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# A current of 1.0 A exists in a copper wire of cross-section 1.0 mm^2. Assuming one free electron per atom calculate the drift speed of the free electrons in the wire. The density of copper is 9000 kg/m^3.

Jitender Pal
6 years ago
Sol. i = 1 A, A = 1 mm2 = 1 * 10^–6 m^2 F’ cu = 9000 kg/m^3 Molecular mass has N base 0 atoms = m Kg has (N base 0/M * m) atoms = N base 0AI9000/36.5 * 10^-3 No.of atoms = No.of electrons n = No.of electrons/Unit volume = N base 0Af/mAI = N base 0f/M = 6 * 10^23 * 9000/63.5 * 10^-3 i = V base d n A e. ⇒ V base d = i/nAe = 1/6 * 10^23 *9000/63.3 * 10^-3 *10^-6 *1.6 * 10^-19 = 63.5 * 10^-3/6 * 10^23 *9000 * 10^-6 * 1.6 * 10^-19 = 63.5 * 10^-3/6 * 9 * 1.6 * 10^26 *10^-19 * 10^-6 = 63.5 * 10^10-3/6 * 9 * 1.6 * 10 = 63.5 * 10^-3/6 * 9 *16
Ashok kumar sharma
28 Points
2 years ago
1. Density of Cu=9×103kg/m3 (mass of 1 m3 of Cu) 6.0×1023 atoms has a mass = 63×103kg \ Number of electrons per m3are               =6.0×102363×10-3×9×103=8.5×10​28​​$Density of Cu=9�103kg/m3 (mass of 1 m3 of Cu) 6.0�1023 atoms has a mass = 63�103kg \ Number of electrons per m3are =6.0�102363�10%u22123�9�103=8.5�1028 Now drift velocity =vd=ineA =1.18.5�1028�1.6�10%u221219�%u03C0�(0.5�10%u22123)2 =0.1�10%u22123m/sec$  Now drift velocity =vd=ineA =1.18.5×1028×1.6×10−19×π×(0.5×10−3)2 =0.1×10−3m/sec