Maxwell's equations , Concept of displacement current and Hertz experiment Questions in English

(D) magnetic field is produced by a moving charge (current) as described by the Biot-Savart law.
Additionally,a changing electric field (displacement current) also produces a magnetic field,as described by the Maxwell-Ampere law.
Therefore,both a moving charge and a changing electric field can produce a magnetic field.

(C) According to Maxwell's $EM$ theory,a time-varying electric field produces a displacement current,which in turn acts as a source of a magnetic field.
This is represented by the Maxwell-Ampere law: $\oint B \cdot dl = \mu_0 (I_c + I_d)$,where $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current.
Therefore,a changing electric field gives rise to a magnetic field.

(D) Maxwell's equations describe the fundamental laws of electricity and magnetism.
These equations explain how electric and magnetic fields are generated and altered by each other,as well as by electric charges and currents.
Therefore,they encompass the principles of both electricity and magnetism.

(C) Heinrich Rudolf Hertz was a German physicist who first conclusively proved the existence of electromagnetic waves,which were previously theorized by James Clerk Maxwell.
He used an oscillating circuit to produce and detect these waves in his laboratory.
Therefore,the correct answer is $C$.

(B) The displacement current $I_D$ is given by the formula $I_D = \varepsilon_0 \frac{d\phi_E}{dt}$.
Since the electric field $E = \frac{V}{d}$ and the electric flux $\phi_E = E \cdot A$,we have $I_D = \varepsilon_0 \frac{A}{d} \frac{dV}{dt}$.
Given: $\varepsilon_0 = 8.854 \times 10^{-12} \, F/m$,$V = 400 \, V$,$A = 60 \, cm^2 = 60 \times 10^{-4} \, m^2$,$d = 2 \, mm = 2 \times 10^{-3} \, m$,and $\Delta t = 10^{-6} \, s$.
Substituting these values:
$I_D = \frac{8.854 \times 10^{-12} \times 60 \times 10^{-4} \times 400}{2 \times 10^{-3} \times 10^{-6}}$
$I_D = \frac{8.854 \times 60 \times 4 \times 10^{-12} \times 10^{-4} \times 10^2}{2 \times 10^{-9}}$
$I_D = \frac{2124.96 \times 10^{-14}}{2 \times 10^{-9}} = 1062.48 \times 10^{-5} = 1.06248 \times 10^{-2} \, A$.
Thus,the correct option is $B$.

(A) In the ionosphere,the interaction of electromagnetic waves with the plasma leads to two types of currents: the conduction current (space current) and the displacement current.
According to Maxwell's equations,the displacement current density is given by $J_d = \epsilon_0 \frac{\partial E}{\partial t}$.
For a harmonic wave $E = E_0 \sin(\omega t)$,the displacement current is $J_d = \epsilon_0 \omega E_0 \cos(\omega t) = \epsilon_0 \omega E_0 \sin(\omega t + \pi / 2)$.
The conduction current (space current) $J_c$ in a plasma is related to the electric field by $J_c = \sigma E$,where $\sigma$ is the conductivity. In the ionosphere,the motion of electrons is dominated by inertia,making the current lag behind the electric field by $\pi / 2$.
Thus,the conduction current $J_c$ is proportional to $\sin(\omega t - \pi / 2)$.
The phase difference between the displacement current $(\sin(\omega t + \pi / 2))$ and the conduction current $(\sin(\omega t - \pi / 2))$ is $(\pi / 2) - (-\pi / 2) = \pi \, rad$.

(D) An electron carries an electric charge. $A$ stationary electron creates no magnetic field. An electron moving at a constant velocity generates a steady magnetic field,but a constant magnetic field does not produce an electric field. An electron moving with a changing velocity (i.e.,accelerating) generates a changing magnetic field,which in turn produces a changing electric field. This continuous cycle of changing electric and magnetic fields propagates through space as an electromagnetic wave.

(C) Maxwell's equations are a set of four partial differential equations that provide a comprehensive description of classical electromagnetism.
They describe how electric and magnetic fields are generated by charges,currents,and by each other.
Therefore,these equations are derived from the fundamental laws of electricity and magnetism.

(B) In Hertz's experiment,two metal rods were connected to an induction coil.
These rods act as a capacitor because they store electrical energy in the electric field between them when a high potential difference is applied.
When the potential difference exceeds the breakdown voltage of the air gap,a spark is produced,and the system oscillates,generating electromagnetic waves.

