Question

A wire of area of cross-section 1mm^2 carries a current of 2A. If number of electrons per unit volume of wire is 10^29, then the drift speed of electrons in the wire is?

Answer 1

Given :

• Area of Cross-section of wire(A)=1mm²
• Current (i) = 2A
• Electron density (N) = ${\rm 10^{29}}$

To Find :

• Drift speed ${\rm (v_d)}$

Formula Used :

$\bullet\underline{\boxed{\sf i = NeAv_d}}$

Solution :

✪ Convert area of cross-section from mm² to m²

$\implies{\sf A = 1mm^2}$

$\implies{\bf A = 10^{-6}m^2 }$

______________________________

$\implies{\sf i =NeAv_d }$

e = 1.6 × ${\rm 10^{-19}\:C}$

$\implies{\sf 2 = 10^{29} \times 1.6 \times 10^{-19} \times 10^{-6} \times v_d}$

$\implies{\sf v_d = \dfrac{2}{1.6 \times 10^{-19} \times 10^{29} \times 10^{-6}} }$

$\implies{\sf v_d = \dfrac{2}{1.6 \times 10^4}}$

$\implies{\sf v_d = 1.25 \times 10^{-4}}$

$\implies{\sf v_d = 0.125 \times 10^{-3}}$

$\implies{\bf v_d = 0.125 \: mm/s }$

Answer :

Drift speed of electron in the wire is 0.125 mm/s