Lecture 19
Flux Linkage and Inductance
Sections: 8.10
Homework: See homework file
LECTURE 19
slide 1
Flux Linkage in Coils
• the sum of all fluxes piercing the surfaces bounded by all
turns (the total flux “linking” the turns)
• ideal scenario: each turn creates the same flux Φ and this flux
“links” all turns (no leakage)
  N , Wb
I
1
• if flux density B is uniform inside the coil the
flux linkage is
2
B  NI    N 2 I

  NBS , Wb
N
• this is the coil’s self-flux linkage, i.e., the
flux links to the current which generates it
LECTURE 19
  BS
  N
slide 2
Self and Mutual Flux Linkage
• flux linkage may be self-flux linkage and mutual-flux linkage
• self-flux linkage links current which it is due to
• every coil has self-flux linkage
• mutual-flux linkage is due to
currents in other inductors
I1
B1
• mutual-flux linkage implies
two coupled coils
LECTURE 19
1
I2 1
2
2


1  11  12 ,  2   22   21
11  N111  N12 I1
12  N112  N1 N 2 I 2
 22  N 2  22  N 22 I 2
 21  N 2  21  N 2 N1I1
B2
11
N1
12
 21
slide 3
N2
 22
Self and Mutual Inductance
• self inductance L shows the amount of self-flux linkage due to unit
current
 self
L
, H = Wb/A
I
• mutual inductance M shows the amount of mutual flux linkage
due to unit current producing this mutual flux
 mutual
M
,H
I
• self inductance (or simply inductance) is defined for a single
inductor
• mutual inductance is defined for a pair of inductors
LECTURE 19
slide 4
Example – Self Flux Linkage and Inductance of Toroid
N I
H 
, H/m, 1     2 (see L16)
2
h 2


0 1
C
B
NIh   2 
 H  d  dz  
ln  

2  1 
2
0
1
B
N 2 Ih   2 
ln   , Wb
  N  
2
 1 
N 2h  2 
ln   , H
L
2
 1 
• if toroid is thin, field is mostly uniform
N I
NIA
N 2A
H 
 
 L
20
20
20
d av
LECTURE 19
w
I

z
h
I
B
0  0.5( 1   2 )
A  h (  2  1 )



w
slide 5
Homework
Find the inductance of a toroid consisting of 200 turns. Its crosssection is of width and height w = h = 1 cm and its inner radius is
ρ1 = 1.5 cm. The toroid has a ferromagnetic core whose relative
permeability is μr = 400.
[Ans.: L ≈ 16.3 mH]
LECTURE 19
slide 6
Example for Mutual Flux Linkage and Mutual Inductance
• two coaxial solenoids (assume uniform field is in cross-sections)
solenoid #1: radius a, length D1, N1 turns a
solenoid #2: radius b, length D2, N2 turns
b  a, D2  D1
b
• solenoid #2 is smaller than solenoid #1 and is positioned in its
middle
N 1 I1
N1 I1 b2
2
H1 
  21   H1   b  
D1
D1
N1 N 2 I1 b2
  21  N 2  21  
D1
 M 21
N1 N 2 b2
 21


,H
I1
D1
LECTURE 19
slide 7
Physical Significance of Flux Linkage: Faraday's Law
Why is the flux linkage important rather than the flux itself? The
answer is given by Faraday’s law of EM induction
d

 e, V
dt
If the magnetic flux changes in time, electromotive force e (V) is
induced in each turn of the coil. Each turn acts like an AC voltage
source. These equivalent voltage sources are connected in series,
thus producing an overall voltage at the coil’s terminations as
d
d
E  Ne   N
, V

dt
dt
It is the rate of change in time of the flux linkage, which determines
the induced voltage of a coil:
d
dI
dI k
V  E 
 L , V or V j  M jk
dt
dt
dt
LECTURE 19
slide 8
Inductance per Unit Length: Coaxial Cable
linkage per unit length determines inductance per unit length

L 
, H/m
I
I c b
0 I
, T (Ampère’s law)
2
0 I


d   d   B d  
d
2
B 
H
b
0 I
0 I  b 
ln   , Wb/m
d 
2
2  a 
a
  
L2 
0  b 
ln   , H/m
2  a 
LECTURE 19
a
slide 9
Inductance per Unit Length: Parallel-plate Line
w
• express flux linkage PUL in terms of H field
    B (
hl )   H ( hl )     Hh
area
l
• express H field in terms of current I using
Ampère’s law
C
 H  dL  Hw  I
h

h
 0 , H/m
I
w
LECTURE 19
I
H
• find inductance PUL Lꞌ
L 
Ampère’s
contour
slide 10
Obtaining Inductance from Capacitance Expressions
• compare the capacitance and inductance PUL expressions for a
coaxial line

L 
ln  b / a  , H/m
2
2
C 
, F/m
ln(b / a )
• and for a parallel-plate line
w
h


C


L 
h
w
• it can be shown that for any infinitely long TL structure
 LC   
the inductance PUL formula can be obtained from the reciprocal
of the capacitance formula where ε is replaced by 1/μ
LECTURE 19
slide 11
Inductance per Unit Length: Twin-Lead Line
C 

2
h

h


ln      1 
r

r


, F/m
if h  r
2
 
h
h
0  h
0  2 h 



0

ln     1 
arccosh   , H/m L 
 L 
ln  
 r
  r 
 
r
r


y
h
h
P
1
s
B
I
A
s
Az  0
r
2
0
equipotentials Az
LECTURE 19
r
x B
I
A
slide 12
You have learned:
that the inductance of a structure is proportional to the flux
linkage
that the flux linkage is proportional not only to the number of turns
generating the flux but also to the number of turns intercepting this
flux
that the self inductance of a coil is proportional to N2
that the mutual inductance of a pair of coils is proportional to N1N2
how to calculate the inductance per unit length of transmission lines
LECTURE 19
slide 13