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The number density of free electrons in a copper conductor is estimated at `8.5 xx 10^(28) m^(-3)`. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is `2.0 xx 10^(-6)m^(2)` and it is carrying a current of 3.0A. |
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Answer» Number density of free electrons in a copper conductor, `n = 8.5 xx 10^(28) m^(-3)` Length of the copper wire, `l = 3.0` m Area of cross-section of the wire, `A = 2.0 xx 10^(-6) m^(2)` Current carried by the wire, `I = 3.0A`, which is given by the relation, `I = nAeV_(d)` where, e = Electric charge `= 1.6 xx 10^(-19)C` `V_(d) = "Drift velocity" = ("Length of the wire" (l))/("Time taken to cover" (t))` `I = nAe(l)/(t)` `t = (nAel)/(I)` `= (3 xx 8.5 xx 10^(28) xx 2xx 10^(-6) xx 1.6 xx 10^(-19))/(3.0)` `= 2.7 xx 10^(2)s` Therefore, the time taken by an electron to drift from one end of the wire to the other is `2.7 xx 10^(4)s`. |
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