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  • The space between two coaxial cylinders, whose radii are   <img alt="\mathrm{a\, \& \, b(a<b)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Ba%5C%2C%20%5C%26%20

The space between two coaxial cylinders, whose radii are   \mathrm{a\, \& \, b(a<b)} is filled with a conducting medium. The specific conductivity of the medium is \mathrm{\sigma}.Then resistance between the cylinders in radial direction is: \mathrm{(L>>b)}\mathrm{L=} Length of the cylinders:  


 

Option: 1

\mathrm{\frac{\ln (b/a)}{2 \pi \sigma\, L}}


Option: 2

\mathrm{\frac{ln\, b}{2\pi \sigma L}}


Option: 3

\mathrm{\frac{\ln L}{2 \pi \sigma(b-a)}}


Option: 4

\mathrm{\frac{\ln (b-a)}{2 \pi \sigma(b-a)}}


Answers (1)

best_answer

From ohm's law:
\vec{J}=\sigma \vec{E}- \text { (1) }

Assuming radial current density \vec{J} becomes
\vec{J}=\frac{I}{2 \pi r L} \hat{r} \text { for } a<r<b
\Rightarrow \vec{E}=\frac{\vec{J}}{\sigma}=\frac{I}{2 \pi \sigma r L} \hat{r}

Here we have used the assumption that L>>b so that \vec{E} and \vec{J}are in cylindrically symmetric form.
potential difference across the medium is thus

V_{a b}=-\int_{b}^{a} \vec{E}(r) \cdot \overrightarrow{d r}=-\frac{1}{2 \pi \sigma L} \cdot \int_{a}^{b} \frac{d r}{r}
V_{a b}=\frac{1}{2 \pi \sigma L} \cdot \ln (b / a).

The resistance
R_{a b}=\frac{V_{a b}}{\ell}=\frac{\ln (b / a)}{2 \pi \sigma L}.

Posted by

rishi.raj

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