Answer:
A) ρ=[tex]1.74x10^{26}[/tex]
B) μ=[tex]1.68x10^{29}\frac{electron}{m^3} [/tex]
C) v=[tex]1.03x10^{-3}\frac{m}{s} [/tex]
D)e=[tex]8.99x10^-9[/tex]
Explanation:
A)
The magnetic field can be find knowing the current is the charge per second
β= [tex]\frac{14eC*s}{1.6x10^{-19}C}\\[/tex]
β= 8.75x10^{19}e*s
Electron density
ρ=[tex]\frac{8.75x10^{19}}{\pi*0.400x10^{-3}m} = 1.74x10^{26}[/tex]
B)
μ= [tex]\frac{7.86}{56.2}\frac{g}{cm^3} \frac{mol}{g}*6.022x10^{23}\frac{molecules}{mol} *2 \frac{electron}{molecule}[/tex]
μ[tex]=1.68x10^{23} \frac{electron}{cm^3}= 1.68x10^{29} \frac{electron}{m^3}[/tex]
C)
The drift speed using last information found
[tex]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} )[/tex]
D)
To compared the random thermal motion and the current's drift speed
[tex]e=\frac{1.03x10^-3}{1.15x10^5} = 8.99x10^-9[/tex]
