91Ó°ÊÓ

The wire has a square cross-section with each side measuring 2.3 mm. Convert this measurement into meters: \[ 2.3 ext{ mm} = 2.3 imes 10^{-3} ext{ m} \]The area \( A \) is then given by:\[ A = ( ext{side})^2 = (2.3 imes 10^{-3} ext{ m})^2 = 5.29 imes 10^{-6} ext{ m}^2 \]

Current density \( J \) is defined as the current \( I \) divided by the cross-sectional area \( A \):\[ J = \frac{I}{A} = \frac{3.6 ext{ A}}{5.29 imes 10^{-6} ext{ m}^2} \approx 6.81 imes 10^5 ext{ A/m}^2 \]

Ohm's Law in terms of current density is \( J = \sigma E \), where \( \sigma \) is the conductivity and \( E \) is the electric field. Conductivity can be expressed as \( \sigma = n e \mu \), where \( n \) is the density of electrons, \( e \) is the elementary charge \( (1.6 imes 10^{-19} ext{ C}) \), and \( \mu \) is mobility:First, find the mobility using the relationship between resistivity \( \rho \), where \( \rho = 1/\sigma \) and standard values:For copper, \( \rho \approx 1.68 \times 10^{-8} \text{ Ohm} \cdot \text{m} \) Thus, \( \sigma = 1/\rho \approx 5.96 imes 10^7 \text{ S/m} \).Now using \( J = \sigma E \) we get:\[ E = \frac{J}{\sigma} = \frac{6.81 \times 10^5}{5.96 \times 10^7} \approx 0.0114 \text{ V/m} \]

The drift velocity \( v_d \) is given by the relation \( J = n e v_d \), solving for \( v_d \):\[ v_d = \frac{J}{n e} = \frac{6.81 \times 10^5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19}} \approx 5.01 \times 10^{-4} \text{ m/s} \]

Using the drift velocity \( v_d \) and wire length \( L = 4.0 \text{ m} \), time \( t \) is:\[ t = \frac{L}{v_d} = \frac{4.0}{5.01 \times 10^{-4}} \approx 7984.03 \text{ s} \]