Section2-2 - Slide Presentations for ECE 329, Introduction...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Slide Presentations for ECE 329, Introduction to Electromagnetic Fields, to supplement “Elements of to Engineering Electromagnetics, Sixth Edition” Edition” by by 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 Nannapaneni Narayana Rao 2.2 The Surface Integral 2.2-3 The Surface Integral Flux of a vector crossing a surface: B an ∆S B ∆S B an α Flux = (B)(∆ S) an Flux = 0 Flux = ( B cos α ) ∆S = B ∆S cos α = B • ∆S a n = B • ∆S ∆S 2.2-4 Normal anj αj ∆ Sj S Bj Flux = = j =1 n j =1 ∑ ∆ψ j ∑B j • ∆S j n In the limit n → ∞ , Flux, ψ = ∫S B • dS = Surface integral of B over S. 2.2-5 = ∫S B • dS Surface integral of B D2.4 over the closed surface S. z 2 y A = x ( ax + a y ) x = 0, a n = ± a x (a) A = 0, A • d S = 0 ∫ A • dS = 0 ∫ A • dS =0 x 2 2.2-6 (b) A = 2 ( ax + a y ) x = 2, an = ± ax z 2 y d S = ± dy dz ax A • dS = ± 2 dy dz 2 x 2 2 2 ∫ A • dS = ± ∫y =0 ∫z =0 2 dy dz = ± 8 ∫ A • dS = 8 2.2-7 z (c) A = x ( ax + a y ) y = 0, an = ± a y 2 d S = ± dz dx a y A gd S = ± x dx dz y 2 x 2 2 ∫ A • dS = ± ∫x =0 ∫z =0 x dx dz = ± 4 ∫ A • dS =4 2.2-8 (d) From (c), A • dS = ± x dx dz 2– x 2 ∫ A • dS = ± ∫x =0 ∫z =0 x dx dz 2 = ± ∫0 x (2 – x ) dx z 2 x+z=2 y x 2 4 =± 3 4 ∫ A • dS = 3 ...
View Full Document

This note was uploaded on 01/22/2011 for the course ECE 329 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.

Page1 / 8

Section2-2 - Slide Presentations for ECE 329, Introduction...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online