r 4 = 4.186 = = 3 6 2 3

2


July 2008


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Exam

le 2 (Contd)( v ) On Volume ABCDE

GH dv = r 2 sin drd d

3 to 5, 60 to 90 , 45 to 60

Volume = dv = 5 2 3

5

r = 3 = 60 = 45

90

60

2 sin d

d d 60 45

=

d

sin d d 60

90

r 90 60 = [ cos ]60 [ ]45 3 3 49 = = 4.276 363Com


July 2008

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105/197

Line integralsLine integral is defined as any integral that is to be evaluated along a line. Aline indicates a

ath along a curve in s

ace. b a A d l

ep

esents a line integ

al whe

e each element o

length d l on the cu

ve is multiplied acco

ding to scala

dot p

oduct

ule by the local value o

A and then these p

oducts a

e summedto get the value o

the integ

al. Let A be a vecto

ield in space and ab a cu

ve

om point a to point b. Let the cu

ve ab is subdivided in to in

initesimallysmall vecto

elements dl 1 , dl 2 , dl 3 ,........, dl

Let the dot p

oducts A1. d l 1 , A 2 . d l 2 , A 3 . d l 3 , ........, A

. d l

a

e taken whe

e A1, A2 , A3 ,........, A

a

e the value o

the vecto

ield at the junction pointso

the vecto

elements dl 1 , dl 2 , dl 3 ,........, dl

b Then the sum o

these p

oducts A

d l

along the enti

e length o

the a cu

ve is known as the line integ

al o

A along the cu

ve ab.

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Line integ

alsZ

A

A3 A2 A1

bdl

dl 4 dl 2 dl 3

A

a

dl 1

Y

X

b a

A dl =a

b

A

d l

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July 2008


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Line integ

alsa

A

dl

A

b

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July 2008


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Line integ

alsAs an example i

ep

esents the

o

ce acting on a moving pa

ticle along a cu

ve ab, then the line integ

al o

ove

the path desc

ibed by the pa

ticle

ep

esents the wo

k done by the

o

ce in moving the pa

ticle

om a to b. The line integ

al a

ound a closed cu

ve is called closed line integ

al

a

A dl

b

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Su

ace integ

alsConside

a vecto

ield A continuous in a

egion o

space containing a smooth su

ace S. The su

ace integ

al o

A th

ough S can be de

ined as =

s

A dS

ASurface S

an

dS

Compiled b

: MKP for CEC S5 EC - Jul

2008


110/197

Surface integralsConsider a small incremental surface area on the surface S denoted b

dS. Let an be a unit normal to the surface dS. The magnitude of flux crossing the unit surface normall

is given b

A cos dS

= A andS

= A dSan

= A dS

Where dS denote the vector area having magnitude e

ual to dS and whose directionis in the direction of the unit normal. d S = d Sa n The total flux crossing the surface is obtained b

integrating A d S over the surface of interest.

=

s

A dS

Compiled b


2008


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Surface integralsFor a closed surface defining a volume the surface integral becomes closed surface integral and is denoted b

=

S

A dS

It represents the net outward flow of flux from surface S

Compiled b


2008


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Volume integralsLet V be a volume bounded b

the surface S. Let ( x,

, z ) to n elements of volumes dV1 , dV2 , dV3 ,........, dVn In each part let us choose an arbitrar

point ( xi ,

i , zi ) be a function of position defined over V. If the volume V issubdivided in

Then the limit of the sum

called the volume integral of ( x,

, z ) over V and is denoted b

( x ,

, z )dVi i i

i

as n and dVi 0 is

v

dv

Compiled b


2008


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Scalar and vector fieldsLet

R

be a region of space at each point of which a scalar

= (x,y,z) is given, then is called a scala

point

unction and called a scala

ield. Examples: Tempe

atu

e dist

ibution in a medium Dist

ibution o

atmosphe

icp

essu

e in space. given, then v is called a vecto

point

unction and scala

ield. Examples: The velocity o

a moving

luid at any instant G

avitational

o

ce in a

egion.

