Question

A current of 5A a runs through a 12 gauge copper wire (diameter 2.05mm) and a light bulb. (A) how many electrons pass through the bulb every second? (B) what is the current density in the wire? (C) how quickly does a common electron pass through any given point of the wire?

Una corriente de 5.00 A corre a través de un alambre de cobre de calibre 12 (diámetro, 2.05 mm) y de una bombilla. (a) ¿Cuántos electrones pasan por la bombilla cada segundo? (b) ¿Cuál es la densidad de corriente en el alambre? (c) ¿Con qué rapidez un electrón común pasa por cualquier 4. mm/s

Answer 1

Given that Current (I) = 5A Diameter of the wire (d) = 2.05 times 10^-3 n and e = charge of electron = 1.6 times 10^-19 c Let us calculate the cross - sectional area is A = pi d^2/4 = 3.14 times (2.05 times 10^-3)^2/4 A = 3.29 times 10^-5 (a) Now, for t = 1 sec q = I t = 1 times 5 = 5c therefore Number of electron passing through the bulb is N = q/e = 5/1.6 times 10^-19 = 3.12 times 10^-19 therefore N = 3.12 times 10^19 electrons/second (b) Current density is J = I/A = 5A/3.29 times 10^-5 m^2 = 1.51 times 10^6 A/m^2 therefore J = 1.51 times 10^6 A/m^2 \r\n (c) Now in general the free electron density of copper = 8.5 times 10^28 per m^3 So the velocity of electron i.e, the drift velocity v_d = J/ne = 1.51 times 10^6/8.5 times 10^28 times 1.6 times 10^-19
$Area=%u03C0d**2/4; A=3.29�10**-6 Current Density; J=I/A; 1.51�10**6 N=Q/e; 3.12�10**19 Drift velocity, v=0.284mm/sec$