Current Density, Drift Velocity and Mobility Questions in English

(C) The current $I$ in a wire is given by the formula $I = neAv_d$,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 material is the same,$n$ and $e$ remain constant.
The cross-sectional area $A$ is given by $A = \pi r^2$.
For the first wire: $I_1 = ne(\pi r^2)v$.
For the second wire: $r_2 = r/2$ and $v_{d2} = 2v$.
$I_2 = ne(\pi (r/2)^2)(2v) = ne(\pi r^2 / 4)(2v) = (1/2) ne \pi r^2 v$.
Comparing the two,$I_2 = I_1 / 2 = I/2$.

(D) The current $I$ due to a charge carrier with charge $q$,number density $n$,cross-sectional area $A$,and drift velocity $v_d$ is given by $I = n q A v_d$.
For the positive ions $(+2e)$: The charge is $q_1 = +2e$,number density is $n$,and drift speed is $v_1 = v/4$. The current is $I_1 = n(2e)A(v/4) = nevA/2$.
For the negative ions $(-e)$: The charge is $q_2 = -e$,number density is $n$,and drift speed is $v_2 = v$. Since the negative ions move in the direction opposite to the positive ions,their contribution to the conventional current is in the same direction as the positive ions. The magnitude of current is $I_2 = n|q_2|Av_2 = n(e)Av = nevA$.
The total current $I_{total} = I_1 + I_2 = nevA/2 + nevA = 3nevA/2$.

(D) The total current $I$ is given by the integral $I = \int J \cdot dA$. For a cylindrical wire,the area element is $dA = 2\pi x dx$.
$I = \int_{0}^{R/2} J_0 \left( \frac{x}{R} - 1 \right) (2\pi x dx) + \int_{R/2}^{R} J_0 \frac{x}{R} (2\pi x dx)$
$I = 2\pi J_0 \left[ \int_{0}^{R/2} \left( \frac{x^2}{R} - x \right) dx + \int_{R/2}^{R} \frac{x^2}{R} dx \right]$
$I = 2\pi J_0 \left[ \left( \frac{x^3}{3R} - \frac{x^2}{2} \right)_{0}^{R/2} + \left( \frac{x^3}{3R} \right)_{R/2}^{R} \right]$
$I = 2\pi J_0 \left[ \left( \frac{R^3/8}{3R} - \frac{R^2/4}{2} \right) + \left( \frac{R^3}{3R} - \frac{R^3/8}{3R} \right) \right]$
$I = 2\pi J_0 \left[ \left( \frac{R^2}{24} - \frac{R^2}{8} \right) + \left( \frac{R^2}{3} - \frac{R^2}{24} \right) \right]$
$I = 2\pi J_0 \left[ \frac{R^2 - 3R^2}{24} + \frac{8R^2 - R^2}{24} \right] = 2\pi J_0 \left[ -\frac{2R^2}{24} + \frac{7R^2}{24} \right]$
$I = 2\pi J_0 \left( \frac{5R^2}{24} \right) = \frac{5}{12} \pi J_0 R^2$.

(D) The relationship between current $I$,cross-sectional area $A$,and drift velocity $v_d$ is given by $I = n e A v_d$.
Since the area $A = \pi (d/2)^2 = \frac{\pi d^2}{4}$,we can write $I = n e \left( \frac{\pi d^2}{4} \right) v_d$.
For the first wire: $I = n e \left( \frac{\pi d^2}{4} \right) V$.
For the second wire with diameter $d' = d/2$: $I = n e \left( \frac{\pi (d/2)^2}{4} \right) v' = n e \left( \frac{\pi d^2}{16} \right) v'$.
Equating the two expressions for $I$: $n e \left( \frac{\pi d^2}{4} \right) V = n e \left( \frac{\pi d^2}{16} \right) v'$.
Simplifying,we get $\frac{V}{4} = \frac{v'}{16}$,which implies $v' = 4V$.

(C) The current $I$ flowing through a conductor is given by the relation $I = neAv_d$, 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 speed.
Since the current $I$ is steady and $n$ and $e$ are constant, we have $A v_d = \text{constant}$, which implies $v_d \propto \frac{1}{A}$.
From the figure, the cross-sectional area at point $P$ $(A_P)$ is smaller than the cross-sectional area at point $Q$ $(A_Q)$.
Therefore, since $A_P < A_Q$, it follows that $v_P > v_Q$.

(B) The current $i$ flowing through a conductor is related to the drift velocity $v_d$ by the formula:
$i = n A q v_d$
Where:
$i$ = current in amperes
$n$ = number of charge carriers per unit volume (carrier density) = $p$
$A$ = cross-sectional area = $s$
$q$ = charge on each carrier
$v_d$ = drift velocity
Substituting the given variables into the formula:
$i = p \cdot s \cdot q \cdot v_d$
Rearranging to solve for the drift velocity $v_d$:
$v_d = \frac{i}{p s q}$
Thus,the drift velocity is $i/psq$.

