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,

ρ = 9000 kg/m³

Length of the conductor = l

Also,

Mass of copper wire = Volume × density

\[\Rightarrow m = A \times l \times \rho\]

\[ \Rightarrow m = A \times l \times 9000    kg\]

We know that the number of atoms in molecular mass M = N_A

∴ Number of atoms in mass m, N = \[\left( \frac{N_A}{M} \right)m\]

where N_A is known as Avagadro's number and is equal to 6 × 10²³ atoms.

\[\Rightarrow N = \left( \frac{N_A}{M} \right)m\]

\[ \Rightarrow N = \left( \frac{N_A}{M} \right) \times A \times l \times 9000\]

Also, it is given that

No. of free electrons = No. of atoms

Let n be the number of free electrons per unit volume

\[n = \frac{\text{Number  of  electrons}}{\text{Volume}}\]

\[     = \frac{N_A \times A \times l \times 9000}{M \times A \times l}\]

\[     = \frac{N_A \times 9000}{M}\]

\[     = \frac{6 \times {10}^{23} \times 9000}{63 . 5 \times {10}^{- 3}}\]

\[   \therefore i =  V_d nAe\]

\[ \Rightarrow  V_d  = \frac{1}{\frac{6 \times {10}^{23} \times 9000}{63 . 5 \times {10}^{- 3}} \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]

\[ = \frac{63 . 5 \times {10}^{- 3}}{6 \times {10}^{23} \times 9000 \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]

\[ = \frac{63 . 5 \times {10}^{- 3}}{6 \times {10}^{26} \times 9 \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]

\[ = \frac{63 . 5 \times {10}^{- 3}}{6 \times 9 \times 16}\]

\[ = 0 . 073 \times  {10}^{- 3}\text{ m/s} \]

\[ =   0 . 073\text{ mm/s}\]