At a certain potential difference we observe 1.5 amps of current in a uniform wire .5 meters long. At the same potential difference and at the same temperature, what will be the current in an otherwise identical wire .74 meters long?
The potential difference is not given, so call it V. The electric field or potential gradient in the first wire is V / .5 m; in the second wire it is V / .74 m. The potential gradient in the second wire is therefore (V / .74) / (V / .5) = .5 / .74 = .6756 times that in the first.
It follows that the drift velocity of the electrons in the second wire is 1.5 times that in the first. The wires are identical: except for length they each have the same number of current carriers available per unit length. So the current, or the rate at which charges pass a point in the second wire, is 1.5 times that in the first. The current in the second wire is therefore .6756 * 1.5 amps = 1.013 amps.
If the same potential difference V is maintained across two uniform wires of lengths L1 and L2, identical except for length, then the ratio of the second to the first potential gradient is
The ratio of drift velocities will therefore be L1 / L2, equal to the ratio of potential gradients.
Since equal numbers of charge carriers are available per unit length, the ratio of currents will be L1 / L2, equal to the ratio of drift velocities.
The figure below shows two wires of different lengths with the same potential difference V between their ends. The potential gradient E = dV / dx is greater in the shorter wire, so the drift velocity is higher. The higher drift velocity implies a proportionally higher current.
"