Electric Current Questions in English

(A) The unit $Ampere-hour$ $(Ah)$ is derived from the formula for electric charge.
Charge $(Q)$ is equal to the product of current $(I)$ and time $(t)$,expressed as $Q = I \times t$.
Since the unit of current is $Ampere$ $(A)$ and the unit of time is $hour$ $(h)$,the product $Ampere-hour$ represents the total quantity of electricity or electric charge.
Therefore,the correct option is $A$.

(D) We know that the capacitance $C$ of a parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
Since $A$ has dimensions of $L^2$ and $d$ has dimensions of $L$,the dimensions of $\varepsilon_0 L$ are equivalent to the dimensions of capacitance $C$.
Substituting this into the expression for $X$:
$X = \frac{(\varepsilon_0 L) V}{t} = \frac{C V}{t}$.
Since $Q = CV$,we have $X = \frac{Q}{t}$.
The rate of flow of charge $Q$ with respect to time $t$ is defined as current $I$.
Therefore,the dimensions of $X$ are the same as those of current.

(A) The current $I$ is defined as the rate of flow of charge,given by $I = \frac{q}{t}$.
Since the total charge $q$ is the product of the number of electrons $n$ and the elementary charge $e$ $(q = ne)$,we can write $I = \frac{ne}{t}$.
To find the number of electrons per second,we rearrange the formula to solve for $\frac{n}{t}$:
$\frac{n}{t} = \frac{I}{e}$.
Given $I = 4.8\,A$ and $e = 1.6 \times 10^{-19}\,C$,we substitute these values:
$\frac{n}{t} = \frac{4.8}{1.6 \times 10^{-19}} = 3 \times 10^{19}$.
Therefore,the number of electrons flowing per second is $3 \times 10^{19}$.

(C) The electric current $I$ is defined as the rate of flow of charge through a conductor.
It is given by the formula: $I = \frac{q}{t}$
Given:
Charge $q = 4 \, C$
Time $t = 2 \, s$
Substituting the values into the formula:
$I = \frac{4 \, C}{2 \, s} = 2 \, A$
Therefore,the electric current is $2 \, A$.

(C) The current $I$ is defined as the rate of flow of charge,given by the formula $I = \frac{q}{t}$.
Since $q = ne$,where $n$ is the number of electrons and $e$ is the elementary charge $(1.6 \times 10^{-19} \, C)$,we have $I = \frac{ne}{t}$.
Given $n/t = 62.5 \times 10^{18} \, s^{-1}$ and $e = 1.6 \times 10^{-19} \, C$:
$I = (62.5 \times 10^{18}) \times (1.6 \times 10^{-19}) \, A$
$I = 62.5 \times 1.6 \times 10^{-1} \, A$
$I = 100 \times 10^{-1} \, A = 10 \, A$.
Therefore,the current flowing through the wire is $10 \, A$.

(B) The current due to positive ions moving to the right is $i_{(+)} = \frac{n_{(+)} q_{(+)}}{t} = (3.2 \times 10^{18}) \times (2e) / 1 \, s = 6.4 \times 10^{18} \times 1.6 \times 10^{-19} = 1.024 \, A$ (to the right).
The current due to negative ions moving to the left is equivalent to a current of positive charge moving to the right. Thus,$i_{(-)} = \frac{n_{(-)} q_{(-)}}{t} = (3.6 \times 10^{18}) \times (e) / 1 \, s = 3.6 \times 10^{18} \times 1.6 \times 10^{-19} = 0.576 \, A$ (to the right).
The net current is $i_{net} = i_{(+)} + i_{(-)} = 1.024 \, A + 0.576 \, A = 1.6 \, A$ (to the right).

(D) The relationship between charge $(Q)$,current $(I)$,and time $(t)$ is given by the formula: $Q = I \times t$.
Given:
Current $(I)$ = $5 \, A$
Time $(t)$ = $1 \text{ minute} = 60 \, s$
Substituting the values into the formula:
$Q = 5 \, A \times 60 \, s = 300 \, C$.
Therefore,the charge flowing through the conductor is $300 \, C$.

