Coaxial Cable A has twice the length, twice the radius of the inner solid conductor, and twice the radius of the outer cylindrical conductin

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 BA) 4 ln 2B) 2C) 4 ln 4D) 2 ln 2E) 2 ln 4

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Tài Đức 3 years 2021-07-29T19:20:32+00:00 1 Answers 68 views 0

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    2021-07-29T19:22:06+00:00

    Answer:2

    Explanation:

    Given

    Cable has twice the length of cable B

    and twice the radius(both inner and outer) of cable B

    Inductance is given by

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

    Where l=length

    r=inner radius

    R=outer radius

    \mu =Permeability of inner cylinder

    suppose l_a and l_b be the length of cable A and B

    so l_a=2l_b

    L_a=2l_b\times \frac{\mu }{2\pi }\times \ln(\frac{2R}{2r})

    L_a=2l_b\times \frac{\mu }{2\pi }\times \ln(\frac{R}{r})---1

    for wire B

    L_b=l_b\times \frac{\mu }{2\pi }\times \ln(\frac{R}{r})---2

    divide 1 and 2 we get

    \dfrac{L_a}{L_b}=2

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