Question

A uniform silver wire has a resistivity of \(1.54 \times 10^{-8} \, \Omega \, \text{m}\) at room temperature. When an electric field along the wire is \(1 \, \text{V/m}\), the drift velocity is:

• A. \(v_d = 0.5 \, \text{m/s}\)
• B. \(v_d = 1.5 \, \text{m/s}\)
• C. \(v_d = 0.15 \, \text{m/s}\)
• D. \(v_d = 0.05 \, \text{m/s}\)

Hint:

Use the relation τ =m/ρne^2 to simplify drift velocity calculations in conductors with known resistivity and electron density.

Answer 1

The Correct Option is D

The drift velocity is calculated using:
\[v_d = \frac{eE\tau}{m}\]
Using resistivity, $\tau$ can be derived from the relation:
\[\rho = \frac{m}{ne^2\tau}\]
Given $E = 1 \text{ V/m}$, $e = 1.6 \times 10^{-19} \text{ C}$, $\rho = 1.54 \times 10^{-8} \text{ }\Omega\cdot\text{m}$, $n = 5.8 \times 10^{28} \text{ m}^{-3}$, and $m = 9.11 \times 10^{-31} \text{ kg}$:
\[\tau = \frac{m}{\rho ne^2}\]
Substituting into $v_d$:
\[v_d = \frac{eE}{\rho n} = \frac{1.6 \times 10^{-19} \times 1}{1.54 \times 10^{-8} \times 5.8 \times 10^{28}} \approx 0.69 \text{ m/s}\]