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:
For current entering at $A$, the electric field at a distance ' $r$ ' from $A$ is
$\frac{\rho l}{8 \pi r^2}$
$\frac{\rho l}{r^2}$
$\frac{\rho l}{2 \pi r^2}$
$\frac{\rho l}{4 \pi r^2}$
Solution
The current entering at $A$ spreads over a hemispherical surface in the block. At a distance $r$ from $A$, the current spreads over the surface of a hemisphere of radius $r$.
The surface area of a hemisphere is given by $2\pi r^2$. Therefore, the current density $j$ at a distance $r$ from $A$ is given by
$j = \frac{I}{2\pi r^2}$
According to Ohm's law in differential form, the electric field $E$ is given by
$E = \rho j$
Substituting the value of $j$ in the above equation, we get
$E = \rho \left(\frac{I}{2\pi r^2}\right) = \frac{\rho I}{2\pi r^2}$
Therefore, the correct answer is option C, $\frac{\rho I}{2\pi r^2}$.