(B) The current $i$ is defined as the rate of flow of charge. When both positive and negative charges move in opposite directions,their currents add up in the direction of the positive charge flow.
Let $n_+$ be the number of positive ions moving to the right per second and $n_-$ be the number of electrons moving to the left per second.
The current due to positive ions is $i_+ = \frac{n_+ e}{t} = 2.9 \times 10^{18} \times 1.6 \times 10^{-19} \, A = 0.464 \, A$ (towards the right).
The current due to electrons is $i_- = \frac{n_- e}{t} = 1.2 \times 10^{18} \times 1.6 \times 10^{-19} \, A = 0.192 \, A$ (towards the right,because electrons moving to the left constitute a current to the right).
The total current $i = i_+ + i_- = 0.464 \, A + 0.192 \, A = 0.656 \, A \approx 0.66 \, A$ towards the right.

(D) For a steady current $i$ flowing through a conductor,the charge conservation principle implies that the current $i$ must be constant at every cross-section of the conductor.
The current density is given by $j = \frac{i}{A}$,where $A$ is the cross-sectional area. Since $A$ varies along the length,$j$ is not constant.
From Ohm's law in microscopic form,$j = \sigma E$. Since $j$ varies,the electric field $E$ must also vary along the length.
Furthermore,the drift velocity is given by ${v_d} = \frac{j}{ne} = \frac{i}{Ane}$. Since $A$ varies,the drift velocity ${v_d}$ is also not constant.
Therefore,only the current $i$ remains constant along the length of the conductor.

(C) The relationship between current $(I)$,time $(t)$,and charge $(Q)$ is given by the formula: $Q = I \times t$.
Given:
Current $(I)$ = $20\,\mu A = 20 \times 10^{-6}\,A$
Time $(t)$ = $30\,s$
Substituting these values into the formula:
$Q = (20 \times 10^{-6}\,A) \times (30\,s)$
$Q = 600 \times 10^{-6}\,C$
$Q = 6 \times 10^{-4}\,C$
Therefore,the correct option is $C$.

(B) The formula for electric current is $I = \frac{q}{t} = \frac{ne}{t}$,where $I$ is the current,$n$ is the number of electrons,$e$ is the elementary charge,and $t$ is the time.
Given: $I = 1.6 \, mA = 1.6 \times 10^{-3} \, A$,$t = 1 \, s$,and $e = 1.6 \times 10^{-19} \, C$.
Rearranging the formula to solve for $n$: $n = \frac{I \times t}{e}$.
Substituting the values: $n = \frac{1.6 \times 10^{-3} \times 1}{1.6 \times 10^{-19}}$.
$n = 10^{-3} \times 10^{19} = 10^{16}$.
Therefore,the number of electrons flowing per second is $10^{16}$.

(B) The electric current $i$ is defined as the rate of flow of charge,given by $i = \frac{q}{T}$.
Since the frequency $\nu = \frac{1}{T}$,we can write the current as $i = q \times \nu$.
Given:
Charge of an electron $q = 1.6 \times 10^{-19} \ C$
Frequency $\nu = 5 \times 10^{15} \ Hz$
Substituting the values:
$i = (1.6 \times 10^{-19} \ C) \times (5 \times 10^{15} \ Hz)$
$i = 8.0 \times 10^{-4} \ A$
To convert the current into $mA$ (milliampere),we multiply by $10^3$:
$i = 8.0 \times 10^{-4} \times 10^3 \ mA = 0.8 \ mA$.

(C) The equivalent current $i$ produced by a revolving charge is given by the formula $i = q \times \nu$,where $q$ is the charge and $\nu$ is the frequency of revolution.
Given:
Charge of electron $q = 1.6 \times 10^{-19} \text{ C}$
Frequency $\nu = 6.8 \times 10^{15} \text{ rev/s}$
Substituting the values:
$i = (1.6 \times 10^{-19} \text{ C}) \times (6.8 \times 10^{15} \text{ s}^{-1})$
$i = 10.88 \times 10^{-4} \text{ A}$
$i \approx 1.1 \times 10^{-3} \text{ A}$
Therefore,the correct option is $C$.

