Answer:
J = 2.044x10⁶ A/m²
v = 1.50x10⁻⁴ m/s
Explanation:
The current density (J) of the copper wire is giving by:
[tex] J = \frac {I}{A} [/tex]
where I: electric current and A: cross-sectional area of the copper wire
The cross-sectional area of the copper wire can be calculated by:
[tex] A = \frac {\pi d^{2}}{4} = \frac {\pi (1.02 \cdot 10^{-3} m)^{2}}{4} = 8.17 \cdot 10^{-07} m^{2} [/tex]
Substituting the calculated area in the equation (1) we have:
[tex] J = \frac {1.67 A}{8.17 \cdot 10^{-7} m^{2}} = 2.044 \cdot 10^{6} \frac {A}{m^{2}} [/tex]
Hence, the current density is 2.044x10⁶ A/m².
To find the drift speed (v), we need to use the next equation:
[tex] v = \frac {J}{n q} [/tex]
where n: the free-electron density, q: module of the charge of the electron
[tex] v = \frac {2.044 \cdot 10^{6} \frac {A}{m^{2}}}{(8.5 \cdot 10^{28} {m^{-3}}) (1.6 \cdot 10^{-19} C)} [/tex]
[tex] v = 1.50 \cdot 10^{-04} \frac {m}{s} [/tex]
So, the drift speed is 1.50x10⁻⁴ m/s.
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