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)
mn
mn
1.7-10
mv1 nv 2
qE q
B 0
m n
mv1 nv 2
v
mn