(A) The current $i$ in a loop is given by the formula $i = q \times f$,where $q$ is the charge of the electron and $f$ is the frequency of revolution.
Given:
Frequency $f = 6 \times 10^{15} \, \text{Hz}$
Charge of an electron $q = 1.6 \times 10^{-19} \, \text{C}$
Substituting the values:
$i = (1.6 \times 10^{-19} \, \text{C}) \times (6 \times 10^{15} \, \text{Hz})$
$i = 9.6 \times 10^{-4} \, \text{A}$
$i = 0.96 \times 10^{-3} \, \text{A} = 0.96 \, \text{mA}$.

(A) The formula for electric current is $I = \frac{q}{t}$,where $q = ne$.
Given current $I = 16 \ mA = 16 \times 10^{-3} \ A$ and time $t = 1 \ s$.
The charge of an electron $e = 1.6 \times 10^{-19} \ C$.
Substituting the values into the equation: $16 \times 10^{-3} = \frac{n \times 1.6 \times 10^{-19}}{1}$.
Solving for $n$: $n = \frac{16 \times 10^{-3}}{1.6 \times 10^{-19}} = 10 \times 10^{16} = 10^{17}$.
Therefore,$10^{17}$ electrons strike the screen each second.

(B) The current $I$ is given by $I = \frac{q}{t}$,where $q = ne$.
Given $I = 1\, mA = 1 \times 10^{-3}\, A$ and $t = 1\, s$.
The charge of an electron is $e = 1.6 \times 10^{-19}\, C$.
Using the formula $I = \frac{ne}{t}$,we can solve for the number of electrons $n$:
$n = \frac{I \times t}{e} = \frac{1 \times 10^{-3} \times 1}{1.6 \times 10^{-19}}$
$n = \frac{1}{1.6} \times 10^{16} = 0.625 \times 10^{16} = 6.25 \times 10^{15}$.
Therefore,$6.25 \times 10^{15}$ electrons will pass the point in one second.

(A) The electric current $I$ is defined as the rate of flow of net charge through a cross-section.
Electrons moving from right to left constitute a current from left to right.
Protons moving from left to right constitute a current from left to right.
Since both currents are in the same direction,they add up.
$I = I_e + I_p = (n_e e) + (n_p e) = (n_e + n_p)e$
Given $n_e = 3.13 \times 10^{15}$ and $n_p = 3.12 \times 10^{15}$,and $e = 1.6 \times 10^{-19}\,C$.
$I = (3.13 \times 10^{15} + 3.12 \times 10^{15}) \times 1.6 \times 10^{-19}\,A$
$I = (6.25 \times 10^{15}) \times 1.6 \times 10^{-19}\,A$
$I = 10 \times 10^{-4}\,A = 10^{-3}\,A = 1\,mA$.
Since both components of the current are directed from left to right,the total current is $1\,mA$ towards the right.

(A) When a steady current $i$ flows through a conductor,the rate of flow of charge into any segment is equal to the rate of flow of charge out of that segment.
Since the number of electrons entering one end of any segment is exactly equal to the number of electrons leaving the other end,the net charge within any segment of the conductor remains constant.
Because the conductor is electrically neutral to begin with,this net charge is zero.
Therefore,any segment of the conductor has zero net charge.

(C) The relationship between current $i$ and charge $Q$ is given by $i = \frac{dQ}{dt}$.
Therefore,the charge $dQ$ flowing in a small time interval $dt$ is $dQ = i \, dt$.
To find the total charge $Q$ flowing in the first $5$ seconds,we integrate the current with respect to time from $t = 0$ to $t = 5$:
$Q = \int_{0}^{5} i \, dt = \int_{0}^{5} (1.2t + 3) \, dt$
$Q = \left[ \frac{1.2t^2}{2} + 3t \right]_{0}^{5}$
$Q = [0.6t^2 + 3t]_{0}^{5}$
Substituting the limits:
$Q = (0.6 \times 5^2 + 3 \times 5) - (0.6 \times 0^2 + 3 \times 0)$
$Q = (0.6 \times 25 + 15) - 0$
$Q = 15 + 15 = 30\,C$.

