What is the self inductance of a system of co-axial cables carrying current in opposite directions as shown? Their radii are 'a' and 'b' respectively and its length l.

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Solution

The 'B' between the space of the cables is $B = μ_{o} I / 2 π r$
The Ampere's law tells that 'B' outside the cables is zero, as the net current through the amperian loop would be zero.
Taking an element of length l and thickness 'dr', $d ϕ$ through it is
$d Φ = \frac{μ_{o} I}{2 π r} l d r \Rightarrow Φ = \frac{m u_{o} I l}{2 π} \int_{a}^{b} \frac{1}{r} d r = \frac{μ_{o} l I}{2 π} \cdot l n \frac{b}{a}$
$L = \frac{Φ}{I} = \frac{μ_{o} l}{2 π} l n (\frac{b}{a})$ .

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