Question

The number density of free electrons in copper is nearly 8 × 10²⁸ m^-3. A copper wire has its area of cross section = 2 × 10^-6 m² and is carrying a current of 3.2 A. The drift speed of the electrons is _____ ×10^-6 ms^-1

Answer 1

Correct Answer: 125

The drift velocity (\( v_d \)) is given by: \[ I = n e A v_d, \] where:  
\( I = 3.2 \, \text{A} \), 
 \( n = 8 \times 10^{28} \, \text{m}^{-3} \) (number density), 
 \( e = 1.6 \times 10^{-19} \, \text{C} \) (charge of an electron), 
 \( A = 2 \times 10^{-6} \, \text{m}^2 \) (cross-sectional area). 
Rearrange to solve for \( v_d \): \[ v_d = \frac{I}{n e A}. \] 
Substitute the values: \[ v_d = \frac{3.2}{(8 \times 10^{28}) (1.6 \times 10^{-19}) (2 \times 10^{-6})}. \] 
Simplify: \[ v_d = \frac{3.2}{2.56 \times 10^4} = 125 \times 10^{-6} \, \text{ms}^{-1}. \] 
Final Answer: The drift velocity is: \[ \boxed{125 \times 10^{-6} \, \text{ms}^{-1}}. \]