Foundation of Engineering

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Amazing PersonalityStephen William Hawking(Born 8 January 1942) is an English theoretical physicist and cosmologist, whose scientific books and public appearances have made him an academic celebrity.He is severely disabled by a motor neurone disease known as amyotrophic lateral sclerosis (ALS).He is an Honorary Fellow of the Royal Society of Arts , a lifetime member of the Pontifical Academy of SciencesiIn 2009 was awarded the Presidential Medal of freedom, the highest civilian award in the United States.

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Lecture # 1

Electromagnetic Fields

Course OutlineChapter 1 12 (5 & 7 exclusive)Marks DistributionAssignments 10%Quizzes 10%Class Participation 5%Mid Term 25%Final 50%Final Examination Question Paper 8 Review Questions out of 12

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Course BreakupChapter #Final Exam Q. NoChapter Heading11Vector Analysis22Coulombs Law & Electric Field Intensity 33Electric Flux Density, Gauss Law, & Divergence 44Energy & Potential 65,6Dielectrics & Capacitance 87The Steady Magnetic Field 98,9Magnetic Forces, Materials & Inductance

Course BreakupChapter #Final Exam Q. NoChapter Heading1010Time-Varying Fields & Maxwells Equations 1111Transmission Lines1212The Uniform Plane Wave

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Text BookEngineering Electromagnetics 7th Edition William H. Hayt, Jr. & John A. Buck

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Chapter 1Vector Analysis

Electric field Magnetic fieldProduced by the motion of electric charges, or electric current, and gives rise to the magnetic force associated with magnets. Electromagnetic is the study of the effects of charges at rest and charges in motion

Produced by the presence of electrically charged particles, and gives rise to the electric force.

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Scalars & VectorsScalarRefers to a Quantity whose value may be represented by a Single (Positive/Negative) real number.Body falling a distance L in Time t Temperature T at any point in a bowl of soup whose co-ordinates are x, y, z L, t, T, z, y & z are all scalarsMass, Density, Pressure, Volume, Volume resistivity, Voltage

Scalars & VectorsVector A quantity who has both a magnitude and direction in space. Force, Velocity, Acceleration & a straight line from positive to negative terminal of a storage battery

Vector AlgebraAddition

Associative Law:

Distributive Law:

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Vector AlgebraCoplanar vectorsLying in a common plane

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Vector AlgebraSubtractionA B = A + (-B)MultiplicationObeys Associative & Distributive laws (r + s) (A + B) = r(A + B) + s (A + B) = rA + rB + sA + sB

Orthogonal Coordinate SystemsA coordinate system defines a set of reference directions. In a 3D space, a coordinate system can be specified by the intersection of 3 surfaces at each and every point in space.The origin of the coordinate system is the reference point relative to which we locate every other point in space.

Orthogonal Coordinate SystemsA position vector defines the position of a point in space relative to the origin. These three reference directions are referred to as coordinate directions or base vectors, and are usually taken to be mutually perpendicular (orthogonal) . In this class, we use three coordinate systems:CartesianCylindricalSpherical

Rectangular Coordinate SystemIn Cartesian or rectangular coordinate system a point P is represented by coordinates (x,y,z) All the three coordinates represent the mutually perpendicular plane surfaces

The range of coordinates are< x< < y< < z<

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Point Locations in Rectangular Coordinates

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Differential Volume Element

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Orthogonal Vector Components

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Orthogonal Unit Vectors

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Vector Representation in Terms of Orthogonal Rectangular Components

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Vector Expressions in Rectangular Coordinates

General Vector, B: Magnitude of B:Unit Vector in the Direction of B:

Example

Vector Components and Unit VectorsExampleGiven points M(1,2,1) and N(3,3,0), find RMN and aMN.

