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

Step 1: Calculate electric field.
\[ E = \frac{V}{L} = \frac{2}{2} = 1 \, \text{V m}^{-1} \]
Step 2: Use drift current relation.
\[ I = nqAv_d \] \[ v_d = \mu E \] Hence: \[ I = nqA\mu E \]
Step 3: Substitute given values.
\[ n = 5 \times 10^{28}, \quad q = 1.6 \times 10^{-19} \] \[ A = 0.2 \, \text{mm}^2 = 0.2 \times 10^{-6} \, \text{m}^2 \] \[ 1.6 = (5 \times 10^{28})(1.6 \times 10^{-19})(0.2 \times 10^{-6})(\mu)(1) \] \[ \mu = 5 \times 10^{-3} \, \text{m}^2 \text{/V s} \]
Step 4: Compare with given form.
\[ \mu = \alpha \times 10^{-3} \Rightarrow \alpha = 5 \]
Final Answer: \[ \boxed{5} \]