Let ' A ' be the cross-sectional area and ' ' be the specific...

Potential gradient is defined as the potential drop per unit length of the wire.

Let $V$ be the potential difference across a length $L$ of the wire.

By Ohm's law, $V = IR$

The resistance $R$ of the wire is given by $R = \rho \dfrac{L}{A}$

Substituting the value of $R$, we get $V = I \left(\rho \dfrac{L}{A}\right)$

The potential gradient is $\dfrac{V}{L} = \dfrac{I\rho}{A}$

Answer: $\dfrac{I\rho}{A}$