30: Problem 30.32

INTRO:
The terminals of a 0.70 V watch battery are connected by a 70.0-m-long gold wire with a diameter of 0.200 mm .

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PART A:
What is the current in the wire?
Express your answer using three significant figures.

SOLUTION:

Givens/ Conversions: ​

V = 0.70 V
L = 70.0 m
d = 0.200 mm = 2×10^-4 m
also, Ω = V/A
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So.. we have a voltage, a length, a diameter, and the material. 
We know that the cross-sectional area : A = π/4⋅d²
From table (30.2), Gold has a resistivity of: 

ρ = 2.4×10^-8 Ωm​

By using eq. (30.22), we can determine the resistance.

R = ρL/A = ρL/(π/4⋅d²)​

Finally, we know V=IR, by know, that means that I = V/R

or... I = V/[ρL/(π/4⋅d²)] = πVd²/(4ρL)
→ I = π(0.70 V)(2×10^-4 m)²/[4(2.4×10^-8 Ωm)(70.0 m)]
⇒ I = 0.0131 A​

They want it in mA, though. 1 mA = 10^-3 A

⇒ I = 13.1 mA​