There is a current of \(0.21\,\text{A}\) in a copper wire of area of cross section \(10^{-6}\,\text{m}^2\). If the number of electrons per m³ is \(8.4 \times 10^{28}\) and \(e = 1.6 \times 10^{-19}\,\text{C}\), the drift velocity is:

Drift Velocity in Copper Wire – Numerical Solution

Question: There is a current of \(0.21\,\text{A}\) in a copper wire of area of cross section \(10^{-6}\,\text{m}^2\). If the number of electrons per m³ is \(8.4 \times 10^{28}\) and \(e = 1.6 \times 10^{-19}\,\text{C}\), the drift velocity is:

a) \(1.562 \times 10^{-5}\,\text{m/s}\)

b) \(2 \times 10^{-5}\,\text{m/s}\)

c) \(0.64 \times 10^{5}\,\text{m/s}\)

d) \(1 \times 10^{5}\,\text{m/s}\)

Detailed Solution:

The drift velocity \(v_d\) is given by:

\[ v_d = \frac{I}{n\,e\,A} \]

Where:

• \(I = 0.21\,\text{A}\)
• \(n = 8.4 \times 10^{28}\,\text{electrons/m}^3\)
• \(e = 1.6 \times 10^{-19}\,\text{C}\)
• \(A = 10^{-6}\,\text{m}^2\)

Compute the denominator:

\[ n\,e\,A = 8.4 \times 10^{28} \times 1.6 \times 10^{-19} \times 10^{-6} = 13.44 \times 10^{3} = 1.344 \times 10^{4} \]

Thus,

\[ v_d = \frac{0.21}{1.344 \times 10^{4}} \approx 1.562 \times 10^{-5}\,\text{m/s} \]

Correct answer: (a) \(1.562 \times 10^{-5}\,\text{m/s}\)