Question
What is the drift velocity of electrons if the current flowing through a copper wire of $1 \mathrm{~mm}$ diameter is $1.1 \mathrm{~A} ?$ Assume that each atom of copper contributes one electron: (Given : density of $\mathrm{Cu}=9 \mathrm{~g} / \mathrm{cm}^{3}$ and atomic weight of $\mathrm{Cu}=63$ )(a) $0.3 \mathrm{~mm} / \mathrm{s}$(b) $0.5 \mathrm{~mm} / \mathrm{s}$(c) $0.1 \mathrm{~mm} / \mathrm{s}$(d) $0.2 \mathrm{~mm} / \mathrm{s}$
Step 1
The electron density can be calculated using the formula: \[ n = \frac{N_A \cdot \rho}{M} \] where \(N_A\) is Avogadro's number, \(\rho\) is the density of copper, and \(M\) is the atomic weight of copper. Show more…
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Assume that each atom of copper contributes one electron. If the current flowing through a copper wire of $1 \mathrm{~mm}$ diameter is $1.1 \mathrm{~A}$, the drift velocity of electrons will be (Density of $\mathrm{Cu}=9 \mathrm{~g} \mathrm{~cm}^{-3}$, at. wt. of $\mathrm{Cu}=63$ ) (a) $0.3 \mathrm{~mm} / \mathrm{s}$ (b) $0.5 \mathrm{~mm} / \mathrm{s}$ (c) $0.1 \mathrm{~mm} / \mathrm{s}$ (d) $0.2 \mathrm{~mm} / \mathrm{s}$
There is a current of $1.344$ amp in a copper wire whose area of cross-section normal to the length of the wire is $1 \mathrm{~mm}^{2}$. If the number of free electrons per $\mathrm{cm}^{3}$ is $8.4 \times 10^{22}$, then the drift velocity would be (a) $1.0 \mathrm{~mm} / \mathrm{s}$ (b) $1.0 \mathrm{~m} / \mathrm{s}$ (c) $0.1 \mathrm{~mm} / \mathrm{s}$ (d) $0.01 \mathrm{~mm} / \mathrm{s}$
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