(A) James Clerk Maxwell was the first scientist to theoretically predict the existence of electromagnetic waves in $1865$. He unified the laws of electricity and magnetism through his set of four equations,known as Maxwell's equations. These equations implied that a time-varying electric field produces a time-varying magnetic field and vice versa,leading to the propagation of electromagnetic waves in free space.

(C) The displacement current $I_d$ in a capacitor is given by the formula: $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.
Since the electric flux $\Phi_E = E \cdot A = \frac{V}{d} \cdot A = \frac{V \cdot A}{d}$, and the capacitance $C = \frac{\epsilon_0 A}{d}$, we can write the displacement current as $I_d = C \frac{dV}{dt}$.
Given values are $C = 2 \; \mu F = 2 \times 10^{-6} \; F$ and $\frac{dV}{dt} = 10^6 \; V/s$.
Substituting these values into the formula:
$I_d = (2 \times 10^{-6} \; F) \times (10^6 \; V/s) = 2 \; A$.
Therefore, the displacement current is $2 \; A$.

(A) According to the Maxwell-Ampere law,the modified Ampere's law is given by:
$\nabla \times B = \mu_0 \left( J + \epsilon_0 \frac{\partial E}{\partial t} \right)$
Here,the term $\epsilon_0 \frac{\partial E}{\partial t}$ represents the displacement current density.
This equation shows that a time-varying electric field $\frac{\partial E}{\partial t}$ acts as a source for a magnetic field $B$.
Therefore,a changing electric field produces a magnetic field.

(C) Ampere's original circuital law was found to be inconsistent for time-varying electric fields.
James Clerk Maxwell introduced the concept of displacement current,$I_d = \epsilon_0 \frac{d\phi_E}{dt}$,to resolve this inconsistency.
The modified Ampere-Maxwell law states that the line integral of the magnetic field around any closed path is equal to $\mu_0$ times the sum of the conduction current $i$ and the displacement current $I_d$.
Mathematically,this is expressed as: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (i + I_d) = \mu_0 i + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$.

(D) The displacement current $i_d$ is defined as $i_d = \epsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E$ is the electric flux between the plates.
Since the electric flux $\phi_E = \frac{q}{\epsilon_0}$,we have $i_d = \frac{dq}{dt}$.
This means that the displacement current exists whenever the charge $q$ on the capacitor plates is changing with time.
Therefore,the charge on the capacitor is either increasing (during charging) or decreasing (during discharging).
Thus,both $A$ and $B$ are correct.

(C) According to the theory of electromagnetism,a stationary charge produces only an electric field.
$A$ charge moving with a constant velocity produces both an electric field and a magnetic field,but it does not radiate energy.
An accelerating charge (or an oscillating charge) produces time-varying electric and magnetic fields,which propagate through space as electromagnetic waves.
Therefore,an accelerating charge is the source of electromagnetic waves.

(D) Heinrich Rudolf Hertz was a German physicist who first conclusively proved the existence of electromagnetic waves,which were previously theorized by James Clerk Maxwell.
By using a spark-gap transmitter,he demonstrated that electromagnetic waves could be generated and detected in a laboratory setting.
Therefore,$Hertz$ is the one who first produced and experimentally verified the existence of electromagnetic waves.

(A) The displacement current $I_d$ is given by the formula $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.
Since $\Phi_E = E \cdot A$ and $V = E \cdot d$,we have $I_d = C \frac{dV}{dt}$.
Given: Capacitance $C = 2 \ pF = 2 \times 10^{-12} \ F$.
Rate of change of potential $\frac{dV}{dt} = 10^{12} \ V \ s^{-1}$.
Substituting the values: $I_d = (2 \times 10^{-12} \ F) \times (10^{12} \ V \ s^{-1})$.
$I_d = 2 \ A$.

(C) The displacement current $I_D$ in a capacitor is equal to the conduction current flowing through the circuit,which is defined as the rate of change of charge on the plates.
$I_D = \frac{dq}{dt}$
Given $q = q_0 \sin(2\pi nt)$.
Differentiating with respect to time $t$:
$I_D = \frac{d}{dt} [q_0 \sin(2\pi nt)]$
$I_D = q_0 \cdot \cos(2\pi nt) \cdot \frac{d}{dt}(2\pi nt)$
$I_D = 2\pi n q_0 \cos(2\pi nt)$.

(C) According to the Maxwell-Ampere law,the induced magnetic field $B$ is proportional to the displacement current density,which is proportional to the rate of change of the electric field $\frac{dE}{dt}$.
Given that the electric field $E$ varies with time as $E \propto t^2$.
We know that the induced magnetic field $B$ is related to the rate of change of the electric field: $B \propto \frac{dE}{dt}$.
Substituting the given variation: $B \propto \frac{d}{dt}(t^2)$.
Calculating the derivative: $B \propto 2t$.
Therefore,the induced magnetic field $B$ varies with time as $B \propto t$.

