A copper bar 0.5 by 1.0 cm in cross-sectional area and 3.0 m long has a steady p
ID: 2237145 • Letter: A
Question
A copper bar 0.5 by 1.0 cm in cross-sectional area and 3.0 m long has a steady potential of 0.1 V between its ends. The atomic weight of copper is 63.5, and its density is 8.96 g/cm3. The resistivity of copper is 1.7 10-8 ? m. a) Find the current flowing through the bar, the current density, and the electric field strength in the bar. b) How many electrons per second flow past any point of the bar? c) Find the average drift velocity of the electrons, and how long it takes, on average, for an electron to flow from one end to the other end of the bar. You will need to estimate n, the number of free electrons per cubic meter in the bar.Explanation / Answer
A copper bar 0.5 by 1.0 cm in cross-sectional area and 3.0 m long has a steady potential of 0.1 V between its ends. The atomic weight of copper is 63.5, and its density is 8.96 g/cm3. The resistivity of copper is 1.7 10-8 ? m.
a) Find the current flowing through the bar, the current density, and the electric field strength in the bar.
R = rho L/A = 1.7e-8*3/(0.005*0.01) = 0.00102 ohms
current : I = V/R = 0.1/0.00102 = 98.0392 A
current density: J = I/A = 0.00102/(0.005*0.01) = 20.4 A/m2
electric field strength: E = V/L = 0.1/3 = 0.0333 V/m
b) How many electrons per second flow past any point of the bar?
n = I t/e = 98.0392 * 1/1.6e-19 = 6.12745 * 10^20 electrons
c) Find the average drift velocity of the electrons, and how long it takes, on average, for an electron to flow from one end to the other end of the bar. You will need to estimate n, the number of free electrons per cubic meter in the bar.
m = density*volume = 8960*(1 m3) = 8960e3 gr/m3
N = 8960e3/63.5 * 6.02e23 = 849436e23 atoms per m3
v = I/nAq = 98.0392/(849436e23*(0.005*0.01)*1.6e-19) = 1.44e-4 m/s