Current Density, Drift Velocity and Mobility Questions in English

(N/A) The drift velocity $v_d$ of an electron in the presence of an external electric field $E$ is given by the relation: $v_d = -\frac{eE}{m} \tau$,where $e$ is the charge of an electron,$m$ is the mass of the electron,and $\tau$ is the relaxation time.
Taking the magnitude,we have $|v_d| = \frac{eE}{m} \tau$.
Mobility $\mu$ is defined as the magnitude of drift velocity per unit electric field: $\mu = \frac{|v_d|}{E}$.
Substituting the expression for $|v_d|$,we get $\mu = \frac{eE\tau}{mE} = \frac{e\tau}{m}$.
The $SI$ unit of mobility is $\frac{m/s}{V/m} = m^2 V^{-1} s^{-1}$.

(A) When $n$ electrons move with drift velocity $v_{d}$ in a conductor of cross-sectional area $A$,the electric current $I$ is given by:
$I = n A v_{d} e$
By definition,mobility $\mu$ is the ratio of drift velocity to the applied electric field $E$:
$\mu = \frac{v_{d}}{E}$
Therefore,the drift velocity can be expressed as:
$v_{d} = \mu E$
Substituting this into the current equation:
$I = n A (\mu E) e$
Rearranging the equation to solve for mobility $\mu$:
$\mu = \frac{I}{n A E e}$
The $SI$ unit of mobility is calculated as:
$\mu = \frac{A}{m^{-3} \times m^{2} \times (V/m) \times C} = \frac{A}{V \cdot C} = m^{2} V^{-1} s^{-1}$

(A) In the absence of an external electric field,the free electrons in a conductor move randomly in all possible directions due to thermal energy.
Because the motion is completely random,for every electron moving with a velocity $\vec{v}$,there is,on average,another electron moving with a velocity $-\vec{v}$.
Consequently,the vector sum of the velocities of all $N$ electrons in the conductor is zero.
Therefore,the average velocity $\vec{v}_{avg} = \frac{1}{N} \sum_{i=1}^{N} \vec{v}_i = 0$.

(N/A) $1$. Relaxation Time $(\tau )$: It is defined as the average time interval between two successive collisions of an electron with the positive ions in a conductor. It is typically of the order of $10^{-14} \ s$.
$2$. Drift Velocity $(v_d)$: It is defined as the average velocity acquired by the free electrons in a conductor in the opposite direction of the externally applied electric field. It is given by the formula $v_d = -\frac{eE\tau}{m}$,where $e$ is the charge of an electron,$E$ is the electric field,$\tau$ is the relaxation time,and $m$ is the mass of the electron.

(N/A) Mobility $(\mu)$ is defined as the magnitude of the drift velocity $(v_d)$ per unit electric field $(E)$.
Mathematically, $\mu = \frac{v_d}{E}$.
The $SI$ unit of drift velocity is $m/s$ and the $SI$ unit of electric field is $V/m$ or $N/C$.
Therefore, the $SI$ unit of mobility is $\frac{m/s}{V/m} = m^2 V^{-1} s^{-1}$ or $m^2 / (V \cdot s)$.

(N/A) The motion of a charge across a junction is generally not momentum-conserving.
When a charge carrier (such as an electron) moves through a conductor,it experiences collisions with the lattice ions,which exert external forces on the charge.
According to Newton's second law,the rate of change of momentum of a system is equal to the net external force acting on it.
Since the lattice ions exert a force on the charge carriers to maintain a steady drift velocity $v_{d} = \frac{eE\tau}{m}$,the momentum of the charge carrier is not conserved because the external force from the lattice is non-zero.
Furthermore,at a junction,the redistribution of charge creates localized electric fields that further alter the momentum of the charge carriers.

(N/A) Relaxation time $\tau$ is defined as the average time interval between two successive collisions of an electron with the lattice ions.
$1$. Dependence on $E$ field: When an external electric field $E$ is applied,the drift velocity $v_d$ acquired by electrons is very small (order of $10^{-3} \ m/s$) compared to their random thermal velocity (order of $10^5 \ m/s$). Since the collision frequency is primarily determined by the random thermal motion,the relaxation time $\tau$ remains nearly independent of the applied $E$ field. This constancy of $\tau$ ensures that the current density $J$ is directly proportional to $E$ $(J = \sigma E)$,which is the microscopic form of Ohm's law.
$2$. Dependence on $T$: As temperature $T$ increases,the random thermal velocity of electrons increases significantly. This leads to more frequent collisions between electrons and lattice ions,thereby decreasing the relaxation time $\tau$. According to the relation $\rho = \frac{m}{n e^2 \tau}$,since $\tau$ decreases with an increase in temperature,the resistivity $\rho$ of the conductor increases.

(B) In metals,the outer shell electrons are loosely bound to the nucleus. These electrons are called free electrons or valence electrons. Due to thermal energy at room temperature,these electrons detach from their parent atoms and move randomly throughout the crystal lattice. When an external electric field is applied,these free electrons drift in a specific direction,creating an electric current. Therefore,free electrons are the charge carriers responsible for the conductivity of metals.

