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

To find the mobility of the charge carrier, we start by using the formula for electric current: \(I = nAqv_d\), where:

• \(I\) is the current (1.2 A)
• \(n\) is the charge density (\(7.5 \times 10^{28} \, \text{m}^{-3}\))
• \(A\) is the cross-sectional area (\(0.2 \, \text{mm}^2 = 0.2 \times 10^{-6} \, \text{m}^2\))
• \(q\) is the charge of an electron (\(1.6 \times 10^{-19} \, \text{C}\))
• \(v_d\) is the drift velocity

Rearranging gives the drift velocity: \[v_d = \frac{I}{nAq} = \frac{1.2}{(7.5 \times 10^{28})(0.2 \times 10^{-6})(1.6 \times 10^{-19})}\]Calculating: \[v_d \approx 5 \times 10^{-5} \, \text{m/s}\]The mobility \(\mu\) is related to the drift velocity and electric field \(E\) by \(\mu = \frac{v_d}{E}\). The electric field \(E\) is calculated as: \[E = \frac{\text{Potential Difference}}{\text{Length}} = \frac{2}{2} = 1 \, \text{V/m}\]Thus, \(\mu = \frac{5 \times 10^{-5}}{1} = 5 \times 10^{-5} \, \text{m}^2/\text{V}\cdot\text{s} = 5 \times 10^{-4} \, \text{SI units}\) Therefore, \(x = 5\). This value, \(x = 5\), lies within the specified range of (5,5).