FieldFunction, which specifies a particular quantity everywhere in the regionTwo types: Vector Field has a direction feature pertaining to it e.g. Gravitational field in space and Scalar Field has only magnitude e.g. Temperature

The Dot Product

Commutative Law:

Vector Projections Using the Dot Product

B a gives the component of Bin the horizontal direction(B a) a gives the vector component of B in the horizontal direction

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Operational Use of the Dot Product

GivenFind

where we have used:

Note also:

The three vertices of a triangle are located at A(6,1,2), B(2,3,4), and C(3,1,5). Find: (a) RAB; (b) RAC; (c) angle

BAC at vertex A; (d) the vector projection of RAB on RAC.

Drill Problem

Solution

Continued

Cross Product

Operational Definition of the Cross Product in Rectangular Coordinates

Therefore:

OrBegin with:

where

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Operational Definition of the Cross Product in Rectangular Coordinates

The Cross ProductExample

Given A = 2ax3ay+az and B = 4ax2ay+5az, find AB.

Circular Cylindrical CoordinatesPoint P has coordinatesSpecified by P(z)The coordinate represents a cylinder of radius with z axis as its axis. The coordinate (the azimuthal angle) is measured from x axis in xy plane. Z is same as in Cartesian coordinatesThe range of coordinates are 0 < 0 < 2 < z <

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Orthogonal Unit Vectors in Cylindrical Coordinates

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Relationship between (x, y, z) and (r, f, z).

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Point P and unit vectors in the cylindrical coordinate system.

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Differential Volume in Cylindrical Coordinates

dv = dddz

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Point Transformations in Cylindrical Coordinates

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The Cylindrical Coordinate System

Dot products of unit vectors in cylindrical and rectangular coordinate systems

Transform the vector B into cylindrical coordinates:

Example

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Transform the vector B into cylindrical coordinates

Start with:

Problem

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Then:

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Finally:

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The r coordinate represents a sphere of radius r centered at origin . The coordinate represents the angle made by the cone with z-axis. The coordinate is the same as cylindrical coordinate.The range of coordinates are 0 r < 0

0 < 2Point P has coordinatesSpecified by P(r)Spherical Coordinates

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Point P and unit vectors in spherical coordinates.

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Relationships between space variables (x, y, z), (r, q, f), and (r, f, z).

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Differential Volume in Spherical Coordinates

dv = r2sindrdd

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Dot Products of Unit Vectors in the Spherical and Rectangular Coordinate Systems

The Spherical Coordinate SystemExample

Given the two points, C(3,2,1) and D(r = 5, = 20, = 70), find: (a) the spherical coordinates of C; (b) the rectangular coordinates of D.

Example: Vector Component Transformation

Transform the field, , into spherical coordinates and components

Chapter 2Coulombs Law & Electric Field Intensity

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The Experimental Law of CoulombCoulomb law states that the force between two very small objects separated in vacuum or free space by a distance which is large compared to their size is Proportional to the charge on each and Inversely Proportional to the square of the distance between them.

Chapter 2

The Experimental Law of CoulombIn SI Units, the quantities of charge Q are measured in coulombs (C), the separation R in meters (m), and the force F should be Newtons (N).This will be achieved if the constant of proportionality k is written as:

Chapter 2

The Experimental Law of CoulombThe permittivity of free space is measured in farads per meter (F/m), and has the magnitude of:

The Coulombs law is now:The force F acts along the line joining the two charges. It is repulsive if the charges are alike in sign and attractive if the are of opposite sign.

The Experimental Law of CoulombIn vector form, Coulombs law is written as:

F2 is the force on Q2, for the case where Q1 and Q2 have the same sign, while a12 is the unit vector in the direction of R12, the line segment from Q1 to Q2.

Chapter 2

The Experimental Law of CoulombChapter 2ExampleA charge Q1 = 3104 C at M(1,2,3) and a charge of Q2 = 104 C at N(2,0,5) are located in a vacuum. Determine the force exerted on Q2 by Q1.