(N/A) Current in the circuit $I = \frac{V}{R} = \frac{6}{6} = 1 \, A$.
Number density of electrons $n = 10^{29} \, m^{-3}$.
Length of the wire $l = 10 \, cm = 0.1 \, m$.
Area of cross-section $A = (1 \, mm)^2 = 10^{-6} \, m^2$.
Volume of the wire $V_{vol} = A \times l = 10^{-6} \times 0.1 = 10^{-7} \, m^3$.
Total number of electrons $N = n \times V_{vol} = 10^{29} \times 10^{-7} = 10^{22}$.
Drift velocity $v_d = \frac{I}{nAe} = \frac{1}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}} = \frac{1}{1.6 \times 10^4} = 0.625 \times 10^{-4} \, m/s = 6.25 \times 10^{-5} \, m/s$.
Total kinetic energy $K.E. = N \times \frac{1}{2} m_e v_d^2 = 10^{22} \times 0.5 \times 9.1 \times 10^{-31} \times (6.25 \times 10^{-5})^2$.
$K.E. = 4.55 \times 10^{-9} \times 39.06 \times 10^{-10} \approx 1.78 \times 10^{-17} \, J$.
$(b)$ Power dissipated $P = I^2 R = (1)^2 \times 6 = 6 \, W$.
Since $P = \frac{E}{t}$,the time scale $t = \frac{E}{P} = \frac{1.78 \times 10^{-17}}{6} \approx 2.97 \times 10^{-18} \, s \approx 3 \times 10^{-18} \, s$.

(C) The mobility $\mu$ of a charged particle is defined as the ratio of its drift velocity $v_d$ to the applied electric field $E$.
Formula: $\mu = \frac{v_d}{E}$
Given:
$v_d = 7.5 \times 10^{-4} \, m s^{-1}$
$E = 3 \times 10^{-10} \, V m^{-1}$
Calculation:
$\mu = \frac{7.5 \times 10^{-4}}{3 \times 10^{-10}}$
$\mu = 2.5 \times 10^{-4 - (-10)}$
$\mu = 2.5 \times 10^{6} \, m^{2} V^{-1} s^{-1}$

(C) The drift speed $v_{d}$ is given by the formula $i = neAv_{d}$,where $n$ is the number density of free electrons.
First,calculate the number density $n$:
$n = \frac{\text{Density} \times N_{A}}{\text{Molar Mass}} = \frac{2.7 \times 10^{3} \, kg/m^{3} \times 6.022 \times 10^{23} \, mol^{-1}}{27 \times 10^{-3} \, kg/mol} \approx 6.022 \times 10^{28} \, m^{-3}$.
Using $n \approx 6 \times 10^{28} \, m^{-3}$ for simplicity.
Now,rearrange the current formula for $v_{d}$:
$v_{d} = \frac{i}{neA}$.
Substitute the given values:
$i = 10 \, A$,$n = 6 \times 10^{28} \, m^{-3}$,$e = 1.6 \times 10^{-19} \, C$,$A = 4 \times 10^{-6} \, m^{2}$.
$v_{d} = \frac{10}{(6 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (4 \times 10^{-6})}$
$v_{d} = \frac{10}{38.4 \times 10^{3}} = \frac{10}{38400} \approx 2.6 \times 10^{-4} \, m/s$.

(B) The current $i$ through a cross-sectional area element $dA$ is given by $di = J(r) dA$.
For a cylindrical wire,the area element at a radial distance $r$ with thickness $dr$ is $dA = 2 \pi r dr$.
Substituting the given current density $J(r) = J_{0}(1 - \frac{r}{R})$,we get:
$di = J_{0}(1 - \frac{r}{R}) (2 \pi r dr) = 2 \pi J_{0} (r - \frac{r^{2}}{R}) dr$.
To find the total current $i$ from $r = 0$ to $r = \frac{R}{4}$,we integrate:
$i = \int_{0}^{R/4} 2 \pi J_{0} (r - \frac{r^{2}}{R}) dr = 2 \pi J_{0} [\frac{r^{2}}{2} - \frac{r^{3}}{3R}]_{0}^{R/4}$.
Substituting the limits:
$i = 2 \pi J_{0} [\frac{(R/4)^{2}}{2} - \frac{(R/4)^{3}}{3R}] = 2 \pi J_{0} [\frac{R^{2}}{32} - \frac{R^{3}}{192R}] = 2 \pi J_{0} [\frac{R^{2}}{32} - \frac{R^{2}}{192}]$.
Finding a common denominator:
$i = 2 \pi J_{0} [\frac{6R^{2} - R^{2}}{192}] = 2 \pi J_{0} [\frac{5R^{2}}{192}] = \frac{5 J_{0} \pi R^{2}}{96}$.

(B) Given: Current $i = 10\, A$,Area $A = 5\, mm^{2} = 5 \times 10^{-6}\, m^{2}$,Drift velocity $v_{d} = 2 \times 10^{-3}\, m/s$,and charge of an electron $e = 1.6 \times 10^{-19}\, C$.
We know the relation between current and drift velocity is $i = neAv_{d}$.
Rearranging for the number density $n$,we get $n = \frac{i}{eAv_{d}}$.
Substituting the values: $n = \frac{10}{1.6 \times 10^{-19} \times 5 \times 10^{-6} \times 2 \times 10^{-3}}$.
$n = \frac{10}{16 \times 10^{-28}} = 0.625 \times 10^{28} = 625 \times 10^{25}\, m^{-3}$.