R

is

Let R be a

egion o

space at each point o

which a vecto

v = v(x, y, z) is

R

is called a

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July 2008


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DEL Ope

ato

The del ope

ato

is the vecto

di

e

ential ope

ato

and is denoted by . In Ca

tesian coo

dinates

= ax + a y + az x y zThe vecto

di

e

ential ope

ato

is not a vecto

in itsel

, but when it ope

ateson a scala

unction the

esult is a vecto

. This ope

ation is use

ul in de

iningThe g

adient o

a scala

V The dive

gence o

a vecto

.A The cu

l o

a vecto

A 2The Laplacian o

a scala

V

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o

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July 2008


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DEL Ope

ato

in cylind

ical coo

dinatesThe exp

ession

o

del ope

ato

in othe

coo

dinate systems can be obtained using the t

ans

o

mation e

uations de

ived ea

lie

.

= x +y2

2

x = cos = cos x

= sin + y

y tan = x y = sin sin equation (1) cos

Substituting in =

ax + a y + az x y z

sin cos a x + sin a y + a z (2) = cos +Com


July 2008


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DEL O

erator in cylindrical coordinatescos sin = ( cosax + sina y ) + ax + a y + z az + z az (4)

a

1 = a + a + a z (5) z 1 = a + a + az Compiled by: MKP

o

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July 2008

a


117/197

DEL Ope

ato

in sphe

ical coo

dinatesThe exp

ession

o

del ope

ato

in othe

coo

dinate systems can be obtained using the t

ans

o

mation e

uations de

ived ea

lie

. x =

sin cos

= x 2 + y 2 + z2 = tan 1 y x y =

sin sin 2 2 x +y z =

cos = tan 1 z cos cos si cos sin cos = sin sin + + e

uation (1) y

s

x + a y + az Substituting in = a x y z

Compiled by: MKP fo



118/197

DEL Ope

ato

in sphe

ical coo

dinates cos cos sin = sin cos + ax r r

cos sin cos sin + sin sin + + + cos r r az = ( sin cos a x + sin sin a y + cos a z )

+ cos cos cos sin sin sin cos ax + az +

Com


July 2008


119/197

DEL O

erator in s

herical coordinates = ( sin cos a x + sin sin a y + cos a z )

1 1 x + cos sin sin a z ) + + ( cos cos a ( sin a x + cos

a

a 1 1

+ + a a a (5) =

1 1 a

+ a + a

a

=

1 1 a

+ a + a

sin Compiled by: MKP

o

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July 2008


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G

adient o

a scala

ieldZ

V3 V2 = V1 + V V1Y

V

P 1dl X

G

P 1

A Level Su

ace

V

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July 2008


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G

adient o

a scala

ieldThe g

adient o

a scala

ield V is a vecto

that

ep

esents the magnitude and di

ection o

the maximum space

ate o

inc

ease o

V. Let V be a scala

ield andlet V1,V2 and V3 be contou

s on which V is constant. Conside

the di

e

ence inthe

ield dV between points P1 and P2dV = V V V dz dx + dy + x y z

V V V = ax + ay + a z ( dxa x + d

a

+ dza z )

z x + az = G y z x

Then dV = G dl

= G cos dlCompiled b


2008


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Gradient of a scalar fielddV Then = G cos dl dV is maximum when = 0 i.e., when dl is in the direction of GdldV dl =GMAX

Magnitude of G is e

ual to the maximum space rate of change of V Direction of Gis along the maximum space rate of change of V

G is defined as the gradient of V and is denoted b

grad Vgrad V =V= V V V ax + ay + az x y zCompiled by: MKP

o

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July 2008


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G

adient o

a scala

ield

o

cylind

ical coo

dinates,V 1 V V g

ad V =V= a + a + az z

o

sphe

ical coo

dinates,V 1 V 1 V g

ad V =V= a

+ a + a

sin

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o

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July 2008


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G

adient o

a scala

ield

Impo

tant Relations

(V + U ) = V + U (VU ) = V U + U V V U V

V U U = U 2n n 1

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o

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July 2008


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G

adient o

a scala

ield

Impo

tant Points1. Magnitude o

V is e

ual to the maximum space

ate o

change o

V 2. Di

ectiono

V is along the maximum space

ate o

change o

V 3. V at any point is pe

pendicula

to the constant V su

ace that passes

th

ough that point.4. I

A = V , V is called the scala

potential o

A 5. The p

ojection o

V in thedi

ection o

a given unit vecto

a is V a

It is called the di

ectional de

ivative o

V along a It indicates the

ate o

change o

V in the di

ection o

a

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July 2008


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Example 1

ind g

ad when = 3x 2 y y 3 z 2

ind the di

ectional de

ivative in the di

eo

3a x + 4a y + 12a z at (2,

1,0)

