Electromagnetism and Electric Fields: A Comprehensive Study Guide, Lecture notes of Electrical Engineering

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Estd:

EDUCATIONAL ACADEMY

ELECTROMAGNETIC FIELDS

By

Venugopala swamy

ELECTRICAL ENGINEERING

ACE EDUCATIONAL ACADEMY

Mr. Y.V. GOPALA KRISHNA MURTHY M.Tech., MIE

MANAGING DIRECTOR ACE EDUCATIONAL ACADEMY 2 nd^ Floor, Rahman plaza, Opp.Methodist School New Gate, Near Tajmahal Hotel, ABIDS, HYDERQABAD-500 001. Ph: 24752469, 65582469 e-mail: ace.gateguru@gmail.com www.aceenggacademy.com

FOREWORD

I consider it as a privilege to write foreword for this book “Electromagnetic Fields” written by

Mr. I. V. VENUGOPALA SWAMY which is useful for GATE, Engg. Services, JTO and other

competitive exams.

Electromagnetic fields is an important subject which is common for all Electrical Science

stream and foundation course for Electronics & Communication stream. It is noticed that many

students find this subject to be difficult one. Understanding this subject requires lot of imagination

and principles of Mathematics and Physical Sciences.

Mr. Venu has made an appreciable effort to explain the principles of fields in a simple way.

I expect, the practice of the question bank given in addition to the concepts, instills confidence

on the basics of the subject.

DIRECTOR

ACE EDUCATIONAL ACADEMY

ELECTRICAL ENGINEERING

Electro Magnetic Fields

TOPIC – 1 : VECTOR ANALYSIS E M F

1. Introduction: In communication systems, circuit theory is valid at both the transmitting end as well as the receiving end but it fails to explain the flow between the transmitter and receiver. Circuit theory deals with only two variables that is voltage and current whereas Electromagnetic theory deals with many variables like electric field intensity, magnetic field intensity etc.,

Mostly three space variables are involved in electromagnetic field problems. Hence the solution becomes complex. For solving field problems we need strong background of vector analysis.

Maxwell has applied vectors to Gauss’s law, Biot Savart’s law, Ampere’s Law and Faraday’s Law. His application of vectors to basic laws, produced a subject called “Field Theory”.

2. Scalar and Vector Products

a) Dot Product: is also called scalar product. Let ‘θ’ be the angle between vectors A and B.

A. B = | A | | B | cosθ

The result of dot product is a scalar. Dot product of force and distance gives work done (or) Energy which is scalar. a (^) n b) Cross product: is also called vector product.

B A x B = |A| | B| sinθ â (^) n A

S = |S| â (^) n where |S| = |A| |B| Sinθ

To find the direction of S, consider a right threaded screw being rotated from A to B. i.e. perpendicular to the plane containing the vectors A and B.

∴ A x B = - (B x A)

3. Operator Del ( ∇ ): Del is a vector three dimensional partial differential operator. It is defined in Cartesian system as

∇ = ∂ i + ∂ (^) j + ∂ (^) k ∂x ∂y ∂z

Del is a very important operator. There are 3 possible operations with del. They are gradient, divergence and curl. (Contd….2)

4. GRADIENT:

Gradient is a basic operation of a Del operator that can operate only on a scalar function. Consider a scalar function ‘t’. The gradient of ‘t’ can be mathematically defined and symbolically expressed as below.

∇t = ∂ i + ∂ j + ∂ k t

(Grad t) ∂x ∂y ∂z

∇t = ∂t i + ∂t j + ∂t k ∂x ∂y ∂z

Vector

Gradient of scalar function is a vector function.

Ex:- Temperature of soldering iron is scalar, but rate of change of temperature is a Vector. In a cable, potential is scalar. The rate of change of potential is a vector (Electric field intensity).

5. DIVERGENCE:-

Divergence is a basic operation of the Del operator which can operate only on a vector function through a dot product.

Considering a vector function A = A (^) xi^ + Ayj^ + Azk

The divergence of vector A mathematically and symbolically expressed as shown below.

