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What is the average drift speed of conduction electrons in a copper wire of cross-sectional area $1.0 \times 10^{–7}m^2$ carrying a current of $1.5 A$. Assume that each copper atom contributes roughly one conduction electron. The density of copper is $ 9.0 \times 10^3 kg/m^3 $ , and its atomic mass is $ 63.5 u.$

$\begin {array} {1 1} (A)\;1.9mm/s & \quad (b)\;1.1 mm/s \\ (C)\;2.9 mm/s & \quad (D)\;2.2 mm/s \end {array}$


1 Answer

The direction of drift velocity of conduction electrons is opposite to the electric field direction,
i.e., electrons drift in the direction of increasing potential. The drift speed$v_d$ is given by
$ v_d =( I / neA)$
Now $ e=1.6 \times 10^{-19}C,A= 1.0 \times 10^{-7}m^2, I =1.5A$
The density of conduction electrons, $n$ is equal to the number of atoms per cubic meter
(assuming one conduction electron per $ Cu$ atom as is reasonable from its valence electron count of one).
A cubic meter of copper has a mass of $9.0 \times 10^3 \: kg$. Since $6.0 \times 10^{23}$ copper atoms have a mass of $ 63.5 g,$
$ n = \large\frac{6 \times 10^{23}}{63.5}$ $ \times 9 \times 10^{6} = 8.5 \times 10^{28}$
on substituting in the formula for$v_d$, one gets the final answer as $1.1 mm/s$
Ans : (B)


answered Feb 14, 2014 by thanvigandhi_1
edited Sep 4, 2014 by thagee.vedartham

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