Paragraph: Consider a block of conducting material of resistivity ' $\rho$ ' shown in the figure. Current…

Paragraph: Consider a block of conducting material of resistivity ' $\rho$ ' shown in the figure. Current 'l' enters at 'A' and leaves from ' $\mathrm{D}$ '. We apply superposition principle to find voltage ' $\Delta \mathrm{V}$ ' developed between ' $\mathrm{B}$ ' and ' $\mathrm{C}$ '. The calculation is done in the following steps: (i) Take current 'l' entering from 'A' and assume it to spread over a hemispherical surface in the block. (ii) Calculate field $E(r)$ at distance ' $r$ ' from $A$ by using Ohm's law $E=\rho j$, where $j$ is the current per unit area at ' $r$ '. (iii) From the ' $r$ ' dependence of $E(r)$, obtain the potential $V(r)$ at $r$. (iv) Repeat (i), (ii) and (iii) for current 'l' leaving ' $D$ ' and superpose results for ' $A$ ' and ' $D$ '.
Question: $\Delta \mathrm{V}$ measured between $\mathrm{B}$ and $\mathrm{C}$ is
  1. $\frac{\rho l}{\pi a}-\frac{\rho l}{\pi(a+b)}$
  2. $\frac{\rho l}{a}-\frac{\rho l}{(a+b)}$
  3. $\frac{\rho l}{2 \pi a}-\frac{\rho l}{2 \pi(a+b)}$
  4. $\frac{\rho l}{2 \pi(a-b)}$

Solution

Choosing $A$ as origin, $ \begin{gathered} E=\rho j=\rho \frac{1}{2 \pi r^2} \\ V_C-V_B=-\frac{\rho l}{2 \pi} \int_a^{(a+b)} \frac{1}{r^2} d r=\frac{\rho l}{2 \pi}\left[\frac{1}{(a+b)}-\frac{1}{a}\right] \\ V_B-V_C=\frac{\rho l}{2 \pi}\left[\frac{1}{a}-\frac{1}{(a+b)}\right] \end{gathered} $

Asked in: JEE Main 2008

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