Current Density, Drift Velocity and Mobility Questions in English

(C) The relationship between drift velocity $v_d$ and current $i$ is given by the formula $i = n e A v_d$,where $n$ is the number density of electrons,$e$ is the charge of an electron,and $A$ is the area of cross-section.
From this,the drift velocity is $v_d = \frac{i}{n e A}$.
In the first case,$v_1 = v = \frac{i}{n e A}$.
In the second case,the current is $i' = 2i$ and the area is $A' = 2A$.
Therefore,the new drift velocity $v_2$ is $v_2 = \frac{i'}{n e A'} = \frac{2i}{n e (2A)} = \frac{i}{n e A}$.
Comparing the two,we get $v_2 = v_1 = v$.

(B) The drift velocity $(v_d)$ of electrons in a conductor is given by the relation $v_d = \frac{I}{neA}$,where $I$ is the current,$n$ is the number density of electrons,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
For typical metallic conductors,the drift velocity of electrons is very small,typically in the range of $10^{-4} \, m/s$ to $10^{-5} \, m/s$.
Converting this to $cm/s$: $10^{-4} \, m/s = 10^{-4} \times 10^{2} \, cm/s = 10^{-2} \, cm/s$.
Therefore,the order of drift velocity is $10^{-2} \, cm/s$.

(B) The number of atoms per unit volume $(n)$ is given by $n = \frac{\text{Density} \times N_A}{\text{Atomic Weight}}$.
Given: Density $\rho = 9 \times 10^3 \ kg/m^3$,Atomic weight $M = 63 \times 10^{-3} \ kg/mol$,Avogadro number $N_A = 6.023 \times 10^{23} \ atoms/mol$.
$n = \frac{9 \times 10^3 \times 6.023 \times 10^{23}}{63 \times 10^{-3}} \approx 8.6 \times 10^{28} \ electrons/m^3$.
The drift velocity $v_d$ is given by $v_d = \frac{I}{neA}$,where $I = 1.1 \ A$,$e = 1.6 \times 10^{-19} \ C$,and $A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 \ m^2$.
$v_d = \frac{1.1}{8.6 \times 10^{28} \times 1.6 \times 10^{-19} \times 3.14 \times 0.25 \times 10^{-6}}$.
$v_d \approx \frac{1.1}{1.08 \times 10^4} \approx 1.01 \times 10^{-4} \ m/s = 0.1 \ mm/s$.

(A) The resistance $R$ of a conductor is given by the formula $R = \frac{ml}{ne^2 \tau A}$,where $m$ is the mass of the electron,$l$ is the length,$n$ is the electron density,$e$ is the charge,$A$ is the cross-sectional area,and $\tau$ is the relaxation time.
From this relation,we see that $R \propto \frac{1}{\tau}$.
When the temperature of a conductor increases,the thermal velocity of the free electrons increases.
This leads to more frequent collisions between the electrons and the lattice ions.
Consequently,the average time between two successive collisions,known as the relaxation time $\tau$,decreases.
Since $R$ is inversely proportional to $\tau$,a decrease in $\tau$ results in an increase in the resistance $R$.

(C) For a steady electric current to flow through a conducting wire,there must be a potential difference across its ends.
According to the relation $E = V/L$,where $E$ is the electric field,$V$ is the potential difference,and $L$ is the length of the wire,an electric field must be present inside the conductor.
This electric field is responsible for the drift velocity of the charge carriers (electrons).
Since the current flows along the length of the wire,the electric field must also be directed along the length of the wire,i.e.,parallel to it.
Therefore,the correct option is $C$.

(B) In the absence of an external electric field,the free electrons in a metallic block are in random thermal motion.
According to the kinetic theory of gases,the root mean square velocity $(V_{rms})$ of these electrons is given by the formula $V_{rms} = \sqrt{\frac{3KT}{m}}$,where $K$ is the Boltzmann constant,$T$ is the absolute temperature,and $m$ is the mass of an electron.
From this expression,it is clear that $V_{rms} \propto \sqrt{T}$.
Therefore,the mean velocity (specifically the root mean square thermal velocity) is proportional to $\sqrt{T}$.

(C) The formula for drift velocity $(v_d)$ is given by $v_d = \frac{I}{nAe}$.
Given:
Current $(I)$ = $20 \, A$
Number density of electrons $(n)$ = $10^{29} \, m^{-3}$
Area of cross-section $(A)$ = $10^{-6} \, m^2$
Charge of an electron $(e)$ = $1.6 \times 10^{-19} \, C$
Substituting these values into the formula:
$v_d = \frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}$
$v_d = \frac{20}{1.6 \times 10^{29-6-19}}$
$v_d = \frac{20}{1.6 \times 10^4}$
$v_d = 12.5 \times 10^{-4} \, m/s = 1.25 \times 10^{-3} \, m/s$.

