Calculate Electron Mobility in a Cylindrical Conductor

This solution details the steps to find the electron mobility ($\mu$) in a conductor based on its electrical properties and dimensions.

Given Parameters

• Length, $L = 2 \text{ m}$
• Area, $A = 0.2 \text{ mm}^2 = 0.2 \times 10^{-6} \text{ m}^2 = 2 \times 10^{-7} \text{ m}^2$
• Current, $I = 1.6 \text{ A}$
• Voltage, $V = 2 \text{ V}$
• Electron concentration, $n = 5 \times 10^{28} \text{ m}^{-3}$
• Electron charge, $e = 1.6 \times 10^{-19} \text{ C}$
• Mobility expression, $\mu = \alpha \times 10^{-3} \text{ m}^2\text{/V.s}$

Calculation Steps

Step 1: Calculate Electric Field ($E$)

The electric field inside the conductor is calculated using the voltage and length.

$ E = \frac{V}{L} = \frac{2 \text{ V}}{2 \text{ m}} = 1 \text{ V/m} $

Step 2: Calculate Current Density ($J$)

Current density is the current per unit area.

$ J = \frac{I}{A} = \frac{1.6 \text{ A}}{2 \times 10^{-7} \text{ m}^2} = 0.8 \times 10^{7} \text{ A/m}^2 = 8 \times 10^{6} \text{ A/m}^2 $

Step 3: Calculate Electron Mobility ($\mu$)

The relationship between current density, electron concentration, charge, and mobility is given by $J = n e \mu E$. We rearrange this to solve for $\mu$.

$ \mu = \frac{J}{n e E} $

Substitute the calculated and given values:

$ \mu = \frac{8 \times 10^{6} \text{ A/m}^2}{(5 \times 10^{28} \text{ m}^{-3}) \times (1.6 \times 10^{-19} \text{ C}) \times (1 \text{ V/m})} $

$ \mu = \frac{8 \times 10^{6}}{8 \times 10^{9}} \text{ m}^2\text{/V.s} $

$ \mu = 1 \times 10^{-3} \text{ m}^2\text{/V.s} $

Step 4: Determine the value of $\alpha$

The calculated mobility is $\mu = 1 \times 10^{-3} \text{ m}^2\text{/V.s}$. The problem states the mobility is $\alpha \times 10^{-3} \text{ m}^2\text{/V.s}$. Comparing these two expressions, we find:

$ \alpha = 1 $

Final Answer

The value of $\alpha$ is 1.