Answer:
a) 1273.23 A/m^2
b) 7.19*10^-5 m/s
c) 236881.7 Ohms
Explanation:
(a) To find the current density you use the following formula:
[tex]J=\frac{I}{A}=\frac{I}{\pi r^2}[/tex]
I: current in the wire
A: cross area of the wire
r: radius of the wire
[tex]J=\frac{5A}{\pi(1.25*10^{-3}m)^2}=1273.23\frac{A}{m^2}[/tex]
(b) The electron drift speed is given by:
[tex]v_d=\frac{I}{nqA}=\frac{I}{nq\pi r^2}[/tex]
n: number of conduction electrons per m^3
q: charge of the electron = 1.6*10^-19C
The number of free electrons is calculated by using:
[tex]n=\frac{\rho N_A}{M}\\\\n=\frac{(9*10^{3}kg/m^3)(6.22*10^{23})}{63.54*10^{-3}}=8.5*10^{28}[/tex]
Next, you replace the values of the parameters in the equation for vd:
[tex]v_d=\frac{5A}{(8.85*10^{28}m^{-3})(1.6*10^{-19}C)\pi (1.25*10^{-3}m)^2}\\\\v_d=7.19*10^{-5}\frac{m}{s}[/tex]
(c) The conductivity is given by:
[tex]\sigma=\frac{L}{RA}[/tex]
You first calculate R:
[tex]R=VI=(0.86V)(5A)=4.3\Omega[/tex]
Next, replace for sigma:
[tex]\sigma=\frac{50m}{(4.3\Omega)(\pi (1.25*10^{-3}m)^2)}=236881.7\Omega^{-1}m^{-1}[/tex]
