A copper wire of area of cross-section $2 × 10^{-6} m^2$ and electron density $10^{28} m^3$ carries a current of 1.6 A. Time taken by an electron to drift from one end of a 1.0 m long wire to its other end will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $2 × 10^3 s$
Given:
Area, $A = 2 \times 10^{-6} \, m^{2}$
Electron density, $n = 10^{28} \, m^{-3}$
Current, $I = 1.6 \, A$
Length, $l = 1.0 \, m$
Charge of an electron, $e = 1.6 \times 10^{-19} \, C$
Drift velocity:
$v_d = \frac{I}{n e A}$
Substituting values:
$v_d = \frac{1.6}{(10^{28})(1.6 \times 10^{-19})(2 \times 10^{-6})}$
$v_d = \frac{1.6}{3.2 \times 10^{3}}$
$v_d = 5 \times 10^{-4} \, m/s$
Time taken to drift:
$t = \frac{l}{v_d} = \frac{1}{5 \times 10^{-4}} = 2 \times 10^{3} \, s$
∴ Time taken by an electron = 2000 s