(B) The electric field $E$ between the plates of a parallel plate capacitor is given by $E = \frac{Q}{\varepsilon_0 A}$,where $Q$ is the charge on the capacitor at time $t$.
For a surface of area $A' = A/2$ placed parallel and symmetrically between the plates,the electric flux $\phi_E$ through this area is $\phi_E = E \cdot A' = \left( \frac{Q}{\varepsilon_0 A} \right) \left( \frac{A}{2} \right) = \frac{Q}{2\varepsilon_0}$.
The displacement current $i_d$ is defined as $i_d = \varepsilon_0 \frac{d\phi_E}{dt}$.
Substituting the expression for $\phi_E$,we get $i_d = \varepsilon_0 \frac{d}{dt} \left( \frac{Q}{2\varepsilon_0} \right) = \frac{1}{2} \frac{dQ}{dt}$.
Since the charging current $i = \frac{dQ}{dt}$,the displacement current is $i_d = \frac{i}{2}$.

(B) The conduction current density is given by $J_c = \sigma E$,where $E = E_0 \sin(\omega t - kx)$. Thus,the magnitude is $J_c = \sigma E_0$.
The displacement current density is given by $J_d = \varepsilon \varepsilon_0 \frac{\partial E}{\partial t}$.
Since $E = E_0 \sin(\omega t - kx)$,we have $\frac{\partial E}{\partial t} = E_0 \omega \cos(\omega t - kx)$.
Thus,the magnitude of displacement current density is $J_d = \varepsilon \varepsilon_0 E_0 \omega$.
The ratio of conduction current density to displacement current density is $\frac{J_c}{J_d} = \frac{\sigma E_0}{\varepsilon \varepsilon_0 E_0 \omega} = \frac{\sigma}{\varepsilon \varepsilon_0 \omega}$.

(D) The displacement current is given by the formula $I_d = \epsilon_0 A \frac{dE}{dt}$.
Since $\epsilon_0$ and $A$ are constants, the displacement current $I_d$ is directly proportional to the rate of change of the electric field, i.e., the slope of the $E-t$ graph $(\frac{dE}{dt})$.
We calculate the slope $(\frac{dE}{dt})$ for different intervals:
$1$. For $t = 0$ to $2 \ \mu s$: Slope = $\frac{1.0 - 0.5}{2 - 0} = 0.25 \ (N/C)/\mu s$.
$2$. For $t = 2$ to $10 \ \mu s$: Slope = $\frac{-1.0 - 1.0}{10 - 2} = \frac{-2.0}{8} = -0.25 \ (N/C)/\mu s$.
$3$. For $t = 10$ to $13 \ \mu s$: Slope = $\frac{0 - (-1.0)}{13 - 10} = \frac{1.0}{3} \approx 0.33 \ (N/C)/\mu s$.
The magnitude of the slope is greatest in the interval $t = 10$ to $13 \ \mu s$, where the slope is $0.33$. Among the given options, $t = 12 \ \mu s$ lies within this interval.

(D) The Ampere-Maxwell law is a modification of Ampere's circuital law to include the displacement current,which accounts for time-varying electric fields.
The generalized Ampere-Maxwell law is given by:
$\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$
Where:
$1. \oint \vec{B} \cdot d\vec{l}$ is the line integral of the magnetic field around a closed loop.
$2. i_c$ is the conduction current.
$3. \epsilon_0 \frac{d\phi_E}{dt}$ is the displacement current $(i_d)$.
Thus,the expression is $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_{\text{enclosed}} + \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$.

(B) When an insulator (dielectric) plate enters the capacitor,the capacitance $C$ increases because $C = \frac{K \epsilon_0 A}{d}$.
Since the capacitor is connected to a battery,the potential difference $V$ remains constant.
The charge on the capacitor $Q = CV$ increases as the plate enters.
This increase in charge requires a current to flow from the battery to the capacitor plates,which corresponds to a charging current in the external circuit.
Inside the capacitor,the displacement current $I_d$ is equal to the conduction current in the external circuit,so it flows in the direction that supports the charging process (clockwise).
As the insulator plate leaves the capacitor,the capacitance $C$ decreases,causing the charge $Q$ on the capacitor to decrease.
This results in a discharge current flowing out of the capacitor,meaning the displacement current now flows in the opposite direction (anticlockwise).
Therefore,the displacement current first flows clockwise and then anticlockwise.