Solution :

ax + ay + az x y z (3x 2 y y 3 z 2 ) (3x 2 y y 3 z 2 ) (3x 2 yaz x y z g

ad = = ( 6 xy ) ax + 3x 2 3 y 2 z 2 a y + 2 y 3 z az g

ad

(

)

(

)

At (2,

1,0) = ( 12 ) ax + (12 ) a y

Di

ectional de

ivative a = ( 12ax + 12a y ) = ( 12ax + 12a y )( 0.23ax + 0.31a y + 0.92az

( 3a

x + 4a y + 12az

)

= 0.96

)

9 + 16 + 144

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o

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July 2008


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Example 2

ind g

ad V when V = 10

sin 2 cos

Solution :V 1 V 1 V g

ad V =V= a

+ a + a

sin V = 10sin 2 cos a

+ 10sin 2 cos a 10sin sin a

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July 2008


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Dive

gence o

a vecto

The dive

gence o

a vecto

uantity A at a given point P is the outwa

d

lux pe

unit volume ove

a closed inc

emental su

ace as the volume sh

inks about P.S

divA = A = lim v 0

A S

S

v A dS is the net out

low o

lux o

a vecto

ield A

om

a closed su

ace S

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July 2008


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Dive

gence o

a vecto

in Ca

tesian coo

dinatesZ

dydxP( x0 , y0 , z0 )

dz

1

2Y

XCompiled by: MKP

o

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July 2008


130/197

Dive

gence o

a vecto

in Ca

tesian coo

dinates To evaluate the dive

gence o

a vecto

ield A at point P( x0 , y0 , z0 )

i

st const

uct a di

e

ential volume a

ound point P

The closed su

ace integ

al o

S

A is obtained as+TOP

A dS =

(

RONT

+

BACK

+LE

T

+

RIGHT

+

BOTTOM

)

Ax about P isP

A th

ee dimensional Taylo

s se

ies expansion o

Ax Ax ( x, y , z ) = Ax ( x0 , y0 , z0 ) + ( x x0 ) x Ax + ( y y0 ) y P

Ax + ( z z0 ) z P

+ highe

o

de

te

ms

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Dive

gence o

a vecto

in Ca

tesian coo

dinates

o

the

ont side x = x0 + dx , A = Ax a x , dS = dydza x 2

dx Ax RONT A S = Ax ( x0 , y0 , z0 ) + 2 x P dydz + higher order terms

or the back side x = x0

dx , A = Ax ( a x ) , dS = dydz ( a x ) 2

dx Ax BACK A S = Ax ( x0 , y0 , z0 ) 2 x P dydz + higher order termAx

RONT A S + BACK A S = x y z x + higher or er termsP

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Divergence of a vector in Cartesian coor inatesSimilarly

LEFT

A S +

RIGHT

A S = x y z

Ay yP

+ higher or er terms

TOP

A S +

BOTTOM Az A S = x y z z

+ higher or er te rm sP

Ax S A S = x y z x

Az + x y z + x y z y P z P

Ay

+ higher or

er termsP

Ax Ay Az S A S = x + y + z v + highe o de te ms PSubstituting in v 0

lim

S

A S vCompiled by: MKP

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Dive

gence o

a vecto

in Ca

tesian coo

dinatesAx Ay Az + + S A dS = lim x y z lim v 0 v 0 v vP

v

Ax Ay Az = + + x y z

P

Since highe

o

de

te

ms vanish as v 0

Dive

gence o

A at P ( x0 , y0 , z0 ) in Ca

tesian coo

dinates is Ax Ay Az A = + + x y z

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Divergence of a vector in other coordinatesIn cylindrical coordinates1 1 A Az A = ( A ) + + z

