Question

Coaxial Cable A has twice the length, twice the radius of the inner solid conductor, and twice the radius of the outer cylindrical conducting shell of coaxial Cable B. What is the ratio of the inductance of Cable A to that of Cable B? a. 4ln2 O b.2 ?. o d. 2In2 e.21n4

Answer 1

Indutance of a co-axial cable is

$L =l\times\frac{\mu}{2\pi}\ln\left ( \frac{R}{r} \right )$

where

= length,

$\mu$ = permeability of the inner solid conductor,

R = radius of the outer cylinder

= radius of the inner solid cylinder conductor

So, inductance of cable A is

$L_A =l_A\times\frac{\mu}{2\pi}\ln\left ( \frac{R_A}{r_A} \right ) = L =2l_B\times\frac{\mu}{2\pi}\ln\left ( \frac{2R_B}{2r_B} \right ) \\ \\ \Rightarrow L_A = 2L_B\\ \\ \Rightarrow \frac{L_A}{L_B}=2.$