Consider a wire of length 4 m and cross-sectional area 1 mm^2 carrying a current 2 A. If each cubic metre of the material contains 10^29 free electrons, find the average time taken by an electron to cross the length of the wire.
Hrishant Goswami
12 Years agoGrade 10
1 Answer
Deepak Patra
12 Years ago
Sol. ℓ = 4 m, A = 1 mm^2 = 1 * 10^–6 m^2
I = 2 A, n/V = 10^29, t = ?
i = n A V base d e
⇒ e = 10^29 * 1 * 10^–6 * V base d * 1.6 * 10^–19
⇒ V base d = 2/10^29 *10^-6 *1.6 *10^-19
= 1/0.8 * 10^4 = 1/8000
t = ℓ/V base b = 4/1/8000 = 4 * 8000
= 32000 = 3.2 * 10^4 sec.
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