Electric Field IntensityLet us consider one charge, say Q1, fixed in position in space.Now, imagine that we can introduce a second charge, Qt, as a unit test charge, that we can move around.We know that there exists everywhere a force on this second charge This second charge is displaying the existence of a force field.

Chapter 2

Electric Field IntensityThe force on it is given by Coulombs law as:

Writing this force as a force per unit charge gives:Chapter 2

Vector Field, Electric Field Intensity

Electric Field IntensityWe define the electric field intensity as the vector of force on a unit positive test charge.Electric field intensity, E, is measured by the unit Newtons per coulomb (N/C) or volts per meter (V/m).

Chapter 2

Electric Field IntensityThe field of a single point charge can be written as:

aR is a unit vector in the direction from the point at which the point charge Q is located, to the point at which E is desired/measured.

Chapter 2

Electric Field IntensityFor a charge which is not at the origin of the coordinate, the electric field intensity is:

Chapter 2

Electric Field IntensityThe electric field intensity due to two point charges, say Q1 at r1 and Q2 at r2, is the sum of the electric field intensity on Qt caused by Q1 and Q2 acting alone (Superposition Principle).

Chapter 2

Electric Field IntensityExampleA charge Q1 of 2C is located at at P1(0,0,0) and a second charge of 3C is at P2(1,2,3). Find E at M(3,4,2).

Chapter 2

Example 2-2

Solution

Assignment

Chapter 2Field Due to a Continuous Volume Charge DistributionIf we now visualize a region of space lled with a tremendous number of charges separated by minute distances, we see that we can replace this distribution of very small particles with a smooth continuous distribution described by a volume charge density, just as we describe water as having a density of 1 g/cm3(gram per cubic centimeter) even though it consists of atomic- and molecular-sized particles.

Chapter 2Field Due to a Continuous Volume Charge DistributionWe denote the volume charge density by v, having the units of Coulombs per cubic meter (C/m3).The small amount of charge Q in a small volume v is

Chapter 2Field Due to a Continuous Volume Charge DistributionWe may define v mathematically by using a limit on the above equation:

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

Example 2.3

Solution

Chapter 2Field Due to a Continuous Volume Charge DistributionExampleFind the total charge inside the volume indicated by v = 4xyz2, 0 2, 0 /2, 0 z 3. All values are in SI units.

Chapter 2Field Due to a Continuous Volume Charge Distribution

Chapter 2Coulombs Law and Electric Field IntensityField Due to a Continuous Volume Charge DistributionThe contributions of all the volume charge in a given region, let the volume element v approaches zero, is an integral in the form of:

Assignment

Field of a Line ChargeNow we consider a filament like distribution of volume charge density. It is convenient to treat the charge as a line charge of density L C/m.Let us assume a straight-line charge extending along the z axis in a cylindrical coordinate system from to +.We desire the electric field intensity E at any point resulting from a uniform line charge density L.

Chapter 2

Field of a Line ChargeChapter 2

Field of a Line ChargeThe incremental field dE only has the components in a and az direction, and no a direction. Why?The component dEz is the result of symmetrical contributions of line segments above and below the observation point P.Since the length is infinity, they are canceling each other dEz = 0.

Chapter 2

SymmetrySymmetry should always be considered rst in order to determine two specic factors: with which coordinates the eld does not vary, which components of the eld are not presentIf z and = constant and vary line charge appears same from every angle (azimuthal symmetry)If and = constant and vary z line charge still recedes into infinite distance in both directions (axial symmetry)If and z = constant and vary Field become weaker as increases (Coulombs Law) Field varies with on Chapter 2

Field of a Line ChargeThe component dE exists, and from the Coulombs law we know that dE will be inversely proportional to the distance to the line charge, .

Chapter 2

Field of a Line ChargeTake P(0,y,0),

Chapter 2

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Field of a Line Charge

Chapter 2

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Field of a Line ChargeNow let us analyze the answer itself:

The field falls off inversely with the distance to the charged line, as compared with the point charge, where the field decreased with the square of the distance.

