When a 1.0-m length of metal wire is connected to a $1.5 - V$ battery, a current of 8.0${~mA}$ flows through it. What is the diameter of the wire? The resistivity of the metal is 2.24 × 10^-8 Ω ∙ m. A) 12 μm B) 6.0 μm C) 24 μm D) 2.2 μm

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The resistance of the wire is given by:$$R = \frac{ρL}{A}$$where:* ρ is the resistivity of the metal (2.24 × 10^-8 Ω ∙ m)* L is the length of the wire (1.0 m)* A is the cross-sectional area of the wire (πr²)The current through the wire is given by:$$I = \frac{V}{R}$$where:* I is the current (8.0 mA = 8.0 × 10^-3 A)* V is the voltage (1.5 V)Substituting the given values into the equations above, we get:$$\frac{2.24 × 10^{-8} Ω ∙ m × 1.0 m}{πr^2} = \frac{1.5 V}{8.0 × 10^{-3} A}$$Solving for r, we get:$$r = 6.0 μm$$Therefore, the diameter of the wire is 2r = 12 μm.

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