Question

You are presented with several wires made of the same conducting material. The radius and drift speed are given for each wire in terms of some unknown units \rm r and \rm v. Rank the wires in order of decreasing electron current.Rank from most to least electron current.a) radius=3r, drift speed=1vb) radius=4r, drift speed=0.5vc) radius=1r, drift speed=5vd) radius=2r, drift speed=2.5v

Answer 1

$\begin{array}{ccc}&\text{Radius} & \text{Drift Speed}\\d) & 2 \; r &2.5 \; v\\a) & 3 \; r &1 \; v\\b) & 4 \; r &0.5 \; v\\c) & 1 \; r &5 \; v\\\end{array}$

Explanation

$I = v \cdot A \cdot n \cdot q$,

where

• $I$ is the current;
• $v$ is the drift speed;
• $A$ is the cross-section area of the wire,
• $n$ is the number of charge carrier per unit volume, and
• $q$ is the charge on each charge carrier.

Area of a circular cross-section:

$A = \pi \cdot r^{2}$,

where

• $r$ is the radius of the wire.

$n$ and $q$ are the same for all four samples, for they are made out of the same material.

As a result, $I$ of each wire is directly proportional to $v \cdot r^{2}$ where the value of $\pi \cdot n \cdot q$ is constant.

For each of the four wires:

$\begin{array}{ccc|c}\\& r & v &I \propto v\cdot r^{2}\\a) & 3 & 1 & 9\\b) & 4 & 0.5 & 8\\c) & 1 & 5 & 5\\d) & 2 & 2.5 & 10\\\end{array}$.

How do the four wires rank by their current?

d > a > b > c.