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Question
Consider a wire of length 4 m and cross-sectional area 1 mm2 carrying a current of 2 A. If each cubic metre of the material contains 1029 free electrons, find the average time taken by an electron to cross the length of the wire.
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Solution
Given:-
Current through the wire, i = 2 A
Length of the wire, l = 4 m
Area of cross section, A = 1 mm2 = 1 × 10–6 m2
Number of electrons per unit volume, n = 1029
We know:-
\[i = nA V_d e\]
\[ \Rightarrow V_d = \frac{i}{nAe}\]
\[ = \frac{2}{{10}^{29} \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]
\[ \Rightarrow V_d = \frac{1}{8000} m/s,\]
where Vd is the drift speed.
Let t be the time taken by an electron to cross the length of the wire.
\[\Rightarrow t = \frac{\text{Length of the wire}}{\text{Drift speed}}\]
\[ = \frac{l}{V_d}\]
\[ \therefore t = \frac{4}{\frac{1}{8000}}\]
\[ = 32 \times {10}^3 s\]
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