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No Slide Title Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jor[.]
Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of Engineering Electromagnetics, Sixth Edition” by Nannapaneni Narayana Rao Edward C Jordan Professor of Electrical and Computer Engineering University of Illinois at Urbana-Champaign, Urbana, Illinois, USA Distinguished Amrita Professor of Engineering Amrita Vishwa Vidyapeetham, Coimbatore, Tamil Nadu, India 1.1 Vector Algebra 1.1-3 (1) Vectors (A) vs Magnitude and direction Ex: Velocity, Force Scalars (A) Magnitude only Ex: Mass, Charge (2) Unit Vectors have magnitude unity denoted by symbol a with subscript A A1a1 A2 a A3a aA A A12 A22 A32 Useful for expressing vectors in terms of their components 1.1-4 (3) Dot Product is a scalar A A A • B = AB cos B B Useful for finding angle between two vectors A • B cos AB A A1a1 A2 a A3 a B B1a1 B2 a B3a A1 B1 A2 B2 A3 B3 A1 A22 A32 B12 B22 B32 1.1-5 (4) Cross Product is a vector B A A B = AB sin an right hand screw A B an is perpendicular to both A and B Useful for finding unit vector perpendicular to two vectors A B A B an AB sin A B 1.1-6 where a1 A B A1 B1 a2 A2 B2 a3 A3 B3 (5) Triple Cross Product A (B C) is a vector A (B C) B (C A) C (A B) in general 1.1-7 (6) Scalar Triple Product A • B C B • C A C • A B A1 A2 A3 B1 B2 B3 C1 C2 C3 is a scalar 1.1-8 Volume of the parallelepiped Area of base Height A × B C a n A × B C C A × B A B ×C A×B A×B an C B A 1.1-9 D1.2 A = 3a1 + 2a2 + a3 B = a1 + a2 – a3 C = a1 + 2a2 + 3a3 (a) A + B – 4C = (3 + – 4)a1 + (2 + – 8)a2 + (1 – – 12)a3 = – 5a2 – 12a3 A B – 4C 25 144 13 1.1-10 (b) A + 2B – C = (3 + – 1)a1 + (2 + – 2)a2 + (1 – + 3)a3 = 4a1 + 2a2 – 4a3 Unit Vector 4a1 2a – 4a = 4a1 2a – 4a = (2a1 a – 2a ) 1.1-11 (c) A • C = 1 + + = 10 a1 a (d) B C 1 a3 –1 = (3 2)a1 (–1 – 3)a (2 – 1)a = 5a1 – 4a2 + a3 1.1-12 (e) A • B C 1 –1 = 15 – + = Same as A • (B C) = (3a1 + 2a2 + a3) • (5a1 – 4a2 + a3) = 5 + (–4) + 1 = 15 – + = 1.1-13 P1.5 D A E B C Common Point D= B–A A + D = B) ( E= C–B B + E = C) ( D and E lie along a straight line 1.1-14 D× E 0 B A × C B 0 B × C A × C B × B A × B 0 A × B + B ×C + C× A = What is the geometric interpretation of this result? 1.1-15 Another Example Given a1 × A a 2a3 a × A a1 2a3 (1) (2) Find A A =C a 2a3 × a1 2a3 a1 a a3 C C 2a1 2a a3 2 1.1-16 To find C, use (1) or (2) a1 C 2a1 2a2 a3 a2 2a3 C 2a3 a2 a2 2a3 C 1 A 2a1 2a2 a3