E field line of dipoles

University Electromagnetism:
Electric field of a line of dipoles ( 3 directions)

1. Line of dipoles 1
Electrical dipoles
on a thin long line
© Frits F.M. de Mul

2. Line of dipoles 2
Electrical dipoles
on a thin long line
Available:
Thin line, infinitely long,
homogeneously filled with dipoles,
each with dipole moment p [Cm]
Question:
Calculate E-field in arbitrary points
around the line

3. Line of dipoles 3
Electrical dipoles
on a thin long line
•Analysis and symmetry
•Approach to solution
•Calculations
•Conclusions

4. Line of dipoles 4
Analysis and Symmetry (1)
1. Cylinder: infinitely long and thin
2. Distribution of dipoles:
n dipoles / meter; homogeneous;
all directions uniform
each: dipole moment p [Cm]
3. Coordinate axes: X,Y,Z
Z-axis = symm. axisY
X
Z
e
z
4. Cylinder symmetry:
e
r
j
all points at equal r are equivalent,
even if at different z or j

5. Line of dipoles 5
Analysis and Symmetry (2)
p
Dipole :
q
r
P
e
r
e
q
E
r
3
0
4
cos2
r
p
E
r
pe
q
=
E
q
3
04
sin
r
p
E
pe
q
q
=
Assume: field components of dipole field at
some distance in point P àre known.

6. Line of dipoles 6
Analysis and Symmetry (3)
Possible directions of the dipoles:
X
Z
e
z
e
r
j
Y
X
Y
X
Z
e
z
e
r
j
Y
Y
X
Z
e
z
e
r
j
Z

7. Line of dipoles 7
Approach to solution
Z
Y
P
y
P
Choose: Point P at
distance y
P
from axis.
Use symmetry:
r
+
r
-
Dipole elements at
+z and -z will contribute
symmetrically
dp in dz at +z
dp in dz at -z
+z
-z
O
dp = n p dz
n = dipole density
(per meter)

8. Line of dipoles 8
Calculations (1): Dipoles // X-axis
P
dp in dz at +z
dp in dz at -z
Z
+z
-z
y
P
r
+
r
-
O
Y
X
Case 1:
all dipoles // +X-axis
0
4
cos2
2
3
0
==
r
dp
dE
r
pe
q
dE
q +
=dE
q-
3
0
3
0
24
sin
2
r
dp
r
dp
dE
pepe
q
q
==
==> E // -X-axis
E
result
q = 90
0

==>
e
q +
=e
q-
q

9. Line of dipoles 9
Calculations (2a): Dipoles // Y-axis
dp in dz at -z
P
dp in dz at +z
Z
+z
-z
O
Y
X
q
+
q
-
Case 2:
all dipoles // +Y-axis
e
q +
e
r+
e
q -
e
r- dE
r
~ cos q
dE
q
~ sin q
dipole +z -z

>0
dE
r+
>0
dE
r-
dE
q +
>0
dE
q
-
>0
resulting vector :
E
result
==> E // Y-axis
Find directions of the
field contributions dE

10. Line of dipoles 10
Calculations (2b): Dipoles // Y-axis
P
Z
+z
-z
O
Y
X
dE
q +
dE
q
-q
+
q
-
e
q -
e
q +
e
r-
e
r+
dE
r-
dE
r+
E
result
r
q
q
q
dp = n.p.dz ; q
+
=q
-
=q
Y-components:
dE
r+
and dE
r-
: each:
qq
pe
coscos2
4
3
0
÷
÷
ø
ö
ç
ç
è
æ
r
dznp
dE
q+
and dE
q -
: each:
qq
pe
sinsin
4
3
0
÷
÷
ø
ö
ç
ç
è
æ
-
r
dznp

11. Line of dipoles 11
Calculations (3a): Dipoles // Z-axis
P
dp in dz at +z
dp in dz at -z
Z
+z
-z
O
Y
X
q
+
q
-
Case 3:
all dipoles // +Z-axis
e
q +
e
r+
e
q -
e
r- dE
r
~ cos q
dE
q
~ sin q
dipool + -

<0
dE
r+
>0
dE
r-
dE
q +
>0
dE
q
-
>0
resulting vector :
dE
result
==> E // Z-axis

12. Line of dipoles 12
Calculations (3b)
P
Z
+z
-z
O
Y
X
dE
q +
dE
q
-
q
+
q
-
e
q -
e
q +
e
r-
e
r+
dE
r-
dE
r+
dE
result
r
r
q
-
p/2-q
-
dp = n p dz
q
+
= p - q
-
Z-components:
dE
r+
and dE
r-
: each:
--
÷
÷
ø
ö
ç
ç
è
æ
qq
pe
coscos2
4
3
0
r
dznp
dE
q+
and dE
q -
: each:
--
÷
÷
ø
ö
ç
ç
è
æ
- qq
pe
sinsin
4
3
0
r
dznp

13. Line of dipoles 13
Conclusions Conclusions
X
Z
e
z
e
r
j
Y
dipole direction: dE - contributions:
X x
edE
3
0
2r
dznp
pe
-=
Y ( )
y
edE qq
pe
22
3
0
sincos2
4
-=
r
dznp
Z ( )
z
edE qq
pe
22
3
0
sincos2
4
-=
r
dznp
(q and r are measured from dipole to P )
the end