Section 1 6 TRƯỜNG ĐIỆN TỪ

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 1.6 The Magnetic Field 1.6-3 The Magnetic Field acts to exert force on charge when it is in motion Fm qv B B Fm v q B = Magnetic flux density vector Alternatively, since charge in motion constitutes current, magnetic field exerts forces on current dFm elements B dFm I dl B I dl 1.6-4 N N–m Units of B: = Am Am Wb = =T m Sources: Currents; Time-varying electric field 1.6-5 Ampère’s Law of Force I1 a21 dl2 I2 a dl1 12 R  0 I d l × a 21  d F1 I1d l1 ×   4 R   I1d l1 × B  0 I1 d l1 × a12  d F2 I d l ×    R   I d l × B1 1.6-6 Magnetic field due to a current element (Biot-Savart Law)  I dl a R B 4 R2 Note B  sin  B R I dl  aR R P B B circular to the axis of the current element 0  Permeability of free space = 4 10 –7 H m 1.6-7 Ex I d l I dx a x  a y A situated at  1,  2,    Find B at  2,  1,  0 I d l × a R 0 I d l × R B  4 R 4 R3 R   since a R   R  R =   1 a x      a y     a z a x  a y  a z 1.6-8 0 I dx  a x  a y  ×  a x  a y  a z  B= 4   0 I dx  ax  a y   12 3 1.6-9 Current Distributions I (a) Filamentary Current I (A) Thin wire (b) Surface Current Surface current density, JS (A/m) JS w I   JS  w max 1.6-10 (c) Volume Current Density, J (A/m2) area A J I   J    A max 1.6-11 P1.44 z a2 P(r, , z) 2  dz aR z – z z I r y 1 x a1 r 0 dz a z × a R dB  2  4 r   z  z     0 I dz sin  a  2  4 r   z  z     1.6-12 z a B  z a1 0 I  4 r z  z cot   r dB d z  sin  a 1     z  z  r    0 I  cosec 2 sin  d  a     4 r cosec  0 I 2   cos     1 a 4 r 0 I   cos 1  cos   a 4 r 1.6-13 For infinitely long wire, a1  – , a2  , 1  ,    0 I B a 2r 1.6-14 Magnetic Field Due to an Infinite Plane Sheet of Uniform Surface Current Density This can be found by dividing the sheet into infinitely long strips parallel to the current density and using superposition, as in the case of finding the electric field due to an infinite plane sheet of uniform surface charge density Instead of going through this procedure, let us use analogy To this, we first note the following: 1.6-15 (a) Point Charge P ar Current Element E R P I dl R B ar Q Q E a 4 R R 0 I dl a R B 4R 1.6-16 (b) Line Charge z L0 r Line Current z I P ar E r=0 L0 E ar 2 r r P B ar r=0 0 I B a 2r 0 I  a z a r 2r 1.6-17 Then, (c) Sheet Charge S0 P Sheet Current JS an E S0 E an 2 P B an 0 B JS a n