Question

A) A wire made from iron with a cross-section of diameter 0.800 mm carries a current of 14.0 A. Calculate the "areal current density"; in other words, how many electrons per square meter per second flow through this wire?
B) The density of iron is 7.86 g/cm3, and its atomic mass is 56.2. Assuming each iron atom contributes two mobile electrons to the metal, what is the number density of free charges in the wire, in electrons/m3?
C) Use your results to calculate the drift speed (i.e., the average net speed) of the electrons in the wire.
D) Due to thermal motion, the electrons at room temperature are randomly traveling to and fro at 1.15×105 m/s, even without any current. What fraction is the current's drift speed, compared to the random thermal motion?

Answer 1

Answer:

A) ρ=$1.74x10^{26}$

B) μ=$1.68x10^{29}\frac{electron}{m^3}$

C) v=$1.03x10^{-3}\frac{m}{s}$

D)e=$8.99x10^-9$

Explanation:

A)

The magnetic field can be find knowing the current is the charge per second

β= $\frac{14eC*s}{1.6x10^{-19}C}$

β= 8.75x10^{19}e*s

Electron density

ρ=$\frac{8.75x10^{19}}{\pi*0.400x10^{-3}m} = 1.74x10^{26}$

B)

μ= $\frac{7.86}{56.2}\frac{g}{cm^3} \frac{mol}{g}*6.022x10^{23}\frac{molecules}{mol} *2 \frac{electron}{molecule}$

μ$=1.68x10^{23} \frac{electron}{cm^3}= 1.68x10^{29} \frac{electron}{m^3}$

C)

The drift speed using last information found

$v= \frac{J}{n*q} \\v= \frac{14A}{\pi*((0.40x10^-3)^2)*1.68x10^29*1.60x10^-19)} = 1.03x10^-3(\frac{m}{s} )$

D)

To compared the random thermal motion and the current's drift speed

$e=\frac{1.03x10^-3}{1.15x10^5} = 8.99x10^-9$