Paragraph: 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 '$$D$$'.
Question: For current entering at $$A$$, the electric field at a distance '$$r$$' from $$A$$ is

When the current $$I$$ enters the block at point $$A$$ it spreads radially through the conducting material. Because the current can occupy only the half-space inside the block, the current lines fill a hemispherical surface of radius $$r$$ centred at $$A$$.

Area of a complete sphere of radius $$r$$ is $$4\pi r^{2}$$. Hence the area of the hemisphere that actually carries current is

$$A_{\text{hemi}} = 2\pi r^{2}\,.$$

Step (i) Current density $$j(r)$$ is the current per unit area on this hemispherical surface:

$$j(r) = \frac{I}{A_{\text{hemi}}} = \frac{I}{2\pi r^{2}}\,.$$

Step (ii) Ohm’s law in microscopic form relates the electric field $$E$$ and the current density $$j$$ through the resistivity $$\rho$$:

$$E(r) = \rho\, j(r)\,.$$

Substituting the expression for $$j(r)$$, we obtain

$$E(r) = \rho \,\frac{I}{2\pi r^{2}}\,.$$

Thus the magnitude of the electric field at a distance $$r$$ from point $$A$$ is

$$E(r) = \frac{\rho I}{2\pi r^{2}}\,.$$

Therefore the correct option is Option C which is: $$\dfrac{\rho I}{2\pi r^{2}}$$.