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If I have a uniform, infinitely thin current. The relationship between inductance and current by definition is:

$\phi_{B} = L I$

Where inductance($L$) the proportionality constant between a wires current, and the magnetic flux that it causes on itself.

This definition only works for infinitely thin currents and B closed loops

since an open loop would not even have a surface attached to it this definition cannot be valid( I'm guessing)

$L$

Another formula derived, is that the total amount of work I would have to do against the back emf is to generate a current $I_{0}$ is

$W =(1/2) LI_{0}^2$

In griffiths, Using the magnetic vector potential and a few "hand wavy " arguments to change this to volume currents

We can also say that this amount of energy is also stored in the fields as.

$\int_{V} \frac{1}{2\mu_0}\vec{|B|}^2$ dv

Comparing the two expressions

$\int_{V} \frac{1}{2\mu_0}\vec{|B|}^2 dv = (1/2) LI_{0}^2$

Clearly

$L = \frac{2}{I_{0}^2}\int_{V} \frac{1}{2\mu_0}\vec{|B|}^2 dv $

Here in textbooks ,like griffiths, they use this formula to derive the inductance for a straight volume current wire. Where I_{0} is taken to be the flux of current density perpendicular to the length of the wire

So, my question is this

What are the steps in this derivation that specifically allow for inductance to be calculated for straight wires using this method. As the starting formula

$\phi_{B} = L I$

Only makes sense when talking about CLOSED loops and thin currents.

And 2: In his transition to thin currents to volume currents, how can we be sure what $I_{0}$ is? As this $I_{0}$ is derived in the context of thin currents. So its generalisation to volume currents doesn't make sense to say that $I_{0}$ is a specific flux integral. As we aren't talking about a specific "current enclosed by a surface" so we cannot say $I_{0}$ is J.da for a perpendicular surface to length of the wire. Aka for a thin wire there is only 1 current. But in volume currents the current flowing through a surface depends on what surface I pick

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  • $\begingroup$ Unfortunately, inductance per unit length of an infinitesimal wire is undefined; it blows up logarithmicly in 1/r. $\endgroup$
    – Whit3rd
    Commented Dec 17, 2021 at 2:32
  • $\begingroup$ Theoretically no current would ever be present in a infinitely thin wire due to infinite work being needed against the back emf thenM. And how does the definition of inductance relate to volume currents, which path should I take? What is the current, as current is measured with respect to some cross-sectional surface ? The outer path, the middle? And what about open paths, what is the definition of inductance then? It definitely can't be the ratio between a current and a flux as no surface can be attached to an open loop $\endgroup$ Commented Dec 17, 2021 at 2:45
  • $\begingroup$ I would be very interested in someone deriving inductance in the context of volume currents first. Instead of doing it for thin currents and then making a generalisation $\endgroup$ Commented Dec 17, 2021 at 2:47
  • $\begingroup$ Currents in a thick wire, ignoring skin effect, are easy to solve for magnetic field inside, and outside, the wire. One just has to start with assuming uniform current density (Ohm's law dominating, that's mainly what happens). Usual inductors, though, use loops/helices/solenoids, and the math only simplifies for long solenoids. $\endgroup$
    – Whit3rd
    Commented Dec 17, 2021 at 5:17
  • $\begingroup$ Not really what I'm asking. My main question is what OPEN surface i attach to an open loop in the definition of inductance, as by definition I need change in flux on itself . I know its easy to solve for the B field inside a wire. $\endgroup$ Commented Dec 17, 2021 at 13:22

1 Answer 1

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Sorry for my poor english. My native language is french. I only try to answer part 2 of the question.

The definition of the self-inductance from the flux is tricky because it is well known that the flux diverges for a wire-shaped circuit. The problem is not too serious because in elementary courses one usually calculates only the self-inductance of a solenoid or a torus, modeled in reality by surface layers of current for which there is no divergence of the field near the circuit. And in complicated cases, one returns to the energy definition without worrying too much about the coherence of the various definitions.

For the following, I limit myself to a single circuit (in the general case, the self-inductance also depends on the presence of other circuits which modify the current distribution).

We can be more precise, starting from the magnetic energy to arrive at the definition of the proper flux.

$E_m=\iiint{\frac{{\vec{B}}^2}{2\mu_0}d\tau}=\iiint{\vec{j}.\vec{A} d\tau}$

Most often, it is possible to decompose the closed circuit into elementary closed tubes of current $(K)$, each of which has a current $dI_K$ flowing through it. The volume integral is then decomposed into an integral on each tube and an integral along a tube, with the substitution $\vec{j}d\tau=\vec{j}dSdl=dI\vec{dl}$ enter image description here $E_m=\int_{all\ tubes\ K}\int_{K}{{\rm dI}_K\vec{\ A}.{\vec{dl}}_K}=\int_{all\ tubes\ K}{{\rm dI}_K\int_{K}{\vec{\ A}.{\vec{dl}}_K}}$

We can (without any divergence problem) define the magnetic field flux through the contour of the elementary tube : $\int_{K}{\vec{\ A}.{\vec{dl}}_K}=\iint{\vec{\ B}.{\vec{dS}}_K}=\Phi_K$

And so we have : $E_m=\int_{all\ tubes\ K}{\Phi_KdI}_K$

We can define the total flux through the non-wire circuit by:

$\Phi=\frac{1}{I}\int_{all\ tubes\ K}{\Phi_KdI}_K$

As $I=\int_{all\ tubes\ K}{\rm dI}_K$ this total flux is an average over the elementary current tubes.

By construction, this definition of the flux is consistent with that of the magnetic energy $E_m=\frac{1}{2}I\Phi$

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  • $\begingroup$ Is your answer for a non thin wire circuit? As I'm not sure i follow your tube method. For a non thin wire, the flux doesn't diverge , as once you get to the radius the B field is linear with radius, but also, what Is the flux of a wire with thickness? Which line do I pick? One that goes through the center? The edge of the wire? And in this case what is the current of this line? As the definition of inductance only works for thin currents $\endgroup$ Commented Dec 18, 2021 at 18:37
  • $\begingroup$ Yes, this is the definition of the flux for a non thin wire circuit. I added a diagram to try to be more clear. $\endgroup$ Commented Dec 19, 2021 at 8:24

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