Answer:
[tex]r = 0.93 \times 10^{-3} m[/tex]
Explanation:
As we know that electrical power consumed by a wire is given as
[tex]P = i^2 R[/tex]
here we know that power per unit length is given as
[tex]P_l = \frac{P}{L} = i^2 \frac{R}{L}[/tex]
also we know that resistance depends on length and area
so we have
[tex]R = \rho \frac{L}{A}[/tex]
so we have
[tex]P_l = i^2(\frac{\rho}{A})[/tex]
now we have
[tex]P_l = 2 W/m[/tex]
i = 18 A
so we have
[tex]2 = 18^2 (\frac{1.68 \times 10^{-8}}{\pi r^2})[/tex]
[tex]r = 0.93 \times 10^{-3} m[/tex]
