Question

The area of cross-section of a copper wire is $4 \times 10^{-7} m^2$ and the number of electrons per cubic meter in copper is $8 \times 10^{28}$. If the wire carries a current of 6.4 A, then the drift velocity of the electrons (in $10^{-3} m/s$) is

• A. 0.25
• B. 2.5
• C. 0.125
• D. 1.25

Hint:

Drift velocity is proportional to current and inversely proportional to number density, charge, and cross-sectional area.

Answer 1

The Correct Option is D

Drift velocity \(v_d\) is given by: \[ I = n e A v_d \Rightarrow v_d = \frac{I}{n e A} \] Where: \(I = 6.4\, A\), \(n = 8 \times 10^{28} \, m^{-3}\), \(e = 1.6 \times 10^{-19} C\), \(A = 4 \times 10^{-7} m^2\) Calculate: \[ v_d = \frac{6.4}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 4 \times 10^{-7}} = \frac{6.4}{5.12 \times 10^{3}} = 1.25 \times 10^{-3} m/s \]