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