(D) When a resistor is heated due to the flow of current, its temperature increases.
$1$. Resistance $(R)$ and resistivity $(\rho)$ are temperature-dependent properties of a conductor. As temperature increases, the relaxation time of electrons decreases, causing both $R$ and $\rho$ to change.
$2$. Drift speed $(v_d)$ is given by $v_d = \frac{eE\tau}{m}$, where $\tau$ is the relaxation time. Since $\tau$ changes with temperature, the drift speed also changes.
$3$. The number of free electrons $(n)$ is a property of the material itself (the number of valence electrons per unit volume). This value remains constant regardless of the temperature changes within the operating range of the resistor.
Therefore, the number of free electrons does not change. The correct option is $D$.

(D) When a steady current $I$ flows through a wire of non-uniform cross-section,the current $I$ remains constant at every cross-section of the wire.
$1$. The charge $q$ crossing a given cross-section in a time interval $t$ is given by $q = I \times t$. Since $I$ is constant,the charge crossing is independent of the cross-section.
$2$. The free-electron density $n$ is a property of the material of the wire and is independent of the geometry or cross-section of the wire.
$3$. The drift velocity $v_d$ is related to current by $I = n e A v_d$. Since $I$ and $n$ are constant,$v_d$ must change if the cross-sectional area $A$ changes.
Therefore,both the charge crossing in a given time interval and the free-electron density are independent of the cross-section.

(B) The current $I$ in a conductor is given by $I = n e A v_d$,where $n$ is the electron density,$e$ is the elementary charge,$A$ is the cross-sectional area,and $v_d$ is the drift speed.
From Ohm's law,$V = I R$,where $R = \rho \frac{l}{A}$.
Substituting $I$ and $R$ into the equation: $V = (n e A v_d) \times (\rho \frac{l}{A}) = n e v_d \rho l$.
Rearranging for resistivity $\rho$: $\rho = \frac{V}{n e v_d l}$.
Given values: $V = 5 \ V$,$l = 0.1 \ m$,$n = 8 \times 10^{28} \ m^{-3}$,$v_d = 2.5 \times 10^{-4} \ m/s$,and $e = 1.6 \times 10^{-19} \ C$.
Substituting these values: $\rho = \frac{5}{8 \times 10^{28} \times 1.6 \times 10^{-19} \times 2.5 \times 10^{-4} \times 0.1}$.
$\rho = \frac{5}{3.2 \times 10^5} = 1.5625 \times 10^{-5} \ \Omega m \approx 1.6 \times 10^{-5} \ \Omega m$.

(A) The current density $J$ is defined as $J = I/A$,where $I$ is the current and $A$ is the cross-sectional area.
Since the current $I$ flowing through the series conductor $PQ$ is constant,the current density $J$ is inversely proportional to the cross-sectional area $A$ $(J \propto 1/A)$.
From the graph,the current density is higher at $P$ and decreases towards $Q$.
This implies that the cross-sectional area $A$ must be smaller at $P$ and larger at $Q$.
Therefore,the strip is narrower at $P$ than at $Q$.

(B) The current $i$ flowing through the resistor is constant throughout its length because it is a series circuit.
The relationship between current $i$, current density $J$, and drift velocity $V_d$ is given by:
$i = J A$
$J = n e V_d$
Substituting $J$ into the current equation:
$i = (n e V_d) A$
Since $i$, $n$, and $e$ are constants, we have:
$A V_d = \text{constant}$
$V_d \propto \frac{1}{A}$
Since the area $A = \pi r^2$, where $r$ is the radius, we have $V_d \propto \frac{1}{r^2}$.
In the middle part of the cylinder, the radius $r$ is larger, so the area $A$ is larger.
Therefore, the drift velocity $V_d$ must be smaller in the middle part of the cylinder compared to the ends.
This corresponds to the graph where the value of $V_d$ decreases in the region with the larger radius. Graph $B$ shows this behavior.

(A) The resistance $R$ of a wire is given by the formula $R = \frac{\rho l}{A} = \frac{m l}{n e^2 \tau A}$,where $m$ is the mass of an electron,$l$ is the length,$A$ is the cross-sectional area,$e$ is the charge of an electron,$n$ is the free electron density,and $\tau$ is the relaxation time.
Given that $l$,$A$,$m$,$e$,and $\tau$ are constant,the resistance $R$ is inversely proportional to the free electron density $n$,i.e.,$R \propto \frac{1}{n}$.
It is given that $n \propto T$,therefore $R \propto \frac{1}{T}$.
This relationship $R = \frac{k}{T}$ (where $k$ is a constant) represents a rectangular hyperbola,which corresponds to the graph shown in option $A$.