(B) The microscopic form of Ohm's law states that the current density $j$ is proportional to the electric field $E$ applied across a conductor.
Mathematically,this is expressed as $j = \sigma E$,where $\sigma$ is the electrical conductivity.
Since conductivity $\sigma$ is the reciprocal of resistivity (specific resistance) $\rho$,we have $\sigma = 1/\rho$.
Substituting this into the equation,we get $j = E/\rho$.
Rearranging the terms to solve for the electric intensity $E$,we obtain $E = j\rho$.
Therefore,the correct relation is $E = j\rho$ (or $E = jk$ if $k$ represents resistivity).

(C) The formula for drift velocity is given by $v_d = \frac{I}{nAe}$.
Given:
$I = 1.344 \, A$
$A = 1 \, mm^2 = 10^{-6} \, m^2$
$n = 8.4 \times 10^{22} \, \text{electrons}/cm^3 = 8.4 \times 10^{28} \, \text{electrons}/m^3$
$e = 1.6 \times 10^{-19} \, C$
Substituting these values into the formula:
$v_d = \frac{1.344}{(8.4 \times 10^{28}) \times (10^{-6}) \times (1.6 \times 10^{-19})}$
$v_d = \frac{1.344}{8.4 \times 1.6 \times 10^3}$
$v_d = \frac{1.344}{13.44 \times 10^3} = 0.1 \times 10^{-3} \, m/s$
$v_d = 0.1 \, mm/s$.

(A) The resistance $R$ of a conductor is given by $R = \rho \frac{l}{A}$.
From the microscopic model of current,the resistivity $\rho$ is given by $\rho = \frac{m}{ne^2 \tau}$.
Substituting the expression for $\rho$ into the resistance formula,we get:
$R = \left( \frac{m}{ne^2 \tau} \right) \frac{l}{A} = \frac{ml}{ne^2 \tau A}$.
Thus,the correct option is $A$.

(B) Current density $J$ is defined as the current per unit cross-sectional area,given by $J = \frac{I}{A} = \frac{I}{\pi r^2}$.
Since the copper wire and the iron wire are connected in series,the same amount of current $I$ flows through both wires,i.e.,$I_{Cu} = I_{Fe} = I$.
The ratio of the current density in the copper wire $(J_{Cu})$ to the iron wire $(J_{Fe})$ is given by:
$\frac{J_{Cu}}{J_{Fe}} = \frac{I / (\pi r_{Cu}^2)}{I / (\pi r_{Fe}^2)} = \frac{r_{Fe}^2}{r_{Cu}^2}$.
Given $r_{Cu} = 1\, mm$ and $r_{Fe} = 3\, mm$,we substitute these values:
$\frac{J_{Cu}}{J_{Fe}} = \frac{(3\, mm)^2}{(1\, mm)^2} = \frac{9}{1} = 9:1$.
Thus,the ratio of the current density is $9:1$.

(C) The drift velocity ${v_d}$ of electrons in a conductor is given by the formula: ${v_d} = \frac{eE\tau}{m}$,where $E$ is the electric field,$\tau$ is the relaxation time,$e$ is the charge of an electron,and $m$ is the mass of an electron.
Since the electric field $E = \frac{V}{l}$,we can substitute this into the equation:
${v_d} = \frac{eV\tau}{ml}$.
Here,$V$ is the potential difference,$l$ is the length of the wire,$e$ is the elementary charge,$m$ is the electron mass,and $\tau$ is the relaxation time.
None of these parameters $(V, l, e, m, \tau)$ depend on the diameter $d$ of the wire.
Therefore,changing the diameter $d$ does not affect the drift velocity ${v_d}$ of the electrons.
Thus,the drift velocity remains unchanged.

(B) The electric current $I$ in a conductor is given by the relation $I = nAev_d$, where $n$ is the number density of free electrons, $A$ is the cross-sectional area, $e$ is the magnitude of the electron charge, and $v_d$ is the drift speed.
Although the drift speed $v_d$ is very small (typically $10^{-4} \, m/s$) and the charge $e$ is also very small $(1.6 \times 10^{-19} \, C)$, the number density $n$ of free electrons in a conductor is extremely high (typically $10^{28} \, to \, 10^{29} \, m^{-3}$).
This very high value of $n$ compensates for the small values of $v_d$ and $e$, allowing for the flow of a significant amount of current through the conductor.

