Ncert Exercises & Solutions

A long straight cable of length l is placed symmetrically along the z-axis and has radius a(<<l). The cable consists of a thin wire and a co-axial conducting tube. An alternating current $I (t) = I_{o} \sin (2 πνt)$ 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:

$E (s, t) = μ_{0} I_{0} νcos (2 πνt) \ln (\frac{s}{a}) \hat{k} .$

(i) Calculate the displacement current density inside the cable.
(ii) Integrate the displacement current density across the cross-section of the cable to find the total displacement current $I^{d} .$
(iii) Compare the conduction current $I_{0}$ with the displacement current $I_{0}^{d} .$

Hint: The displacement current density depends on the variation of the electric field between the plates of the capacitor.

(i) Step 1 : Given, the induced electric field at a distane r from the wire inside the cable is :

\vec{E} (s, t) = μ_{0} I_{0} νcos (2 πνt) \ln (\frac{s}{a}) \hat{k}

Now, displacement current density,

J_{d} = ε_{0} \frac{d \vec{E}}{dt} = ε_{0} \frac{d}{dt} [μ_{0} I_{0} νcos (2 πνt) \ln (\frac{s}{a}) \hat{k}]

= ε_{0} μ_{0} I_{0} ν \frac{d}{dt} [\cos 2 πνt] In (\frac{s}{a}) \hat{k}

= \frac{1}{c^{2}} I_{0} ν^{2} \times 2 π [- \sin 2 πνt] In (\frac{s}{a}) \hat{k}

= \frac{ν^{2}}{C^{2}} 2 {πI}_{0} \sin 2 πνt In (\frac{a}{s}) \hat{k} [∴ \ln (\frac{s}{a}) = - lln (\frac{a}{s})]

= \frac{1}{λ^{2}} 2 {πI}_{0} In (\frac{a}{s}) \sin 2 πνt \hat{k}

= \frac{2 {πI}_{0}}{λ^{2}} In (\frac{a}{s}) \sin 2 πνt \hat{k}

Step 2 : I_{d} = \int J_{d} sdsdθ = \int_{s = 0}^{a} J_{d} sds \times \int_{0}^{2 π} dθ = \int_{s = 0}^{a} J_{d} sds \times 2 π

= \int_{s = 0}^{a} [\frac{2 π}{λ^{2}} I_{0} \log_{e} (\frac{a}{s}) sds \times \sin 2 πνt] \times 2 π

= {(\frac{2 π}{λ})}^{2} I_{0} \int_{s = 0}^{a} \ln (\frac{a}{s}) sds \times \sin 2 πνt

= {(\frac{2 π}{λ})}^{2} I_{0} \int_{s = 0}^{a} In (\frac{a}{s}) \frac{1}{2} d (s^{2}) . \sin 2 πνt

= \frac{a^{2}}{2} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 πνt \int_{s = 0}^{a} In (\frac{a}{s}) . d {(\frac{s}{a})}^{2}

= \frac{a^{2}}{4} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 πνt \int_{s = 0}^{a} In {(\frac{a}{s})}^{2} . d {(\frac{s}{a})}^{2}

= - \frac{a^{2}}{4} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 πνt \int_{s = 0}^{a} In {(\frac{s}{a})}^{2} . d {(\frac{s}{a})}^{2}

= - \frac{a^{2}}{4} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 πνt \times (- 1) [∴ \int_{s = 0}^{a} In {(\frac{s}{a})}^{2} d {(\frac{s}{a})}^{2} = - 1]

∴ I_{d} = \frac{a^{2}}{4} {(\frac{2 π}{λ})}^{2} I_{0} \sin 2 πνt

= {(\frac{πa}{λ})}^{2} I_{0} \sin 2 πνt

Step 3 : The displacement current,

I_{d} = {(\frac{πa}{λ})}^{2} I_{0} \sin 2 πνt = {I_{0}}^{d} \sin 2 πνt

Here,

{I_{0}}^{d} = {(\frac{πa}{λ})}^{2} I_{0}

∴ \frac{{I_{0}}^{d}}{I_{0}} = {(\frac{aπ}{λ})}^{2}

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