Question

The number of free electrons per cm of copper wire is \(2\times10^{21}\). The average drift speed of the electrons is \(0.25\ \text{mm/s}\). The current flowing is

• A. \(0.8\ \text{A}\)
• B. \(8\ \text{A}\)
• C. \(80\ \text{A}\)
• D. \(800\ \text{A}\)

Hint:

If electrons per unit length and drift speed are given, first find electrons crossing per second: \[ N=(\text{electrons per cm})\times(\text{drift speed in cm/s}) \] Then use \(I=Ne\).

Answer 1

The Correct Option is B

Concept:
Electric current is the amount of charge flowing through a cross-section of a conductor per unit time. \[ I=\frac{Q}{t} \] If \(N\) electrons cross a section per second, then current is: \[ I=Ne \] where: \[ e=1.6\times10^{-19}\ \text{C} \]

Step 1: Write the given data.
Number of free electrons per cm length of wire is: \[ 2\times10^{21} \] Average drift speed of electrons is: \[ 0.25\ \text{mm/s} \] Convert drift speed into cm/s: \[ 1\ \text{mm}=0.1\ \text{cm} \] So: \[ 0.25\ \text{mm/s}=0.25\times0.1\ \text{cm/s} \] \[ 0.25\ \text{mm/s}=0.025\ \text{cm/s} \]

Step 2: Find the number of electrons crossing per second.
Since there are: \[ 2\times10^{21} \] electrons in \(1\ \text{cm}\) length of wire. In \(1\) second, electrons drift through: \[ 0.025\ \text{cm} \] Therefore, number of electrons crossing any section per second is: \[ N=2\times10^{21}\times0.025 \] \[ N=2\times10^{21}\times\frac{25}{1000} \] \[ N=2\times10^{21}\times\frac{1}{40} \] \[ N=5\times10^{19} \]

Step 3: Find the current.
Current is: \[ I=Ne \] Substitute: \[ N=5\times10^{19} \] and: \[ e=1.6\times10^{-19}\ \text{C} \] \[ I=(5\times10^{19})(1.6\times10^{-19}) \] \[ I=5\times1.6 \] \[ I=8\ \text{A} \] Hence, the correct answer is: \[ \boxed{(B)\ 8\ \text{A}} \]