Question

On applying 10 V across the two ends of a 100 cm long copper wire, the average drift velocity (in cm/s) in the wire is (rounded off to two decimal places).

Hint:

Even with large current, drift velocity is very small because electron density is extremely high. Electrons respond slowly as a group even though individual thermal speeds are huge.

Answer 1

Correct Answer: 4.2

Step 1: Data given.
\[ V = 10 \, \text{V}, L = 100 \, \text{cm}, n = 8.43 \times 10^{22} \, \text{cm}^{-3}, \] \[ \rho = 1.67 \times 10^{-6} \, \Omega \, \text{cm}, e = 1.6 \times 10^{-19} \, \text{C} \]

Step 2: Electric field.
\[ E = \frac{V}{L} = \frac{10}{100} = 0.1 \, \text{V/cm} \]

Step 3: Current density.
Ohm’s law in terms of current density: \[ J = \frac{E}{\rho} \] \[ J = \frac{0.1}{1.67 \times 10^{-6}} \approx 5.99 \times 10^4 \, \text{A/cm}^2 \]

Step 4: Drift velocity.
\[ J = n e v_d \Rightarrow v_d = \frac{J}{n e} \] \[ v_d = \frac{5.99 \times 10^4}{(8.43 \times 10^{22})(1.6 \times 10^{-19})} \] \[ = \frac{5.99 \times 10^4}{1.3488 \times 10^4} \approx 0.4449 \, \text{cm/s} \] Final Answer:
\[ \boxed{0.45 \, \text{cm/s}} \]