30: Problem 30.19

INTRO:
The electric field in a 1.0mm×1.0mm square aluminum wire is 2.8×10⁻² V/m .
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PART A:
What is the current in the wire?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
E = 2.8×10⁻² V/m
b = h = 1.0 mm = 0.001 m

∴ A = 1×10^-6 m²​

using table 30.2... 

σ_Al = 3.5×10⁷ 1/(Ωm)​

Using eq. (30.17) & eq. (30.13)... 
J = I/A = σE

→ I = σEA

= (3.5×10⁷ 1/(Ωm))(2.8×10⁻² V/m)(1×10^-6 m²)
⇒ I = 0.98 A

* note 1 V/Ω = A ( since V=IR ) *​