(B) The relationship between current $I$ and charge $Q$ is given by $I = \frac{dQ}{dt}$,which implies $dQ = I \, dt$.
To find the total charge $Q$ flowing between $t = 2 \, s$ and $t = 3 \, s$,we integrate the current with respect to time:
$Q = \int_{2}^{3} I \, dt = \int_{2}^{3} (2t + 3t^2) \, dt$
$Q = \left[ \frac{2t^2}{2} + \frac{3t^3}{3} \right]_{2}^{3} = \left[ t^2 + t^3 \right]_{2}^{3}$
$Q = (3^2 + 3^3) - (2^2 + 2^3)$
$Q = (9 + 27) - (4 + 8)$
$Q = 36 - 12 = 24 \, C$.

(A) The electric power $P$ produced by the generator is given by $P = 250\,kW = 250,000\,W$.
The voltage $V$ at which the power is transmitted is $V = 10,000\,V$.
The relationship between power,voltage,and current is given by the formula $P = VI$.
To find the current $I$,we rearrange the formula: $I = \frac{P}{V}$.
Substituting the given values: $I = \frac{250,000\,W}{10,000\,V} = 25\,A$.
Therefore,the current in the cable is $25\,A$.

(D) The formula for electric charge $q$ is given by $q = I \times t$, where $I$ is the current and $t$ is the time in seconds.
Given:
Current $I = 0.5\,A$
Time $t = 1\,hour = 3600\,s$
Substituting the values into the formula:
$q = 0.5\,A \times 3600\,s = 1800\,C$.
Therefore, the total charge passing through the bulb is $1800\,C$.

(A) scalar quantity is defined as a physical quantity that has only magnitude and no direction.
Electric current has both magnitude and direction,but it does not follow the vector laws of addition (like the parallelogram law).
Instead,it follows the algebraic rules of addition.
Therefore,electric current is classified as a scalar quantity.
In contrast,velocity,force,and acceleration are vector quantities because they possess both magnitude and direction and obey vector addition laws.

(B) The relationship between current $I$ and charge $Q$ is given by $I = \frac{dQ}{dt}$,which implies $dQ = I \ dt$.
To find the total charge $Q$ passing through the cross-section between time $t_1 = 2 \ s$ and $t_2 = 3 \ s$,we integrate the current with respect to time:
$Q = \int_{2}^{3} I \ dt = \int_{2}^{3} (2t + 3t^2) \ dt$
Performing the integration:
$Q = [t^2 + t^3]_{2}^{3}$
$Q = (3^2 + 3^3) - (2^2 + 2^3)$
$Q = (9 + 27) - (4 + 8)$
$Q = 36 - 12 = 24 \ C$.
Thus,the total charge passed is $24 \ C$.

(C) Given: Current $I = 5\,A$, Resistance $R = 10\, \Omega$, Time $t = 4\, \text{minutes} = 4 \times 60 = 240\,s$.
First, calculate the total charge $Q$ using the formula $Q = I \times t$:
$Q = 5\,A \times 240\,s = 1200\,C$.
Next, calculate the number of electrons $n$ using the quantization of charge formula $Q = n \times e$, where $e = 1.6 \times 10^{-19}\,C$:
$n = \frac{Q}{e} = \frac{1200}{1.6 \times 10^{-19}} = \frac{12000}{1.6} \times 10^{18} = 7500 \times 10^{18} = 75 \times 10^{20}$.
Thus, the number of electrons is $75 \times 10^{20}$ and the charge is $1200\,C$.

