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Compare the drift speed obtained above with (i) Thermal speed of copper atoms at ordinary temperatures. (ii) Speed of propagation of electric field along the conductor which causes the drift motion |
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Answer» (i) At a temperature T, the thermal speed of a copper atom of mass is obtained from `(1)/(2) M_(e) V_(rms)^(2) = [(3)/(2)] K_(B) T` `implies V_(rms) = sqrt((3K_(B) T)/(M_(e)))` Here `K_(B) = 1.38 xx 10^(-23)` `T = 300 K, M_(e)) = 9.1 xx 10^(-31) kg` `implies V_("rms") = sqrt((3 xx 1.38 xx 10^(-23) xx 300)/(9.1 xx 10^(-31)))` `= sqrt((1.38 xx 9 xx 10^(10))/(9.1))` ltBrgt `= sqrt(1.36) xx 10^(5) m//s` `:.` Thermal speed, `V_("rms") = 1.17 xx 10^(5) m//s` `(V_(d))/(V_(rms)) = (1.225 xx 10^(-3))/(1.17 xx 10^(5)) = 1.047 xx 10^(-8)` `:.` drift speed of electron`(V_(d)) = 1.047 xx 10^(-8)` `= 10^(-8)` times of thermal speed at ordinary temperature. (ii) The electric field travels along conductor with speed of EMW. `C = 3 xx 10^(8) m//s` `V_(d) = 1.25 xx 10^(-3) m//s` `(V_(d))/(C ) = (1.225 xx 10^(-3))/(3 xx 10^(8))` `V_(d) =0.408 xx 10^(-11) C` `:.` Drift speed is, in comparision of C, extermely smallar by a factor of `10^(-11)`. |
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