The resistance of a conductor is given by:
[tex]R= \frac{\rho L}{A} [/tex]
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area
We can use this formula to solve both parts of the problem.
a) The length of the copper wire is L=1.0 m. Its diameter is d=0.50 mm, so its radius is
[tex]r= \frac{d}{2}=0.25 mm=0.25 \cdot 10^{-3} m [/tex]
And its cross-sectional area is
[tex]A=\pi r^2 = \pi (0.25 \cdot 10^{-3}m)^2 = 1.96 \cdot 10^{-7} m^2[/tex]
The copper resistivity is [tex]\rho=1.68 \cdot 10^{-8} \Omega m[/tex], therefore the resistance of this piece of wire is
[tex]R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.0 m)}{1.96 \cdot 10^{-7} m^2}= 8.57 \cdot 10^{-2} \Omega[/tex]
b) The length of this piece of iron is L=10 cm=0.10 m. Its cross-sectional size is L=1.0 mm=0.001 m, so its cross-sectional area is
[tex]A=L^2 = (0.001 m)^2 =1 \cdot 10^{-6}m^2[/tex]
The iron resistivity is [tex]\rho = 9.71 \cdot 10^{-8} \Omega m[/tex], therefore the resistance of this piece of wire is
[tex]R= \frac{\rho L}{A}= \frac{(9.71 \cdot 10^{-8} \Omega m)(0.10 m)}{1.0 \cdot 10^{-6} m^2}=9.71 \cdot 10^{-3} \Omega [/tex]