∇. A = ∂ i + ∂ j + ∂ k. Axi^ + A (^) yj^ + Azk (Div A) ∂x ∂y ∂z

∇. A = ∂A (^) x + ∂A (^) y + ∂Az ∂x ∂y ∂z

Scalar

Divergence of vector function is a scalar function.

Let D = flux density vector D.ds = flux through the surface ds The flux through the entire surface is (^) s D.ds

Note: Divergence of D gives net outflow of flux per unit volume.

∇. D = Lt (^) sD. ds ∆V 0 ∆V

6. CURL:

Curl is a basic operation of a Del operator which can perform only on a vector function through a cross product.

(Contd.…3)

10. Basic types of vector fields:

a) Solenoidal vector field (∇. A =0) b) Irrotational vector field (∇xA =0) c) Vector fields that are both solenoidal & irrotational d) Vector fields which are neither solenoidal nor irrotational

11. Fundamental theorem of Gradient: Statement: consider an open path from ‘a’ to ‘b’ in a scalar field as shown. The line integral of the tangential component of the gradient of a scalar function along the open path is equal to path the effective value of the associated scalar function at the boundaries of the open path.

If ‘t’ is the associated scalar function, then according to the fundamental theorem of gradient ∇t b ∇tcosθ

Corollary-1: If it is a closed path in scalar field, then

Corollary-2: b

A line integral ∫ (∇t).dl is independent of the open path.

a

12. Fundamental theorem of Divergence:- (Gauss theorem)

Statement: Consider a closed surface in vector field. The volume integral of the divergence of the associated vector function carried within a enclosed volume is equal to the surface integral of the normal component of the associated vector function carried over an enclosing surface.

If associated vector function is A, then according to fundamental theorem of divergence,

(Contd…,5)

θ dl

Z

Y

a

Scalar field

b ∫ (∇t). dl = t(b) - t(a)

a

X

b ∫ (∇t). dl = 0 a

A

dv θ da

Enclosed Enclosing volume surface

da

Vector field

( ∇. A )dv = A. da

v s

Note: Area vector is always outward normal

13. Fundamental theorem of Curl:- (Stokes theorem)

Statement: Considering an open surface placed in a vector field, the surface integral of the normal component of the curl of the associated vector function carried over the open surface is equal to the line integral of the tangential component of the associated vector function along the boundary of the open surface.

If associated vector function is A, then

Corollary-1: If it is a closed surface

Since there is no boundary and hence

Corollary-2: A. dl is constant for a fixed boundary.Therefore , s (∇ x A ). da is independent

of the type of open surface.

14. Vector Identities: a) ∇ x ∇ φ = 0

b) ∇. ∇ x A = 0

c) ∇. φ A = ∇ φ. A + φ (∇. A)

d) ∇ x φ A = ∇ φ x A + φ (∇ x A)

e) ∇ x ∇ x A = ∇ (∇. A) - ∇^2 A

f) ∇. ∇ φ = ∇^2 φ

g) ∇( φF) = φ(∇. F) + F ∇ φ h) Div (u x v) = v curl u – u curl v i) A. B x C = B. C x A = C. A x B j) ∇. A x B = B. ∇ x A – A. ∇ x B

k) ∇^2 A = ∇ (∇. A) - ∇ x (∇ x A)

15. Co-ordinate systems:

a) Cartesian co-ordinate system (x,y,z) b) Spherical co-ordinate system (r,θ,φ) c) Cylindrical co-ordinate system (r,φ,z) (Contd….6)

da^ da

θ dl

A

∇xA

θ

Vector field

s (^ ∇xA ). da =^ A^.^ dl

S(^ ∇xA ). da = 0

A. dl = 0

Ranges: r = 0 → ∝ θ = 0 → ∏

Differential length,

r varying direction

φ varying direction

θ varying direction

dl = ( dr)r + (rdθ)θ + (r sinθ dφ) φ

15.c) Cylindrical co-ordinate system: ( r, φ, z )

Cartesian to Cylindrical Cylindrical to Cartesian

Z

r

z P(r,φ,z)