(A) Given:
Conductivity $\sigma = 5 \times 10^{7} \, S/m$
Radius $r = 0.5 \, mm = 5 \times 10^{-4} \, m$
Electric field $E = 10 \, mV/m = 10 \times 10^{-3} \, V/m = 10^{-2} \, V/m$
Using Ohm's law in microscopic form,current density $J = \sigma E$:
$J = (5 \times 10^{7}) \times (10^{-2}) = 5 \times 10^{5} \, A/m^{2}$
Current $i = J \times A = J \times (\pi r^{2})$:
$i = (5 \times 10^{5}) \times \pi \times (5 \times 10^{-4})^{2}$
$i = 5 \times 10^{5} \times \pi \times 25 \times 10^{-8}$
$i = 125 \times 10^{-3} \, \pi \, A$
$i = 125 \, \pi \, mA$
Given that $i = x^{3} \, \pi \, mA$,we have:
$x^{3} = 125$
$x = \sqrt[3]{125} = 5$

(B) The current $I$ is given by the dot product of current density $\vec{J}$ and area vector $\vec{A}$: $I = \vec{J} \cdot \vec{A} = J A \cos(\theta)$.
Given $I = 5\; A$,$A = 0.04\; m^2$,and $\theta = 60^{\circ}$.
Substituting the values: $5 = J \times 0.04 \times \cos(60^{\circ})$.
Since $\cos(60^{\circ}) = 0.5$,we have $5 = J \times 0.04 \times 0.5 = J \times 0.02$.
Thus,$J = \frac{5}{0.02} = 250\; A/m^2$.
Using Ohm's law in vector form,$\vec{E} = \rho \vec{J}$,the magnitude is $E = \rho J$.
$E = (44 \times 10^{-8}) \times 250 = 11000 \times 10^{-8} = 11 \times 10^{-5}\; V/m$.

(B) In a series circuit,the current $i$ flowing through the conductor remains constant at every cross-section.
Since $i = n e A v_{d}$,where $n$ is the number density of electrons,$e$ is the charge,$A$ is the cross-sectional area,and $v_{d}$ is the drift velocity,we have $v_{d} = \frac{i}{n e A}$.
As we move from $P$ to $Q$,the radius $r$ decreases,so the cross-sectional area $A = \pi r^{2}$ decreases. Since $i$ is constant,the drift velocity $v_{d}$ must increase.
From the relation $v_{d} = \frac{e E \tau}{m}$,where $E$ is the electric field,$\tau$ is the relaxation time,and $m$ is the mass of the electron,we see that $v_{d} \propto E$. Therefore,as $v_{d}$ increases,the electric field $E$ also increases.
Since the current $i$ is constant throughout the conductor,the electron current does not change.
Thus,as we move from $P$ to $Q$,the drift velocity of electrons increases and the electric field increases.

(C) Given: Length $L = 10 \, m$,Radius $r = \frac{10^{-2}}{\sqrt{\pi}} \, m$,Resistance $R = 10 \, \Omega$,Electric field $E = 10 \, V/m$.
First,calculate the cross-sectional area $A$:
$A = \pi r^{2} = \pi \left( \frac{10^{-2}}{\sqrt{\pi}} \right)^{2} = \pi \cdot \frac{10^{-4}}{\pi} = 10^{-4} \, m^{2}$.
We know that current density $J = \sigma E$,where $\sigma$ is conductivity.
Since $\sigma = \frac{1}{\rho}$ and $R = \rho \frac{L}{A}$,we have $\rho = \frac{RA}{L}$.
Thus,$\sigma = \frac{L}{RA}$.
Substituting this into the expression for $J$:
$J = \left( \frac{L}{RA} \right) E = \frac{E \cdot L}{R \cdot A}$.
Plugging in the values:
$J = \frac{10 \times 10}{10 \times 10^{-4}} = \frac{100}{10^{-3}} = 10^{5} \, A/m^{2}$.

(A) The current density $J$ is given as $4 \times 10^{6} \; A \cdot m^{-2}$.
The radius of the wire is $R = 4 \; mm = 4 \times 10^{-3} \; m$.
The current $I$ through an area $A$ is given by $I = \int J \cdot dA$.
For the outer portion between radial distance $\frac{R}{2}$ and $R$,the area is $A' = \pi R^{2} - \pi \left(\frac{R}{2}\right)^{2} = \pi R^{2} - \frac{\pi R^{2}}{4} = \frac{3}{4} \pi R^{2}$.
Substituting the values:
$I = J \times A' = 4 \times 10^{6} \times \frac{3}{4} \pi R^{2}$.
$I = 4 \times 10^{6} \times \frac{3}{4} \times \pi \times (4 \times 10^{-3})^{2}$.
$I = 3 \times 10^{6} \times \pi \times 16 \times 10^{-6}$.
$I = 48 \; \pi \; A$.

