When 5V potential difference is applied …
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Naveen Sharma 8 years, 5 months ago
Given : V = 5V
Lenght of wire l = 0.1m
Vd = 2.5 *10-4 m/s
Electron Density n = 0.8 * 1028 m-3
We know the drift velocity and current are related by the formula
i = neAVd ................. (1) where n is electron density, e is charge on electron, A is area of cross section and Vd is drift velocity.
also i = V/R and R = ρl/A where ρ is resitivity l is length of conductor
So
i = VA/ρl
put value of i in equation (1) we get
VA/ρl = neAVd
or ρ = V/nelVd
Put values of all,
ρ= 5 / 0.1 * 8 * 1028 * 1.6 * 10-19 * 2.5 * 10-4
after solving we get, 1.5625 * 10-5
Approx 1.6 * 10-5 Ωm
1Thank You