NCERT Solutions

We need to determine the total time required for a single electron to travel from one end of a $3.0\,\text{m}$ long copper wire to the other, given the current and physical properties of the conductor.

Concept: Drift velocity

The drift velocity ($v_d$) is the average velocity that free electrons attain in a conductor due to an electric field. It is related to the electric current ($I$) by the following equation:

$$I = n e A v_d$$

where $n$ is the free electron density, $e$ is the elementary charge, and $A$ is the cross-sectional area. The time ($t$) taken to traverse a length ($L$) at this constant average speed is:

$$t = \frac{L}{v_d}$$

Given:

• Number density of electrons, $n = 8.5 \times 10^{28}\,\text{m}^{-3}$
• Length of the wire, $L = 3.0\,\text{m}$
• Area of cross-section, $A = 2.0 \times 10^{-6}\,\text{m}^2$
• Current, $I = 3.0\,\text{A}$
• Charge of an electron, $e = 1.6 \times 10^{-19}\,\text{C}$

Solving:

1. First, we express the drift velocity $v_d$ from the current equation:

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

2. Next, we substitute this expression into the formula for time:

$$\begin{aligned} t &= \frac{L}{v_d} \\ &= \frac{L}{\left( \frac{I}{n e A} \right)} \\ &= \frac{n e A L}{I} \end{aligned}$$

3. Now, we substitute the given numerical values into the derived equation:

$$\begin{aligned} t &= \frac{(8.5 \times 10^{28}\,\text{m}^{-3}) \cdot (1.6 \times 10^{-19}\,\text{C}) \cdot (2.0 \times 10^{-6}\,\text{m}^2) \cdot (3.0\,\text{m})}{3.0\,\text{A}} \\ &= (8.5 \times 1.6 \times 2.0) \times 10^{28 - 19 - 6} \times \frac{3.0}{3.0}\,\text{s} \\ &= 27.2 \times 10^3\,\text{s} \\ &= 27200\,\text{s} \end{aligned}$$

4. To express the result in a more intuitive unit, we convert seconds to hours:

$$\begin{aligned} t_{\text{hours}} &= \frac{27200\,\text{s}}{3600\,\text{s}\cdot\text{h}^{-1}} \\ &\approx 7.555\ldots\,\text{h} \end{aligned}$$

Rounding to two significant figures as per the input data:

$$t \approx 2.7 \times 10^4\,\text{s} \approx 7.6\,\text{h}$$

The time taken by an electron to drift from one end of the wire to the other is approximately $2.7 \times 10^4\,\text{s}$ or $7.6\,\text{h}$.

$$2.7 \times 10^4\,\text{s}$$