(C) The current $I$ in a wire is given by the relation $I = neAv_d$,where $n$ is the number density of free electrons,$e$ is the charge of an electron,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.
Since the wire has a circular cross-section,$A = \pi r^2$.
Therefore,$I = ne(\pi r^2)v_d$.
This implies $I \propto r^2 v_d$.
For the first wire,$I_1 = I$,$r_1 = r$,and $v_{d1} = V$.
For the second wire,$r_2 = r/2$ and $v_{d2} = 2V$.
Taking the ratio: $\frac{I_2}{I_1} = \frac{r_2^2 v_{d2}}{r_1^2 v_{d1}} = \frac{(r/2)^2 (2V)}{r^2 V} = \frac{(r^2/4) (2V)}{r^2 V} = \frac{2}{4} = \frac{1}{2}$.
Thus,$I_2 = I_1 / 2 = I/2$.

(C) The drift velocity $v_d$ of electrons in a conductor is given by the formula $v_d = \frac{I}{neA}$,where $I$ is the current,$n$ is the number density of electrons,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
Since $I$,$n$,and $e$ are constant for both sections,we have $v_d \propto \frac{1}{A}$.
The area of cross-section $A$ is given by $A = \pi r^2 = \pi (d/2)^2 = \frac{\pi d^2}{4}$,which implies $A \propto d^2$.
Therefore,$v_d \propto \frac{1}{d^2}$.
For sections $P$ and $Q$,we have $\frac{v_P}{v_Q} = \left( \frac{d_Q}{d_P} \right)^2$.
Given $d_P = d$ and $d_Q = d/2$,we get $\frac{v_P}{v_Q} = \left( \frac{d/2}{d} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Thus,$v_P = \frac{1}{4}v_Q$.

(B) The drift velocity $v_d$ is given by the formula $v_d = \frac{I}{neA}$,where $I$ is the current,$n$ is the number density of free electrons,$e$ is the charge of an electron,and $A$ is the cross-sectional area of the wire.
From this relation,it is clear that $v_d$ depends on the current $(I)$,the number density of free electrons $(n)$,and the cross-sectional area $(A)$.
However,the drift velocity does not depend on the length of the wire $(L)$ for a macroscopic conductor.
Therefore,the correct option is $B$.

(B) The formula for drift velocity $(V_d)$ is given by $V_d = \frac{I}{neA}$.
Given:
Current $(I)$ = $40 \ A$
Number density of electrons $(n)$ = $10^{29} \ m^{-3}$
Area of cross-section $(A)$ = $10^{-6} \ m^2$
Charge of an electron $(e)$ = $1.6 \times 10^{-19} \ C$
Substituting these values into the formula:
$V_d = \frac{40}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}$
$V_d = \frac{40}{1.6 \times 10^{29-19-6}}$
$V_d = \frac{40}{1.6 \times 10^4}$
$V_d = 25 \times 10^{-4} \ m/s = 2.5 \times 10^{-3} \ m/s$.

(C) The formula for drift velocity $(V_d)$ is given by $V_d = \frac{I}{nAe}$.
Given:
$I = 5.4 \, A$
$n = 8.4 \times 10^{28} \, m^{-3}$
$A = 10^{-6} \, m^2$
$e = 1.6 \times 10^{-19} \, C$
Substituting the values:
$V_d = \frac{5.4}{8.4 \times 10^{28} \times 10^{-6} \times 1.6 \times 10^{-19}}$
$V_d = \frac{5.4}{8.4 \times 1.6 \times 10^3}$
$V_d = \frac{5.4}{13.44 \times 10^3} \approx 0.4017 \times 10^{-3} \, m/s$
$V_d \approx 0.4 \, mm/s$.

(A) Current,wattless current,and power are scalar quantities because they do not follow the laws of vector addition.
Current density $\vec{J} = \frac{I}{A} \hat{n}$ is defined as the current per unit area perpendicular to the flow of charge,and it is a vector quantity directed along the flow of positive charge.

