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The number density of free electrons in ...

The number density of free electrons in a copper conductor is estimated is `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.

Text Solution

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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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