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?

Question

Quizplus · Accepted Answer

Inductance is proportional to the natural logarithm of the ratio of the outer radius to the inner radius. 
L ∝ ln(b/a)

For cable A: 
Let the inner radius be r1 A and outer radius be r2 A. 
Length = 2L and r1 A = 2r1 B and r2 A = 2r2 B

So, b/a for cable A = (2r2 A)/(2r1 A) = r2 A/r1 A 
And b/a for cable B = r2 B/r1 B

So, the ratio of their inductances L_A/L_B = ln(r2 A/r1 A)/ln(r2 B/r1 B) 
= ln[(2r2 B)/(2r1 B)]/ln(r2 B/r1 B) 
= ln2/ln2 
= 2

Hence, the ratio of the inductance of Cable A to that of Cable B is 2.

Coaxial Cable a Has Twice the Length,twice the Radius of the Inner