(A) The relationship between current $I$ and charge $q$ is given by $I = \frac{dq}{dt}$.
Therefore,the charge $dq$ is $dq = I \, dt = (3t^2 + 2t + 5) \, dt$.
To find the total charge $q$ passing through the cross-section from $t = 0$ to $t = 2$ seconds,we integrate the expression:
$q = \int_{0}^{2} (3t^2 + 2t + 5) \, dt$
$q = [t^3 + t^2 + 5t]_{0}^{2}$
Substituting the limits:
$q = (2^3 + 2^2 + 5(2)) - (0^3 + 0^2 + 5(0))$
$q = (8 + 4 + 10) - 0 = 22 \, C$.
Thus,the total charge is $22 \, C$.

(D) The charge flowing per minute is given by $Q = I \times t$.
Given $I = 3.2 \ A$ and $t = 60 \ s$,we have $Q = 3.2 \times 60 = 192 \ C$.
Each copper ion $(Cu^{2+})$ requires $2$ electrons to be deposited at the cathode.
The charge of one electron is $e = 1.6 \times 10^{-19} \ C$.
Therefore,the charge required per ion is $2e = 2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19} \ C$.
The number of ions $n$ deposited is $n = \frac{Q}{2e}$.
$n = \frac{192}{3.2 \times 10^{-19}} = 60 \times 10^{19} = 6 \times 10^{20}$ ions.

(A) The relationship between current $i$ and charge $q$ is given by $i = \frac{dq}{dt}$.
Therefore,$dq = i \cdot dt = (2 + 3t) \cdot dt$.
To find the total charge $Q$ passed in $10 \text{ s}$,we integrate from $t = 0$ to $t = 10 \text{ s}$:
$Q = \int_{0}^{10} (2 + 3t) \cdot dt$
$Q = [2t + \frac{3t^2}{2}]_{0}^{10}$
Substituting the limits:
$Q = (2(10) + \frac{3(10)^2}{2}) - (0 + 0)$
$Q = 20 + \frac{3 \times 100}{2}$
$Q = 20 + 150 = 170 \text{ C}$.

(B) The current is defined as the rate of flow of charge: $I = \frac{dQ}{dt} = 4 - 0.08t$.
To find the total charge $Q$ passing through the cross-section in $50 \ s$,we integrate the current with respect to time from $t = 0$ to $t = 50 \ s$:
$Q = \int_{0}^{50} (4 - 0.08t) dt = [4t - \frac{0.08t^2}{2}]_{0}^{50} = [4(50) - 0.04(50)^2] = 200 - 0.04(2500) = 200 - 100 = 100 \ C$.
Using the quantization of charge formula $Q = ne$,where $e = 1.6 \times 10^{-19} \ C$ is the charge of an electron:
$n = \frac{Q}{e} = \frac{100}{1.6 \times 10^{-19}} = 6.25 \times 10^{20}$ electrons.

(B) The formula for electric current is $I = \frac{q}{t} = \frac{Ne}{t}$,where $I$ is the current,$N$ is the number of electrons,$e$ is the elementary charge,and $t$ is the time.
Given: $I = 8 \ A$,$t = 1 \ \text{minute} = 60 \ s$,and $e = 1.6 \times 10^{-19} \ C$.
Rearranging the formula to solve for $N$: $N = \frac{I \times t}{e}$.
Substituting the values: $N = \frac{8 \times 60}{1.6 \times 10^{-19}}$.
$N = \frac{480}{1.6 \times 10^{-19}} = 300 \times 10^{19} = 3 \times 10^{21}$ electrons.

(A) Consider a point $A$ on the circle. The electron passes through this point in every revolution. The number of revolutions per second made by the electron is given by the frequency $f$.
$f = \frac{v}{2\pi r} = \frac{4 \times 10^6}{2\pi \times 10 \times 10^{-2}} = \frac{2}{\pi} \times 10^7 \, \text{rev/sec}$.
The electric current $I$ is defined as the charge passing through a point per unit time:
$I = \frac{q}{t} = f \times e$
$I = \left( \frac{2}{\pi} \times 10^7 \right) \times (1.6 \times 10^{-19}) \, A$
$I = \frac{3.2}{\pi} \times 10^{-12} \, A$
Since $\pi \approx 3.14$,we get $I \approx 1.019 \times 10^{-12} \, A \approx 1 \times 10^{-12} \, A$.