(C) The current $I$ through an area element $dA$ is given by $I = \int J \cdot dA$.
For a cylindrical wire,the area element of a ring at radius $r'$ with thickness $dr'$ is $dA = 2 \pi r' dr'$.
Given $J = 1.0 \times 10^{6} \, A/m^{2}$ and $r = 4.0 \, mm = 4.0 \times 10^{-3} \, m$.
The current $I$ through the region between $r/2$ and $r$ is:
$I = \int_{r/2}^{r} J (2 \pi r') dr' = 2 \pi J \int_{r/2}^{r} r' dr'$
$I = 2 \pi J \left[ \frac{(r')^{2}}{2} \right]_{r/2}^{r} = \pi J \left[ r^{2} - \left( \frac{r}{2} \right)^{2} \right]$
$I = \pi J \left[ r^{2} - \frac{r^{2}}{4} \right] = \pi J \left( \frac{3r^{2}}{4} \right)$
Substituting the values $J = 10^{6} \, A/m^{2}$ and $r = 4 \times 10^{-3} \, m$:
$I = \pi \times 10^{6} \times \frac{3}{4} \times (4 \times 10^{-3})^{2}$
$I = \pi \times 10^{6} \times \frac{3}{4} \times 16 \times 10^{-6} = 12 \pi \, A$
Comparing this with $x \pi \, A$,we get $x = 12$.

(B) The drift velocity $v_d$ is given by the formula: $v_d = \frac{e \tau E}{m} = \frac{e \tau}{m} \left( \frac{\Delta V}{\ell} \right)$.
$1$. Effect of temperature: As temperature increases,the collision frequency of electrons increases,which causes the relaxation time $\tau$ to decrease. Since $v_d \propto \tau$,the drift velocity decreases. Thus,statement $A$ is correct and $E$ is incorrect.
$2$. Effect of length: From the formula $v_d = \frac{e \tau \Delta V}{m \ell}$,it is clear that $v_d \propto \frac{1}{\ell}$. Thus,statement $D$ is correct.
$3$. Effect of area: The drift velocity is related to current by $I = n e A v_d$. If the current $I$ is constant,then $v_d = \frac{I}{n e A}$,implying $v_d \propto \frac{1}{A}$. However,in a general conductor where potential difference $\Delta V$ is fixed,$v_d = \frac{e \tau \Delta V}{m \ell}$,which is independent of the cross-sectional area $A$. Thus,statement $B$ is incorrect.
$4$. Effect of potential difference: Since $v_d = \frac{e \tau \Delta V}{m \ell}$,the drift velocity is directly proportional to the applied potential difference $\Delta V$. Thus,statement $C$ is incorrect.
Therefore,statements $A$ and $D$ are correct.

(B) Given: Length $l = 1\,m$,Current $i = 1\,A$,Area $A = 2.0\,mm^{2} = 2.0 \times 10^{-6}\,m^{2}$,Resistivity $\rho = 1.7 \times 10^{-8}\,\Omega\,m$,Charge $e = 1.6 \times 10^{-19}\,C$.
First,calculate the resistance $R$ of the wire:
$R = \frac{\rho l}{A} = \frac{1.7 \times 10^{-8} \times 1}{2.0 \times 10^{-6}} = 0.85 \times 10^{-2}\,\Omega$.
The potential difference $V$ across the wire is given by Ohm's Law:
$V = iR = 1 \times 0.85 \times 10^{-2} = 0.85 \times 10^{-2}\,V$.
The electric field $E$ in the wire is:
$E = \frac{V}{l} = \frac{0.85 \times 10^{-2}}{1} = 0.85 \times 10^{-2}\,V/m$.
The force $F$ experienced by an electron is:
$F = eE = 1.6 \times 10^{-19} \times 0.85 \times 10^{-2} = 1.36 \times 10^{-21}\,N$.
Expressing this in terms of $10^{-23}$:
$F = 136 \times 10^{-23}\,N$.
Thus,$x = 136$.

(D) The current $I$ is related to the charge density $\rho$,cross-sectional area $A$,and velocity $v$ by the formula $I = \rho A v$.
Given:
$I = 500 \,\mu A = 500 \times 10^{-6} \,A$
$v = 3 \times 10^7 \,m/s$
$A = 1.50 \,mm^2 = 1.50 \times 10^{-6} \,m^2$
Rearranging the formula to solve for charge density $\rho$:
$\rho = \frac{I}{A v}$
Substituting the values:
$\rho = \frac{500 \times 10^{-6}}{(1.50 \times 10^{-6}) \times (3 \times 10^7)}$
$\rho = \frac{500}{1.50 \times 3 \times 10^7}$
$\rho = \frac{500}{4.5 \times 10^7} \approx 111.11 \times 10^{-9} \approx 1.11 \times 10^{-7} \,C/m^3$.
Wait,re-evaluating the calculation:
$\rho = \frac{500 \times 10^{-6}}{1.5 \times 10^{-6} \times 3 \times 10^7} = \frac{500}{4.5 \times 10^7} = \frac{500}{45,000,000} = \frac{5}{450,000} = \frac{1}{90,000} \approx 1.11 \times 10^{-5} \,C/m^3$.
Thus,the value is close to $10^{-5} \,C/m^3$.