(B) The drift velocity $v_d$ of free electrons in a conductor is given by the formula: $v_d = \frac{i}{neA}$,where $i$ is the current,$n$ is the number density of electrons,$e$ is the charge of an electron,and $A$ is the cross-sectional area of the conductor.
Since the conductor is a wire,the cross-sectional area $A = \pi r^2$,where $r$ is the radius.
Thus,$v_d = \frac{i}{ne\pi r^2}$.
From this,we can 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 $r_1 = r$.
Let the new drift velocity be $v_2$,new current be $i_2 = 2i$,and new radius be $r_2 = 2r$.
Using the proportionality $v_d \propto \frac{i}{r^2}$,we get:
$\frac{v_2}{v_1} = \frac{i_2}{i_1} \times \left( \frac{r_1}{r_2} \right)^2$
$\frac{v_2}{v} = \frac{2i}{i} \times \left( \frac{r}{2r} \right)^2$
$\frac{v_2}{v} = 2 \times \left( \frac{1}{2} \right)^2 = 2 \times \frac{1}{4} = \frac{1}{2}$
Therefore,$v_2 = \frac{v}{2}$.

(A) The formula for drift velocity is given by $v_d = \frac{I}{nAe}$.
Here,$I = 8 \, A$,$n = 8 \times 10^{28} \, m^{-3}$,and $e = 1.6 \times 10^{-19} \, C$.
The cross-sectional area $A$ of the square wire is $(2.0 \, mm)^2 = (2.0 \times 10^{-3} \, m)^2 = 4 \times 10^{-6} \, m^2$.
Substituting these values into the formula:
$v_d = \frac{8}{8 \times 10^{28} \times 4 \times 10^{-6} \times 1.6 \times 10^{-19}}$
$v_d = \frac{8}{51.2 \times 10^3} = \frac{1}{6.4} \times 10^{-3} \approx 0.156 \times 10^{-3} \, m/s$.

(C) When a potential difference is applied across the ends of a linear metallic conductor,an electric field is established inside the conductor.
This electric field exerts a force on the free electrons,causing them to move.
However,due to frequent collisions with the positive ions of the conductor,the electrons do not accelerate continuously.
Instead,they acquire a small,constant average velocity known as the drift velocity,directed from the lower potential end to the higher potential end of the conductor.

(C) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.
Here,$I = 100\,A$,$n = 10^{28}\,m^{-3}$,and $e = 1.6 \times 10^{-19}\,C$.
The area of cross-section $A = \pi r^2 = \pi (d/2)^2 = \pi (0.01)^2 = \pi \times 10^{-4}\,m^2$.
Substituting these values into the formula:
$v_d = \frac{100}{10^{28} \times 1.6 \times 10^{-19} \times \pi \times 10^{-4}}$
$v_d = \frac{100}{1.6 \times \pi \times 10^5} \approx \frac{100}{5.026 \times 10^5} \approx 1.989 \times 10^{-4}\,m/s$.
Thus,the drift velocity is nearly $2 \times 10^{-4}\,m/s$.

(B) From Ohm's law in microscopic form,the current density $J$ is directly proportional to the electric field $E$ applied across a conductor.
The relationship is given by the equation: $J = \sigma E$,where $\sigma$ is the electrical conductivity of the material.
Rearranging this equation to solve for conductivity $\sigma$,we get:
$\sigma = \frac{J}{E}$
Thus,the correct relation is $\sigma = J/E$.

(D) The total current density $J$ is the sum of the current density due to electrons $(J_e)$ and the current density due to positive ions $(J_i)$.
$J = J_e + J_i = n e v_e + n e v_i = n e (v_e + v_i)$
Given:
$n = 5 \times 10^7\, cm^{-3} = 5 \times 10^7 \times 10^6\, m^{-3} = 5 \times 10^{13}\, m^{-3}$
$e = 1.6 \times 10^{-19}\, C$
$v_e = 0.4\, m/s$
$J = 4\, \mu A/m^2 = 4 \times 10^{-6}\, A/m^2$
First,calculate the current density due to electrons:
$J_e = n e v_e = (5 \times 10^{13}) \times (1.6 \times 10^{-19}) \times 0.4 = 8 \times 10^{-6} \times 0.4 = 3.2 \times 10^{-6}\, A/m^2$
Now,find the current density due to ions:
$J_i = J - J_e = 4 \times 10^{-6} - 3.2 \times 10^{-6} = 0.8 \times 10^{-6}\, A/m^2$
Using $J_i = n e v_i$,we solve for $v_i$:
$v_i = \frac{J_i}{n e} = \frac{0.8 \times 10^{-6}}{(5 \times 10^{13}) \times (1.6 \times 10^{-19})} = \frac{0.8 \times 10^{-6}}{8 \times 10^{-6}} = 0.1\, m/s$.

