Question

find the self inductance per unit length of a long straight conductor circular ctoss section​

Answer 1

(Let us consider the "long straight conductor" to be a wire.)

Now, we need to keep in mind that;

Inductance is associated with the flux linkage (λ)

Consider a coil consisting of $N$ number of turns, and it is linked by flux Φ due to current $I$;

Then,

Inductance (L) = Flux Linkage ($N$Φ) ÷ Current ($I$)

However, for the transmission line $N$ = 1; the value of Φ only has to be calculated. With that, we get the transmission line inductance.

1. Calculating the Inductance of a Single Conductor

Let us suppose that - a conductor is carrying current $I$ through its length $l$, with $x$ as the internal variable radius of the conductor and $r$ the conductor's original radius.

Now, the cross-sectional area with respect to radius $x$ is $\pi x^2$ units², and current $I_x$ flowing through this particular cross-sectional portion.

Then, the value of $I_x$ can be expressed using the following equation;

$\frac{I_x}{I} = \frac{\pi x^2}{\pi r^2 }$ ⇒ $I_x = I\frac{\pi x^2}{\pi r^2} = I \frac{x^2}{r^2}$

(Refer to the cross sectional diagram below, in grey-yellow-orange colours)

Now, we will look into the small thickness $dx$ with 1 m length of the conductor, in which $H_x$ signifies the magnetizing force due to current $I_x$ around the area $\pi x^2$;

∴ $H_x = \frac{I_x}{2\pi x} = \frac{I}{2\pi r^2}$ × $x (A/m)............[I _x = I\frac{x^2}{r^2]}$

Then, the magnetic flux density, i.e. $B_x$ is equal to μ$H_x$,

in which μ = permeability of the conductor.

But then again, µ = µo × µr

Now, if the relative permeability of this conductor is considered to be µr= 1;

Then, µ = µo;

Thus, $B_x$ = μo × $H_x$ = μo × $\frac{I}{2\pi r^2}$ × $x$ =  μo ÷ $2\pi$ $(\frac{xI}{r^2})$  Tesla

Then, dφ for the strip $dx$;

dφ = $B_xdx =$ μo ÷ $2\pi$ $(\frac{xI}{r^2})dx$   Wb/m  

Now, it is inferred that the entire cross-sectional area does not enclose the flux stated above. The ratio of the cross-sectional area within the circle having radius $x$ to the total cross-section of the conductor viewed as a fractional turn that can link the flux. Henceforth, the flux linkage will be;

dλ = $\frac{\pi x^2}{\pi r^2 }$dΦ =  μo ÷ $2\pi$ $(\frac{x^3I}{r^4})$     Wb/m

Now, the total flux linkage for the conductor which is 1 m long and with radius $r$ can be derived via;

λ = $\int\limits^r_0$ dλ =  (μo ÷ $8\pi$)$I$   Wb/m

∴ Internal inductance = $L_i_n_t_e_r_n_a_l =$ λ ÷ $I$ = (μo ÷ $8\pi$)   Henry/m

2. Looking for the External Inductance due to External Magnetic Flux of a Conductor

Let us suppose that because of the skin effect conductor, current $I$ is concentrated near the conductor's surface. Assuming that he distance $y$ is derived from the conductor centre, which is making the conductor's external radius.

(refer to the monochromatic diagram below)

In the diagram,

$H_y$ = magnetizing force

$B_y$ = magnetic field density at $y$ distance per unit length

So,

$H_y = \frac{I}{2\pi y}$   A/m

and, $B_y =$ μo × $H_y$ = μo ÷ $2\pi$ × $(\frac{I}{y})$   Tesla

Now, we consider that dφ is exists within the thickness dy from $D_1$ to $D_2$

for 1 m. Then,

dφ = $B_ydy$ =  μo ÷ $2\pi$ × $(\frac{I}{y})dy$  Wb/m

∵ $I$ is assumed to flow in the conductor's surface,

∴ dλ = dφ

(refer to the formula below)

Moving on;

However, the following must be taken into consideration; the flux linkage from conductor's surface to any external distance (r to D)

∴ λ = (μo ÷ $2\pi$)$I$ × $ln(\frac{D}{r})$  Wb/m

∴ External Inductance = $L_e_x_t_e_r_n_a_l$ = (μo ÷ $2\pi$)$I$ × $ln(\frac{D}{r})$  Henry/m

∴ Total Inductance = $L_i_n_t_e_r_n_a_L +$$L_e_x_t_e_r_n_a_l$ = (μo ÷ $8\pi$) + (μo ÷ $2\pi$) × $ln(\frac{D}{r})$  

= (μo ÷ $2\pi$)$I$ × $[\frac{1}{4} + ln(\frac{D}{r})]$ Henry/m  

(the last portion of the answer has been attached below, because they are not enough symbols here to type it all down)

[∵  0.7788r = Geometric Mean Radius]

Hopes this helps you.

Answer 2

Explanation:

Inductance is closely related to the flux linkage, denoted by λ. Suppose a coil with N number of turn is linked by flux Φ due to current I, then,

But for transmission line N = 1. We have to calculate only the value of flux Φ, and hence, we can get the transmission line inductance.

Calculation of Inductance of Single Conductor

Calculation of Internal Inductance due to Internal Magnetic Flux of a Conductor

Suppose a conductor is carrying current I through its length l, x is the internal variable radius of the conductor and r is the original radius of the conductor. Now the cross-sectional area with respect to radius x is πx^2 square – unit and current Ix is flowing through this cross-sectional area. So the value of Ix can be expressed in term of original conductor current I and cross-sectional area πr^2 square – unit

Now consider small thickness dx with the 1m length of the conductor, where Hx is the magnetizing force due to current Ix around the area πx^2.

And magnetic flux density Bx = μHx, where μ is the permeability of this conductor. Again, µ = µo*µr.

If it is considered that the relative permeability of this conductor µr= 1,

then µ = µo. Hence, here Bx= μo*Hx.

Using Ampere's law for one of the wires to obtain the magnetic field in terms of current I and the variable radius rdue to one of the wire

μ

0

I=∮

B

.

dl

=∫

0

2π

Br dθ

μ

0

I=B(2πr)

⟹B=

2πr

μ

0

I

Note that because of the symmetry of the setup, the total magnetic flux (ϕ) will be two times generated by one of the wires. Then, we continue to use the relation between self inductance and flux

ϕ

total

=2∫

A

B

.

dA

=2∫

a

d−a

(

2πr

μ

0

I

)l dr

ϕ

total

=

π

μ

0

Il

ln(

a

d−a

)

Inductance, L=

I

ϕ

total

=

π

μ

0

l

ln(

a

d−a

)

Inductance per unit length , ϕ

total

=

π

μ

0

ln(

a

d−a

)

dφ for small strip dx is expressed by

Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of the cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux. Therefore the flux linkage is

Now, the total flux linkage for the conductor of 1m length with radius r is given by

Hence, the internal inductance is

External Inductance due to External Magnetic Flux of a Conductor

Let us assume, due to skin effect conductor current I is concentrated near the surface of the conductor. Consider, the distance y is taken from the center of the conductor making the external radius of the conductor.

Hy is the magnetizing force and By

is the magnetic field density at y distance per unit length of the conductor.

Let us assume magnetic flux dφ is present within the thickness dy from D1 to D2

for 1 m length of the conductor as per the figure.

As the total current I is assumed to flow in the surface of the conductor, so the flux linkage dλ is equal to dφ.

But we have to consider the flux linkage from conductor surface to any external distance, i.e. r to D