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 1.7 Lorentz Force Equation 1.7-3 Lorentz Force Equation Fe qE Fm qv × B Fm B F = Fe Fm F = q E + v × B For a given B, to find E, F E – v B q One force is sufficient q E v Fe 1.7-4 D1.21 B0 B ax 2a y 2az Find E for which acceleration experienced by q is zero, for a given v F q E v × B 0 E = v × B (a) v = v0 a x a y a z v0 B0 E= ax a y a z × a x 2a y 2a z v0 B0 a y az 1.7-5 (b) v = v0 2ax a y 2az v0 B0 E 2ax a y 2az × a x 2a y 2az v0 B0 2ax 2a y az (c) v = v0 along y z 2 x v0 ax 2a y a z v0 E ax 2a y 2a z × a x 2a y 2a z 0 1.7-6 For a given E, to find B, F v B – E q One force not sufficient Two forces are needed F1 v1 B – E C1 q F2 v B – E C2 q C1 × C2 v1 × B × v × B v1 × B B v2 v1 × B v B = C1 v2 B 1.7-7 C2 C1 B C1 • v provided C1 • v2 0, which means v2 and v1 should not be collinear (since C1 • v1 = 0) 1.7-8 P1.54 For v = v1 or v = v2, test charge moves with constant velocity equal to the initial value It is to be shown that for mv1 nv v , where m + n 0, m n the same holds qE qv1 B 0 (1) qE qv B 0 (2) qE qv B 0 (3) 1 3 v1 v × B = 2 3 v2 v × B = 1.7-9 Both v1 v and v2 v are collinear to B v1 v k v2 v v1 – kv v 1– k mv1 mv m n n for k = – m Alternatively, m n (1) (2) mn mn 1.7-10 mv1 nv qE q B 0 m n mv1 nv v mn