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= (I)/(A)= (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~ (A)/(Vm)` of the aluminium, the electric field is given by

`E= (J)/(\sigma)= ( 6.63\cdot10^5 ~A/m^2)/(3.6\cdot10^7~ (A)/(Vm) ) = 0.018~V/m`