Question

Consider a block of conducting material of resistivity $\rho$ shown in the figure. Current l enters A and B 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:
(a) Take current l entering from A and assume it to spread over a hemispherical surface in the block.
(b) 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.
(c) From the r dependence of E(r), obtain the potential V(r) at r.
(d) Repeat (i), (ii) and (iii) for current l leaving D superpose results for A to D.

$\Delta V$ measured between B and C is

• A. $\frac{\rho L}{\pi a}-\frac{\rho L}{\pi (a+b)}$
• B. $\frac{\rho L}{a}-\frac{\rho L}{(a+b)}$
• C. $\frac{\rho L}{2\pi a}-\frac{\rho L}{2\pi (a+b)}$
• D. $\frac{\rho L}{2\pi (a+b)}$

Answer 1

The correct option is A

$\frac{\rho L}{\pi a}-\frac{\rho L}{\pi (a+b)}$

Let j be the current density . then $j\times 2\pi r^2=l\Rightarrow j=\frac{I}{2\pi r^2}\Delta V_{BC}=-{\int }_{a+b}^{a}Ed\overline{r}=-{\int }_{a+b}^a\frac{\rho l}{2\pi r^2}dr=\frac{\rho l}{2\pi a}-\frac{\rho l}{2\pi (a+b)}$

On applying superposition as mentioned we get total $\Delta V_{BC}=2\times \Delta V_{BC}=\frac{\rho L}{\pi a}-\frac{\rho L}{\pi (a+b)}.$