In s

herical coordinates1 2 1 1 A A = 2 ( r Ar ) + ( A sin ) + r r r sin r sin

Com


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Divergence of a vector

Physical significanceZ

yx

vy z F1P ( x, y , z )

v y + yF2Y

XCompile by: MKP for CEC S5 EC July 2008


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Physical significance Consi er the motion of a flui having velocity V = v x a x + v y a y + v z a z at a point P(x,y,z). Consi er a small parallelepipe with e ges x , y an z parallel to the axes enclosing the point P. The mass of flui entering through the face F1 per unit time is given by velocity x area v y x z The mass of flui flowingout through face F2 is v y + y x z By Taylors theorem (neglecting higher or er term)

v y + y = v y +

v y

y v y v y + y x z = v y + y x z y Com


July 2008

y


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Physical significanceThe net ecrease in the mass of flui flowing across these two faces is

v y v y v y + y y x z v y x z = y x y z

Similarly by considering the other two faces we get the total decrease in the mass of fluid inside the

arallele

i

ed

er unit time

v x v y vz + + Rate of loss of fluid

er unit volume is x y z The above

uantity is defined as the divergence of fluid velocity at the

oint Pand is denoted by div V or VCompiled by: MKP

o

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July 2008

v x v y vz x + y + z x y z


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Physical significanceEXAMPLES Divergence of the velocity of water in a container after the outlet haso

ened is zero because water is an incom

ressible fluid. Volume of water entering and leaving different regions of the closed surface is e

ual. When the valveon a steam boiler is o

ened there is a net outward flow of steam for each elemental volume. In this case the divergence has a

ositive value. It indicates a source of vector

uantity at that

oint. When an evacuated glass bulb is broken there is a sudden inrush of air and there is a net inward flow of air for each elemental volume. In this case the divergence has a negative value. It indicates a sink of that vector

uantity.

Com


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Physical significance

ZERO DIVERGENCE

Com


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Physical significance

POSITIVE DIVERGENCE

NEGATIVE DIVERGENCE

SOURCE

SINK

Com


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Exam

lesDetermine the divergence of the vector fields (i ) P = x 2 yza x + xza zP = 2 ( x yz ) + y ( 0 ) + z ( xz ) x Py Pz P P = x + + x y z

P = 2 xyz + x

(ii ) Q = sin a + 2 za + z cos a z1 1 Q = ( 2 sin ) + ( 2 z ) + z ( z cos ) Q = 2sin + cos

1 Q 1 Q Q = ( Q ) + + zz

Com


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Exam

les1 (iii ) T = 2 cos ar + r sin cos a + cos a

1 1 T = ( r T ) + r2

2

r

sin ) +

1 T r sin

1 1 1 r sin 2 cos ) + cos ) T = 2 ( cos ) + ( ( r r r sin

sin

T = 2 cos cos

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Gausss Dive

gence theo

emThe integ

al o

the no

mal component o

any vecto

ield ove

a closed su

ace is e

ual to the integ

al o

the dive

gence o

this vecto

ield th

oughout the volume enclosed by the closed su

ace. The total outwa

d

lux o

a vecto

ield Ath

ough the closed su

ace S is the same as the volume integ

al o

the dive

gence o

A

S

A dS = AdVV

Volume integ

als a

e easie

to evaluate than su

ace integ

als. Using dive

gencetheo

em we can conve

t su

ace integ

al to a volume integ

al and then easily evaluate it.

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Dive

gence theo

emP

oo

Ax Ay Az A = + + x y z

V

A dV =

Ax Ay Az dxdydz + + V x y z

Ay Ax Az dxdydz + dxdydz + dxdydz A dV = V V x V y V z Ay Ax A dxdy= V x y zCompiled by: MKP

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Dive

gence theo

emAx x dx = Ax

Ay y

dy = Ay

Az z dz = Az

V

A dV = Ax dydz + Ay dxdz + Az dxdy dxdy = dS y

dydz = dS x

dxdz = dS z= A dS

V A dV = Ax dS x + Ay dS y + Az dS zS

A dS = A dVV

S

A dS = A dVVCompiled by: MKP

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Dive

gence theo

em

ExplanationZClosed su

ace S

VY

XCompiled by: MKP

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Dive

gence theo

em

ExplanationLet the volume V bounded by the su

ace S is subdivided in to a numbe

o

elementa

y volumes V The

lux dive

ging

om each such cell ente

s o

conve

ges on theadjacent cells unless the cell contains a po

tion o

the oute

su

ace. As a

esult the dive

gence o

the

lux density th

oughout the volume leads to the same

esult as dete

mining the net

lux c

ossing the enclosing su

ace.