(B) When current $I$ flows through a conductor of varying cross-section,the current remains constant through every cross-section due to the continuity of charge.
Since $I = jA$,where $j$ is the current density and $A$ is the cross-sectional area,we have $j_1 A_1 = j_2 A_2$.
As the area decreases $(A_2 < A_1)$,the current density must increase $(j_2 > j_1)$.
Also,current density is related to drift velocity $v_d$ by $j = n e v_d$. Since $n$ and $e$ are constant,the drift velocity $v_d$ increases as the area decreases.
From Ohm's law in microscopic form,$j = \frac{E}{\rho}$,where $E$ is the electric field and $\rho$ is the resistivity of the material.
Since $j$ increases and $\rho$ is constant for a given material,the magnitude of the electric field $E$ must increase in the narrowing region.
Therefore,the correct option is $(b)$.

(B) Given the resistance formula $R = \frac{m_e v}{n e^2 L^2}$.
From the de-Broglie wavelength relation,$\lambda = \frac{h}{p} = \frac{h}{m_e v}$.
Given $\lambda = L$,we have $L = \frac{h}{m_e v}$,which implies $m_e v = \frac{h}{L}$.
Substituting this into the resistance formula: $R = \frac{(h/L)}{n e^2 L^2} = \frac{h}{n e^2 L^3}$.
Since $n = \frac{N}{V} = \frac{N}{L^3}$ (where $N$ is the number of electrons),we get $R = \frac{h}{N e^2}$.
For maximum resistance,the number of electrons $N$ must be minimum. The minimum number of electrons in a metallic sample is $N = 1$.
Thus,$R_{max} = \frac{h}{e^2} = \frac{6.63 \times 10^{-34}}{(1.60 \times 10^{-19})^2}$.
$R_{max} = \frac{6.63 \times 10^{-34}}{2.56 \times 10^{-38}} \approx 2.59 \times 10^4 \, \Omega$.
This value is closest to $10^4 \, \Omega$.

(A) The relationship between current $I$ and drift velocity $V_d$ is given by the formula: $I = n e A V_d$.
Here,$I = 10 \,A$,$n = 9 \times 10^{28} \,m^{-3}$,$e = 1.6 \times 10^{-19} \,C$,and $A = 1 \,cm^2 = 10^{-4} \,m^2$.
Substituting the values into the equation:
$10 = (9 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (10^{-4}) \times V_d$.
$10 = (9 \times 1.6 \times 10^{28-19-4}) \times V_d$.
$10 = (14.4 \times 10^5) \times V_d$.
$V_d = \frac{10}{14.4 \times 10^5} = \frac{1}{1.44} \times 10^{-5} \approx 0.6944 \times 10^{-5} = 6.94 \times 10^{-6} \,m/s$.

(B) The mobility $\mu$ of electrons is defined as the ratio of drift velocity $v_d$ to the electric field $E$.
Given:
Potential difference $V = 5 \,V$
Length $l = 10 \,cm = 0.1 \,m$
Drift velocity $v_d = 2.5 \times 10^{-4} \,m/s$
The electric field $E$ is given by $E = \frac{V}{l} = \frac{5}{0.1} = 50 \,V/m$.
Now,calculate mobility $\mu$:
$\mu = \frac{v_d}{E} = \frac{2.5 \times 10^{-4}}{50}$
$\mu = \frac{2.5}{50} \times 10^{-4} = 0.05 \times 10^{-4} = 5 \times 10^{-6} \,m^2 V^{-1} s^{-1}$.

(A) The current $I$ in a conductor is given by the relation $I = n e A v_d$,where $n$ is the number density of electrons,$e$ is the charge of an electron,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.
Since both wires are of the same material,$n$ and $e$ are constant.
Thus,$v_d = \frac{I}{n e A} = \frac{I}{n e (\pi r^2)}$.
Given the ratio of radii $r_A : r_B = 1 : 2$,the ratio of areas is $A_A : A_B = r_A^2 : r_B^2 = 1^2 : 2^2 = 1 : 4$.
Given the ratio of currents $I_A : I_B = 4 : 1$.
The ratio of drift velocities is $\frac{v_{dA}}{v_{dB}} = \frac{I_A}{A_A} \times \frac{A_B}{I_B} = \left(\frac{4}{1}\right) \times \left(\frac{4}{1}\right) = \frac{16}{1}$.
Therefore,the ratio is $16:1$.

(D) The number of free electrons per unit volume $(n)$ is given by:
$n = \frac{N_A \times \rho}{M} = \frac{6.023 \times 10^{26} \times 10.5 \times 10^3}{108} \approx 5.86 \times 10^{28} \, m^{-3}$.
Using the formula for drift velocity $v_d = \frac{I}{n e A}$:
$v_d = \frac{20}{(5.86 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (3.14 \times 10^{-6})}$.
$v_d = \frac{20}{29.45} \times 10^{-4} \, m/s \approx 6.798 \times 10^{-4} \, m/s$.

