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  • Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity and a constant potential differenceV is maintained between them by a battery. Then the c

Two long coaxial and conducting cylinders of radius a and b are separated by a material of conductivity and a constant potential differenceV is maintained between them by a battery. Then the current per unit length of the cylinder flowing from one cylinder to the other is

Option: 1

\mathrm{\frac{4 \pi \sigma}{\ln (b / a)} V}


Option: 2

\mathrm{\frac{4 \pi \sigma}{(b / a)} V}


Option: 3

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


Option: 4

\mathrm{\frac{2 \pi \sigma}{(b+a)} V}


Answers (1)

best_answer

\mathrm{E=\frac{\lambda}{2 \pi \varepsilon_0 r r^{\prime}},~ where~ \lambda}  is the linear charge density of the inner cylinder
And    \mathrm{V=\int_a^b E d l=\frac{\lambda}{2 \pi \varepsilon_0} \ln \left(\frac{b}{a}\right)} 
Now,   \mathrm{I=\int \vec{J} \cdot d \vec{A}=\sigma \int \vec{E} \cdot d \vec{A}}
\mathrm{ =\sigma \int \frac{\lambda}{2 \pi \varepsilon_0 r} 2 \pi r d r }
Current per unit length will be

\mathrm{ I=\frac{\sigma \lambda}{\varepsilon_0} }
From Eq. (i), we get
\mathrm{ I=\frac{2 \sigma \pi \varepsilon_0}{\varepsilon_0 i n(b / a)} v=\frac{2 \pi \sigma}{\ln (b / a)} v }

Alternatively,

\mathrm{ I=\frac{V}{R} \Rightarrow R=\int_{x=a}^b \frac{1}{\sigma} \frac{d x}{2 \pi x 1}=\frac{1}{2 \pi r} \ln \left(\frac{b}{a}\right) \\ }
\mathrm{ \therefore I=\frac{2 \pi \sigma V}{\operatorname{In}(b / a)} }

Posted by

Divya Prakash Singh

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