A 4.0 m length of gold wire is connected to a 1.5 V battery,and a current of 4.0 mA flows through it.What is the diameter of the wire? (The resistivity of gold is 2.44 × 10^-8 Ω·m. )

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The resistance of the wire is given by:$$R = \frac{ho L}{A}$$where:* R is the resistance in ohms (Ω)* ρ is the resistivity in ohm-meters (Ω·m)* L is the length in meters (m)* A is the cross-sectional area in square meters (m²)The cross-sectional area of a wire is given by:$$A = \frac{\pi d^2}{4}$$where:* d is the diameter in meters (m)Substituting the given values into the first equation, we get:$$4.0 	imes 10^{-3} = \frac{2.44 	imes 10^{-8} 	imes 4.0}{A}$$Solving for A, we get:$$A = 4.88 	imes 10^{-11} 	ext{ m}^2$$Substituting this value into the second equation, we get:$$d = \sqrt{\frac{4A}{\pi}} = 1.7 	imes 10^{-5} 	ext{ m} = 17 \mu	ext{m}$$Therefore, the diameter of the wire is 17 μm.

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