Question

A coaxial cable consists of alternating coaxial cylinders of conducting and insulating material. Coaxial cabling is the primary type of cabling used by the cable television industry and is also widely used for computer networks such as Ethernet, on account of its superior ability to transmit large volumes of the electrical signals with minimum distortion. Like all other kinds of cables, however, coaxial cables also have some self-inductance that has undesirable effects, such as producing some distortion and heating. Consider a long coaxial cable made of two coaxial cylindrical conductors that carry equal currents I in opposite directions (see figure). The inner cylinder is a small solid conductor of radius a. The outer cylinder is a thin-walled conductor of outer radius b, electrically insulated from the inner conductor. Calculate the inductance per unit length Ll of this coaxial cable.

Answer 1

Answer:

inductance per unit length is  $\frac{L}{l} = \frac{\mu_o}{2 \pi} ln (\frac{b}{a} )$

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

          ∮ $(\= B \ \= dl) = \mu_o I$

=>       $B (2 \pi r) = \mu_o I$

=>      $B = \frac{\mu_o I }{2 \pi r }$      

   The magnetic flux on the coaxial cable can be mathematically represented as

            $\phi = \int\limits^{r_2}_{r_1} {\= B \cdot \= da} \, dx$  

          $\phi = \int\limits^{\frac{b}{2} }_{\frac{a}{2} } { \frac{\mu_o I }{2 \pi r } * l * dr } \,$  

        $\phi = \frac{\mu_o I \ * \ l }{2 \pi r } [ln [\frac{b}{2} - ln [\frac{a}{2} ] ]]$  

        $\phi = \frac{\mu_o I \ * \ l }{2 \pi r } ln [\frac{b}{a} ]$  

Now the emf induced in the coaxial cable is mathematically represented as

            $\epsilon = \frac{d \phi }{dt} = L \frac{dI}{dt}$

=>             $\frac{d \phi }{dt} = L \frac{dI}{dt}$

=>            $\int\limits {\phi} \, = L \int\limits {dI} \,$

=>            $\phi = L I$

substituting for  $\phi$

               $\frac{\mu_o I \ * \ l }{2 \pi r } ln [\frac{b}{a} ] = LI$

dividing through by $l$

         $\frac{L}{l} = \frac{\mu_o}{2 \pi} ln (\frac{b}{a} )$