(A) The drift velocity $V_{d}$ is given by the formula $V_{d} = \frac{eE}{m}\tau$,where $e$ is the charge of an electron,$E$ is the electric field,$m$ is the mass of the electron,and $\tau$ is the relaxation time.
Since $E = \frac{V}{L}$,where $V$ is the applied voltage and $L$ is the length of the conductor,we can write $V_{d} = \frac{eV}{mL}\tau$.
In this problem,the applied voltage $V$,length $L$,material (which determines $\tau$),and charge/mass of the electron remain constant.
Therefore,the drift velocity $V_{d}$ is independent of the area of cross-section $A$.
Thus,the new drift velocity remains $V_{d}$.

(B) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.
Here,$I = 2\,A$,$n = 2.0 \times 10^{28}\,m^{-3}$,$e = 1.6 \times 10^{-19}\,C$,and $A = 25.0\,mm^2 = 25.0 \times 10^{-6}\,m^2$.
Substituting these values into the formula:
$v_d = \frac{2}{(2.0 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (25.0 \times 10^{-6})}$
$v_d = \frac{2}{80 \times 10^3 \times 10^{-6}}$
$v_d = \frac{2}{80 \times 10^{-3}}$
$v_d = \frac{2}{0.08} = 25\,ms^{-1}$.
Since the question asks for the value in terms of $\times 10^{-6}\,ms^{-1}$,we express $25$ as $25 \times 10^{-6} \times 10^6$ (Note: The calculation yields $25 \times 10^{-6}\,ms^{-1}$ based on the provided parameters).

(A) Given:
Number density of electrons,$n = 8 \times 10^{28} \ m^{-3}$
Area of cross-section,$A = 2 \times 10^{-6} \ m^2$
Current,$I = 3.2 \ A$
Charge of an electron,$e = 1.6 \times 10^{-19} \ C$
The formula for drift velocity $(v_d)$ is given by:
$I = n e A v_d$
Rearranging for $v_d$:
$v_d = \frac{I}{n e A}$
Substituting the values:
$v_d = \frac{3.2}{(8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2 \times 10^{-6})}$
$v_d = \frac{3.2}{25.6 \times 10^3}$
$v_d = 0.125 \times 10^{-3} \ m/s$
$v_d = 125 \times 10^{-6} \ m/s$
Thus,the drift speed is $125 \times 10^{-6} \ m/s$.

(B) When an electric field $E$ is applied to a metallic conductor,the free electrons experience an electrostatic force $F = -eE$ in the direction opposite to the electric field.
Since the electric field points from higher potential to lower potential,the electrons are accelerated towards the higher potential.
However,due to continuous collisions with the positive ions of the metallic lattice,the electrons do not move in straight lines but follow random,curved paths between successive collisions.
On average,they exhibit a net drift velocity towards the higher potential region.
Thus,the motion of free electrons between collisions is along curved paths.

(D) The mobility $\mu$ of charge carriers is defined as the ratio of drift velocity $v_d$ to the electric field $E$.
Formula: $\mu = \frac{v_d}{E}$
Since the electric field $E$ is given by $E = \frac{V}{\ell}$,where $V$ is the potential difference and $\ell$ is the length of the wire,we have:
$\mu = \frac{v_d}{V/\ell} = \frac{v_d \cdot \ell}{V}$
Given values: $v_d = 0.5\ m/s$,$\ell = 2\ m$,$V = 200\ V$.
Substituting these values into the formula:
$\mu = \frac{0.5 \times 2}{200} = \frac{1}{200} = 0.005\ m^2 V^{-1} s^{-1}$
$\mu = 5 \times 10^{-3}\ m^2 V^{-1} s^{-1}$

(A) The current $I$ flowing through the wire is constant at all cross-sections.
Current density is defined as $J = \frac{I}{A}$,where $A$ is the cross-sectional area.
Since $I$ is constant,$J \propto \frac{1}{A}$.
From the figure,the area at $A$ is smaller than the area at $B$ $(A_A < A_B)$,therefore $j_A > j_B$.
According to Ohm's law in microscopic form,$\vec{J} = \sigma \vec{E}$,where $\sigma$ is the conductivity of the material.
Since the material is homogeneous and isotropic,$\sigma$ is constant throughout the wire.
Thus,$E = \frac{J}{\sigma}$,which implies $E \propto J$.
Since $j_A > j_B$,it follows that $E_A > E_B$.

(B) Given:
Length of conductor $L = 40 \ cm = 0.4 \ m$
Potential difference $V = 10 \ V$
Drift velocity $v_d = 5 \times 10^{-6} \ m/s$
Electric field $E$ is given by $E = \frac{V}{L} = \frac{10}{0.4} = 25 \ V/m$.
Mobility $\mu$ is defined as the ratio of drift velocity to the electric field: $\mu = \frac{v_d}{E}$.
Substituting the values: $\mu = \frac{5 \times 10^{-6}}{25} = 0.2 \times 10^{-6} = 2 \times 10^{-7} \ m^2 \ V^{-1} \ s^{-1}$.

