Answer to Question #154882 in Electric Circuits for Nethononda

Question

A wire 4.00 m long and 6.00 mm in diameter has a resistance of 15.0 mΩ.

A potential difference of 23.0 V is applied between the ends.

2.1. What is the current in the wire?

2.2. What is the magnitude of the current density?

2.3. Calculate the resistivity of the wire material.

2.4. Using the resistivity constants table, identify the 

Answer 1

1. By Ohm's law we get $I = \frac{U}{R} = \frac{23}{15\cdot10^{-3}}\approx 1.5\cdot10^3A$ .
2. If we suppose that the current density $j$ is uniform, we get by definition $jS =I$ , where $S$ is a section area. As section is circular, $S = \pi r^2 = \pi (d/2)^2$. Thus we have $j = \frac{I}{S}=\frac{1.5\cdot10^3}{\pi \cdot3^2\cdot10^{-6}}\approx 5.3\cdot10^7 A/m^2$ .
3. By definition of resistivity $R = \frac{\rho l}{S}$, where $\rho$ is material resistivity, $l$ is wire length. Thus we find $\rho = R\frac{S}{l} = \frac{15\cdot10^{-3}\cdot\pi\cdot3^2\cdot10^{-6}}{4}\approx 1.06 \cdot10^{-7} \Omega\cdot m$
4. Using the resistivity table from Wikipedia ( Electrical resistivity and conductivity - Wikipedia ) we find that this material should be platinum.