This preview shows pages 1–3. Sign up to view the full content.
Internal Inductance of Conductor in H/m:
L
int
4
π
⋅
10
7
−
⋅
8
π
⋅
H
m
⋅
=
1
2
10
7
−
⋅
H
m
⋅
=
Internal Inductance of Conductor in H/m:
L
int
λ
int
I
=
1
8
µ
o
π
⋅
=
Result of calculation of internal Flux
Linkage per unit length in z direction:
λ
int
1
8
µ
o
⋅
I
π
⋅
=
Integrate internal Flux Linkage (per unit length)
over the complete range of the conductor radius.
λ
int
0
r
x
µ
o
I
⋅
x
3
⋅
2
π
⋅
r
4
⋅
⌠
⌡
d
=
λ
int
1
20000000
henry
m
⋅
I
⋅
=
→
Flux at radial position x is linked with Current
I
by a ratio of
x
2
/
r
2
:
d
λ
int
x
2
r
2
d
Φ
⋅
=
µ
o
I
⋅
x
3
⋅
2
π
⋅
r
4
⋅
dx
⋅
=
Calculate Flux per unit length of conductor:
d
Φ
B
x
dx
⋅
=
Calculate
B
x
from
H
x
:
B
x
µ
o
H
x
⋅
=
µ
o
I
⋅
x
⋅
2
π
⋅
r
2
⋅
=
Substitute
I
x
into Formula for
H
x
:
H
x
I
x
2
r
2
⋅
2
π
⋅
x
⋅
=
Ix
⋅
2
π
⋅
r
2
⋅
=
Solve for
I
x
:
I
x
I
x
2
r
2
⋅
=
I
x
I
π
x
2
⋅
π
r
2
⋅
=
Calculate Ratio of
I
x
and
I
considering constant
current Density:
Solve for
H
x
:
H
x
I
x
2
π
⋅
x
⋅
=
Solution for cylindrical conductor:
2
π
⋅
x
⋅
H
x
⋅
I
x
=
Start with Ampere's Law:
l
H
tan
⌠
⌡
dI
enclosed
=
Dr. M. Giesselmann, Feb272008:
Internal Inductance of Conductors:
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentComplete internal and external Flux Linkage
between wire with radius r and return wire at distance
D
:
λ
p
1
2
10
7
−
⋅
I
⋅
210
7
−
⋅
I
⋅
ln
D
r
⋅
+
=
unit length in z direction between Distances
D
1
and
D
2
from reference point:
L
12
,
7
−
⋅
ln
D
2
D
1
⋅
=
Result of calculation of external Flux Linkage per unit
length in z direction between Distances
D
1
and
D
2
from
reference point:
λ
,
7
−
⋅
I
⋅
ln
D
2
D
1
⋅
=
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 GIESSELMANN

Click to edit the document details