Section 1 7 TRƯỜNG ĐIỆN TỪ

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 
3
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 

3
 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 

3
v0 B0 2ax  2a y  az 
(c) v = v0 along y  z 2 x

v0
  ax  2a y  a z 
3
v0
E   ax  2a y  2a z × a x  2a y  2a z 
3

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 2 B 
– E C2
q

C1 × C2 v1 × B × v 2 × B 

 v1 × B  B  v2  v1 × B  v 2 B

=  C1  v2 B


1.7-7

C2 C1
B 
C1 • v 2
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 2
v
, where m + n 0,
m n
the same holds.

qE  qv1 B 0

(1)

qE  qv 2 B 0

(2)

qE  qv B 0

(3)

1  3  v1  v × B = 0
2   3  v2  v × B = 0


1.7-9

Both  v1  v  and  v2  v  are collinear to B.
  v1  v  k  v2  v 
v1 – kv 2
v
1– k
mv1  mv 2

m n

n
for k = –
m

Alternatively,
m
n
(1) 
 (2) 

mn
mn

1.7-10

mv1  nv 2 
qE  q 
B 0
 m  n 
mv1  nv 2
v 
mn