(A) When an insulator (dielectric) plate enters the capacitor,the capacitance $C$ increases. Since $Q = CV$,the charge on the plates increases. This increase in charge causes a change in the electric field between the plates. As the plate enters,the electric flux increases,leading to a displacement current in the direction of the electric field (from positive plate $A$ to negative plate $B$). As the plate leaves,the capacitance decreases,the charge decreases,and the electric flux decreases,causing the displacement current to flow in the opposite direction (from $B$ to $A$). Therefore,it first flows from $A$ to $B$ and then from $B$ to $A$.

(A) The displacement current $I_d$ in a capacitor is defined as the rate of change of charge on the plates,which is equal to the conduction current $I_c$ flowing into the capacitor.
Mathematically,$I_d = I_c = \frac{dq}{dt}$.
Given that the rate of loss of charge is $\frac{dq}{dt} = 1.8 \times 10^{-8}\, C/s$.
Therefore,the magnitude of the displacement current is $I_d = 1.8 \times 10^{-8}\, A$.

(D) According to the Maxwell-Ampere law,the magnetic field $B$ induced by a time-varying electric field is given by the integral form: $\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.
For a cylindrical region of radius $R$ with a uniform electric field $E$ changing with time,we consider an Amperian loop of radius $r$ $(r < R)$ centered on the axis.
The electric flux through this loop is $\Phi_E = E \cdot \pi r^2$.
Substituting this into the Maxwell-Ampere law: $B(2\pi r) = \mu_0 \epsilon_0 \frac{d}{dt}(E \pi r^2)$.
Since $E$ is uniform in space,$\frac{dE}{dt}$ is constant with respect to $r$.
$B(2\pi r) = \mu_0 \epsilon_0 \pi r^2 \frac{dE}{dt}$.
Solving for $B$,we get $B = \frac{\mu_0 \epsilon_0 r}{2} \frac{dE}{dt}$.
Thus,the magnitude of the magnetic field $B$ is proportional to $r$.

(D) The displacement current $I_D$ in a capacitor is given by the formula $I_D = C \frac{dV}{dt}$,where $C$ is the capacitance and $\frac{dV}{dt}$ is the rate of change of potential difference.
Given,$I_D = 2\,A$ and $C = 1\,\mu F = 10^{-6}\,F$.
Substituting the values into the formula:
$2 = 10^{-6} \times \frac{dV}{dt}$
$\frac{dV}{dt} = \frac{2}{10^{-6}} = 2 \times 10^6\,V/s$.
Thus,the potential difference must be changed at the rate of $2 \times 10^6\,V/s$.

(D) The magnetic field between the plates of a parallel plate capacitor is determined by the displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.
In this scenario,the medium between the plates is slightly conducting,which means there is a conduction current $I_c = -\frac{dq}{dt} = 2.5\times10^{-8}\, A$ flowing between the plates.
According to the modified Ampere-Maxwell law,the total current $I_{total} = I_c + I_d$ is the source of the magnetic field.
For a parallel plate capacitor,the conduction current $I_c$ and the displacement current $I_d$ are equal in magnitude but opposite in direction (or effectively cancel each other out in the context of the magnetic field generated between the plates in this specific idealized setup).
However,in a standard parallel plate capacitor where the plates are circular and the current is uniform,the magnetic field at a distance $r$ from the axis is given by $B = \frac{\mu_0 I r}{2\pi R^2}$.
Since the problem does not specify a radial distance $r$ and asks for the magnetic field 'between the plates' generally,and given the symmetry of the conduction current flowing through the medium,the magnetic field is zero at the center or if we consider the net current enclosed by an Amperian loop to be zero due to the continuity of current,the result is zero.

(C) The displacement current $(i_d)$ is a quantity that appears in the modified Ampere's law to account for the magnetic field produced by a time-varying electric field.
The formula for displacement current is defined as:
$i_d = \varepsilon_0 \frac{d\phi_E}{dt}$
Where:
$\varepsilon_0$ is the permittivity of free space.
$\phi_E$ is the electric flux.
$\frac{d\phi_E}{dt}$ is the rate of change of electric flux with respect to time.

(B) The displacement current $I_D$ in a capacitor is equal to the rate of change of charge on the plates,which is the conduction current $I_C$.
Given the charge $q = q_0 \cos(2\pi \nu t)$.
The displacement current is given by $I_D = \frac{dq}{dt}$.
$I_D = \frac{d}{dt} [q_0 \cos(2\pi \nu t)]$.
Using the chain rule for differentiation,$\frac{d}{dt} \cos(kt) = -k \sin(kt)$.
$I_D = q_0 \cdot (-2\pi \nu) \sin(2\pi \nu t)$.
$I_D = -q_0 \, 2\pi \nu \sin(2\pi \nu t)$.

