Question

What is the magnitude of the applied electric field inside an aluminum wire of radius 1.0 mmmm that carries a 6.0-AA current? [[ σaluminumσaluminum = 3.6 ×× 1077 A/(V⋅m)A/(V⋅m) ]]

Express your answer with the appropriate units.

Answer 1

We know that current density in a wire is given by:

J = I/A

Also, Current density is given by:

J = *E

So,

I/A = *E

E = I/(A*)

= conductivity of aluminum = 3.6*10^7 A/V.m

A = Cross-sectional Area of wire = pi*r^2

r = radius of wire = 1.0 mm = 1.0*10^-3 m

I = current through wire = 6.0 A

E = Electric field inside wire = ?

So, Using given values:

E = 6.0/(pi*(1.0*10^-3)^2*3.6*10^7)

E = 0.053 V/m

Let me know if you've any query.