Lecture Notes On Electromagnetic Field, PDF

Electromagnetic Field Dr. Wafaa Rady  Course Grades 30 % Midterm [Week 7] 10% Assignments / Participation [ LC &TU] 20% Quizzes 40% Final Exam  Text Book  Lecture Notes  Engineering Electromagnetics 8th Edition William H. Hayt...

Electromagnetic Field Dr. Wafaa Rady  Course Grades 30 % Midterm [Week 7] 10% Assignments / Participation [ LC &TU] 20% Quizzes 40% Final Exam  Text Book  Lecture Notes  Engineering Electromagnetics 8th Edition William H. Hayt 2012 - ISBN 978-0-07-338066-7  Electric and Magnetic Field Electric Field Magnetic Field Electric Field Intensity E Magnetic Field Intensity H Electric Flux Density D = 𝜺𝑬 Magnetic Flux Density 𝑩 = 𝝁𝑯 , 𝝁o= 1.26x10-6 Electric Flux 𝜓 = ‫𝑫 ׭‬. 𝒅𝒔 Magnetic Flux 𝜑 = ‫𝑩 ׭‬. 𝒅𝒔 𝑑𝜑 emf = - volt 𝑑𝑡 Charge 𝑄 = ‫𝑫 ׭‬. 𝒅𝒔, 𝑄 = ‫𝝆 ׭‬. 𝒅𝒗 Current 𝐼 = ‫𝑱 ׭‬. 𝒅𝒔  Vector Analysis 1- Vector Components in Rectangular Coordinate System A point is located by giving its x, y, and z coordinates. These are, respectively, the distances from the origin to the intersection of perpendicular lines dropped from the point to the x, y, and z axes. Example The location of points P(1, 2, 3) and Q(2,−2, 1). P(1, 2, 3)  x= 1, y=2, z = 3 Q(2,−2, 1)  x= 2, y=-2, z = 1 To describe a position vector in the rectangular coordinate system: rP = ax + 2ay + 3az rQ = 2ax - 2ay + az  Vector Analysis To describe a position vector in the rectangular coordinate system: rP = ax + 2ay + 3az rQ = 2ax - 2ay + az The desired vector from P(1, 2, 3) to Q(2,−2, 1) is therefore: RPQ = rQ − rP = (2 − 1)ax + (−2 − 2)ay + (1 − 3)az RPQ = ax − 4ay − 2az  Vector Analysis 2- Vector Magnitude and Unit vector Any vector B then may be described by B = Bx ax + By ay + Bz az. The magnitude of B written |B| unit vector in the direction of the vector B is: or simply B, is given by  Vector Analysis 3- Vector Multiplication A. Dot Product Given two vectors A and B, the dot product, or scalar product, is defined as the product of the magnitude of A, the magnitude of B, and the cosine of the smaller angle between them, Note: Because the sign of the angle does not affect the cosine term. Finding the angle between two vectors in three-dimensional space is often a job we would prefer to avoid, and for that reason the definition of the dot product is usually not used in its basic form. A more helpful result is obtained by considering two vectors whose rectangular components are given, such as A = Ax ax + Ay ay + Az az and B = Bx ax + By ay + Bz az. A· B yields the sum of nine scalar terms Because the angle between two different unit vectors of the rectangular coordinate system is 90◦, we then have SCALAR Example  Gauss’s Law and Divergence Relation between electric flux density and volume charge density: 𝜕𝐷𝑥 𝜕𝐷𝑦 𝜕𝐷Ζ 𝛻.𝐷 = + + 𝜕𝑥 𝜕𝑦 𝜕Ζ  Gauss’s Law and Divergence Example: 𝜕𝐷𝑥 𝜕𝐷𝑦 𝜕𝐷Ζ 𝑎) 𝛻. 𝐷 = + + 𝜕𝑥 𝜕𝑦 𝜕Ζ = 2𝑦Ζ − 2𝑥 + 0 𝛻. 𝐷 𝑎𝑡 𝑝(2,3,−1) = 2 ∗ 3 ∗ −1 − 2 ∗ 2 = −10 𝐶/𝑚3 Practice Yourself  complete solving b and c  Vector Analysis 3- Vector Multiplication B. Cross Product Given two vectors A and B, we now define the cross product, or vector product, of A and B. The cross product A × B is a vector; the magnitude of A × B is equal to the product of the magnitudes of A, B, and the sine of the smaller angle between A and B; the direction of A×B is perpendicular to the plane containing A and B Note: B×A = −(A×B) Because the sign of the angle affects the sine term. For two vectors whose rectangular components are given, such as A = Ax ax + Ay ay + Az az and B = Bx ax + By ay + Bz az. Ax B yields the sum of nine vector terms Vector  Curl write the point form of Ampere’s law Example Calculate the value of the vector current density in rectangular coordinates at PA(2, 3, 4) if H = x2z ay − y2x az; Hx = 0 , Hy= x2z , Hz =− y2x J= (-2yx –x2) ax – (-y2 -0) ay + ( 2xz -0)az at PA(2, 3, 4) Ans. −16ax + 9ay + 16az A/m2;  Gradient Obtain Electric Field from Potential Function The operation on V by which E is obtained is known as the gradient, and the gradient of a scalar field T is defined as  Energy and Potential Example  Vector Analysis 4- OTHER COORDINATE SYSTEMS: CYLINDRICAL COORDINATES CYLINDRICAL CIRCULAR COORDINATES A. Point to Point Transformation CIRCULAR COORDINATES  Vector Analysis 4- OTHER COORDINATE SYSTEMS: CYLINDRICAL CIRCULAR COORDINATES B. Vector to Vector Transformation A = Ax ax + Ay ay + Az az A = Aρ aρ + Aφ aφ + Az az A = Ar ar + Aθ a θ + Aφaφ A =A.a k k Ak = A. ak Ak = A. ak A ρ =( Ax ax + Ay ay + Az az ). a ρ Ar =( Ax ax + Ay ay + Az az ). a r A φ =( Ax ax + Ay ay + Az az ). a φ Aθ =( Ax ax + Ay ay + Az az ). a θ A z =( Ax ax + Ay ay + Az az ). a z Aφ =( Ax ax + Ay ay + Az az ). a φ Thank You