$(A)$ The drift velocity $v_d$ is given by the formula $v_d = \frac{eE\tau}{m}$, where $E = \frac{V}{L}$.
Thus, $v_d = \frac{eV\tau}{mL}$.
Since all rods are made of copper and are at the same temperature, the relaxation time $\tau$ and the mass of electron $m$ are constant.
Therefore, $v_d \propto \frac{V}{L}$.
For rod $(A)$: $v_A \propto \frac{V}{L} = 1 \cdot (V/L)$.
For rod $(B)$: $v_B \propto \frac{2V}{2L} = 1 \cdot (V/L)$.
For rod $(C)$: $v_C \propto \frac{2V}{3L} = 0.67 \cdot (V/L)$.
Comparing the values, we get $v_A = v_B > v_C$.

(B) The current $I$ in a conductor is given by $I = neAv_d$,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.
The total number of electrons $N$ in a conductor of length $l$ is $N = nAl$.
The total momentum $P$ of the electrons is $P = Nmv_d$,where $m$ is the mass of an electron.
Substituting $N = nAl$,we get $P = (nAl)mv_d$.
From the current formula,$nAv_d = \frac{I}{e}$.
Substituting this into the momentum equation: $P = l \cdot (nAv_d) \cdot m = l \cdot \frac{I}{e} \cdot m$.
Since the specific charge $S$ is defined as the ratio of charge to mass,$S = \frac{e}{m}$,therefore $\frac{m}{e} = \frac{1}{S}$.
Thus,the total momentum $P = \frac{Il}{S}$.

(A) The current $I$ flowing through a conductor is given by the formula $I = n_e e A v_d$,where $n_e$ is the number of free electrons per unit volume,$e$ is the charge of an electron,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.
Since the two rods are connected in series,the same current $I$ flows through both of them.
For the left rod: $I = (2n) e A v_L$
For the right rod: $I = (n) e (2A) v_R$
Equating the two expressions for current:
$(2n) e A v_L = (n) e (2A) v_R$
$2 n e A v_L = 2 n e A v_R$
Dividing both sides by $2 n e A$,we get:
$v_L = v_R$
Therefore,the ratio $\frac{v_L}{v_R} = 1$.

(B) The current $I$ passing through a cross-section is given by the integral of current density $J$ over the area $A$: $I = \int J \cdot dA$.
For a cylindrical shell of radius $r$ and thickness $dr$,the elemental area is $dA = 2\pi r \cdot dr$.
Substituting the given current density $J = \frac{J_0}{r^2}$ and the elemental area $dA$ into the integral,we get:
$I = \int_{a}^{2a} \left( \frac{J_0}{r^2} \right) (2\pi r \cdot dr)$
$I = 2\pi J_0 \int_{a}^{2a} \frac{1}{r} \cdot dr$
$I = 2\pi J_0 [\ln r]_{a}^{2a}$
$I = 2\pi J_0 (\ln(2a) - \ln(a))$
$I = 2\pi J_0 \ln\left( \frac{2a}{a} \right)$
$I = 2\pi J_0 \ln 2$.

(C) The formula for current is $I = n e A v_d$,where $I$ is the current,$n$ is the number density,$e$ is the elementary charge,$A$ is the cross-sectional area,and $v_d$ is the drift velocity.
Given: $I = 1.7\, A$,$d = 1.02\, mm \implies r = 0.51 \times 10^{-3}\, m$,$n = 8.5 \times 10^{28}\, m^{-3}$,$e = 1.6 \times 10^{-19}\, C$.
The area $A = \pi r^2 = \pi \times (0.51 \times 10^{-3})^2 \approx 8.17 \times 10^{-7}\, m^2$.
Rearranging for $v_d$: $v_d = \frac{I}{n e A}$.
Substituting the values: $v_d = \frac{1.7}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (\pi \times (0.51 \times 10^{-3})^2)}$.
$v_d = \frac{1.7}{8.5 \times 1.6 \times \pi \times 0.2601 \times 10^{-7}} \approx \frac{1.7}{11.12} \times 10^{-3} \approx 0.15 \times 10^{-3}\, m/s$.
Thus,$v_d = 0.15\, mm/s$.

(B) In conductors,as the temperature increases,the thermal velocity of free electrons increases.
This leads to more frequent collisions between electrons and the lattice ions.
Consequently,the average time between two successive collisions,known as the relaxation time $(\tau)$,decreases.
Since resistivity $(\rho)$ is inversely proportional to the relaxation time $(\rho = \frac{m}{ne^2\tau})$,an increase in temperature causes the relaxation time to decrease.

