Use a swept-volume argument to obtain I=Anvq and define every symbol. A conductor has one carrier population of charge magnitude 2e, n=5.0 × 10²³ m⁻³, A=2.0 × 10⁻⁷ m² and v=3.0 × 10⁻³ m s⁻¹. Calculate current using e=1.60 × 10⁻¹⁹ C. Explain how direction changes if the carriers are negative.

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I=(2.0×10⁻⁷)(5.0×10²³)(3.0×10⁻³)(3.20×10⁻¹⁹)=9.6×10⁻⁵ A. For negative carriers conventional current is opposite their drift direction. Identify A in m², n in m⁻³, v as mean drift speed in m s⁻¹ and q in C. Use distance vΔt and swept volume AvΔt. Multiply by n for number and q for charge, then divide by Δt. Use q=2e, not e, to obtain 9.6×10⁻⁵ A. Explain conventional direction relative to carrier sign.

Derive and interpret Anvq

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