Calculation of Total Inductance of a Straight Conductor

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Calculation of Total Inductance of a Straight
Conductor of Finite Length
PRIYANKA PATEL
C.U. Shah Science College, Ashram Road, Ahmedabad 380014
E-mail: jghosh04@yahoo.com
ABSTRACT
This article presents calculation of total inductance of a straight
conductor. Surprisingly, it is not easy to find an expression for the
inductance of a straight piece of conductor. In this article, the total
inductance of a straight conductor is calculated using the Biot-Savart
law. The result is very useful as a building block for more complex
structures, and also because the inductance of a conductor is important
in high frequency or high speed electrical circuits.
Introduction
The pioneer experiments of Faraday on the
induction of a current in a conductor when the
magnetic flux linked with the conductor is
changed led to two very important laws in
physics. First is the Faraday’s Law and the
second is the Lenz’s Law. Faraday’s law states
that the induced electro motive force (e.m.f.) in
a circuit, when the magnetic field around it is
changed, is equal to the rate of variation of the
number of lines of magnetic induction linked
with the circuit. Lenz’s law states that the
Physics Education • July − September 2009
direction of the induced current due to the
induced electro motive force is such as to try to
stop the action taking place, i.e. to oppose the
variation of the magnetic flux linked with the
circuit. The magnetic flux is made to increase
positively, so the e.m.f. opposes this change.
This phenomenon is called the electromagnetic
induction. Consequences of this phenomenon
of electromagnetic induction are the effects of
self-induction and mutual induction.
The coefficient of self-inductance, or the
self-inductance L, is defined as the flux linked
with the circuit when unit current flows in it.
193
Elaborating further, when an electric current I
flowing round a circuit it produces a magnetic
field and hence a magnetic flux Φ through the
circuit. The ratio of the magnetic flux to the
current is called the inductance, or more
accurately self-inductance of the circuit. The
term was coined by Oliver Heaviside in
February 1886. It is customary to use the
symbol L for inductance, possibly in honour of
the physicist Heinrich Lenz. The quantitative
definition of the inductance is therefore L=ΦI,
where I is the current in the curcuit. It follows
that the SI units for inductance are webers per
ampere. In honour of Joseph Henry, the unit of
inductance has been given the name henry (H):
1H = 1Wb/A. Hence when I =1 Ampere, L=Φ.
Calculation of Inductance of a Straight
Conductor
Although the inductances of circuit elements
not associated with magnetic materials are
independent of the value of the current and
dependent only on the geometry of the system,
it is only in the simplest cases that these
constants can be calculated exactly.
Surprisingly, it is not easy to find an
expression for the inductance of a straight
piece of conductor of a finite length. And yet
this result is very useful as a building block for
more complex structures, and also because the
inductance of a conductor is important in high
frequency or high speed electrical circuits.
If we consider the conductor in isolation we
ignore the question of how the current gets to
the conductor. But that current, however it is
delivered, will affect the flux, which is
developed in the vicinity (outside) of the
conductor and also inside the conductor. In this
report we calculated the flux developed per
unit current outside and inside the conductor
separately and define it as external and internal
inductance respectively. The total selfinductance or the inductance of the straight
194
conductor is then given by adding the external
and internal inductances.
(a)
ance
Calculation of External Induct-
We defined the external inductance of the
straight conductor as the flux developed
outside the conductor due to unit current
passing through the conductor in the region
bounded by lines perpendicular to the
beginning and end of the conductor. We
derived the formula in the simplest possible
manner, using the law of Biot and Savart in the
differential form as it gives a better physical
view of the various problems considered.
Let AB in Figure 1 is a straight conductor of
length ‘l’. Let I current pass through the
conductor. First we use the Biot-Savart’s law
to determine the magnetic field at a given point
P outside the conductor due to constant current
I. Let dl’ be a small length element of the
conductor.
