Answer to Question #91879 in Electric Circuits for Priskilla

Question

A 6.0m long wire having a diameter of 2.0 mm has a resistance of 12.2 m ohm. A potential difference of 23V is applied between the ends. Calculate the current in wire, the magnitude of the current density and the resistivity of the wire material

Answer 1

According to the Ohm's law, the current in a wire is

$I = \frac{V}{R}$

Substituting the numerical values, we obtain:

$I = \frac{23 \, V}{12.2 \cdot 10^{-3} \, \Omega} \approx 1.89 \cdot 10^3 \, A$

The current density is a current flowing through a unit cross-sectional area of a wire:

$j = \frac{I}{S} = \frac{4 I}{\pi d^2}$

Substituting the numerical values, we obtain:

$j = \frac{4 \cdot 1.89 \cdot 10^{3}}{3.14 \cdot 2^2 \cdot 10^{-6}} \approx 6.0 \cdot 10^8 \frac{A}{m^2}$

The resistivity of a wire material is:

$R = \rho \frac{l}{S} \, \Rightarrow \, \rho = \frac{R S}{l} = \frac{R \pi d^2}{4 l}$

Substituting the numerical values, we obtain:

$\rho = \frac{12.2 \cdot 10^{-3} \cdot 3.14 \cdot 2^2}{4 \cdot 6} \approx 6.3 \cdot 10^{-3} \, \frac{Ohm \cdot mm^2}{m}$