(C) Free electron theory:
In metals,most atoms have loosely bound electrons in their outermost orbits,known as free electrons,as they have a tendency to detach from their parent atom and move freely.
The flow of electric charge occurs due to the movement of these electrons. In the presence of an external electric field,all the free electrons flow in a direction opposite to the field,thus constituting an electric current.
As the electrons move through the metal conductor,they frequently collide with atoms,other electrons,or impurities within the lattice structure.
These collisions result in the transfer of kinetic energy from the electrons to the atoms of the conductor,which increases the vibrational energy of the atoms,thereby generating heat in the conductor.

(D) When an electric potential difference is applied across a metallic wire,the conduction electrons are accelerated by the electric field.
As these electrons move through the lattice,they frequently collide with the vibrating atoms (or ions) of the metal.
During these collisions,the electrons transfer a portion of their kinetic energy to the atoms,causing the atoms to vibrate with greater amplitude.
This increase in the vibrational energy of the atoms manifests as an increase in the temperature of the wire,which is observed as the production of heat.
Therefore,the correct reason for the production of heat is the collisions of the conduction electrons with the atoms of the metallic wire.

(B) The current $i$ is given as $1 \mu A = 10^{-6} \text{ A}$.
The cross-sectional area $A = 0.5 \text{ mm}^2 = 0.5 \times 10^{-6} \text{ m}^2$.
The velocity of the protons $v = 3 \times 10^4 \text{ m/s}$.
In one second,the length of the beam segment is $L = v \times t = 3 \times 10^4 \times 1 = 3 \times 10^4 \text{ m}$.
The volume of this segment is $V = A \times L = (0.5 \times 10^{-6} \text{ m}^2) \times (3 \times 10^4 \text{ m}) = 1.5 \times 10^{-2} \text{ m}^3$.
The charge flowing in one second is $Q = i \times t = 10^{-6} \text{ C} \times 1 \text{ s} = 10^{-6} \text{ C}$.
Charge density $\rho = \frac{Q}{V} = \frac{10^{-6}}{1.5 \times 10^{-2}} = \frac{1}{1.5} \times 10^{-4} = 0.666 \times 10^{-4} = 6.66 \times 10^{-5} \text{ C/m}^3$.
Thus,the correct option is $B$.

(C) In metallic conductors,electric current is primarily due to the drift of free electrons under the influence of an external electric field.
In semiconductors,electric current is due to the drift of both free electrons and holes under the influence of an external electric field.
Therefore,electric current in both metallic conductors and semiconductors involves the drift of electrons.
Thus,the correct option is $(c)$.

(B) In the absence of an electric field,the free electrons in a conductor move in random directions with random velocities,and their paths between collisions are straight lines.
However,in the presence of an external electric field $E$,the electron experiences an electric force $F = -eE$.
This force causes the electron to undergo constant acceleration $a = -eE/m$.
Due to this constant acceleration,the path of the electron between two successive collisions becomes parabolic,not a straight line.

(C) The drift velocity $v_d$ of electrons is very small (typically $10^{-3} \, m/s$).
However,the electric field is established throughout the entire length of the wire at the speed of light $(c \approx 3 \times 10^8 \, m/s)$ as soon as the switch is closed.
This electric field exerts a force on all free electrons in the wire simultaneously,causing them to drift and establishing a current throughout the circuit almost instantaneously.
Therefore,the bulb glows immediately because the electric field propagates rapidly,not because individual electrons travel from the switch to the bulb.

(A) The drift velocity $v_d$ is given by the formula $v_d = \frac{eV\tau}{ml}$,where $e$ is the charge of an electron,$V$ is the potential difference,$\tau$ is the relaxation time,$m$ is the mass of the electron,and $l$ is the length of the wire.
From the formula,we can see that $v_d \propto V$ and $v_d \propto \frac{1}{l}$.
If the potential difference $V$ is doubled,the drift velocity $v_d$ becomes $v_d' = \frac{e(2V)\tau}{ml} = 2v_d$.
Therefore,the drift velocity is doubled if the potential difference $V$ is doubled.

(B) According to Ohm's law in vector form,the current density $J$ is related to the electric field $E$ by the equation $J = \sigma E$,where $\sigma$ is the electrical conductivity of the material.
Rearranging this equation to solve for the electric field,we get $E = J / \sigma$.
Since the conductivity $\sigma$ is a constant for a given material at a constant temperature,we can conclude that $E \propto J$.
Therefore,the electric field is directly proportional to the current density.

