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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.
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Solution
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 = NA
∴ Number of atoms in mass m, N = \[\left( \frac{N_A}{M} \right)m\]
where NA 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
\[n = \frac{\text{Number of electrons}}{\text{Volume}}\]
\[ = \frac{N_A \times A \times l \times 9000}{M \times A \times l}\]
\[ = \frac{N_A \times 9000}{M}\]
\[ = \frac{6 \times {10}^{23} \times 9000}{63 . 5 \times {10}^{- 3}}\]
\[ \therefore i = V_d nAe\]
\[ \Rightarrow V_d = \frac{1}{\frac{6 \times {10}^{23} \times 9000}{63 . 5 \times {10}^{- 3}} \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]
\[ = \frac{63 . 5 \times {10}^{- 3}}{6 \times {10}^{23} \times 9000 \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]
\[ = \frac{63 . 5 \times {10}^{- 3}}{6 \times {10}^{26} \times 9 \times {10}^{- 6} \times 1 . 6 \times {10}^{- 19}}\]
\[ = \frac{63 . 5 \times {10}^{- 3}}{6 \times 9 \times 16}\]
\[ = 0 . 073 \times {10}^{- 3}\text{ m/s} \]
\[ = 0 . 073\text{ mm/s}\]
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