The magnetic field dB, at point P due to
current I in a small length element of the
conductor dl’, according to bio-savart’s law is
given by (Figure1)
dB =
μ 0 I dl × r
4π r 3
(1)
where, dl = length of small element, r =
distance from point P to small element,
μ0=permeability constant = 4π × 10−7 Tm/A.
Let us assume that the conductor is placed
along X-axis and P is any point on Y-axis then
the direction of dBwill be along Z-axis. If we
represent the direction of X, Y and Z axis by
iˆ , ĵ and k̂ respectively. Thus the magnitude
of the magnetic field at point P is
B=
μ0l
[sin θ 2 + sin θ1 ]
4π h
(2)
Physics Education • July − September 2009
θ3
dl
P
dh
θ
θ1
θ2
r2
r1
r
h
Q
I
A
I
O
B
dl’
x
l
Figure 1: Magnetic field calculation at point P outside the conductor due to current I
passing through the conductor AB of length l.
Equation (2) gives the magnetic field at a
particular point P due to the current I passing
though a conductor of length l. Now, the
magnetic flux dΦ in a differential area at a
fixed distance h from the conductor, i.e. dΦ =
B dS = B dl. dh is given by
d Φ = BdS =
μ0l
[sin θ 2 + sin θ1 ]dldh
4π h
To obtain the total flux over all of the area
we integrate the above equation over the length
of the conductor and then over the distance
from the conductor to infinity. Integrating over
the length of the conductor gives
l
μ0 I
l
∫ Bdl = 4π h ∫ (sin θ
1
0
+ sin θ 2 ) dl
(3)
l
∫ Bdl =
0
μ0 I ⎡ l 2 + h 2 ⎤
− 1⎥
⎢
h
2π h ⎢⎣
⎥⎦
Now, we integrate the above equation over
the distance from the edge of the conductor to
infinity. Let the diameter of the conductor is d
then we integrate the above equation form d/2
to ∞ to obtain the total magnetic flux outside
the conductor. Hence the total flux is given by:
Φ=
∞
l
∫ ∫ Bdl dh =
d /2 0
∞
∞
⎤
μ0 l ⎡
l 2 + h2
dh − ∫ 1dh ⎥
⎢l / 2 ∫
2π ⎣⎢ d / 2
h
d /2
⎦⎥
0
Physics Education • July − September 2009
195
∞
Φ=
current, i.e. I = 1 in the above equation.
Therefore the external inductance of a straight
conductor of finite length is given by
l
∫ ∫ Bdl dh =
d /2 0
∞
∞
⎡
⎤
l 2 + h2
dh − ∫ 1dh ⎥
⎢l / 2 ∫
h
⎢⎣ d / 2
⎥⎦
d /2
⎡ ⎧⎛ 2l ⎞
⎫
d⎤
Lext = 2l ⎢ ln ⎨⎜ ⎟ (1 + x) ⎬ − x + ⎥ × 10−7 H
d
2
l⎦
⎭
⎣ ⎩⎝ ⎠
(4)
Hence,
Φ=
∞
⎡ ⎧⎛ 2l ⎞
⎫
d⎤
Lext = 2l ⎢ ln ⎨⎜ ⎟ (1 + x) ⎬ − x + ⎥ × 102 nH
2l ⎦
⎭
⎣ ⎩⎝ d ⎠
l
∫ ∫ Bdl dh = =
where
d /2 0
⎞
μ0 I ⎡ ⎛ 4l 2 + d 2
⎢l ln ⎜
+l⎟
⎟
2π ⎢ ⎜⎝
2
⎠
( 2l )
x = 1+ d
⎣
2
(b)
Calculation of Internal
Inductance
d d1
⎤
− l ln +
4l 2 + d 2 ⎥
2 22
⎦
From the definition of inductance, we
obtain the external inductance by putting unit
I
r
P
I1
h
l
A
B
Figure 2: Magnetic field calculation at point P inside the conductor due to current I passing through
the conductor AB of length l.
We know that when a current, I, passes
through a conductor, the internal energy stored
1
in the conductor is equal to Lm I 2 . Where, Lm
2
is the internal inductance of the conductor.
