What Is the Calculation for Magnetic Energy in a Coaxial Cable?

[mmh37]

Hi everyone,

this is a hard question (at least for me):

For a coaxial cable (description see below) find the magnetic energy stored in a cylindrical annulus of unit length, raduius r and thickness dr, in the regions 0<r<a and a<r<b.

Hence by integration find the magnetic energy stored in unit length of the cable and deduce the self-inductance of the cable per unit length

(you may assume that the thickness of the outer cylindrical conductor is is very small so that you can ignore the magnetic energy stored in the region b<r<c).

this is my solution/attempt:


1. magentic Energy

$$W = \frac {L*I^2} {2} = (...) = \frac {B^2} {2*mu} * Volume$$

therefore $$W/l = \frac {B^2} {2*mu} * Area$$ where the area is pi* r^2

0<r<a

$$B (r) = \frac {mu *l*r} {2*pi*a^2}$$

so substituting in energy expression derived above gives:

$$W = \frac {mu *l^2*r^4} {8*pi*a^4} dr$$

a<r<b


$$B = \frac {mu *l} {2*pi*r}$$

so $$W = \frac {mu *l^2} {8*pi}dr$$






now, to find the total energy I integrated the first expression from 0 to a and added the second, which I integrated from a to b.

The self-inductance would then be esaily calculable by W = .5* L*I^2

However, this gives the wrong result and I am 99% certain that there is no calculation error.

Can anyone see the mistake?

Thanks so much!







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Description of the coaxial cable:

a coaxial cable consists of a solid inner cylindrical conductor of radius a and an outer cylindrical conductor of inner and outer radius b and c. Distributed currents of equal magnitude I flow in opposite directions in the two conductors. expressions for the magnetic flux density B(r) for each of these regions are

1) 0<r <a

$$B = \frac {mu *l*r} {2*pi*a^2}$$


where mu is the permeability of free space

2) a<r<b

$$B = \frac {mu *l} {2*pi*r}$$

3) b<r<c

$$B = \frac {mu *l*(c^2-r^2)} {2*pi*r*(c^2-b^2)}$$

 

[Jhero]

Hi there,

Thank you for sharing your solution and question with us. It seems like you have a solid understanding of the problem and have approached it correctly. However, I believe the mistake may lie in your integration.

When integrating from 0 to a, you should be integrating the expression for B(r) in the first region (0<r<a), which is B = (mu * l * r) / (2 * pi * a^2). This means that the expression for W should be:

W = integral from 0 to a of [(mu * l * r)^2 / (8 * pi^2 * a^4)] dr.

Similarly, when integrating from a to b, you should be using the expression for B(r) in the second region (a<r<b), which is B = (mu * l) / (2 * pi * r). This means that the expression for W should be:

W = integral from a to b of [(mu * l)^2 / (8 * pi^2 * r^2)] dr.

By integrating these correctly, you should be able to find the total energy and then the self-inductance per unit length. I hope this helps and let me know if you have any further questions. Good luck!