Section 1 7 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

1.7 Lorentz Force Equation

1.7-3 Lorentz Force Equation

For a given B, to find E,

E F

q – v B

 One force is sufficient

Fm

q

B

E

v

Fe

e

m

q q q







F = F F

F = E + v×B

B

B a a a

q

F E v×B

E = v×B

Find E for which acceleration experienced by q is

zero, for a given v

0

0 0

0 0

3

v

v B

v B

 

0

0 0

0 0

3

v

v B

v B

 

0 0 0

along 2

2 3

3 0

v v

 



v =

(b)

(c)

For a given E, to find B,

One force not sufficient Two forces are needed.

v B F

q – E

v1 B F1

q – E C1

v2 B F2

q – E C2





= C v B



 B C2 C1

C1 • v2

provided C1 • v2 0, which means v2 and v1 should not be collinear (since C1 • v1 = 0)



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

the same holds.

v  mv1  nv2

m  n , where m + n 0,

qE  qv1 B 0

qE  qv2 B 0

qE  qv B 0

(1) (2) (3)

1 2

1 3

2 3

v v ×B = 0

v v ×B = 0

v  v1 – kv2

1 – k

 mv1  mv2

m  n for k = – n m

Alternatively,

(1)  m

m  n  (2)  n m  n 

Both and are collinear to k

qE  q mv1  nv2

m  n





 B 0

 v  mv1  nv2

m  n