Question

A copper wire of length 1 m and uniform cross sectional area 5 × 10^-7 m² carries a current of 1 A. Assuming that there are 8 × 10²⁸ free electrons per m³ in copper, how long will an electron take to drift from one end of the wire to the other?

• A. 0.8 × 10³ S
• B. 1.6 × 10³ S
• C. 3.2 × 10³ S
• D. 6.4 × 10³ S

Answer 1

The Correct Option is D

Step 1: Calculate the drift velocity ($v_d$) of electrons.

The drift velocity is related to the current density ($J$), number density of electrons ($n$), and charge of an electron ($e$) by the formula:

$J = n e v_d$

Current density $J = \frac{I}{A}$, where $I$ is the current and $A$ is the cross-sectional area. So, $\frac{I}{A} = n e v_d$

$v_d = \frac{I}{n e A}$

Step 2: Calculate the drift velocity using given values.

$v_d = \frac{1 \ A}{(8 \times 10^{28} \ m^{-3}) \times (1.6 \times 10^{-19} \ C) \times (5 \times 10^{-7} \ m^{2})}$

$v_d = \frac{1}{8 \times 1.6 \times 5 \times 10^{28 - 19 - 7}} \ m/s$

$v_d = \frac{1}{64 \times 10^{2}} \ m/s = \frac{1}{6400} \ m/s$

Step 3: Calculate the time ($t$) taken to drift the length of the wire.

Time = Distance / Speed

$t = \frac{L}{v_d} = \frac{1 \ m}{\frac{1}{6400} \ m/s} = 6400 \ s = 6.4 \times 10^{3} \ s$

The correct answer is (D) $6.4 \times 10^3\,\text{s}$.

Answer 2

The drift velocity ($v_d$) of electrons in a conductor is given by:

$v_d = \frac{I}{neA}$

where:

• $I$ is the current,
• $n$ is the number density of free electrons,
• $e$ is the elementary charge ($1.6 \times 10^{-19}\,\text{C}$), and
• $A$ is the cross-sectional area.

$v_d = \frac{1\,\text{A}}{(8 \times 10^{28}\,\text{m}^{-3})(1.6 \times 10^{-19}\,\text{C})(5 \times 10^{-7}\,\text{m}^2)} \approx 1.56 \times 10^{-4}\,\text{m/s}$

The time taken ($t$) to drift across the length ($l$) of the wire is:

$t = \frac{l}{v_d} = \frac{1\,\text{m}}{1.56 \times 10^{-4}\,\text{m/s}} \approx 6.4 \times 10^3\,\text{s}$

The correct answer is (D) $6.4 \times 10^3\,\text{s}$.