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QPages 198-203: Q8: HC Verma: Concepts of Physics - Volume 2 Electric Current in Conductors Pages 198-203: Q8 Solution: Consider a wire of length 4 m and cross-sectional area carrying …

Consider a wire of length 4 m and cross-sectional area 1 mm21 \text{ mm}^2 carrying a current of 2 A. If each cubic metre of the material contains 102910^{29} free electrons, find the average time taken by an electron to cross the length of the wire.

Step-by-Step Verified Solution

**Answer:** t=3.2times104t = 3.2 \\times 10^4 s. **Step-by-Step Explanation:** Given: l=4l = 4 m, A=10−6A = 10^{-6} m_, i=2i = 2 A, n=1029n = 10^{29} m__. vd=fracinAe=frac21029times10−6times1.6times10−19=frac18000v_d = \\frac{i}{nAe} = \\frac{2}{10^{29} \\times 10^{-6} \\times 1.6 \\times 10^{-19}} = \\frac{1}{8000} m/s. t=fraclvd=4times8000=32000t = \\frac{l}{v_d} = 4 \\times 8000 = 32000 s [citation:4][citation:6][citation:9].

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