Answer:
The current density is [tex]J = 2.04 * 10^{6} A /m^2[/tex]
The drift velocity is [tex]v_d = 1.5 * 10^{-4} m/s[/tex]
Explanation:
From the question we are told that
The nominal diameter of the wire is [tex]d = 1.02 mm= \frac{1.02}{1000} = 0.00102 \ m[/tex]
The current carried by the wire is [tex]I = 1.67 A[/tex]
The power rating of the lamp is [tex]P = 200 W[/tex]
The density of electron is [tex]n = 8.5 * 10^{28} \ e/m^3[/tex]
The current density is mathematically represented as
[tex]J = \frac{I}{A}[/tex]
Where A is the area which is mathematically evaluated as
[tex]A = \pi \frac{d^2}{4}[/tex]
Substituting values
[tex]A = 3.142 * \frac{(1.02 * 10^{-3})^2 }{4}[/tex]
[tex]A = 8.0*10^{-4}m^2[/tex]
So
[tex]J = \frac{1.67}{8.0*10^{-4}}[/tex]
[tex]J = 2.04 * 10^{6} A /m^2[/tex]
The drift velocity is mathematically represented as
[tex]v_d = \frac{J}{ne}[/tex]
Where e is the charge on one electron which has a value [tex]e = 1.602 *10^{-19} C[/tex]
So
[tex]v_d =\frac{2.04 * 10^6 }{8.5 *10^{28} * 1.6 * 10^{-19}}[/tex]
[tex]v_d = 1.5 * 10^{-4} m/s[/tex]