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Example 1T

Determine the flux of D = 2 cos 2 a + zsin a ove

the su

ace o

the cylinde, = 4 Z Ve

i

y dive

gence theo

em .

S

Y

X

B

Compiled b


2008


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Example 1If is the flux through the given surface = T + B + S

For T , z = 1, dS = d d a z

T = D dS = B = D dS =

4

=0 =0

2

2cos 2 a + zsin a ) d d a z = 0 (

o

B , z = 0, dS = d d ( a z )4

=0 =0

2 2cos 2 a + zsin a ) d d ( a z ) = 0 (

o

S , = 4, dS = d dza

S = D dS =

1

z =0 =0

2

2cos 2 a + zsin a ) d dza (Compiled by: MKP

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Example 1S =

z =0

1

2 =0

64 cos 2 d dz2

64 2 = 64 cos d = = 0 1 + cos2 d =0 22

= 32

2

=0

1d = 64

A

lying divergence theorem

S

D S = D dVV

1 1 D = ( D ) + D + z Dz = 1 1 z sin 3 cos 2 ) + ( Compiled by: MKP

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Example 1 D = 3 cos 2 +4

1

z cos 1

1 2 V D dV = =0 =0 z =0

3 cos + z cos

d d dz

4 3 2 2

=

2

4z 2 cos d = 64 z cos + =0 2 02 2

=0 z =0

( 64 cos + 4 z cos )d dz

1 21

=

2

=0

( 64 cos + 2 cos ) d2

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Example 1=2

=0

64 cos d + 2

2

=0

2 cos d

64 2 = =0 (1 + cos 2 ) d 2= 32 2

=02

d +

2 =0

cos 2d

= 32

=0

d = 64


July 2008


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Exam

le 2 Verify divergence theorem for the flux of D = x a x + xya y + yza z for the volume of cube with 1unit for each side. The cube is situated in the first octant ofthe coordinate system with one corner on the origin. Z2

1E

A

DB

C

Y

1H G

1

XCom iled by: MKP for CEC S5 EC

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Exam

le 2 D = x 2 a x + xya y + yza z

S

D S = D dVV

S

D dS =

(

RONT

+1

BACK

+

LE

T

+

RIGHT

+

TOP

+

BOTTOM

)and x = 1

RONT

D dS =

1

y =0 z =0

x 2a x + xya y + yza z ) dydza x (

=BACK


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1 1

0 0

dydz =

1and x = 0

D dS =

1

y =0 z =0

1

x 2a x + xya y + yza z ) dydz ( a x ) (

= 0

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Example 2LE

T

D dS =

1

z =0 x =0

(x a1 2

x

+ xya y + yza z ) dxdz ( a y )

and y = 0

= 0

RIGHT

D dS =

1

z =0 x =0

(x a1 2

x

+ xya y + yza z ) dxdza y

1 1

and y = 1

1 = xdxdz = 0 0 2

TOP

D dS =

1

y =0 x =0

1

x 2 a x + xya y + yza z ) dxdya z ( =1 1 0 0


157/197

and z = 1

ydxdy =

1 2Compiled by: MKP

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Example 2BOTTOM

D dS =

1

y =0 x =0

(x a1 2

x

+ xya y + yza z ) dxdy ( a z )

and z = 0

= 0

S

D dS = 1 +

1 1 + = 2 2 2

D x D y D z D = + + x y z

2 = ( x ) + ( xy ) + ( yz ) = 3 x + y x y z= 2

V

DdV = S V

1

0

(3x + y )dxdydz0 0

1

1

D dS = D dV

Thus dive

gence theo

em is ve

i

ied .Compiled by: MKP

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Cu

l o

a vecto

The cu

l o

a vecto

A is an axial o

otational vecto

whose magnitude is the maximum ci

culation (closed line integ

al) o

A pe

unit a

ea as the a

ea tends to ze

o and whose di

ection is the no

mal di

ection o

the a

ea when the a

ea iso

iented so as to make the ci

culation maximum. The ci

culation o

a vecto

ield A a

ound a closed path L is the integ

al A dL

L

Cu

l A =

A = lim

S 0

A l L an S M AX

Where the area S is bounded by the cu

ve L an is the unit vecto

no

mal to the su

ace S and is dete

mined by

ight hand

ule.Compiled by: MKP

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Cu

l o

a vecto

Conside

the di

e

ential a

ea in the yz plane.X

ddz

c

Y

a

dy

bCompiled by: MKP

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Cu

l o

a vecto

Closed line integ

al o

vecto

A a

ound abcd is obtained as below:

l

A dl =

(

ab

+

bc

+

cd

+

da

) A dl

(1)+ ( y y0 )P

Ay ( x, y, z ) = Ay ( x0 , y0 , z0 ) + ( x x0 )