(A) The current $dI$ through a thin cylindrical shell of radius $r$ and thickness $dr$ is given by $dI = J \cdot dA$,where $dA = 2 \pi r dr$.
Substituting the given current density $J = J_0 \frac{r}{R}$:
$dI = (J_0 \frac{r}{R}) \cdot (2 \pi r dr) = \frac{2 \pi J_0}{R} r^2 dr$.
To find the total current $I$,we integrate from $r = 0$ to $r = R$:
$I = \int_{0}^{R} \frac{2 \pi J_0}{R} r^2 dr = \frac{2 \pi J_0}{R} \int_{0}^{R} r^2 dr$.
$I = \frac{2 \pi J_0}{R} \left[ \frac{r^3}{3} \right]_{0}^{R} = \frac{2 \pi J_0}{R} \cdot \frac{R^3}{3} = \frac{2}{3} J_0 (\pi R^2)$.
Since the cross-sectional area $A = \pi R^2$,the total current is $I = \frac{2}{3} J_0 A$.

(A) The drift velocity $V_{d}$ of electrons in a conductor is given by the formula: $V_{d} = \frac{eE\tau}{m}$.
Since the electric field $E$ is related to the potential difference $V$ across a wire of length $L$ by $E = \frac{V}{L}$,we can substitute this into the equation:
$V_{d} = \frac{e}{m} \left( \frac{V}{L} \right) \tau$.
From this expression,it is clear that $V_{d} \propto V$.
Therefore,if the potential difference $V$ is doubled,the drift velocity $V_{d}$ will also be doubled.

(A) The formula for drift velocity is $I = neAv_d$, where $I$ is the current, $n$ is the number density of electrons, $e$ is the elementary charge, $A$ is the cross-sectional area, and $v_d$ is the drift velocity.
Given values:
$I = 0.2 \, A$
$n = 0.8 \times 10^{23} \, \text{electrons}/cm^3 = 0.8 \times 10^{29} \, \text{electrons}/m^3$
$e = 1.6 \times 10^{-19} \, C$
$A = 0.01 \, cm^2 = 10^{-6} \, m^2$
Rearranging for $v_d$:
$v_d = \frac{I}{neA}$
$v_d = \frac{0.2}{(0.8 \times 10^{29}) \times (1.6 \times 10^{-19}) \times (10^{-6})}$
$v_d = \frac{0.2}{1.28 \times 10^4}$
$v_d = 0.156 \times 10^{-4} \, m/s = 1.56 \times 10^{-5} \, m/s$.

(B) For a steady current flowing through a conductor,the current $I$ remains constant throughout the wire regardless of the cross-sectional area,due to the principle of conservation of charge.
The current density $J$ is defined as $J = I / A$,where $A$ is the cross-sectional area.
Since the wire has a non-uniform cross-section,$A$ varies along the length of the wire.
Therefore,because $I$ is constant and $A$ varies,the current density $J$ must also vary along the wire.
Thus,only the current is constant.

(A) The relationship between current $I$ and drift velocity $v_d$ is given by the formula: $I = n e A v_d$.
Here,$I = 1.5 \, A$,$n = 9 \times 10^{28} \, m^{-3}$,$e = 1.6 \times 10^{-19} \, C$,and $A = 5 \, mm^2 = 5 \times 10^{-6} \, m^2$.
Substituting the values into the formula:
$1.5 = (9 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (5 \times 10^{-6}) \times v_d$.
$1.5 = (9 \times 1.6 \times 5) \times 10^{28-19-6} \times v_d$.
$1.5 = 72 \times 10^3 \times v_d$.
$v_d = \frac{1.5}{72 \times 10^3} = \frac{1.5}{72} \times 10^{-3} \, m/s$.
$v_d \approx 0.0208 \times 10^{-3} \, m/s = 0.0208 \, mm/s$.
Thus,the value of $v$ is close to $0.02 \, mm/s$.

(D) The resistivity $\rho$ of a conductor is given by the formula: $\rho = \frac{m_e}{n e^2 \tau}$.
Given values:
$n = 8.5 \times 10^{28} \ m^{-3}$
$\tau = 25 \ fs = 25 \times 10^{-15} \ s$
$m_e = 9.1 \times 10^{-31} \ kg$
$e = 1.6 \times 10^{-19} \ C$
Substituting these values into the formula:
$\rho = \frac{9.1 \times 10^{-31}}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19})^2 \times (25 \times 10^{-15})}$
$\rho = \frac{9.1 \times 10^{-31}}{8.5 \times 10^{28} \times 2.56 \times 10^{-38} \times 25 \times 10^{-15}}$
$\rho = \frac{9.1 \times 10^{-31}}{544 \times 10^{-25}}$
$\rho \approx 0.0167 \times 10^{-6} \ \Omega m \approx 1.67 \times 10^{-8} \ \Omega m$.
Comparing this with the given options,the approximate value is $10^{-8} \ \Omega m$.

