Inductance/derivation of self inductance

The mutual inductance by circuit "i" on circuit "j" is given by the double integral "Neumann formula":$M\_\{ij\}\; =\; frac\{mu\_0\}\{4pi\}\; oint\_\{C\_i\}oint\_\{C\_j\}\; frac\{mathbf\{ds\}\_icdotmathbf\{ds\}\_j\}$so that the mutual inductance is a purely geometrical quantity independent of the current in the circuits.

elf inductance

In the self inductance case $C\_\{i\}=C\_\{j\}$. Therefore 1/R gets singularand the finite radius of the wire and the distribution of the current in thewire must be taken into account. A generic formula for the self inductance $M$ is available provided that the length $l$ of the wire is much larger thanits radius $a$::$M\; =\; M\_\{ii\}\; =\; frac\{mu\_0\}\{4pi\}\; left\; (\; oint\_\{C\}oint\_\{C\text{'}\}\; frac\{mathbf\{ds\}cdotmathbf\{ds\}\text{'}\}d^\{3\}xd^\{3\}x^\{prime\; \}\; ight\; )\_.$The integral over z (from $-Delta$ to $Delta$) may be performed using$mathbf\{R\}\_\{xx^\{prime\; ^\{2\}=N^\{2\}+z^\{2\}$, $N^\{2\}=r^\{2\}+r^\{prime2\}-2rr^\{prime\; \}cos\; left(\; phi\; ight)$,:$M\_\{Delta\; \}=mu\; \_\{0\}lint\; dphi\; drdr^\{prime\; \}rfleft(\; r\; ight)\; r^\{prime\}fleft(\; r^\{prime\; \}\; ight)\; ln\; left(\; frac\{Delta\; \}\{N\}+sqrt\{1+left(Delta\; /N\; ight)\; ^\{2\; ight)\; .$Because of $Nll\; Delta$ the logarithm may be replaced by:$ln\; left(\; frac\{Delta\; \}\{N\}+sqrt\{...\}\; ight)\; simeq\; ln\; left(\; frac\{2Delta\; \}\{N\}\; ight)\; =ln\; left(\; frac\{2Delta\; \}\{a\}\; ight)\; -frac\{1\}\{2\}ln\; frac\{r^\{2\}+r^\{prime\; 2\}-2rr^\{prime\; \}cos\; left(\; phi\; ight)\; \}\{a^\{2.$In the realistic cases $fleft(\; r\; ight)=const$ and $fleft(\; r\; ight)=constcirc\; delta\; left(\; r-a\; ight)$ the remaining integrals now eitherare trivial or may be looked up in an integral table. The result is $M\_\{Delta\; \}=\; left(\; mu\; \_\{0\}l\; /\; 2pi\; ight)\; left(\; ln\; left(\; 2Delta\; /a\; ight)+Y\; ight)$. Finally, the self inductance cannot depend on the arbitrarylength $Delta$, which is the lower integration limit for the curveintegral and the upper integration limit for $M\_\{Delta\; \}$. Therefore thecontribution proportional to $lnleft(\; Delta\; ight)$ cancels against acontribution from the curve integral, so that one now also may choose $Delta\; =a/2$ to simplify the calculation.