Question

A current of 1.0 A exists in a copper wire of cross-section 1.0 mm². Assuming one free electron per atom, calculate the drift speed of the free electrons in the wire. The density of copper is 9000 kg m^–3.

Answer 1

Given:
Current, i = 1 A
Area of cross-section, A = 1 mm² = 1 × 10^–6 m²
Density of copper, $\rho$ = 9000 kg/m³
Length of the conductor = l
Also,
Mass of copper wire = Volume × density
$$\begin{gathered} \Rightarrow m\mathit{=}A\times l\times \rho \\ \mathit{\Rightarrow }m\mathit{=}A\times l\times 9000\mathrm{kg} \end{gathered}$$
We know that the number of atoms in molecular mass M = N_A
∴ Number of atoms in mass m, N = $\left(\frac{{\mathrm{N}}_{\mathrm{A}}}{M}\right)m$,
where N_A is known as Avagadro's number and is equal to 6 × 10²³ atoms.
$$\begin{gathered} \Rightarrow N=\left(\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}\right)m \\ \Rightarrow N=\left(\frac{{\mathrm{N}}_{\mathrm{A}}}{\mathrm{M}}\right)\times A\times l\times 9000 \end{gathered}$$
Also, it is given that
No. of free electrons = No. of atoms
Let n be the number of free electrons per unit volume

$$\begin{gathered} n=\frac{\mathrm{Number}\mathrm{of}\mathrm{electrons}}{\mathrm{Volume}} \\ =\frac{{\mathrm{N}}_{\mathrm{A}}\times A\times l\times 9000}{M\times A\times l} \\ =\frac{{\mathrm{N}}_{\mathrm{A}}\times 9000}{M} \\ =\frac{6\times {10}^{23}\times 9000}{63.5\times {10}^{-3}} \\ \therefore i\mathit{=}{V}_{\mathrm{d}}nAe \\ \Rightarrow V_d=\frac{1}{{\displaystyle \frac{6\times {10}^{23}\times 9000}{63.5\times {10}^{-3}}\times {10}^{-6}\times 1.6\times {10}^{-19}}} \\ =\frac{63.5\times {10}^{-3}}{6\times {10}^{23}\times 9000\times {10}^{-6}\times 1.6\times {10}^{-19}} \\ =\frac{63.5\times {10}^{-3}}{6\times {10}^{26}\times 9\times {10}^{-6}\times 1.6\times {10}^{-19}} \\ =\frac{63.5\times {10}^{-3}}{6\times 9\times 16} \\ =0.073\times {10}^{-3}\mathrm{m}/\mathrm{s} \\ =0.073\mathrm{mm}/\mathrm{s} \end{gathered}$$