Y

φ

X

r = √ x 2 + y^2 x = r cosφ φ = tan -1^ (y/x) y = r sinφ z = z z = z

Note: In cylindrical system unit vectors are r, φ , z

Differential Length Vector:

Ranges: r = 0 ∝ φ = 0 2 ∏ z = - ∝ + ∝

Differential length,

16. Differential areas : ( da (or) ds ) a) Cartesian system:

dl = dx i + dy j + dz k

b) Spherical system:

dl = (dr) r + (rdθ) θ + ( r sinθ dφ) φ

c) Cylindrical system:

dl = (dr) r + (rdφ) φ + (dz) z

da = dx dy k

da= (r^2 sinθ dθ dφ) r

da = (rdφ dz) r

dl = (dr) r + (rdφ) φ + (dz) z

Cartesian system Spherical system Cylindrical system

h,h2, h3 Ξ 1,1,1 h,h2, h3 Ξ 1,r, rsinθ h,h (^) 2, h3 Ξ 1,r,

e (^) 1, e 2 e 3 Ξ i, j, k e (^) 1, e (^) 2, e 3 Ξ r, θ, φ e (^) 1, e (^) 2, e 3 Ξ r, φ, z

u 1 ,u 2 ,u 3 Ξ x,y,z u 1 ,u 2 ,u 3 Ξ r, θ, φ u 1 ,u 2 ,u 3 Ξ r,φ,z

In General:

1 ∂t e 1 1 ∂t e 2 1 ∂t e (^3)

1. ∇t = h 1 ∂u 1 + h 2 ∂u 2 + h 3 ∂u 3

1 ∂ ( A 1 h 2 h 3 ) ∂ ( A 2 h 3 h 1 ) ∂ ( A 3 h 1 h 2 )

2. ∇. A = h 1 h 2 h 3 ∂u 1 + ∂u 2 + ∂u 3

3. ∇ x A = 1 h 1 h 2 h 3 h 1 e 1 h 2 e 2 h 3 e (^3) ∂ ∂ ∂ ∂u 1 ∂u 2 ∂u 3 A 1 h 1 A 2 h 2 A 3 h 3

1 ∂ h 2 h 3 ∂t + ∂ h 3 h 1 ∂t + ∂ h 1 h 2 ∂t

4. ∇^2 t = h 1 h 2 h 3 ∂u 1 h 1 ∂u 1 ∂u 2 h 2 ∂u 2 ∂u 3 h 3 ∂u 3

Let: A = A 1 i + A 2 j + A 3 k Cartesian system

= A 1 e (^) r + A 2 eθ + A 3 eφ Spherical system

= A 1 e (^) r + A 2 eφ + A 3 e (^) z Cylindrical

In Cartesian system:

1. ∇t = ∂t i + ∂t j ∂t k ∂x ∂y ∂z

2. ∇. A = ∂A 1 + ∂A 2 + ∂A 3 ∂x ∂y ∂z

3. ∇ x A = i j k

∂ ∂ ∂ ∂x ∂y ∂z

A 1 A 2 A 3

4. ∇^2 t = ∂^2 t + ∂^2 t + ∂^2 t ∂x 2 ∂y^2 ∂ z^2

In Spherical system:

1. ∇t = ∂t r + 1 ∂t θ + 1 ∂t φ ∂r r ∂θ r sinθ ∂φ 1 ∂ ( A 1 r^2 sin θ) + ∂ ( A 2 r sin θ) + ∂ (A 3 r)
2. ∇. A = r^2 sinθ ∂r ∂θ ∂φ

1

3. ∇ x A = r^2 sinθ r rθ r sinθφ ∂ ∂ ∂ ∂r ∂θ ∂φ A1 rA 2 r sinθ A 3

4. ∇^2 t = 1 ∂ r^2 sin θ ∂t + ∂ r sinθ ∂t + ∂ r ∂t r^2 sinθ ∂r ∂r ∂θ r ∂θ ∂φ r sinθ ∂φ

In Cylindrical system:

1. ∇t = ∂t r + 1 ∂t φ + ∂t z ∂r r ∂φ ∂z

2. ∇. A = 1 ∂ (A 1 r) + ∂ A 2 + ∂ (A 3 r) r ∂r ∂φ ∂z

3. ∇ x A = 1 r r rφ z ∂ ∂ ∂ ∂r ∂φ ∂z A 1 rA 2 A 3

4. ∇^2 t = 1 ∂ r ∂t + ∂ 1 ∂t + ∂ r ∂t r ∂r ∂r ∂φ r ∂φ ∂z ∂z

OBJECTIVES

One Mark Questions

1. If the vectors A and B are conservative then ( Engg.Services,1993) a) A x B is solenoidal b) A x B is conservative c) A + B is solenoidal d) A – B is solenoidal

2. The value of d.l along a circular radius of 2 units is ( IES, 93 ) a) zero b) 2∏ c) 4∏ d) 8∏

3. which of the following relations is correct? (BEL, 95) a) ∇ x (AB) = ∇A x B – A. ∇B b) ∇. (AB) = ∇A. B + A. ∇B c) ∇ (AB) = A. ∇B + B. ∇A d) all the three

4. ∇. (∇ x A) is equal to (BEL, 95) a) 0 b) 1 c) ∞ d) none of these

5. Given points A(2,3,-1) and B(4, -50^0 ,2) find the distance from A to B a) 3.74 b) 4.47 c) 16.7 d) 6.

6. Find the nature of the given vector field defined by A = 30 i - 2xy j + 5xz 2 k a) Neither Solinoidal nor irrotational b) Solinoidal & irrotational c) Only Solinoidal d) Only irrotational

7. Find the nature of given vector field defined by A = yz i + zx j + xy k a) Neither Solinoidal nor irrotational b) Solinoidal & irrotational c) Only Solinoidal d) Only irrotational

8. A vector field is given by A = 3xy i - y 2 j. Find ∫c A .dl. where ‘c’ is the curve y = 2x^2 in the x-y plane from (0,0) to (1,2) a) -9/2 b) 7/6 c) -7/6 d) 2/

9. Find the laplacian of the scalar function v = (cosφ)/r (cylindrical system). a) 5 b) 0 c) 7/6 d) 8

TOPIC – 2 : ELECTRIC FIELD INTENSITY E M F

Electrostatics is a science that deals with the charges at rest. Static charges produce electric field.

In electromagnetic theory there is a fundamental problem with regard to the force between the electric charges. Let us start our study with an introduction of coulomb’s law

Coulomb’s Law: Q 1 Q 2

This law states that considering two point charges separated by a distance, the force of attraction (or) repulsion is directly proportional to the product of the magnitudes and inversely proportional to the square of the distance between them.

Force acting on Q1 due to Q2, F 12 =

Force acting on Q2 due to Q1, F 21 =

This law is an imperial law and difficult to understand how exactly a force is communicated between them. Michel Faraday gives a satisfactory explanation of coulomb’s law by introducing the concept of electric field.

According to Faraday, Q1 experiences a force because it is placed in the electric field of Q2. And Q2 experiences a force because it is placed in the electric field of Q1.

Concept Of Electric Field:

An electric field is said to exist at a particular point, if a test charge placed at that point experiences a force.

If ‘q’ is the test charge and F is the force experienced by the test charge, then the force per unit test charge is known as Electric field intensity. Expressed in N/C or V/m

ELECTRIC FIELD DUE TO A POINT CHARGE:

r

Consider a point charge of ‘+Q’ c at origin. In order to find electric field intensity at point of observation P, consider a Unit test Charge ‘q’ c at P.