This energy is stored in form of magnetic
energy, which is given by B2/2μ0 per unit
volume. The total energy stored can be
obtained by integrating over total volume of
the conductor. Equating this energy to ½ LmI2,
we can calculate the internal inductance of the
conductor.
We know that to a good approximation the
total magnetic field due to a current I, at any
196
Physics Education • July − September 2009
point close to the surface of the conductor with
its length, l >>> than its radius, r, is given by,
μI
(5)
B= 0
2π
where, μ0= permeability and r = radius.
μ0 I 2 l
B12
=
dV
∫ 2μ0
16π
We know this energy is equal to
The magnetic field, B1, at any given point P
inside the cylindrical conductor, at a distance h
from the axis of the conductor is given by
Hence,
μ0 I1
2π h
(6)
where, the current I1 is passing through the
conductor of radius h.
If the total current, I, passes through the
conductor, then the current I1 can be given in
terms of I as,
⎛h⎞
∴I1 = I ⎜ ⎟
⎝r⎠
1
Lm I 2 , hence
2
μ I 2l
B2
1
Lm I 2 = ∫ 1 dV = 0
2
2 μ0
16π
B
B1 =
(9)
2
(7)
Because I=Jπr2 and I1=Jπh2
(c)
Lin =
μ0 I 4π × 10= 7 × l 1
=
= × 10−7
8π
8π
2
Lin =
1
× 10−7 H
2
Total Inductance
Total inductance of the straight conductor is
the sum of internal & external inductance and
is given by:
Substituting Eq. (7) in Eq. (6), we have
B1=
B
μ0 ⎛ h ⎞
I⎜ ⎟
2π h ⎝ r ⎠
⎡ ⎧ ⎛ ⎛ 2l ⎞
d⎫ l⎤
⎞
LTotal = ⎢ 2l ⎨ln ⎜ ⎜ ⎟ (1 + x) ⎟ − x + ⎬ + ⎥
2l ⎭ 2 ⎥⎦
⎢⎣ ⎩ ⎝ ⎝ d ⎠
⎠
−7
×10 H
2
Hence
μ I 2 h2
B12
= 0 2 4
2 μ0 8π r
(8)
where l = length of the conductor, d = diameter
( 2l )
of the conductor and x = 1 + d
The volume element of the conductor of radius h is
given by 2πhldh. The energy in this volume is
B12
(2π hldh)
2 μ0
2
.
Conclusion
Integrating this equation with respect to h
from zero and r (= d/2, Figure 2), the total
energy inside the conductor can be obtained
=
μ0 I 2
2π h l dh h 2
8π 2 r 4 ∫
Physics Education • July − September 2009
197
We generally associate inductance with a loop
or coil of conductor. However, even a straight
piece of conductor, has some self-inductance or
inductance. The inductance of a straight
conductor of finite length is calculated starting
from the first principle. It is not easy to find an
expression for the inductance of a straight
piece of conductor of a finite length. The
results are very useful and can be used as a
building block for more complex structures.
Further, because the inductance of a conductor
is important in high frequency or high speed
electrical circuits, the formula derived will be
very useful in rapid calculation of the same.
Acknowledgements
First, I would like to thank Vikram A. Sarabhai
Community Science Center, Ahmedabad for
giving me this opportunity to work at Institute
for Plasma Research as a part of Ad. B.Sc.
(Physics) Program. I am very much thankful to
Dr. Joydeep Ghosh and Mr. Bhooshan
Paradkar for providing proper guidance during
the course of this work and also for helping me
in many ways. I am very much thankful to the
Director, I.P.R. for allowing me to use the
institute’s facilities. I am very thankful to Dr.
Raghvan Rangrajan (Physical Research
Laboratory, Ahmedabad), for giving me such a
great opportunity.
Reference
D. J. Griffith, Introduction to Electrodynamics (3rd
ed.), (Engelwoods Cliff: Prentice Hall) 1998
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