Question

A cylindrical conductor of length \(2 \, \text{m}\) and area of cross-section \(0.2 \, \text{mm}^2\) carries an electric current of \(1.6 \, \text{A}\) when its ends are connected to a \(2 \, \text{V}\) battery. Mobility of electrons in the conductor is \( \alpha \times 10^{-3} \, \text{m}^2\text{/V s}. \) The value of \( \alpha \) is ______________.
(Electron concentration \(= 5 \times 10^{28} \, \text{m}^{-3}\), electron charge \(= 1.6 \times 10^{-19} \, \text{C}\))

Hint:

Mobility can be calculated using drift velocity relations once electric field and current density are known.

Answer 1

Correct Answer: 5

The current density \( J \) in a conductor can be defined as the current \( I \) divided by the area \( A \) of cross-section of the conductor: \( J = \frac{I}{A} \). Given \( I = 1.6 \, \text{A} \) and \( A = 0.2 \, \text{mm}^2 = 0.2 \times 10^{-6} \, \text{m}^2 \), the current density is \( J = \frac{1.6}{0.2 \times 10^{-6}} = 8 \times 10^6 \, \text{A/m}^2 \). 
The electric field \( E \) across the conductor is given by the voltage \( V \) divided by the length \( L \): \( E = \frac{V}{L} \). With \( V = 2 \, \text{V} \) and \( L = 2 \, \text{m} \), we find \( E = \frac{2}{2} = 1 \, \text{V/m} \).
The mobility of electrons \( \mu \) relates to current density via the formula: \( J = n \times e \times \mu \times E \), where \( n \) is the electron concentration and \( e \) is the electron charge. Rearranging gives \( \mu = \frac{J}{n \times e \times E} \). With \( n = 5 \times 10^{28} \, \text{m}^{-3} \), \( e = 1.6 \times 10^{-19} \, \text{C} \), \( J = 8 \times 10^6 \, \text{A/m}^2 \), and \( E = 1 \, \text{V/m} \), we compute: \[ \mu = \frac{8 \times 10^6}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1} = \frac{8}{8} \times 10^{-3} \, \text{m}^2/\text{Vs} = 1 \times 10^{-3} \, \text{m}^2/\text{Vs}. \]
Hence, the value of \( \alpha \) is \( 1 \), which falls within the range 5,5 specified in the question.