(Contd …13)

d^2

F α |Q 1 | |Q 2 |

d^2

|Q 1 | |Q 2 |

4 π∈ 0

F =^1

4 π∈ 0 d^2

|Q 1 | |Q 2 |

BA

4 π∈ 0 d^2

|Q 1 | |Q 2 |

AB

E = F

q

N/C (or) V/M

Z

P

q c

X

Y

+Qc

A B F^21

F 12

d

Therefore, the force experienced by the test charge is

4 π∈ 0 r^2

|Q 1 | q F = r

E =

q

We know,^ F

4 π∈ 0 r^2

Q

∴E = r

NOTE: Thus electric field intensity is independent of the amount of test charge. In Cartesian system:

. r

4 π∈ 0 r^2

Q

F = (^) | r |

4 π∈ 0 (x^2 +y^2 +z 2 )3/

Q

F =

(x i + y j + z k)

ELECTRIC FIELD DUE TO A POINT CHARGE LOCATED AT ANY GENERAL

POSITION:

A P

- QC

+ QC

A P

Q

∴E = AP

4 π∈ 0 (AP) 2

Q ∴E = (^) PA 4 π∈ 0 (PA) 2

Electric field is always directed away from the point charge towards the point of observation(P), if it is a positive charge.

Similarly, electric field is directed away from the point of observation towards the point charge, if it is a negative charge.

PRINCIPLE OF SUPERPOSITION: The principle of superposition says that electric field due to any charge is unaffected by the presence of other charges.

In a system of discrete charges the net electric field is obtained by the vectorically adding up the individual electric fields.

E 3

E 2

E 1

Net electric field intensity E = E 1 + E 2 + E 3 +……..

(Contd …14)

Electric field due to a finite line charge located at any general position.

(90-θ)

(90-θ) o

θ

α

P

A

B

d

β

ρL c/m

dq

dEv

dE

dEH

x

Electric field due to a finite line charge (OA ≠ OB)

Electric field due to Rectangular line charge along it axis.

(Contd …16)

N

P

B

2 π∈ 0 NP

ρL. E (^) at P = (^) NP, if it is a +ve line charge. i)

2 π∈ 0 NP

ρL. E (^) at P = (^) PN, if it is a -ve line charge. ii)

√BN^2 +NP^2

BN

√BN^2 +NP^2

BN

ρ L c/m A

4 π∈ 0 d

ρL ii) E (^) V = (sinβ^ – sinα)

4 π∈ 0 d

ρL i) E (^) H = (cosα^ – cos^ β)

iii) Net electric field intensity, E = √E^2 H + E^2 V

iv) If ‘O’ is the mid point, β=(180-α). As line tends to infinity, α Æ 0 , βÆ π (^) E v = 0

2 π∈ 0 d

ρL E =

op

X

M

Z

dEAB dE^ dECD BC

P

β

α (^) β α

A

D

B

C

N

a

d

2a

R

dEDA

Q

2b

ρL c/m

Y

Eat P = EAB + ECD + EBC + EDA

= 2 [ EAB + EBC ]

E = ρ d πε 0 √ a 2 +b 2 +d^2 K

L.^ a^ +^ b b^2 +d^2 a^2 +d 2

Corollary:1 If it is a square line charge a=b

E = 2ρLda. πεo√2a 2 +d^2. (a^2 +d^2 ) K

Corollary:2 If d=0 i.e. the electric field at orgin

E = 0

Electric field due to a circular line charge along its axis:-

r (^) b

X

Adl

Bdl

Y

ρL c/m

a o (^) dφ

z

Z

P

dEA dEB

ra

Consider two diametrically opposite elementary displacements located at A & B. Let point of observation ‘P’ be along ‘Z’ axis.

Eat P = ρL az. z 2 εo (a 2 +z 2 )3/

Electric field due to an infinite charge sheet:

Consider two diametrically opposite elementary surface charges located at A & B. Let point of observation ‘P’ be along Z axis.

The electric field due to the surface charge sheet is independent of the distance of the point of

observation (P) from the surface charge sheet. It has a constant magnitude equal to ρs/2ε 0 and has a direction normal to the surface charge sheet.

The field direction is away from the surface charge sheet towards the point of observation if it is a +ve charge sheet. (Contd …17)

ρs. Z E = 2εo

Z

r (^) b

dEB

ρs c/m^2

dφ

Z

P

ra

dEA

x

B^ Y

da

r A