(B) The mobility $\mu$ is defined as the ratio of drift velocity $V_d$ to the electric field $E$: $\mu = \frac{V_d}{E}$.
From Ohm's law in vector form,$E = \rho J$,where $\rho$ is resistivity and $J$ is current density.
Current density $J = \frac{I}{A} = \frac{I}{\pi r^2}$.
Given: $I = 5\, A$,$\rho = 1.7 \times 10^{-8}\, \Omega \, m$,$r = 5\, mm = 5 \times 10^{-3}\, m$,and $V_d = 1.1 \times 10^{-3}\, m/s$.
First,calculate the electric field $E$:
$E = \rho \times \frac{I}{\pi r^2} = 1.7 \times 10^{-8} \times \frac{5}{\pi \times (5 \times 10^{-3})^2} = 1.7 \times 10^{-8} \times \frac{5}{\pi \times 25 \times 10^{-6}} = \frac{1.7 \times 10^{-8} \times 5}{78.54 \times 10^{-6}} \approx 1.08 \times 10^{-3}\, V/m$.
Now,calculate mobility $\mu$:
$\mu = \frac{1.1 \times 10^{-3}}{1.08 \times 10^{-3}} \approx 1.01\, m^2/Vs$.
Rounding to the nearest given option,the correct value is $1.0\, m^2/Vs$.

(C) The current $I$ flowing through any cross-section of a conductor in a steady state is constant,so $I$ is the same at all points.
Using the relation $I = neA V_d$,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 speed.
Since $I$,$n$,and $e$ are constant,we have $V_d \propto \frac{1}{A}$.
As the cross-sectional area $A$ decreases along the wire from $A$ to $B$,the drift speed $V_d$ must increase.
Therefore,the drift speed at point $A$ is less than the drift speed at point $B$,i.e.,$(V_d)_A < (V_d)_B$.

(A) Ohm's law states that the current $I$ flowing through a conductor is directly proportional to the potential difference $V$ applied across its ends,provided the physical conditions remain constant. This implies that the current density $J$ is directly proportional to the electric field $E$ $(J = \sigma E)$.
Since the current density is given by $J = n e v_d$,where $n$ is the number density of electrons,$e$ is the charge of an electron,and $v_d$ is the drift velocity,we have $n e v_d = \sigma E$.
This implies $v_d \propto E$.
Therefore,Ohm's law is obeyed when the drift velocity is directly proportional to the electric field.

(C) The relationship between current $I$ and drift velocity $v_d$ is given by $I = neAv_d$,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,$n$ and $e$ remain constant. Thus,$I \propto A v_d$.
The cross-sectional area $A$ is given by $A = \pi (d/2)^2 = \pi d^2 / 4$,so $A \propto d^2$.
Therefore,$I \propto d^2 v_d$.
Since the current $I$ is the same in both cases,we have $d_1^2 v_{d1} = d_2^2 v_{d2}$.
Given $d_1 = d$,$v_{d1} = v_d$,and $d_2 = d/2$,we substitute these values:
$d^2 v_d = (d/2)^2 v_{d2}$
$d^2 v_d = (d^2 / 4) v_{d2}$
$v_{d2} = 4v_d$.

(A) The current $i$ is given by $i = neAv_d$,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.
The total number of electrons $N$ in a wire of length $L$ is $N = nAL$.
Thus,$nA = N/L$.
Substituting this into the current equation: $i = (N/L)ev_d$,which implies $Nv_d = iL/e$.
The total momentum $P$ of all electrons is $P = Nmv_d$,where $m$ is the mass of an electron $(m \approx 9.1 \times 10^{-31}\, kg)$.
Substituting $Nv_d = iL/e$ into the momentum equation: $P = m(iL/e) = (m/e) \times iL$.
Given $m = 9.1 \times 10^{-31}\, kg$,$e = 1.6 \times 10^{-19}\, C$,$i = 16\, A$,and $L = 1\, m$:
$P = (9.1 \times 10^{-31} / 1.6 \times 10^{-19}) \times 16 \times 1$
$P = (9.1 / 1.6) \times 10^{-12} \times 16$
$P = 9.1 \times 10^{-12} \times 10 = 91 \times 10^{-12}\, kg\, m/s$.

(A) The relation between current $I$ and drift velocity $v_d$ is given by the formula:
$I = n e A v_d$
Where:
$I = 10\,A$ (current)
$n = 9 \times 10^{28}\,m^{-3}$ (number density of electrons)
$e = 1.6 \times 10^{-19}\,C$ (charge of an electron)
$A = 1\,cm^2 = 1 \times 10^{-4}\,m^2$ (cross-sectional area)
Rearranging for drift velocity $v_d$:
$v_d = \frac{I}{n e A}$
Substituting the values:
$v_d = \frac{10}{(9 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-4})}$
$v_d = \frac{10}{9 \times 1.6 \times 10^{28-19-4}}$
$v_d = \frac{10}{14.4 \times 10^5}$
$v_d = \frac{10}{1.44 \times 10^6} \approx 6.94 \times 10^{-6}\,m/s$

(A) The current $I$ in a conductor is given by $I = n e A v_d$,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.
The current density $J$ is defined as $J = \frac{I}{A} = n e v_d$.
According to Ohm's law in microscopic form,the current density $J$ is related to the electric field $E$ by the relation $J = \sigma E$,where $\sigma$ is the electrical conductivity.
Equating the two expressions for $J$,we get $n e v_d = \sigma E$.
Since $n$,$e$,and $\sigma$ are constants for a given material at a constant temperature,it follows that $v_d \propto E$.