Ay x Az x

Ay y Az yP

+ ( z z0 )

Ay zP

Az ( x, y, z ) = Az ( x0 , y0 , z0 ) + ( x x0 )

+ ( y y0 )P

+ ( z z0 )P

Az z

P

Along ab dl = dya y and z = z0

dz 2

dz Ay ab A dl = Ay ( x0 , y0 , z0 ) 2 z dy

(a) PCompiled by: MKP for CEC S5 EC

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Curl of a vector Along bc dl = dza z and y = y0 +

dy Az A dl = Az ( x0 , y0 , z0 ) + dz

(b) bc 2 y P Along cd dl = dya y and z = z0 + dz 2

dy 2

dz Ay cd A dl = Ay ( x0 , y0 , z0 ) + 2 z dy

(c) Pdy Along da dl = dza z and y = y0 2

dy Az da A dl = Az ( x0 , y0 , z0 ) 2 y dz

(d) PCompiled by: MKP for CEC S5 EC July 2008


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Curl of a vectorSubstituting e

uations a, b, c and d in (1)Ay Az l A dl = y dydz z dydz

Az Ay = S z y

l

A l S

A A = z y z y

S 0

lim

l

A dl S

Az Ay = y z A A = z y z yBy definition the above e

uation re resents curl of vector about x axis

(

curl A

) (x

=

A

)

x

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Cu

l o

a vecto

y and z components o

the cu

l o

A can simila

ly obtained .

(

cu

l A

(

A A = x z y y x z A A curl A =

A = y x z z y x

) (=

A

)

) (

)

The

esultant cu

l will be the sum o

component cu

ls about x, y,z axes A A A A A A

A = z y ax + x z a y + y x az ax

A= x Ax ay y Ay az z Az

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Cu

l o

a vecto

in othe

coo

dinatesThe exp

ession

o

cu

l in cylind

ical and sphe

ical coo

dinates a

e

a

A= A a

A=

A

a A

a

A

az z Az

sin a

sin ACompiled by: MKP

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Cu

l o

a vecto

Some p

ope

ties The cu

l o

a vecto

ield is anothe

vecto

ield The cu

l o

a scala

ield does not exist

( )

( A

B ) = A ( B ) B ( A)

(VA) = V

A + V

A

A+ B =

A+

B

The dive gence o the cu l o a vecto ield vanishes

A = 0

(

)

The cu

l o

the g

adient o

a scala

ield vanishes.

V = 0Compiled by: MKP

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Cu

l o

a vecto

Z

a

b

d

c

Y

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Cu

l o

a vecto

Z

Y

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Cu

l o

a vecto

Z

a

b

d

cY

Compiled by: MKP

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Cu

l o

a vecto

Z

Y

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Cu

l o

a vecto

Physical inte

p

etation

Z

X

Y

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Cu

l o

a vecto

Physical inte

p

etation

Z Y

X

Rotation o

the paddle wheel along x,y and z axes

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Cu

l o

a vecto

Physical inte

p

etationCu

l Di

ection

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Example 1Dete

mine the cu

l o

the

ollowing vecto

ields

(i ) P = x 2 yza x + xza z

(ii ) Q = sin a + 2 za + z cos a z(iii ) T = 1 cos a

+

sin cos a + cos a 2

Solution : (i ) P = x 2 yza x + xza z ax

P = x Px ay y Py az z Pz Pz Py Px Pz Py Px

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Exam

le 1 P P P P P P

P = z y ax + x z a y + y x az

= ( 0 0 ) ax + ( x 2 y z ) a y + ( 0 x 2 z ) az

= ( x 2 y z ) a y ( x 2 z ) az

(ii ) Q = sin a + 2 za + z cos a z

a

Q = Q

a Q

az z QzCompiled by: MKP

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Example 1 1 Q z Q

Q = z Q Q z + a z Q 1

z 1 2 = sin z a + ( 0 0 ) a + ( 3 2 z cos ) az = ( z sin + ) a + ( 3 z cos ) a 13

z

(iii ) T =

1 cos a

+

sin cos a + cos a 2

a

a

sin a

T =

T

T

sin TCompiled by: MKP

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Example 1

T = 1 1 1 1 T

(rT ) a + ( rT ) Tr a (T s

=

1 1 1 cos (cos sin ) (r sin cos ) ar + (r cos )