(C) The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$,where $E$ is the electric field.
Since $E = \frac{V}{L}$,we have $v_d = \frac{eV\tau}{mL}$.
Here,$e$ is the charge of an electron,$V$ is the potential difference,$\tau$ is the relaxation time,$m$ is the mass of the electron,and $L$ is the length of the wire.
Note that the drift velocity $v_d$ depends only on the potential difference $V$,the length $L$,and the material properties (relaxation time $\tau$).
It does not depend on the diameter $d$ or the cross-sectional area $A$ of the wire.
Therefore,if the diameter $d$ is doubled,the drift velocity $v_d$ remains unchanged.

(B) $1$. The current $I$ flowing through the conductor is constant throughout because it is a series circuit.
$2$. Current density $J = I/A$. Since the cross-sectional area $A_P > A_Q$, the current density $J_P < J_Q$.
$3$. Electric field $E = J/\sigma = I/(A\sigma)$. Since $A_P > A_Q$, $E_P < E_Q$. Thus, the electric field is higher at $Q$ than at $P$.
$4$. The rate of heat production per unit length is $dH/dx = I^2 R/dx = I^2 \rho / A$. Since $A_P > A_Q$, the rate of heat production is lower at $P$ than at $Q$.
$5$. Therefore, the correct statement is that the current density is lower at $P$ than at $Q$ (Option $B$).

(C) When a potential difference $V$ is applied across a metallic conductor,an electric field $E$ is established inside the conductor directed from the higher potential end to the lower potential end.
Since electrons are negatively charged,they experience an electric force $F = -eE$ in the direction opposite to the electric field.
Therefore,the free electrons experience a force directed from the lower potential end to the higher potential end.
Due to frequent collisions with the lattice ions,the electrons do not accelerate indefinitely but instead acquire a constant average velocity known as the drift velocity $(v_d)$ in the direction of the force (i.e.,from lower potential to higher potential).

(A) The drift velocity is given by the formula $\upsilon_d = \frac{e\tau V}{mL}$,where $e$ is the charge of an electron,$\tau$ is the relaxation time,$V$ is the potential difference,$m$ is the mass of the electron,and $L$ is the length of the conductor.
Since the conductors are identical and kept at the same temperature,the relaxation time $\tau$,mass $m$,and length $L$ are constant for both conductors.
Therefore,the drift velocity is directly proportional to the potential difference: $\upsilon_d \propto V$.
Thus,the ratio of the drift velocities is $\frac{\upsilon_{d1}}{\upsilon_{d2}} = \frac{V_1}{V_2}$.
Given the ratio of potential differences is $1 : 2$,the ratio of drift velocities is $1 : 2$.

(C) Given: Current $I = 1.344 \, A$,Area $A = 1 \, mm^2 = 1 \times 10^{-6} \, m^2$,and electron density $n = 8.4 \times 10^{22} \, cm^{-3} = 8.4 \times 10^{28} \, m^{-3}$.
The formula for drift velocity is $v_d = \frac{I}{neA}$.
Substituting the values:
$v_d = \frac{1.344}{(8.4 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-6})}$
$v_d = \frac{1.344}{8.4 \times 1.6 \times 10^{3}}$
$v_d = \frac{1.344}{13.44 \times 10^{3}} = 0.1 \times 10^{-3} \, m/s = 0.1 \, mm/s$.

(C) The relationship between current $I$,drift velocity $v_d$,and cross-sectional area $A$ is given by $I = neAv_d$.
Since $A = \pi r^2$,we have $I = ne(\pi r^2)v_d$,which implies $I \propto r^2 v_d$.
For the first wire: $I_1 = I$,$r_1 = r$,$v_{d1} = V$.
For the second wire: $r_2 = r/2$,$v_{d2} = 2V$.
Taking the ratio: $\frac{I_2}{I_1} = \frac{r_2^2 v_{d2}}{r_1^2 v_{d1}} = \frac{(r/2)^2 (2V)}{r^2 V} = \frac{(r^2/4) (2V)}{r^2 V} = \frac{1}{2}$.
Therefore,$I_2 = I_1 / 2 = I/2$.

(B) Current density $J$ is defined as $J = I / A$,where $I$ is the current and $A$ is the cross-sectional area.
Since the wires are connected in series,the current $I$ flowing through both wires is the same.
The cross-sectional area $A$ is given by $A = \pi r^2$.
For the copper wire $(1)$: $J_1 = I / (\pi r_1^2)$.
For the iron wire $(2)$: $J_2 = I / (\pi r_2^2)$.
The ratio of current densities is $J_1 / J_2 = (I / \pi r_1^2) / (I / \pi r_2^2) = r_2^2 / r_1^2$.
Given $r_1 = 1 \ mm$ and $r_2 = 3 \ mm$,we have:
$J_1 / J_2 = (3)^2 / (1)^2 = 9 / 1$.
Therefore,the ratio is $9 : 1$.