(B) The drift velocity $V_d$ is given by the formula $V_d = \frac{I}{neA}$,where $I$ is the current,$n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area of the conductor.
Since $A = \pi r^2$,we can write $V_d = \frac{I}{ne\pi r^2}$.
From this expression,we see that $V_d \propto \frac{I}{r^2}$.
Let the initial drift velocity be $V_1 = V$,initial current be $I_1 = i$,and initial radius be $r_1 = r$.
When the current and radius are doubled,the new current $I_2 = 2i$ and the new radius $r_2 = 2r$.
The new drift velocity $V_2$ is given by $V_2 = V_1 \times \left( \frac{I_2}{I_1} \right) \times \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values: $V_2 = V \times \left( \frac{2i}{i} \right) \times \left( \frac{r}{2r} \right)^2 = V \times 2 \times \frac{1}{4} = \frac{V}{2}$.

(C) The drift velocity $v_d$ of a charge carrier is directly proportional to the applied electric field $E$,given by $v_d = \mu E$.
Here,$\mu$ is defined as the mobility of the charge carrier.
Therefore,mobility $\mu = \frac{v_d}{E}$.
It represents the magnitude of the drift velocity per unit electric field.

(D) Current density $(J)$ is defined as the current $(I)$ flowing per unit area $(A)$ perpendicular to the direction of flow.
Mathematically,$J = \frac{I}{A}$.
The $SI$ unit of current $(I)$ is Ampere $(A)$ and the $SI$ unit of area $(A)$ is square meter $(m^{2})$.
Therefore,the $SI$ unit of current density is $\frac{A}{m^{2}} = A \ m^{-2}$.
Thus,the correct option is $(D)$.

(C) Mobility $\mu$ is defined as the ratio of drift velocity $v_d$ to the electric field $E$: $\mu = \frac{v_d}{E}$.
The unit of drift velocity $v_d$ is $m/s$.
The unit of electric field $E$ is $N/C$ (or $V/m$).
Substituting the fundamental units:
$1 \ N = 1 \ kg \cdot m/s^2$
$1 \ C = 1 \ A \cdot s$
Therefore,the unit of $E$ is $\frac{kg \cdot m/s^2}{A \cdot s} = \frac{kg \cdot m}{A \cdot s^3}$.
Now,calculating the unit of $\mu$:
$\mu = \frac{m/s}{kg \cdot m / (A \cdot s^3)} = \frac{m}{s} \cdot \frac{A \cdot s^3}{kg \cdot m} = \frac{A \cdot s^2}{kg} = kg^{-1} s^2 A$.
Thus,the correct option is $C$.

(B) The relationship between current $I$ and drift velocity $v_d$ is given by $I = n A v_d e$,where $n$ is the number density,$A$ is the cross-sectional area,and $e$ is the elementary charge.
Since $v_d = \frac{l}{t}$,where $l$ is the length of the wire and $t$ is the time taken,we can write:
$I = n A \left( \frac{l}{t} \right) e$
Rearranging for $t$:
$t = \frac{n A l e}{I}$
Given values:
$n = 8.5 \times 10^{28} \ m^{-3}$
$A = 1.0 \times 10^{-6} \ m^2$
$l = 6 \ m$
$e = 1.6 \times 10^{-19} \ C$
$I = 1.5 \ A$
Substituting these values:
$t = \frac{8.5 \times 10^{28} \times 1.0 \times 10^{-6} \times 6 \times 1.6 \times 10^{-19}}{1.5}$
$t = \frac{81.6 \times 10^3}{1.5} = 54.4 \times 10^3 \ s$
$t = 5.44 \times 10^4 \ s$
Thus,the correct option is $B$.

(B) The current $I$ flowing through a conductor is given by the relation: $I = n A v_d e$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,$v_d$ is the drift velocity,and $e$ is the charge of an electron.
Since the current $I$ is constant throughout the wire,we have $v_d = \frac{I}{n A e}$.
Since $I$,$n$,and $e$ are constant,$v_d \propto \frac{1}{A}$.
Since the cross-section is circular,$A = \pi r^2$,therefore $v_d \propto \frac{1}{r^2}$.
Now,calculating the ratio of drift velocities at points $A$ and $B$:
$\frac{(v_d)_A}{(v_d)_B} = \frac{r_B^2}{r_A^2} = \frac{(r)^2}{(3r)^2} = \frac{r^2}{9r^2} = \frac{1}{9}$.
Thus,the ratio is $\frac{1}{9}$.

(D) Mobility $\mu$ is defined as the ratio of drift velocity $v_d$ to the applied electric field $E$: $\mu = \frac{v_d}{E}$.
The dimensional formula for drift velocity $v_d$ is $[L^1 T^{-1}]$.
The dimensional formula for electric field $E$ is $[M^1 L^1 T^{-3} A^{-1}]$.
Substituting these into the formula: $\mu = \frac{[L^1 T^{-1}]}{[M^1 L^1 T^{-3} A^{-1}]}$.
Simplifying the expression: $\mu = [M^{-1} L^{1-1} T^{-1 - (-3)} A^1] = [M^{-1} L^0 T^2 A^1]$.