1 cos + ( r 2 sin cos ) 2 a r r

=

1 1 1 sin ( cos2 + r sin sin ) a

+ ( 0 cos ) a + 2

sin cos + 2

a

cos2 cos a = + sin ar r r sin

1 + 2cos + 3 a

Com


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Stokes theoremStokes theorem states that the circulation of a vector field A around a closed

ath L is e

ual to the surface integral of the curl of A over the o

en surface Sbounded by L rovided that A and

A a

e continuous on S

L

A dl = (

A) dSS

Closed path L

dS dl

Su

ace S

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Stokes theo

em

The su

ace S is subdivided in to a la

ge numbe

o

cells.

I

the k th cell has su

ace a

ea Sk and is bounded by the path L

L

A dl = A dl = k Lk k

Lk

A dl

S k

S kCompiled by: MKP

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Stokes theo

emThe

e is cancellation on eve

y inte

io

path. So the sum o

line integ

als a

ound Lks is the same as the line integ

als a

ound the bounding cu

ve L. Taking the limit o

the above e

uation

Lk A dl S lim0 S k S k k k By definition the

uantity insidethe k th cell As Sk 0 the summation becomes integ

ation ove

the whole su

ace.

L

A dl = (

A).dS which is stoke

s theo

emS

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Example 1Y

I

cos a + sin a , evaluate A dl a

ound the path shown below. Ve

i

y Sto

emda

5

2

S Sb30

c

60

0

2

5YCompiled by: MKP

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Example 1L

b + A dl = a

c b

+

d c

+

a d

A dl

Along ab = 2 dl = d a

b a

A dl =

30 = 60

sin d

= [ 2 cos ]60 = 30

(

3 1

)

Along bc = 30 dl = d a

c b

A dl =

5 =2

cos d


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Along cd = 5 dl = d a

21 3 = cos 30 = 2 2 4 2

5

d c

A dl =

60 = 30

sin d

= [ 5 cos ]3060

5 = 2

(

3 1

)

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Example 1 Along da = 60 dl = d a

a d

A dl =

2 =5

cos d

21 = cos 60 = 2 5 4 2

2

L

A dl = (

21 3 5 + 3 1 + 4 2

)

(

21 3 1 4

)

L

A dl = 4.941

Using stoke's theorem

L

A dl = (

A).dSS

1 Az A A A + z

A= a z z

A 1 + ( A ) a z a

dS = d d a zCompiled by: MKP

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Example 11

A = ( 0 0 ) a + ( 0 0 ) a + (1 + ) sin a z 1 = (1 + ) sin

S

(

A).dS =

60

=3060

5 =2

1

(1 + ) sin d d az5

=

=30

sin d 60

=2

(1 + )d

2 5

S

= [ cos ]30 + = 4.941 2 2 (

A).dS = 4.941 = A dlL

Stoke s Theo

em is thus ve

i

iedCompiled by: MKP

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Laplacian o

a scala

The Laplacian o

a scala

ield V , w

itten as 2V is the dive

genceo

the g

adient o

V . It is anothe

scala

ield

In Ca

tesian coo

dinates,Laplacian V= V = 2V V V V = ax + a y + az ax + ay + az y z x

2V 2V 2V 2 V= 2 + 2 + 2 x y z

A scala

ield V is said to be ha

monic in a given

egion i

itsLaplacian vanishes in that

egion. 2 V=0 Laplace s EquationCompiled by: MKP for CEC S5 EC - July 2008


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Laplacian of a scalar in other coordinatesIn cylindrical coordinates,

1 V 2 V=

1 2V 2V + 2 2 + z 2

In s

herical coordinates,

1 2 V 1 V 1 2V 2 V= 2 r + 2 sin + 2 2 r sin 2

Com


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La

lacian

35675966 EMT Electromagnetic Theory MODULE I (1)

Documents

Transcript of 35675966 EMT Electromagnetic Theory MODULE I (1)