(C) Let $v_d$ be the drift velocity,$n$ be the number of free electrons per unit volume,$l$ be the length of the conductor,and $A$ be the cross-sectional area of the conductor.
The total number of free electrons in the conductor is given by $N = n(lA)$.
The total charge $q$ contained in the conductor is $q = Ne = n(lA)e$.
The time $t$ taken by the electrons to cross the length $l$ of the conductor with drift velocity $v_d$ is $t = l / v_d$.
The electric current $I$ is defined as the rate of flow of charge: $I = q / t$.
Substituting the values of $q$ and $t$,we get:
$I = \frac{n(lA)e}{l / v_d} = neAv_d$.
Thus,the relation is $I = neAv_d$.

(D) The radius at the left end is $a$ and at the right end is $b$. Thus,the increase in radius over the length $\ell$ is $(b - a)$.
The rate of increase of radius per unit length is $\frac{b - a}{\ell}$.
The increase in radius at a distance $x$ from the left end is $\left( \frac{b - a}{\ell} \right) x$.
Therefore,the radius $r$ at distance $x$ is $r = a + \left( \frac{b - a}{\ell} \right) x$.
The cross-sectional area at this point is $A = \pi r^2 = \pi \left[ a + \left( \frac{b - a}{\ell} \right) x \right]^2$.
Since the current $i$ is constant throughout the conductor,the current density $J$ is given by $J = \frac{i}{A} = \frac{i}{\pi \left[ a + \frac{x(b - a)}{\ell} \right]^2}$.

(C) The mass of $1 \ m^3$ of copper is $8990 \ kg = 8990 \times 10^3 \ g$.
The number of moles in $1 \ m^3$ is $n_{moles} = \frac{8990 \times 10^3}{63.5} \approx 1.415 \times 10^5 \ mol$.
Since each mole contains $N_A = 6.022 \times 10^{23}$ atoms and each atom contributes one free electron,the number density of free electrons $n$ is:
$n = (1.415 \times 10^5) \times (6.022 \times 10^{23}) \approx 8.52 \times 10^{28} \ m^{-3}$.
Using the formula for current $I = neAv_d$,the drift velocity $v_d$ is:
$v_d = \frac{I}{neA}$.
Given $I = 1.34 \ A$,$A = 1.0 \ mm^2 = 10^{-6} \ m^2$,and $e = 1.6 \times 10^{-19} \ C$:
$v_d = \frac{1.34}{(8.52 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (10^{-6})}$
$v_d \approx \frac{1.34}{1.3632 \times 10^4} \approx 0.98 \times 10^{-4} \ m/s$.
Converting to $mm/s$:
$v_d \approx 0.1 \ mm/s$.

(D) Given: Length $l = 10\,cm = 0.1\,m$,Area $A = 1 \times 10^{-4}\,m^2$,Voltage $V = 2\,V$,Mobility $\mu = 0.14\,m^2V^{-1}s^{-1}$,Electron density $n = 1.5 \times 10^{16}\,m^{-3}$,Charge of electron $e = 1.6 \times 10^{-19}\,C$.
The current $I$ is given by $I = neAv_d$,where $v_d$ is the drift velocity.
Since $v_d = \mu E$ and $E = V/l$,we have $v_d = \mu (V/l)$.
Substituting this into the current equation: $I = neA \left( \frac{\mu V}{l} \right)$.
$I = \frac{n e A \mu V}{l} = \frac{(1.5 \times 10^{16}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-4}) \times 0.14 \times 2}{0.1}$.
$I = \frac{1.5 \times 1.6 \times 0.14 \times 2 \times 10^{16-19-4}}{10^{-1}}$.
$I = \frac{0.672 \times 10^{-7}}{10^{-1}} = 0.672 \times 10^{-6} = 6.72 \times 10^{-7}\,A$.

(B) The formula for drift velocity is given by $v_d = \frac{I}{neA}$.
Given:
$I = 24 \, mA = 24 \times 10^{-3} \, A$
$n = 3 \times 10^{23} \, m^{-3}$
$e = 1.6 \times 10^{-19} \, C$
$A = 1 \, cm^2 = 10^{-4} \, m^2$
Substituting these values into the formula:
$v_d = \frac{24 \times 10^{-3}}{3 \times 10^{23} \times 1.6 \times 10^{-19} \times 10^{-4}}$
$v_d = \frac{24 \times 10^{-3}}{4.8 \times 10^{0}}$
$v_d = 5 \times 10^{-3} \, m/s$.

