Question

A wire of iron which is 3.14 m long and has a radius of 0.5 mm is connected between

the terminals of a 5 V battery. If the resistivity of the iron is 10-72.m, so the passing
electric current intensity in the wire equals..
(Take: pi=3.14)

Answer 1

12.5 A

Step-by-step explanation:

You want the current intensity in an iron wire 3.14 m long with a radius of 0.5 mm connected to a 5V battery, given the resistivity of iron is 10^-7 Ω·m.

Resistance

The resistance of the wire is proportional to wire length, and inversely proportional to area:

R = ρl/A

R = (10^-7 Ω·m)(3.14 m)/(3.14·(0.5·10^-3 m)²) = (1/.25)·10^-1 Ω = 0.4 Ω

Current

The current is given by Ohm's law:

I = V/R

I = (5 v)/(0.4 Ω) = 12.5 A

The current intensity is 12.5 amperes.

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Additional comment

The resistivity of iron is 1.0·10⁻⁷ Ω·m, so we presume the 10-72.m in the problem statement is an error in picture-to-text translation.