Chapter 2

Field of a Line ChargeExample D 2.5.Infinite uniform line charges of 5nC/m lie along the (positive and negative) x and y axes in free space. Find E at: (a) PA(0,0,4); (b) PB(0,3,4).

Chapter 2

PAPB is the shortest distance between an observation point and the line charge

Field of a Line ChargeChapter 2

Field of a Sheet of ChargeAnother basic charge configuration is the infinite sheet of charge having a uniform density of S C/m2.The charge-distribution family is now complete: point (Q), line (L), surface (S), and volume (v).

Chapter 2

Field of a Sheet of ChargeLet us examine a sheet of charge above, which is placed in the yz-plane.The plane can be seen to be assembled from an infinite number of line charge, extending along the z axis, from to +.Chapter 2

Field of a Sheet of ChargeFor a differential width strip dy, the line charge density is given by L = Sdy.The component dEz at P is zero, because the differential segments above and below the y axis will cancel each other.

Chapter 2

Field of a Sheet of ChargeThe component dEy at P is also zero, because the differential segments to the right and to the left of z axis will cancel each other.Only dEx is present, and this component is a function of x alone.

Chapter 2

Field of a Sheet of ChargeThe contribution of a strip to Ex at P is given by (l = s dy)

Adding the effects of all the strips,

Chapter 2

Charge per unit length =C

Field of a Sheet of ChargeFact: The electric field is always directed away from the positive charge, into the negative charge.We now introduce a unit vector aN, which is normal to the sheet and directed away from it.

The field of a sheet of charge is constant in magnitude and direction. It is not a function of distance.

Chapter 2

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Evaluation of the E field due to an infinite sheet of charge

Chapter 3Electric Flux Density, Gauss Law,and Divergence

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Faraday ExperimentHe started with a pair of metal spheres of different sizes; The larger one consisted of two hemispheres that could be assembled around the smaller sphere

Faraday Apparatus, Before GroundingThe inner charge, Q, induces an equal and opposite charge, -Q, on the inside surface of the outer sphere, by attracting free electrons in the outer material toward the positive charge. This means that before the outer sphere is grounded, charge +Q resides on the outside surface of the outer conductor.

Faraday Apparatus, Before Grounding

+Q

Faraday Apparatus, After GroundingAttaching the ground connects the outer surface to an unlimited supply of free electrons, which then neutralize the positive charge layer. The net charge on the outer sphere is then the charge on the inner layer, or -Q.

q = 0

ground attachedFaraday Apparatus, After Grounding

Interpretation of the Faraday Experiment

q = 0

Faradays ExperimentFaraday concluded there was a sort of displacement from the charge on the inner sphere through the insulator to the outer sphere and is now known as electric flux. The electric flux originate on a positive charge and terminates on a negative charge and It is independent of the medium. In absence of negative charge it terminates at infinity

Faradays ExperimentThe electric displacement (or electric flux) is equal in magnitude to the charge that produces it, independent of the insulating material and the size of the spheres. The net flux passing normal through a unit surface area is called the electric flux density D. Its direction is normal to the surface area and is a vector field. Electric Flux = Charge (unit is coulomb)

Faradays ExperimentOne coulomb of electric charge give rise to one coulomb of electric fluxFor a point charge Q at the center of a sphere of radius a D = total flux/area =d/dS (magnitude)(C/m2)=Q/4a2

Electric Flux Density

q = 0

q = 0

Vector Field Description of Flux Density

The direction is the outward radial direction At each surface, we would have:

q = 0

D(r)Radially-Dependent Electric Flux Density

rAt a general radius r betweenspheres, we would have:

Expressed in units of Coulombs/m2, and defined over the range (a r b)

Point Charge FieldsIf we now let the inner sphere radius reduce to a point, while maintaining the same charge, and let the outer sphere radius approach infinity, we have a point charge. The electric flux density is unchanged, but is defined over all space:

C/m2 (0 < r