[tex]\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}[/tex]
[tex]I = v \cdot A \cdot n \cdot q[/tex],
where
Area of a circular cross-section:
[tex]A = \pi \cdot r^{2}[/tex],
where
[tex]n[/tex] and [tex]q[/tex] are the same for all four samples, for they are made out of the same material.
As a result, [tex]I[/tex] of each wire is directly proportional to [tex]v \cdot r^{2}[/tex] where the value of [tex]\pi \cdot n \cdot q[/tex] is constant.
For each of the four wires:
[tex]\begin{array}{ccc|c}\\& r & v &I \propto v\cdot r^{2}\\a) & 3 & 1 & 9\\b) & 4 & 0.5 & 8\\c) & 1 & 5 & 5\\d) & 2 & 2.5 & 10\\\end{array}[/tex].
How do the four wires rank by their current?
d > a > b > c.
