Question

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σaluminum = 3.6 × 107 a/(v⋅m) ]?

Answer 1

Hello

1) First of all, since we know the radius of the wire ($r=1.2~mm=0.0012~m$), we can calculate its cross-sectional area
$A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2$

2)  Then, we can calculate the current density J inside the wire. Since we know the current, $I=3~A$, and the area calculated at the previous step, we have
$J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2$

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity $\sigma=3.6\cdot10^7~ \frac{A}{Vm}$ of the aluminium, the electric field is given by
$E= \frac{J}{\sigma}= \frac{ 6.63\cdot10^5 ~A/m^2}{3.6\cdot10^7~ \frac{A}{Vm} } = 0.018~V/m$