(C) The formula for electric current in terms of drift velocity is $i = neAv_d$,where $A = \pi r^2$ is the cross-sectional area.
Thus,$i = ne \pi r^2 v$.
For the new wire,the radius is $r' = r/2$ and the required drift velocity is $v' = 2v$.
The new current $i'$ is given by $i' = ne \pi (r')^2 v'$.
Substituting the values: $i' = ne \pi (r/2)^2 (2v)$.
$i' = ne \pi (r^2/4) (2v) = \frac{1}{2} (ne \pi r^2 v)$.
Since $i = ne \pi r^2 v$,we get $i' = i/2$.

(A) The relationship between current $i$ and drift velocity $v_d$ is given by $i = neA v_d$,where $n$ is the electron density,$e$ is the charge of an electron,and $A$ is the cross-sectional area.
Since the material is the same,$n$ is constant. Thus,$i \propto A v_d$.
Given $A = \pi r^2$,we have $i \propto r^2 v_d$,which implies $v_d \propto \frac{i}{r^2}$.
Therefore,the ratio of drift velocities is $\frac{v_{d1}}{v_{d2}} = \left( \frac{i_1}{i_2} \right) \times \left( \frac{r_2}{r_1} \right)^2$.
Substituting the given values: $\frac{i_1}{i_2} = \frac{4}{1}$ and $\frac{r_1}{r_2} = \frac{1}{2} \Rightarrow \frac{r_2}{r_1} = \frac{2}{1}$.
$\frac{v_{d1}}{v_{d2}} = \left( \frac{4}{1} \right) \times \left( \frac{2}{1} \right)^2 = 4 \times 4 = 16$.
Thus,the ratio is $16 : 1$.

(A) The current density $J$ is given by the formula $J = nqv$,where $n$ is the number density of charge carriers,$q$ is the charge of each carrier,and $v$ is the drift velocity.
Given:
Number density $n = 2 \times 10^8 \text{ ions/cm}^3 = 2 \times 10^8 \times 10^6 \text{ ions/m}^3 = 2 \times 10^{14} \text{ ions/m}^3$.
Charge of a doubly charged ion $q = ze = 2 \times 1.6 \times 10^{-19} \text{ C} = 3.2 \times 10^{-19} \text{ C}$.
Velocity $v = 10^5 \text{ m/s}$.
Substituting these values into the formula:
$J = (2 \times 10^{14}) \times (3.2 \times 10^{-19}) \times 10^5$
$J = 6.4 \times 10^{14-19+5} \text{ A/m}^2$
$J = 6.4 \times 10^0 = 6.4 \text{ A/m}^2$.

(D) The wire $AB$ is uniform,meaning its cross-sectional area $A$ is constant throughout its length.
According to the principle of continuity for steady currents,the current $I$ flowing through any cross-section of a series circuit is the same.
The current density $J$ is defined as the current per unit cross-sectional area,given by the formula $J = I/A$.
Since both the current $I$ and the cross-sectional area $A$ are constant for a uniform wire,the current density $J$ must also be constant at every point along the wire $AB$.
Therefore,the graph of $J$ versus the position along $AB$ will be a horizontal straight line.
This corresponds to the graph shown in option $D$.

(C) The current $I$ flowing through a conductor is given by the formula $I = n e A V_{d}$,where $n$ is the charge carrier density,$e$ is the elementary charge,$A$ is the cross-sectional area,and $V_{d}$ is the drift velocity.
Since the wires are connected in series,the current $I$ flowing through both wires must be the same.
Given that both wires have the same radius $r$,their cross-sectional areas $A$ are also equal.
Therefore,$n_{1} e A V_{d_{1}} = n_{2} e A V_{d_{2}}$.
This simplifies to $n_{1} V_{d_{1}} = n_{2} V_{d_{2}}$,which implies $\frac{V_{d_{1}}}{V_{d_{2}}} = \frac{n_{2}}{n_{1}}$.
Given the ratio of charge carrier densities is $\frac{n_{1}}{n_{2}} = \frac{1}{4}$,we have $\frac{n_{2}}{n_{1}} = \frac{4}{1}$.
Thus,the ratio of drift velocities is $\frac{V_{d_{1}}}{V_{d_{2}}} = 4 : 1$.