The resistivity of gold is $2.44 \times 10 ^{- 8 }\Omega \cdot \mathrm { m }$ at a temperature of 20°C. A gold wire that is 0.50 mm in diameter and 44 cm long carries a current of 380 mA. How many electrons each second pass a given cross section of the wire at 20°C? (e = 1.60 × 10^-19C) A) 2.4 × 10¹⁸ B) 2.4 × 10¹⁷ C) 1.2 × 10²² D) 2.8 × 10¹⁴ E) 6.3 × 10¹⁵

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The current is given by:$$I = nAqv$$where:* n is the number of charge carriers per unit volume* A is the cross-sectional area of the wire* q is the charge of each carrier* v is the drift velocity of the carriersThe resistivity is given by:$$ho = \frac{E}{J}$$where:* E is the electric field* J is the current densityThe electric field is given by:$$E = \frac{V}{d}$$where:* V is the potential difference* d is the distance between the ends of the wireThe current density is given by:$$J = \frac{I}{A}$$Combining these equations, we get:$$ho = \frac{V}{Id}\frac{A}{nAqv}$$Simplifying, we get:$$n = \frac{V}{Iho d}$$Substituting the given values, we get:$$n = \frac{(380 	imes 10^{-3} 	ext{ A})(2.44 	imes 10^{-8} \Omega \cdot 	ext{ m})(0.44 	ext{ m})}{(1.60 	imes 10^{-19} 	ext{ C})}$$$$n = 2.4 	imes 10^{18} 	ext{ m}^{-3}$$The number of electrons passing a given cross section of the wire each second is given by:$$N = nAv$$Substituting the given values, we get:$$N = (2.4 	imes 10^{18} 	ext{ m}^{-3})(\pi (0.25 	imes 10^{-3} 	ext{ m})^2)(380 	imes 10^{-3} 	ext{ m/s})$$$$N = 2.4 	imes 10^{18} 	ext{ s}^{-1}$$Therefore, the correct answer is A.

The Resistivity of Gold Is 2.44×10−8Ω⋅m2.44 \times 10 ^{- 8 }\Omega \cdot \mathrm { m }2.44×10−8Ω⋅m

The Resistivity of Gold Is $2.44 \times 10 ^{- 8 }\Omega \cdot \mathrm { m }$