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

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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)

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