By using Ohm's law, we can find what should be the resistance of the wire, R:
[tex]R= \frac{V}{I}= \frac{1.5 V}{0.40 A} =3.75 \Omega [/tex]
Then, let's find the cross-sectional area of the wire. Its radius is half the diameter,
[tex]r=35 mm=0.35 \cdot 10^{-3} m[/tex]
So the area is
[tex]A=\pi r^2 = \pi (0.35 \cdot 10^{-3} m)^2=3.85 \cdot 10^{-7} m^2[/tex]
And by using the resistivity of the Aluminum, [tex]\rho=2.65 \cdot 10^{-8} \Omega m[/tex], we can use the relationship between resistance R and resistivity:
[tex]R= \frac{\rho L}{A} [/tex]
to find L, the length of the wire:
[tex]L= \frac{RA}{\rho}= \frac{(3.75 \Omega)(3,85 \cdot 10^{-7} m^2)}{2.65 \cdot 10^{-8} \Omega m}=54.48 m [/tex]
