What is the magnitude of the applied electric field inside an aluminum wire of radius 1.5 mm that carries a 5.5-A current?

E = 2.17 x 10⁻² V/m

Explanation:

First we will find out the current density by using the formula:

J = I/A

where.

J = Current Density = ?

I = Current = 5.5 A

A = Cross-Sectional Area = πr² = π(1.5 x 10⁻³ m)² = 7.068 x 10⁻⁶ m²

Therefore,

J = 5.5 A/7.068 x 10⁻⁶ m²

J = 0.778 x 10⁶ A/m²

Now, we calculate the magnitude of applied field:

E = ρJ

where,

E = Magnitude of applied field = ?

ρ = resistivity of Aluminum = 2.8 x 10⁻⁸ Ω.m

Therefore,

E = (2.8 x 10⁻⁸ Ω.m)(0.778 x 10⁶ A/m²)

E = 2.17 x 10⁻² V/m

pe = weightxheight = 2000x1000=2,000,000joules

a.metals can become magnetic if they have many spinning electrons that create magnetic fields.d.only some metals are attracted by magnets; these metals are called ferromagnetic.

d.wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.