You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resulting current through the wire is 17.6
a. what is the resistivity of the wire?
1) First of all, let's find the resistance of the wire by using Ohm's law: [tex]V=IR[/tex] where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have: [tex]R= \frac{V}{I}= \frac{5.70 V}{17.6 A}=0.32 \Omega [/tex]
2) Now that we have the value of the resistance, we can find the resistivity of the wire [tex]\rho[/tex] by using the following relationship: [tex]\rho = \frac{RA}{L} [/tex] Where A is the cross-sectional area of the wire and L its length. We already have its length [tex]L=2.90 m[/tex], while we need to calculate the area A starting from the radius: [tex]A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2[/tex]
And now we can find the resistivity: [tex]\rho = \frac{RA}{L}= \frac{(0.32 \Omega)(1.34 \cdot 10^{-6}m^2)}{2.90m}= 1.48 \cdot 10^{-7}\Omega \cdot m[/tex]
