Question

A metallic conductor of length 2m and cross-sectional area \( 0.2 \, \text{mm}^2 \) carries a steady current of 1.2 A when a potential difference of 2 V is applied across it. (\( e = 1.6 \times 10^{-19} \), charge density = \( 7.5 \times 10^{28} \, \text{m}^{-3} \)) Then the mobility of the charge carrier is \( x \times 10^{-4} \) SI units. Find \( x \).

Hint:

To find the mobility of charge carriers, use the relation \( I = n A e \mu \frac{V}{L} \) and solve for \( \mu \).

Answer 1

Correct Answer: 5

Step 1: Use Ohm's law to find the current.
Ohm's law states that the current \( I \) is related to the potential difference \( V \), the charge density \( n \), the charge of the carrier \( e \), the cross-sectional area \( A \), the length \( L \), and the mobility \( \mu \) by the relation: \[ I = n A e \mu \frac{V}{L}. \] Step 2: Substitute the known values.
Given: - \( I = 1.2 \, \text{A} \), - \( A = 0.2 \, \text{mm}^2 = 0.2 \times 10^{-6} \, \text{m}^2 \), - \( V = 2 \, \text{V} \), - \( L = 2 \, \text{m} \), - \( e = 1.6 \times 10^{-19} \), - \( n = 7.5 \times 10^{28} \, \text{m}^{-3} \). Substitute these values into the equation: \[ 1.2 = (7.5 \times 10^{28}) \times (0.2 \times 10^{-6}) \times (1.6 \times 10^{-19}) \times \mu \times \frac{2}{2}. \] Simplifying this: \[ 1.2 = (7.5 \times 10^{28} \times 0.2 \times 10^{-6} \times 1.6 \times 10^{-19}) \times \mu. \] \[ 1.2 = (2.4 \times 10^{4}) \times \mu. \] Step 3: Solve for \( \mu \).
\[ \mu = \frac{1.2}{2.4 \times 10^{4}} = 5 \times 10^{-4} \, \text{SI units}. \] Thus, \( x = 5 \). Final Answer: \[ \boxed{5}. \]