(A) Given: Area $A = (2 \ mm)^2 = (2 \times 10^{-3} \ m)^2 = 4 \times 10^{-6} \ m^2$. Current $I = 8 \ A$. Number density $n = 8 \times 10^{28} \ m^{-3}$. Charge of electron $e = 1.6 \times 10^{-19} \ C$.
Using the formula for drift velocity: $I = n A v_d e$.
Rearranging for $v_d$: $v_d = \frac{I}{n A e}$.
Substituting the values: $v_d = \frac{8}{8 \times 10^{28} \times 4 \times 10^{-6} \times 1.6 \times 10^{-19}}$.
$v_d = \frac{8}{51.2 \times 10^3} = \frac{8}{51200} \approx 1.56 \times 10^{-4} \ ms^{-1}$.

(A) The mobility of an electron is given by the formula: $\mu = \frac{e \tau}{m}$.
Since the temperature is constant,the relaxation time $\tau$ remains constant,and therefore the mobility $\mu$ remains the same.
The relationship between electric current $I$ and drift velocity $v_d$ is given by $I = n e A v_d$.
From this equation,we can see that $v_d = \frac{I}{n e A}$.
Since $n$,$e$,and $A$ are constants for a given wire,$v_d \propto I$.
Therefore,when the current $I$ is increased,the drift velocity $v_d$ also increases.

(D) Given: Electric field,$E = 3 \times 10^{-10} \text{ Vm}^{-1}$.
Mobility,$\mu = 2.5 \times 10^6 \text{ m}^2 \text{V}^{-1} \text{s}^{-1}$.
The relationship between drift velocity $(v_d)$,mobility $(\mu)$,and electric field $(E)$ is given by the formula:
$v_d = \mu E$
Substituting the given values:
$v_d = (2.5 \times 10^6) \times (3 \times 10^{-10})$
$v_d = 7.5 \times 10^{-4} \text{ m/s}$.
Therefore,the drift velocity is $7.5 \times 10^{-4} \text{ m/s}$.

(D) Given: Length $l = 1 \,m$,Area $A = 5 \times 10^{-7} \,m^{2}$,Current $I = 1 \,A$,Electron density $n = 8 \times 10^{28} \,m^{-3}$,Charge of electron $e = 1.6 \times 10^{-19} \,C$.
The drift velocity $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$.
Substituting the values: $v_{d} = \frac{1}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 5 \times 10^{-7}} = \frac{1}{64 \times 10^{2}} = \frac{1}{6.4 \times 10^{3}} \,m/s$.
The time $T$ taken to drift across the length $l$ is $T = \frac{l}{v_{d}}$.
$T = \frac{1}{1 / (6.4 \times 10^{3})} = 6.4 \times 10^{3} \,s$.

(B) The current $I$ in a conductor is given by the formula $I = neAv_d$, where $n$ is the electron number density, $e$ is the charge of an electron, $A$ is the cross-sectional area, and $v_d$ is the drift velocity.
Even though the drift velocity $v_d$ is very small (typically $10^{-4} \, m/s$) and the charge $e$ is small $(1.6 \times 10^{-19} \, C)$, the number density $n$ of free electrons in a conductor is extremely large, typically of the order of $10^{28} \, m^{-3}$.
Therefore, the product $neAv_d$ results in an appreciably large current.

(B) The mobility $\mu$ of free electrons is defined as the drift velocity per unit electric field,given by the formula:
$\mu = \frac{v_d}{E} = \frac{e \tau}{m}$
Where:
$e$ is the charge of an electron,
$\tau$ is the relaxation time,
$m$ is the mass of an electron.
Since $e$ and $m$ are constants,we have:
$\mu \propto \tau$
Therefore,the mobility of free electrons in a conductor is directly proportional to the relaxation time.

(D) The formula for the drift velocity $(v_{d})$ of electrons is given by:
$v_{d} = \frac{I}{n e A}$
Where:
$I = 2 \,A$ (Current)
$n = 5 \times 10^{26} \,m^{-3}$ (Number density of free electrons)
$e = 1.6 \times 10^{-19} \,C$ (Charge of an electron)
$A = 2 \times 10^{-6} \,m^{2}$ (Cross-sectional area)
Substituting the values:
$v_{d} = \frac{2}{(5 \times 10^{26}) \times (1.6 \times 10^{-19}) \times (2 \times 10^{-6})}$
$v_{d} = \frac{2}{16 \times 10^{26-19-6}} = \frac{2}{16 \times 10^{1}} = \frac{2}{160} = \frac{1}{80} \,ms^{-1}$

(B) The relationship between current $I$ and drift velocity $v_{d}$ is given by the formula: $I = n A e v_{d}$.
Here, $n = 10^{29} \,m^{-3}$ is the electron density, $A = 1 \,mm^{2} = 10^{-6} \,m^{2}$ is the cross-sectional area, $e = 1.6 \times 10^{-19} \,C$ is the charge of an electron, and $I = 20 \,A$.
Rearranging the formula for drift velocity: $v_{d} = \frac{I}{n A e}$.
Substituting the values: $v_{d} = \frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}$.
$v_{d} = \frac{20}{1.6 \times 10^{4}} = \frac{20}{16000} = 1.25 \times 10^{-3} \,ms^{-1}$.