HELP ME OUT PLEASE ASAP

Question

A potential difference of 1.2 V will be applied to a 33 m length of 18-gauge copper wire

with a radius of 0.508 mm. Calculate (I) the current, (II) the magnitude of the current density,

(III) the magnitude of the electric field within the wire, and (IV) the rate at which thermal

energy (power) will appear in the wire. Resistivity for copper is 1.7x10 ^{- 8} Ω.m.

PLEASE SOLVE WITH STEPS.

Answer 1

P = V²/R

Where P = Power

V = Potential difference

R = Resistance of wire

R =ρ l/A

Where ρ= Resistivity of material of wire

l = Length of wire

A = cross sectional area of wire

Also Electric Field = Potential difference /length of wire =V/l

I = current = V*R

So current = I = 1.2/ (ρl/A)

Plug in ρ, l and A values to get Current I

Given ρ = 1.7x10 ^{- 8} Ω.m

l= 33 m and A = ∏r² where ∏ = 3.14

r= 0.508 * 10^-3 m

For Current density = I/Area= Current/Area Amp/m²

Electric field in wire = V/l Volt/m

and Power =V²/ (ρl/A)

Just plug in the values given above