Question

At room temperature, what is the strength of the electric field in a 12 gauge copper wire (diameter 2.05 mm) that is needed to cause a 4.70 A current to flow? Use the resistivity at room temperature for copper rho = 1.72×10−8 Ω⋅m.

Answer 1

Answer:

Electric field will be $2.33\times 10^{-2}V/m$

Explanation:

We have given diameter of the copper d = 2.05 mm

So radius $r=\frac{d}{2}=\frac{2.05}{2}=1.05mm=1.05\times 10^{-3}m$

We know that area $A=\pi r^2=3.14\times (1.05\times 10^{-3})^2=3.461\times 10^{-6}m^2$

Current is given as i = 4.7 A

So current density $j=\frac{i}{A}=\frac{4.70}{3.461\times 10^{-6}}=1.357\times 10^6A/m^2$

Resistivity is given $\rho =1.72\times 10^{-8}ohm-m$

We know that electric field is given by $E=\rho j=1.72\times 10^{-8}\times 1.357\times 10^6=2.33\times 10^{-2}V/m$