Answer:
inductance per unit length is [tex]\frac{L}{l} = \frac{\mu_o}{2 \pi} ln (\frac{b}{a} )[/tex]
Explanation:
The diagram of coaxial cable is shown on the first uploaded image
From the question we are told that
The radius of the inner conductor is a
The current passing through the first cylindrical conductors is = I
The current passing through the first cylindrical conductors is = - I
The radius of the outer conductor is b
According to Ampere's law
∮ [tex](\= B \ \= dl) = \mu_o I[/tex]
=> [tex]B (2 \pi r) = \mu_o I[/tex]
=> [tex]B = \frac{\mu_o I }{2 \pi r }[/tex]
The magnetic flux on the coaxial cable can be mathematically represented as
[tex]\phi = \int\limits^{r_2}_{r_1} {\= B \cdot \= da} \, dx[/tex]
[tex]\phi = \int\limits^{\frac{b}{2} }_{\frac{a}{2} } { \frac{\mu_o I }{2 \pi r } * l * dr } \,[/tex]
[tex]\phi = \frac{\mu_o I \ * \ l }{2 \pi r } [ln [\frac{b}{2} - ln [\frac{a}{2} ] ]][/tex]
[tex]\phi = \frac{\mu_o I \ * \ l }{2 \pi r } ln [\frac{b}{a} ][/tex]
Now the emf induced in the coaxial cable is mathematically represented as
[tex]\epsilon = \frac{d \phi }{dt} = L \frac{dI}{dt}[/tex]
=> [tex]\frac{d \phi }{dt} = L \frac{dI}{dt}[/tex]
=> [tex]\int\limits {\phi} \, = L \int\limits {dI} \,[/tex]
=> [tex]\phi = L I[/tex]
substituting for [tex]\phi[/tex]
[tex]\frac{\mu_o I \ * \ l }{2 \pi r } ln [\frac{b}{a} ] = LI[/tex]
dividing through by [tex]l[/tex]
[tex]\frac{L}{l} = \frac{\mu_o}{2 \pi} ln (\frac{b}{a} )[/tex]
