Q74P - Fundamentals Of Physics

Question

The current density in a wire is uniform and has magnitude $\mathbf{2}\mathbf{.}\mathbf{0}\mathit{x}{\mathbf{10}}^{\mathbf{6}}\mathit{A}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$, the wire’s length is 5.0m, and the density of conduction electrons is  $\mathbf{8}\mathbf{.}\mathbf{49}\mathit{x}{\mathbf{10}}^{28}\mathbf{/}{\mathit{m}}^3$. How long does an electron take (on the average) to travel the length of the wire?

Step-by-Step Solution

Answer

The time required by an electron to travel a wire is $3.4\times {10}^{4}s$.   

Step 1: The given data

1. Current density, $\left|J\right|=2\times {10}^6 A/m^2$
2. Density of the conduction electrons $n=8.49\times {10}^{28}/m^3$
3. Charge on electron $e=1.6\times {10}^{-19} C$
4. Length of wire L = 5m

Step 2: Understanding the concept of the drift velocity

To calculate the time at which an electron travels a wire, we will calculate first its drift speed. For that, we can use the formula for the drift speed in terms of current density and density of electrons, and charge on electrons. To find the time, we can use the relationship between velocity, distance, and time.

 

Formulae:

The drift velocity of electrons due to an uniform current density, $v_d=\left|J\right|l ne$    (i)

The time taken by the electron, $t=L/v_d$                                                        (ii)

Step 3: Calculation of the traveling time of the electron

At first, we can calculate the drift speed using the given data in equation (i) as follows:

$\begin{array}{l}{v}_d=\frac{(2\times {10}^6)}{(8.49\times {10}^{28})\times (1.6\times {10}^{-19})}\\ =147\times {10}^{-4}m/s\end{array}$

Now, average time required to travel taken by an electron is given using the data in equation (ii) as follows:

$$\begin{gathered} t=\frac{5}{\left(1.47\times {10}^{-4}\right)} \\ =3.4\times {10}^{4}s \end{gathered}$$

Hence, the value of the time is $3.4\times {10}^{4}s$.