Question

A long straight cable of length l is placed symmetrically along z-axis and has radius $a(<<l)$. the cable consists of a thin wire and co-axial conducting tube. An alternating current I(t) = $I_0\sin \left(2\pi \nu t\right)$flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is E(s, t) = ${\mu }_0{I}_0\cos \left(2\pi \nu t\right)ln\left(\frac{s}{a}\right)\hat{k}$. , the ratio of conduction current and displacement current is

• A. ${\left(\frac{a\pi }{\pi }\right)}^2$
• B. $\left(\frac{a\pi }{\lambda }\right)$
• C. ${\left(\frac{\lambda }{a\pi }\right)}^2$
• D. $\left(\frac{\lambda }{a\pi }\right)$

Answer 1

The correct option is A ${\left(\frac{a\pi }{\pi }\right)}^2$

$I^d=\int J_{D}sdsd\theta$

$=\frac{2\pi }{{\lambda }^2}{I}_{0}2\pi {\int }_{x=0}^{a}ln\left(\frac{a}{s}\right).sds\sin \left(2\pi vt\right)$

$={\left(\frac{2\pi }{\lambda }\right)}^2{I}_0{\int }_{x=0}^a\frac{1}{2}d{s}^{2}ln\left(\frac{a}{s}\right).\sin \left(2\pi vt\right)$

$=\frac{a^2}{4}{\left(\frac{2\pi }{\lambda }\right)}^2{I}_0{\int }_{x=0}^{a}d{\left(\frac{s}{a}\right)}^{2}ln{\left(\frac{a}{s}\right)}^2.\sin \left(2\pi vt\right)$

$=-\frac{a^2}{4}{\left(\frac{2\pi }{\lambda }\right)}^2{I}_0{\int }_0^{1}ln\xi d\xi .\sin \left(2\pi vt\right)$

$=+{\left(\frac{a}{2}\right)}^2{\left(\frac{2\pi }{\lambda }\right)}^2{I}_0\sin 2\pi vt$

$(\therefore$The integral has value $-1)$

The displacement current

$I^d={(\frac{a}{2},\frac{2\pi }{\lambda })}^2{I}_0\sin 2\pi vt=I_0^4\sin 2\pi vt$

ratio

$\frac{I_0^d}{I_0}={\left(\frac{ax}{\lambda }\right)}^2$.