Q13P - Fundamentals Of Physics

Question

How long does it take electrons to get from a car battery to the starting motor? Assume the current is 300 A and the electrons travel through a copper wire with cross-sectional area $\mathbf{0}\mathbf{.}\mathbf{21}\mathit{c}{\mathit{m}}^{\mathbf{2}}$ and length 0.85 m . The number of charge carriers per unit volume is $\mathbf{8}\mathbf{.}\mathbf{49}\mathbf{\times }{\mathbf{10}}^{\mathbf{28}}{\mathit{m}}^{\mathbf{-}\mathbf{3}}$ .

Step-by-Step Solution

Answer

The time required by the electrons to get from a car battery to the starting motor is $8.1\times {10}^{2}s$ .

Step 1: The given data

a) Current, i = 300A

b) Cross-sectional area of copper wire,$A=0.21c{m}^2 or 0.21\times {10}^{-4}{m}^2$

c) Length of copper wire, $L=0.85m$

d) Concentration of charge carriers, $n=8.49\times {10}^{28}{m}^{-3}$

Step 2: Understanding the concept of the current density

The current density is the current across the unit area at a given point in the conductor. We have to use the relation between drift velocity and current density to find the required time.

 

Formulae:

The current density relation to the drift velocity,  $J=ne{v}_d$                                      ...(i)

Where, v is the speed of protons and n is the concentration of protons and e is the charge value.

The current density for the current flowing through an area, $J=\frac{i}{A}$                        ...(ii)

Where, i is current flowing through the motor, A is the cross sectional area

The speed of a body in motion,  $v_d=\frac{L}{t}$                                                                ...(iii)

Where, L is the length of copper wire and t is the time taken by the electrons to get the motor started

Step 3: Calculation of the required time by the electrons

Substituting the given data and the values of speed and current density from equations (iii) and (ii) in equation (i), we can get the value of the required time taken by the electrons to get the motor started as follows:

$$\begin{gathered} t=\frac{neAL}{i} \\ =\frac{\left(8.49\times {10}^{28}{m}^{-3}\right)\left(1.6\times {10}^{-19}C\right)\left(0.21\times {10}^{-4}{m}^2\right)\left(0.85m\right)}{300A} \\ =\frac{242474.4}{300}s \\ =808.248s \\ =8.082\times {10}^{2}sec \\ \approx 8.1\times {10}^{2}sec \end{gathered}$$

Hence, the value of the required time is $8.1\times {10}^{2}sec.$