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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Physics
Chapter
Electrostatics

BITSAT 2014 · Physics · Electrostatics

Two long coaxial and conducting cylinders of radius a and b are...

Two long coaxial and conducting cylinders of radius a and b\mathrm{b} are separated by a material of conductivity σ\sigma and a constant potential difference V\mathrm{V} is maintained between them, by a battery. Then the current, per unit length of the cylinder flowing from one cylinder to the other is -

Options

  1. A4πσln⁡(b/a)V\frac{4 \pi \sigma}{\ln (b / a)} V
  2. B4πσ(b+a)V\frac{4 \pi \sigma}{(b+a)} V
  3. C2πσln⁡(b/a)V\frac{2 \pi \sigma}{\ln (b / a)} V
  4. D2πσ(b+a)V\frac{2 \pi \sigma}{(b+a)} V

Correct answer

C. 2πσln⁡(b/a)V\frac{2 \pi \sigma}{\ln (b / a)} V

Step-by-step solution

\mathrm{V}=\int_{\mathrma^{\mathrm{b}} \mathrm{E} \cdot \mathrm{d} \ell=\frac{\lambda}{2 \pi \varepsilon_{0} \mathrm{r}} \ln \left(\frac{\mathrm{b}}{\mathrm{a}}\right) Now, I=σ∫E⃗⋅dA→=σ∫λ2πε0r⋅2π\sigma \int \vec{E} \cdot \overrightarrow{d A}=\sigma \int \frac{\lambda}{2 \pi \varepsilon_{0} r} \cdot2 \pi d r=σλε0From(1)\frac{\sigma \lambda}{\varepsilon_{0}} From (1): I=2σπε0ε0ln⁡(b/a)=2πσln⁡(b/a)\frac{2 \sigma \pi \varepsilon_{0}}{\varepsilon_{0} \ln (b / a)}=\frac{2 \pi \sigma}{\ln (b / a)} v
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