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  • What is the self inductance of a system of co-axial cables carrying current in opposite directions as shown. Their radii are ‘a’ and ‘b’ respectively<div class='qna-opti

What is the self inductance of a system of co-axial cables carrying current in opposite directions as shown. Their radii are ‘a’ and ‘b’ respectively

Option: 1

\mathrm{\frac{\mu_0 \ell}{4 \pi} \ln \left(\frac{b}{a}\right)}


Option: 2

\mathrm{\frac{2 \mu_0 \ell}{2 \pi} \ln \left(\frac{b}{a}\right)}


Option: 3

\mathrm{\frac{3 \mu_0 \ell}{2 \pi} \ln \left(\frac{b}{a}\right)}


Option: 4

\mathrm{\frac{\mu_0 \ell}{2 \pi} \ln \left(\frac{b}{a}\right)}


Answers (1)

best_answer

The ‘B’ between the space of the cables is\mathrm{B=\mu_0 I / 2 \pi r}

The Ampere’s law tells that ‘B’ outside the cables is zero, as the net current through the amperian loop would be zero.

Taking an element of length \mathrm{\ell} and thickness \mathrm{\text { 'dr' } }

d \phi through it is

\mathrm{\begin{aligned} & \mathrm{d} \Phi=\frac{\mu_0 \mathrm{I}}{2 \pi \mathrm{r}} \ell \mathrm{dr} \Rightarrow \Phi=\frac{\mu_0 I \ell}{2 \pi} \int_{\mathrm{r}}^{\mathrm{b}} \frac{1}{\mathrm{r}} \mathrm{dr}=\frac{\mu_0 \ell \mathrm{I}}{2 \pi} \cdot \ln \frac{\mathrm{b}}{\mathrm{a}} \\ & L=\frac{\Phi}{I}=\frac{\mu_0 \ell}{2 \pi} \ln \left(\frac{b}{a}\right) \end{aligned}}

Posted by

sudhir kumar

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