Consider a block of conducting material of resistivity $'\rho '$ shown in the figure. Current $'I'$ enters at $'A'$ and leaves from $'D'$. We apply superposition principle to find voltage $'\Delta V'$ developed between $'B'$ and $'C'$. The calculation is done in the following steps: (i) Take current $'I'$ 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 $'I'$ leaving $'D'$ and superpose results for $'A'$ and $'D'.$ $\Delta V$ measured between $B$ and $C$ is

Question

Consider a block of conducting material of resistivity $'\rho '$ shown in the figure. Current $'I'$ enters at $'A'$ and leaves from $'D'$. We apply superposition principle to find voltage $'\Delta V'$ developed between $'B'$ and $'C'$. The calculation is done in the following steps:
(i) Take current $'I'$ 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 $'I'$ leaving $'D'$ and superpose results for $'A'$ and $'D'.$

$\Delta V$ measured between $B$ and $C$ is

${{\rho I} \over {\pi a}} - {{\rho I} \over {\pi \left( {a + b} \right)}}$

${{\rho I} \over a} - {{\rho I} \over {\left( {a + b} \right)}}$

${{\rho I} \over {2\pi a}} - {{\rho I} \over {2\pi \left( {a + b} \right)}}$

${{\rho I} \over {2\pi \left( {a - b} \right)}}$

Solution

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