2.2 Mobility $\left( \mu \right)$

1.1 Current in different situations

2.1 Expression for drift velocity
2.2 Mobility $\left( \mu \right)$

3.1 Temperature dependence of resistance
3.2 Resistivities of different materials
3.3 Limitations of ohm's law

5.1 Combination of cells

6.1 Maximum power transfer theorem
6.2 Kirchhoff's law

9.1 Comparison of emfs of two primary cells.
9.2 Determination of Internal resistance of a cell using potentiometer

10.1 Ammeter
10.2 Voltmeter

Drift Velocity per unit electric field is known as mobility. $$\begin{equation} \begin{aligned} \Rightarrow \mu = \frac{{{v_d}}}{E} \\ \Rightarrow {{\vec v}_d} = \frac{{e\vec E\tau }}{m} \\ \Rightarrow \mu = \frac{{e\tau }}{m} \\\end{aligned} \end{equation} $$

where

$\tau=$ Relaxation time

$m=$ Mass of an electron

$e=$ Charge of an electron

SI unit of mobility $\left( \mu \right)$ is $\frac{{{m^2}}}{{Vs}}$

where,

$m=$ mass of an electron

$V=$ Potential difference

$s=$ Seconds

In a conductor, conduction is due to free electrons.

Current is, $$\begin{equation} \begin{aligned} \Rightarrow I = enA{v_d} \\ \Rightarrow {v_d} = {\mu _e}E \\ \Rightarrow I = enA{\mu _e}E \\\end{aligned} \end{equation} $$

In a semiconductor, both electrons and holes act as current carriers.

The electric current is, $$\begin{equation} \begin{aligned} \Rightarrow I = {I_e} + {I_h} \\ \Rightarrow enA{v_e} + epA{v_h} \\ \Rightarrow enA{\mu _e}E + epA{\mu _h}E\quad (As,\;{v_e} = {\mu _e}E,\;{v_h} = {\mu _h}E) \\ \Rightarrow eA(n{\mu _e} + p{\mu _h})E \\\end{aligned} \end{equation} $$

where,

$n=$ is the number of electrons per unit volume

$p=$ is the number of holes per unit volume

${v_e}=$ is the velocity of an electron

${v_h}=$ is the velocity of holes

${\mu _e}=$ is the mobility of an electron

${\mu _e}=$ is the mobility of holes

$E=$ is the electric field

Question 4. The number density of free electrons in a copper wire is $8.5 \times {10^{28}}{m^{ - 3}}$.

Calculate

(i) Drift speed of the electrons when $1A$ of current exists in a copper wire of cross-section $2m{m^2}$.

(ii) Time is taken by an electron to drift from one end of the wire to its other end. The length of the wire is 1.0 $m$.

Given, $$\begin{equation} \begin{aligned} n = 8.5 \times {10^{28}}{m^{ - 3}} \\ A = 2.0 \times {10^{ - 6}}{m^2} \\ I = 1.0A \\ L = 1.0m \\\end{aligned} \end{equation} $$

(i) $$\begin{equation} \begin{aligned} \Rightarrow I = neA{v_d} \\ \Rightarrow {v_d} = \frac{I}{{nAe}} \\ \Rightarrow \frac{1}{{8.5 \times {{10}^{28}} \times 2 \times {{10}^{ - 6}} \times 1.6 \times {{10}^{ - 19}}}} \\ \Rightarrow 0.036 \times {10^{ - 3}}m{s^{ - 1}} \\\end{aligned} \end{equation} $$

(ii) $$\begin{equation} \begin{aligned} \Rightarrow t = \frac{L}{{{v_d}}} \\ \Rightarrow t = \frac{1}{{0.036 \times {{10}^{ - 3}}}} \\ \Rightarrow 27.7 \times {10^3}s \\\end{aligned} \end{equation} $$

Question 5. A potential difference of $12V$ is applied across a conductor of length $0.24m$. Calculate the drift velocity of electrons, if the electron mobility is $5.6 \times {10^{ - 6}}{m^2}{V^{ - 1}}{s^{ - 1}}$.

Given

$V=6V$ ,

$L=0.24m$

$\mu = 5.6 \times {10^{ - 6}}{m^2}{V^{ - 1}}{s^{ - 1}}$

Drift velocity, $$\begin{equation} \begin{aligned} \Rightarrow {v_d} = \mu E \\ \Rightarrow \mu .\frac{V}{L} \\ \Rightarrow \frac{{5.6 \times {{10}^{ - 6}} \times 12}}{{0.24}} \\ \Rightarrow 2.8m{s^{ - 1}} \\\end{aligned} \end{equation} $$