(C) The drift velocity $(v_d)$ of electrons in a typical metallic conductor carrying a steady current is approximately $10^{-4} \ m/s$.
This value is extremely small compared to the random thermal velocity of electrons,which is of the order of $10^5 \ m/s$ at room temperature.
Therefore,the correct order of magnitude for drift velocity is $10^{-4} \ m/s$.

(B) Given: Current $I = 1.1\, A$,Diameter $d = 1\, mm = 10^{-3}\, m$,Density $\rho = 9 \times 10^{3}\, kg/m^3$,Molar mass $M = 63 \times 10^{-3}\, kg/mol$.
The number of atoms per unit volume $n$ is given by $n = \frac{\rho N_A}{M}$.
Substituting the values: $n = \frac{9 \times 10^{3} \times 6.023 \times 10^{23}}{63 \times 10^{-3}} \approx 8.6 \times 10^{28}\, m^{-3}$.
Since each atom releases one free electron,the electron density $n_e = n = 8.6 \times 10^{28}\, m^{-3}$.
The cross-sectional area $A = \frac{\pi d^2}{4} = \frac{3.14 \times (10^{-3})^2}{4} = 7.85 \times 10^{-7}\, m^2$.
The drift velocity $v_d$ is given by $v_d = \frac{I}{n_e A e}$.
Substituting the values: $v_d = \frac{1.1}{(8.6 \times 10^{28}) \times (7.85 \times 10^{-7}) \times (1.6 \times 10^{-19})}$.
$v_d = \frac{1.1}{1.08 \times 10^{4}} \approx 1.018 \times 10^{-4}\, m/s = 0.1018\, mm/s \approx 0.1\, mm/s$.

(A) The relationship between current $i$ and drift velocity $V_d$ is given by the formula: $i = neAV_d$,where $n$ is the electron density,$e$ is the elementary charge $(1.6 \times 10^{-19}\, C)$,and $A$ is the cross-sectional area.
Given:
$i = 1.6\, A$
$n = 10^{29}\, m^{-3}$
$A = 1\, mm^2 = 10^{-6}\, m^2$
$e = 1.6 \times 10^{-19}\, C$
Rearranging for $V_d$:
$V_d = \frac{i}{neA}$
Substituting the values:
$V_d = \frac{1.6}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-6}}$
$V_d = \frac{1.6}{1.6 \times 10^{29-19-6}}$
$V_d = \frac{1}{10^4} = 10^{-4}\, m/s$.

(C) The current $I$ in a conductor is given by the relation $I = neAv_d$,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 $A = \pi r^2$,the expression becomes $I = ne(\pi r^2)v_d$.
Given that the current $I$ remains constant,we have $v_d \propto \frac{1}{r^2}$.
If the radius is doubled $(r' = 2r)$,the new drift velocity $v_d'$ is:
$v_d' = v_d \times \left(\frac{r}{r'}\right)^2 = v_d \times \left(\frac{r}{2r}\right)^2 = \frac{v_d}{4}$.
Thus,the new drift velocity is $\vec v/4$.

(C) In a conductor,there are a large number of free electrons. When we close the circuit,an electric field is established instantly throughout the conductor at the speed of electromagnetic waves (approximately $3 \times 10^8 \ m/s$).
This electric field exerts a force on all free electrons simultaneously,causing them to drift.
Consequently,the current is established in the entire circuit instantly.
The current does not depend on the time taken for an individual electron to travel from one end of the conductor to the other.
The drift velocity of electrons is actually very small (typically in the order of $10^{-4} \ m/s$).
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) From the microscopic form of Ohm's law,$\vec J = \sigma \vec E$,where $\sigma$ is the conductivity. Since $\sigma$ is a positive scalar,the current density $\vec J$ is always in the direction of the electric field $\vec E$. Thus,the Assertion is correct.
Regarding the Reason,a point charge released from rest in an electrostatic field moves along the electric line of force only if the electric line of force is a straight line. If the electric lines of force are curved,the charge will not follow the path of the line of force because the velocity vector and the force vector (which is tangent to the line of force) will not remain collinear. Thus,the Reason is incorrect.

(A) In the absence of an electric current,the free electrons in a conductor are in a state of random motion,similar to molecules in a gas.
Their average velocity is zero,meaning they do not have any net velocity in a specific direction.
As a result,there is no net magnetic force on the free electrons in a magnetic field.
When a current is passed,the free electrons acquire a drift velocity in a definite direction,and consequently,a magnetic force acts on them (provided the magnetic field has a component perpendicular to the direction of flow).