(A) For a metallic conductor connected to a constant potential difference $V$,the current $I$ flowing through any cross-section must be the same due to the principle of conservation of charge (steady state condition).
Since $I = nAev_d$,where $n$ is the number density of electrons,$A$ is the cross-sectional area,$e$ is the charge of an electron,and $v_d$ is the drift velocity,if $A$ varies,$v_d$ must also vary to keep $I$ constant.
Similarly,current density $J = I/A$ varies with $A$,and since $J = \sigma E$,the electric field $E$ also varies along the conductor.
Therefore,the current $I$ is the only quantity that remains constant.

(B) For a metallic conductor, the total current $I$ flowing through any cross-section must be the same due to the principle of conservation of charge (steady state flow).
Since $I = nAev_d$ and $J = I/A = nev_d$, if the cross-sectional area $A$ varies, the current density $J$ and drift velocity $v_d$ must also vary to keep $I$ constant.
The electric field $E = J/\sigma$ also varies with the cross-sectional area.
Therefore, the current $I$ is the only quantity that remains constant along the conductor.

(C) The effective current $I$ is defined as the net rate of flow of charge. $A$ positive charge moving in one direction is equivalent to a current in that direction,while a negative charge moving in the same direction is equivalent to a current in the opposite direction.
For figure $(i)$: $A$ positive charge moves to the right at $7 \ C/s$. Thus,$I_1 = 7 \ A$.
For figure $(ii)$: $A$ positive charge moves to the right at $4 \ C/s$ and a negative charge moves to the left at $3 \ C/s$. Both contribute to a current to the right. Thus,$I_2 = 4 + 3 = 7 \ A$.
For figure $(iii)$: $A$ positive charge moves to the right at $5 \ C/s$ and a negative charge moves to the left at $2 \ C/s$. Both contribute to a current to the right. Thus,$I_3 = 5 + 2 = 7 \ A$.
For figure $(iv)$: $A$ negative charge moves to the left at $6 \ C/s$ (equivalent to $6 \ A$ to the right) and a positive charge moves to the left at $1 \ C/s$ (equivalent to $1 \ A$ to the left). Thus,$I_4 = 6 - 1 = 5 \ A$.
Comparing the values,we get $I_1 = I_2 = I_3 = 7 \ A$ and $I_4 = 5 \ A$. Therefore,$i = ii = iii > iv$.

(A) Good conductors of electricity are materials that allow electric current to flow through them easily due to the presence of free electrons.
Metals like $Cu$ (Copper),$Ag$ (Silver),and $Au$ (Gold) have a large number of free electrons in their metallic lattice,making them excellent conductors.
$Si$ (Silicon) and $Ge$ (Germanium) are semiconductors.
$Diamond$ is an insulator.
$NaCl$ (Sodium Chloride) is an ionic solid that conducts electricity only in a molten state or aqueous solution,not in its solid state.
Therefore,the set containing $Cu, Ag$,and $Au$ consists entirely of good conductors.

(C) The current $I$ is defined as the rate of flow of charge,$I = \frac{dq}{dt}$. By convention,the direction of current is the direction of flow of positive charge.
$1$. For $\alpha$-particles: Each $\alpha$-particle has a charge of $+2e$. Since $10^{19}$ $\alpha$-particles move to the left per second,the current $I_{\alpha} = (10^{19} \times 2e) \, C/s$ towards the left.
$2$. For protons: Each proton has a charge of $+e$. Since $10^{19}$ protons move to the left per second,the current $I_p = (10^{19} \times e) \, C/s$ towards the left.
$3$. For electrons: Each electron has a charge of $-e$. Since $10^{19}$ electrons move to the right per second,the current $I_e$ is in the opposite direction of the electron flow,i.e.,towards the left. The magnitude is $I_e = (10^{19} \times e) \, C/s$ towards the left.
Total current $I = I_{\alpha} + I_p + I_e = (2e + e + e) \times 10^{19} \, C/s = 4e \times 10^{19} \, C/s$.
Substituting $e = 1.6 \times 10^{-19} \, C$:
$I = 4 \times 1.6 \times 10^{-19} \times 10^{19} \, A = 6.4 \, A$.
Since all components contribute to current towards the left,the total current is $6.4 \, A$ towards the left.

