Self inductance of a very thin conductor, is infinity?

In summary, the problem is that the inductance goes off to infinity as you integrate between the inner conductor and the shell surrounding it. The inductance is proportional to the distance to the central axis. Thicker wires don't make a difference.
  • #1
constfang
10
0
Hi, I was reading some text and came across this problem, the problem is also mentioned in this link from Wiki: http://en.wikipedia.org/wiki/Inductance#Self-inductance
They said that it is because the 1/R now becomes infinite, this is what I'm confused about. From my understand, there would be some point where R=0 while taking the integration, but in the thick conductor case, isn't there also overlapping point that make R=0? how come these points doesn't make the integration infinite? what made a thick wire so different from a very thin wire? and what's the physical meaning of this phenomena? Thank you.
 
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  • #2
I think the problem's even worse than you think it is! The inductance, L, of a length [itex]\ell[/itex] of straight wire is given by
L = [itex]\frac{\mu_{0}\ell}{2\pi}[/itex] ln[itex]\frac{b}{a}[/itex]
in which a is the radius of the conductor and b is... Well, that's the problem: if you put it to infinity, your inductance goes off to infinity (just as badly as if you put a to zero. In fact the 'b' problem is easily understood: by putting b to infinity you're denying the possibility of a return path for the current. Surround the conductor by a return path in the form of a cylindrical shell (making a co-axial cable) and the problem is solved. There's no resultant field outside the cylinder, so you have to integrate only between the inner conductor and the cylinder.

As for your original problem, the field doesn't drop to zero as you penetrate the inner conductor, but is proportional to r, that is the distance to the central axis. You can easily derive this from Ampere's law. I think you'll find that the interior of the conductor contributes a finite amount, L0 to the inductance, namely
L0 = [itex]\frac{\mu_{0}\ell}{4\pi}[/itex].
 
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  • #3
I should have pointed out that (as you can see from my formula for L0) the thickness of the wire makes no difference to L0. Physically, for a given current, if the wire is thinner, the current density is greater, and also the flux density inside the wire, and there is the same amount of flux linked with it – even within the thinner wire.

But making the wire thinner will increase the inductance due to the flux linkage outside the wire, the bit that depends on ln[itex]\frac{b}{a}[/itex].
 
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  • #4
Thank you very much, that cleared out the cloud a little bit for me.
 
  • #5


I can provide some clarification on the concept of self-inductance and its relationship to thin and thick conductors. The self-inductance of a conductor refers to its ability to create an opposing magnetic field when a current is passed through it. This opposing magnetic field is what causes the inductance, or resistance, to the flow of current through the conductor.

In the case of a very thin conductor, the distance between the current-carrying wires is extremely small, leading to a high value of the inductance. This is because the magnetic field produced by the current in one wire is tightly coupled to the other wire, resulting in a strong opposing field. This is why the self-inductance of a very thin conductor is often considered to be infinite.

On the other hand, in a thick conductor, the distance between the current-carrying wires is much larger, resulting in a weaker coupling between the two wires and a lower self-inductance value. While there may still be some overlapping points where R=0, the overall effect is not as significant as in a thin conductor.

The physical meaning of this phenomenon is that the amount of inductance in a conductor is dependent on its size and shape. A thinner conductor will have a higher inductance due to its smaller size and closer proximity of the current-carrying wires. This has practical implications in circuit design and can affect the performance of electronic devices.

In conclusion, the self-inductance of a very thin conductor can be considered infinite due to the close proximity of the current-carrying wires and resulting strong opposing magnetic field. This is a unique characteristic of thin conductors and is important to consider in understanding the behavior of electrical circuits.
 

Related to Self inductance of a very thin conductor, is infinity?

1. What is self inductance and how does it relate to a very thin conductor?

Self inductance is a physical property of a conductor that describes its ability to generate an electric current in response to a changing magnetic field. In a very thin conductor, self inductance is directly proportional to the length of the conductor and inversely proportional to its cross-sectional area.

2. Why is the self inductance of a very thin conductor considered to be infinite?

The self inductance of a very thin conductor is considered to be infinite because as the cross-sectional area of the conductor approaches zero, the magnetic flux produced by the conductor approaches infinity. This results in an infinitely high self-inductance value.

3. What are the implications of a very thin conductor having infinite self inductance?

The main implication of infinite self inductance is that the conductor will have a very strong resistance to changes in current, as any change in current would require an infinite amount of energy to overcome the infinite self-inductance value.

4. Is it possible for a very thin conductor to have a finite self inductance?

No, it is not possible for a very thin conductor to have a finite self inductance. As mentioned before, as the cross-sectional area approaches zero, the self-inductance approaches infinity. Therefore, a very thin conductor will always have an infinite self-inductance value.

5. How does the self inductance of a very thin conductor affect its performance in circuits?

The high self-inductance of a very thin conductor can have a significant impact on the behavior of circuits, as it can cause delays in current flow and distortions in voltage. It can also lead to problems such as electromagnetic interference and unwanted oscillations.

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