$(A)$ The drift speed $v_{d}$ is given by the formula $v_{d} = \frac{I}{neA}$.
Given: $I = 1.5 \; A$, $A = 1.0 \times 10^{-7} \; m^{2}$, $e = 1.6 \times 10^{-19} \; C$.
The number density $n$ is calculated as:
$n = \frac{\text{Density} \times N_{A}}{\text{Atomic Mass}} = \frac{9.0 \times 10^{3} \; kg/m^{3} \times 6.022 \times 10^{23} \; atoms/mol}{63.5 \times 10^{-3} \; kg/mol} \approx 8.5 \times 10^{28} \; m^{-3}$.
Substituting these values:
$v_{d} = \frac{1.5}{8.5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.0 \times 10^{-7}} \approx 1.1 \times 10^{-3} \; m/s = 1.1 \; mm/s$.
$(b) (i)$ Thermal speed of copper atoms at $300 \; K$ is approximately $2 \times 10^{2} \; m/s$. The drift speed is about $10^{-5}$ times smaller than the thermal speed.
$(ii)$ The electric field propagates at the speed of light, $c = 3.0 \times 10^{8} \; m/s$. The drift speed is smaller by a factor of $10^{-11}$ compared to this speed.

(N/A) The electric field is established throughout the circuit almost instantly (at the speed of light),causing a local electron drift at every point. The establishment of a current does not require electrons to travel from one end of the conductor to the other. However,it takes a short time for the current to reach its steady value.
$(b)$ Each 'free' electron accelerates,increasing its drift speed until it collides with a positive ion of the metal. It loses its drift speed after the collision but starts to accelerate again,only to suffer another collision. On average,therefore,electrons acquire a steady drift speed.
$(c)$ This is possible because the electron number density is enormous,approximately $10^{29}\; m^{-3}$.
$(d)$ No. The drift velocity is superposed over the large random thermal velocities of the electrons.
$(e)$ In the absence of an electric field,the paths are straight lines. In the presence of an electric field,the paths are,in general,curved due to the acceleration provided by the field between collisions.

(C) Given:
Number density of free electrons,$n = 8.5 \times 10^{28} \; m^{-3}$
Length of the wire,$l = 3.0 \; m$
Area of cross-section,$A = 2.0 \times 10^{-6} \; m^2$
Current,$I = 3.0 \; A$
Charge of an electron,$e = 1.6 \times 10^{-19} \; C$
The relation for current is $I = nAe v_d$,where $v_d$ is the drift velocity.
Since $v_d = \frac{l}{t}$,we have $I = nAe \frac{l}{t}$.
Rearranging for time $t$,we get $t = \frac{nAe l}{I}$.
Substituting the values:
$t = \frac{(8.5 \times 10^{28}) \times (2.0 \times 10^{-6}) \times (1.6 \times 10^{-19}) \times 3.0}{3.0}$
$t = 8.5 \times 2.0 \times 1.6 \times 10^{(28 - 6 - 19)}$
$t = 27.2 \times 10^3 = 2.72 \times 10^4 \; s$.
Rounding to two significant figures,$t = 2.7 \times 10^4 \; s$.

(N/A) In solid conductors,the absence of electric current can be explained as follows:
$1$. When metallic solids are formed,valence electrons separate from their parent atoms and become free electrons,while positive ions remain fixed in a definite three-dimensional geometric arrangement.
$2$. In the absence of an external electric field,these positive ions oscillate about their mean positions due to thermal energy.
$3$. The free electrons move randomly in the space between these positive ions.
$4$. During their motion,electrons frequently collide with the positive ions,causing their directions of motion to change constantly.
$5$. Because the motion is random,at any given time,the net number of electrons crossing any given cross-section in one direction is equal to the number of electrons crossing in the opposite direction. Thus,the net flow of charge is zero,and no current is formed.

(N/A) Relaxation time,denoted by $\tau$,is defined as the average time interval between two successive collisions of free electrons in a conductor.
When an electric field is applied to a conductor,the free electrons undergo random motion and collide with the positive ions of the lattice.
The time elapsed between two such consecutive collisions is known as the relaxation time.
It is typically of the order of $10^{-14} \ s$.

(A) $1$. Conductors: Materials that allow electric charge to flow through them easily are called conductors. This is because they contain a large number of free electrons that can move freely throughout the material. Examples include metals like copper,silver,and aluminum.
$2$. Non-conductors (Insulators): Materials that do not allow electric charge to flow through them easily are called non-conductors or insulators. This is because they have very few or no free electrons,and the electrons are tightly bound to their respective atoms. Examples include glass,plastic,rubber,and wood.
$3$. Comparison: Conductors have a significantly higher number of free electrons compared to non-conductors. Therefore,conductors possess more free electrons.