(B) The charge $q$ passing through a cross-section is given by the integral of current $I$ with respect to time $t$:
$q = \int_{t_1}^{t_2} I \, dt$
Given $I = 4 + 2t$,$t_1 = 2 \, s$,and $t_2 = 6 \, s$:
$q = \int_{2}^{6} (4 + 2t) \, dt$
$q = [4t + t^2]_{2}^{6}$
$q = (4(6) + (6)^2) - (4(2) + (2)^2)$
$q = (24 + 36) - (8 + 4)$
$q = 60 - 12 = 48 \, C$
Thus,the total charge passed is $48 \, C$.

(C) The volume charge density $\rho$ is defined as the total charge $q$ per unit volume $V$. For a rod of length $d$ and cross-sectional area $A$,the volume is $V = A d$.
Thus,$\rho = \frac{q}{V} = \frac{q}{A d}$,which implies $q = \rho A d$.
Since current $I$ is defined as the rate of flow of charge,$I = \frac{q}{t}$,where $t$ is the time taken for the charge $q$ to pass through a cross-section.
Rearranging for time $t$,we get $t = \frac{q}{I}$.
Substituting the expression for $q$,we get $t = \frac{\rho A d}{I}$.

(A) The electric current $I$ is defined as the rate of flow of charge,which is given by the slope of the $q-t$ graph: $I = \frac{dq}{dt}$.
From the given graph,between $t = 2 \, s$ and $t = 6 \, s$,the charge $q$ on the capacitor plate is constant at $3 \, \mu C$.
Since the charge is constant,the slope of the graph in this interval is zero.
Therefore,the current at $t = 4 \, s$ is $I = 0 \, \mu A$.

(B) The current is given by $i = (2 + 3t) \times 10^{-3} \, A$.
To find the total charge $q$ crossing a section in time $t = 0$ to $t = 60 \, s$,we use the relation $q = \int i \, dt$.
$q = \int_{0}^{60} (2 + 3t) \times 10^{-3} \, dt$.
$q = 10^{-3} \left[ 2t + \frac{3t^2}{2} \right]_{0}^{60}$.
$q = 10^{-3} \left[ 2(60) + \frac{3(60)^2}{2} \right]$.
$q = 10^{-3} [120 + 5400]$.
$q = 10^{-3} \times 5520 = 5.52 \, C$.

(D) The current $I$ is defined as the rate of flow of charge: $I = \frac{dQ}{dt}$.
Given $Q = at - bt^2$,differentiating with respect to $t$ gives $I = a - 2bt$.
$(i)$ Since $I = a - 2bt$,the current decreases linearly with time. Thus,statement $(A)$ is correct.
$(ii)$ The current starts at $I = a$ at $t = 0$ and decreases continuously. It does not reach a maximum after $t = 0$. Thus,statement $(B)$ is incorrect.
$(iii)$ Setting $I = 0$,we get $a - 2bt = 0$,which implies $t = a/2b$. Thus,statement $(C)$ is correct.
$(iv)$ The rate of change of current is $\frac{dI}{dt} = \frac{d}{dt}(a - 2bt) = -2b$. Thus,statement $(D)$ is correct.
Since only statement $(B)$ is incorrect,the correct option is $(D)$.

