Question

A write 4.0 m long and 6.0 mm in a diameter has a resistance of 96 . If a potential difference of 20 V is applied between the ends, (a) what is the current in the wire? (b) what is the current density? (c) what is the resistivity of the wire material? 

Answer 1

(a) By Ohm's law we get

$I=\frac{U}{R} \\ I = \frac{20}{96 \times 10^{-3}} = 208.33 \;A$

(b) 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

$S = \pi r^2 = \pi (\frac{d}{2})^2\\ j = \frac{I}{S}\\ = \frac{208.33}{3.14 \times 9 \times 10^{-6}} \\ = 7.37 \times 10^6 \;A/m^2$

(c) By definition of resistivity $R = \frac{\rho L}{S}$

$\rho$ is material resistivity

L is wire length

$\rho = \frac{RS}{L} \\ = \frac{96 \times 10^{-3} \times 3.14 \times 9 \times 10^{-6}}{4.0} \\ = 6.78 \times 10^{-7} \; \Omega \cdot m$