Given:
The maximum current in the circuit is,
[tex]i=34\text{ A}[/tex]The power per length is,
[tex]\frac{P}{l}=1.6\text{ W/m}[/tex]To find:
The diameter of the copper wire
Explanation:
The power (P) produced by current i, through a copper wire of resistance R and length l is given by,
[tex]\begin{gathered} Pl=i^2R \\ \frac{R}{l}=\frac{P}{i^2} \\ \frac{R}{l}=\frac{1.6}{34\times34} \end{gathered}[/tex]Now,
[tex]\begin{gathered} R=\frac{\rho l}{A} \\ R=\frac{\rho l}{\pi r^2} \end{gathered}[/tex]The resistivity of copper is,
[tex]\rho=1.72\times10^{-8}\text{ ohm.m}[/tex]So, we can write,
[tex]\begin{gathered} \frac{R}{l}=\frac{\rho}{\pi r^2} \\ \frac{1.6}{34\times34}=\frac{1.72\times10^{-8}}{\pi r^2} \\ r^2=\frac{1.72\times10^{-8}\times34\times34}{1.6} \\ r=3.5\times10^{-3}\text{ m} \\ diamer\text{ is,} \\ 2r=7.0\times10^{-3}\text{ m} \end{gathered}[/tex]Hence, the diameter is,
[tex]7.0\times10^{-3}\text{ m}[/tex]