(D) The number of electrons flowing per unit time is given by $\frac{n}{t} = 10^{7} \; s^{-1}$.
The electric current $I$ is defined as the rate of flow of charge,$I = \frac{q}{t}$.
Since $q = ne$,where $e$ is the elementary charge $(e = 1.6 \times 10^{-19} \; C)$,we have $I = \frac{ne}{t} = \left(\frac{n}{t}\right) \times e$.
Substituting the given values: $I = 10^{7} \times (1.6 \times 10^{-19} \; C) = 1.6 \times 10^{-12} \; A$.

(N/A) Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor. When charges (typically electrons in a metal) move in a specific direction due to an applied electric field,an electric current is formed.
Electric charge is of two types:
$(1)$ Positive charge
$(2)$ Negative charge
In nature,electric current can be produced under certain conditions. For example,lightning in the sky is a natural phenomenon where a large amount of charge flows between clouds or between a cloud and the earth. However,this is not a steady flow of charge.
In devices used in our day-to-day life,such as a torch or a battery-operated clock,a steady flow of charge is maintained by a source of electromotive force (like a cell or battery),which results in a constant electric current.

(N/A) An electric current is produced when there is a net flow of electric charge through a cross-section of a conductor over a period of time.
$1$. In a conductor,free electrons are in random motion due to thermal energy,resulting in no net flow of charge.
$2$. When an external electric field $(E)$ is applied across the conductor,the free electrons experience an electric force $(F = -eE)$,causing them to drift in the direction opposite to the applied field.
$3$. This directed motion of charge carriers constitutes an electric current $(I)$.
$4$. Mathematically,current is defined as the rate of flow of charge: $I = \frac{dq}{dt}$,where $dq$ is the net charge passing through a cross-section in time $dt$.

(B) The statement is $False$.
Lightning is an example of a transient or non-steady discharge of electricity.
In a steady flow of charge (direct current),the rate of flow of charge remains constant over time.
Lightning involves a massive,instantaneous discharge of accumulated static electricity between clouds or between a cloud and the ground,which lasts for a very short duration and is not a continuous or steady current.

(N/A) Electric current is defined as the net flow of charge through a cross-section of a conductor per unit time.
Consider a cross-section of a conductor where a net charge $q$ flows in time $t$. For a steady current,the current $I$ is given by $I = \frac{q}{t}$.
When the flow of charge changes with time,the instantaneous current $I$ is defined as the limit of the average current as the time interval approaches zero:
$I = \lim_{\Delta t \rightarrow 0} \frac{\Delta Q}{\Delta t} = \frac{dQ}{dt}$.
The direction of conventional current is taken to be the direction of motion of positive charge. Since positive charges do not flow in metallic conductors,the direction of current is taken opposite to the direction of motion of electrons.
The $SI$ unit of electric current is the Ampere $(A)$,where $1 \text{ Ampere} = 1 \text{ Coulomb/second}$.
One Ampere is defined as the current flowing through a cross-section of a conductor when $1 \text{ Coulomb}$ of charge flows through it in $1 \text{ second}$ ($6.25 \times 10^{18}$ electrons).

(N/A) When the flow of charge is not steady,the electric current at any instant $t$ is defined as the rate of flow of charge through a cross-section of the conductor at that specific instant.
Mathematically,it is expressed as the limit of the ratio of the small amount of charge $\Delta q$ passing through a cross-section in a small time interval $\Delta t$ as $\Delta t$ approaches zero:
$I(t) = \lim_{\Delta t \to 0} \frac{\Delta q}{\Delta t} = \frac{dq}{dt}$
Here,$I(t)$ represents the instantaneous current,and $\frac{dq}{dt}$ is the time derivative of the charge $q$.

(B) By convention,the direction of electric current is defined as the direction in which positive charges would move.
In an external circuit connected to a battery,this corresponds to the flow from the positive terminal to the negative terminal.
Therefore,the conventional current flows from the positive terminal to the negative terminal of the source.

(A) The electrical signals in the human nervous system are generated by the movement of ions across nerve cell membranes.
These signals,known as action potentials,involve very small currents.
The typical magnitude of current flowing in the nerves of the human body is on the order of microamperes $(10^{-6} \ A)$.