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

a cold ice-tea.

pour in room temperature or cold water. cover the pitcher and refrigerate. steep white or green tea for 6 to 8 hours; steep black or oolong tea for 8 to 12 hours.jun 28, 2015

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//