To solve the problem step by step, we will address both parts (a) and (b) separately.
### Part (a): Time taken for an electron to displace from one end of the wire to the other.
1. **Identify the given values:**
- Length of the wire, \( L = 100 \, \text{m} \)
- Cross-sectional area, \( A = 1.0 \, \text{mm}^2 = 1.0 \times 10^{-6} \, \text{m}^2 \)
- Current, \( I = 4.5 \, \text{A} \)
- Resistivity of copper, \( \rho = 1.72 \times 10^{-8} \, \Omega \cdot \text{m} \)
- Density of copper, \( \text{Density} = 8.96 \, \text{g/cm}^3 = 8960 \, \text{kg/m}^3 \)
2. **Calculate the number density of free electrons, \( n \):**
- The molar mass of copper (Cu) is approximately \( 63.5 \, \text{g/mol} \).
- The number of atoms (or free electrons) per mole is Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \).
- The number density \( n \) can be calculated as:
\[
n = \frac{\text{Density}}{\text{Molar mass}} \times N_A = \frac{8960 \, \text{kg/m}^3}{0.0635 \, \text{kg/mol}} \times 6.022 \times 10^{23} \, \text{atoms/mol}
\]
- Calculating \( n \):
\[
n \approx \frac{8960}{0.0635} \times 6.022 \times 10^{23} \approx 8.5 \times 10^{28} \, \text{electrons/m}^3
\]
3. **Use the formula for drift velocity, \( V_D \):**
- The current \( I \) is related to drift velocity by the equation:
\[
I = n \cdot e \cdot A \cdot V_D
\]
- Rearranging for drift velocity \( V_D \):
\[
V_D = \frac{I}{n \cdot e \cdot A}
\]
- Where \( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron).
4. **Calculate the drift velocity \( V_D \):**
\[
V_D = \frac{4.5 \, \text{A}}{(8.5 \times 10^{28} \, \text{electrons/m}^3) \cdot (1.6 \times 10^{-19} \, \text{C}) \cdot (1.0 \times 10^{-6} \, \text{m}^2)}
\]
- Calculate \( V_D \):
\[
V_D \approx \frac{4.5}{(8.5 \times 10^{28}) \cdot (1.6 \times 10^{-19}) \cdot (1.0 \times 10^{-6})} \approx 3.3 \times 10^{-4} \, \text{m/s}
\]
5. **Calculate the time \( t \) taken for an electron to travel the length of the wire:**
- Using the formula \( t = \frac{L}{V_D} \):
\[
t = \frac{100 \, \text{m}}{3.3 \times 10^{-4} \, \text{m/s}} \approx 3.03 \times 10^{5} \, \text{s}
\]
### Part (b): Sum of electrostatic forces acting on all free electrons in the wire.
1. **Calculate the total number of free electrons in the wire:**
- The total volume \( V \) of the wire is given by:
\[
V = A \cdot L = (1.0 \times 10^{-6} \, \text{m}^2) \cdot (100 \, \text{m}) = 1.0 \times 10^{-4} \, \text{m}^3
\]
- The total number of free electrons \( N \) is:
\[
N = n \cdot V = (8.5 \times 10^{28} \, \text{electrons/m}^3) \cdot (1.0 \times 10^{-4} \, \text{m}^3) \approx 8.5 \times 10^{24} \, \text{electrons}
\]
2. **Calculate the total electrostatic force \( F \):**
- The electric field \( E \) in the wire can be calculated using \( E = \frac{V}{L} \) where \( V \) is the potential difference across the wire.
- The potential difference can be calculated using Ohm's law \( V = I \cdot R \), where \( R = \frac{\rho L}{A} \):
\[
R = \frac{1.72 \times 10^{-8} \, \Omega \cdot \text{m} \cdot 100 \, \text{m}}{1.0 \times 10^{-6} \, \text{m}^2} = 1.72 \, \Omega
\]
- Thus, the potential difference \( V \) is:
\[
V = I \cdot R = 4.5 \, \text{A} \cdot 1.72 \, \Omega \approx 7.74 \, \text{V}
\]
- Now, the electric field \( E \) is:
\[
E = \frac{7.74 \, \text{V}}{100 \, \text{m}} = 0.0774 \, \text{V/m}
\]
3. **Calculate the force on a single electron:**
- The force \( F_e \) on a single electron is given by:
\[
F_e = e \cdot E = (1.6 \times 10^{-19} \, \text{C}) \cdot (0.0774 \, \text{V/m}) \approx 1.24 \times 10^{-20} \, \text{N}
\]
4. **Calculate the total force \( F \) acting on all free electrons:**
\[
F = N \cdot F_e = (8.5 \times 10^{24}) \cdot (1.24 \times 10^{-20} \, \text{N}) \approx 1.06 \, \text{N}
\]
### Final Answers:
- (a) The time taken for an electron to displace from one end of the wire to the other is approximately \( 3.03 \times 10^{5} \, \text{s} \).
- (b) The sum of electrostatic forces acting on all free electrons in the wire is approximately \( 1.06 \, \text{N} \).
