## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

V= 0.70 V l= 100 m Resistivity of gold $\rho=2.4\times10^{-8} Ωm$ $r=0.05\times10^{-3}m$ A= $\pi r^{2}=\pi (0.05\times10^{-3}m)^{2}$ Current I= $\frac{V}{R}=\frac{V}{\rho \frac{l}{A}}$ $= \frac{VA}{\rho l}=\frac{0.70 V\times\pi (0.05\times10^{-3}m)^{2}}{2.4\times10^{-8} Ωm\times100m}= 2.3\times10^{-3}A$= 2.3 mA