(N/A) Current density: The electric current density at any point is defined as the amount of electric current flowing per unit cross-sectional area perpendicular to the current at that point. Current density is a vector quantity.
$\overrightarrow{J} = \frac{I}{A} \hat{n}$
$\text{Unit} = A/m^2 = A \cdot m^{-2}$
$\text{Dimensional formula} = [M^0 L^{-2} T^0 A^1]$
Derivation of Ohm's law in vector form:
Consider a conductor of length $l$ and cross-sectional area $A$. Let $E$ be the electric field applied across it. The potential difference $V$ is given by $V = E \cdot l$.
From Ohm's law,$V = I \cdot R$.
We know that resistance $R = \rho \cdot \frac{l}{A}$,where $\rho$ is the resistivity.
Substituting $V$ and $R$ in Ohm's law:
$E \cdot l = I \cdot \left( \frac{\rho \cdot l}{A} \right)$
$E = \left( \frac{I}{A} \right) \cdot \rho$
Since current density $J = I/A$,we have:
$E = J \cdot \rho$
Using conductivity $\sigma = 1/\rho$,we get:
$E = J / \sigma$
$J = \sigma \cdot E$
In vector form,this is written as $\overrightarrow{J} = \sigma \overrightarrow{E}$,which is the vector form of Ohm's law.

(N/A) Ohm's law in its microscopic or vector form relates the current density vector $\vec{J}$ to the electric field vector $\vec{E}$.
According to Ohm's law,the current density $\vec{J}$ at a point in a conductor is directly proportional to the electric field $\vec{E}$ at that point.
Mathematically,this is expressed as:
$\vec{J} = \sigma \vec{E}$
Where:
$\vec{J}$ is the current density vector (measured in $A/m^2$),
$\sigma$ is the electrical conductivity of the material (measured in $\Omega^{-1} m^{-1}$ or $S/m$),
$\vec{E}$ is the electric field vector (measured in $V/m$).
Alternatively,since $\sigma = 1/\rho$,where $\rho$ is the resistivity,the equation can also be written as $\vec{E} = \rho \vec{J}$.

(N/A) $1$. Drift of Electrons: In a conductor,free electrons move randomly due to thermal energy and collide with positive ions. Their average velocity is zero. When an external electric field $E$ is applied,these electrons experience a force $F = -eE$,causing them to drift slowly in the direction opposite to the electric field. This net slow motion is called drift.
$2$. Drift Velocity $(v_d)$: It is defined as the average velocity acquired by free electrons in a conductor under the influence of an external electric field. If $\tau$ is the average relaxation time,then $v_d = -\frac{eE\tau}{m}$.
$3$. Derivation of Current $(I)$: Consider a conductor of cross-sectional area $A$ and length $L$. Let $n$ be the number density of free electrons. The total number of electrons in a volume $A \Delta x$ is $n A \Delta x$. Since $\Delta x = v_d \Delta t$,the total charge $\Delta Q$ passing through the cross-section in time $\Delta t$ is $\Delta Q = (n A v_d \Delta t) e$. The current $I$ is given by $I = \frac{\Delta Q}{\Delta t} = n e A v_d$.

(N/A) Consider a conductor of cross-sectional area $A$ in which an electric field $\vec{E}$ is applied. Due to this field,free electrons experience a force and drift in the opposite direction of $\vec{E}$ with a drift velocity $v_d$.
In a small time interval $\Delta t$,the distance covered by the electrons is $\Delta x = |v_d| \Delta t$.
The volume of the cylindrical element containing these electrons is $V = A \Delta x = A |v_d| \Delta t$.
If $n$ is the number density of free electrons (number of electrons per unit volume),then the total number of electrons $N$ in this volume is $N = n V = n A |v_d| \Delta t$.
The total charge $\Delta Q$ passing through the cross-section in time $\Delta t$ is $\Delta Q = N e = n A e |v_d| \Delta t$,where $e$ is the magnitude of the charge of an electron.
The current $I$ is given by $I = \frac{\Delta Q}{\Delta t} = n A e |v_d|$.
Since current density $J = \frac{I}{A}$,we have $J = \frac{n A e |v_d|}{A} = n e |v_d|$.
In vector form,since electrons drift opposite to the electric field,$\vec{J} = -n e \vec{v}_d$.

(N/A) Mobility $(\mu)$ is defined as the magnitude of the drift velocity $(v_d)$ per unit electric field $(E)$ applied across the conductor.
Mathematically, $\mu = \frac{|v_d|}{E}$.
Since the drift velocity is given by $v_d = \frac{eE\tau}{m}$, where $e$ is the charge, $\tau$ is the relaxation time, and $m$ is the mass of the carrier, substituting this into the mobility equation gives $\mu = \frac{e\tau}{m}$.
$SI$ unit: $\frac{m/s}{V/m} = m^2 V^{-1} s^{-1}$.
Dimensional formula: $[M^{-1} L^0 T^2 A^1]$.