(C) Displacement current $(I_d)$ is defined as the current that comes into existence in addition to the conduction current whenever the electric field and hence the electric flux changes with time. Mathematically, it is given by $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, where $\Phi_E$ is the electric flux. Therefore, it is due to the change in electric flux with time.

(B) The displacement current $I_D$ in a capacitor is given by the formula $I_D = C \frac{dV}{dt}$,where $C$ is the capacitance and $\frac{dV}{dt}$ is the rate of change of potential difference.
Given: $I_D = 1 \, A$ and $C = 2 \, \mu F = 2 \times 10^{-6} \, F$.
Substituting the values into the equation: $1 = (2 \times 10^{-6}) \times \frac{dV}{dt}$.
Solving for $\frac{dV}{dt}$: $\frac{dV}{dt} = \frac{1}{2 \times 10^{-6}} = 0.5 \times 10^{6} \, V/s$.

(A) An oscillating electric dipole consists of charges that are accelerating.
According to the fundamental principles of electromagnetism,an accelerated charge acts as a source of electromagnetic waves.
Since the charges in a dipole oscillation are undergoing acceleration,they necessarily produce electromagnetic waves.
Therefore,the Reason correctly explains why the Assertion is true.

(B) The charge $Q$ on a capacitor is given by $Q = CV$, where $C$ is the capacitance and $V$ is the potential difference.
The conduction current $i_c$ in the connecting wires is the rate of change of charge: $i_c = \frac{dQ}{dt} = C \frac{dV}{dt}$.
Given $C = 20 \; \mu F$ and $\frac{dV}{dt} = 3 \; V/s$, we have $i_c = 20 \; \mu F \times 3 \; V/s = 60 \; \mu A$.
According to Maxwell's modification of Ampere's law, the displacement current $i_d$ between the plates of a capacitor is equal to the conduction current in the connecting wires.
Therefore, $i_d = i_c = 60 \; \mu A$.
Thus, the conduction current is $60 \; \mu A$ and the displacement current is $60 \; \mu A$.

(N/A) Following Oersted's observation, scientists conducted significant research:
In $1864$, the laws governing electricity and magnetism were unified and formulated by James Clerk Maxwell, who subsequently realized that light consists of electromagnetic waves.
Radio waves were discovered by Heinrich Hertz and were later produced by $J. C. Bose$ and $G. Marconi$ by the end of the $19^{th}$ century.
A remarkable scientific and technological progress occurred in the $20^{th}$ century, primarily due to an increased understanding of electromagnetism and the invention of devices for the production, amplification, transmission, and detection of electromagnetic waves.

(A) The time constant of the $CR$ circuit is $\tau = CR = (10^{-9} \; F) \times (10^6 \; \Omega) = 10^{-3} \; s$.
The charge on the capacitor at time $t$ is $q(t) = CV[1 - \exp(-t/\tau)] = (10^{-9} \; F)(2 \; V)[1 - \exp(-t/10^{-3})] = 2 \times 10^{-9} [1 - \exp(-t/10^{-3})] \; C$.
The electric field $E$ between the plates is $E = \frac{q(t)}{\varepsilon_0 A}$,where $A = \pi r_0^2 = \pi(1)^2 = \pi \; m^2$.
Consider a circular loop of radius $r = 0.5 \; m$ passing through point $P$. The electric flux through this loop is $\Phi_E = E \times (\pi r^2) = \frac{q(t)}{\varepsilon_0 A} \times (\pi r^2) = \frac{q(t)}{\varepsilon_0 \pi} \times \pi (0.5)^2 = \frac{q(t)}{4 \varepsilon_0}$.
The displacement current $i_d$ is $i_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 \frac{d}{dt} \left( \frac{q(t)}{4 \varepsilon_0} \right) = \frac{1}{4} \frac{dq}{dt}$.
Since $\frac{dq}{dt} = \frac{d}{dt} [CV(1 - e^{-t/\tau})] = \frac{CV}{\tau} e^{-t/\tau} = \frac{V}{R} e^{-t/\tau}$,at $t = 10^{-3} \; s$ (which is $t = \tau$),$\frac{dq}{dt} = \frac{2 \; V}{10^6 \; \Omega} e^{-1} = 2 \times 10^{-6} e^{-1} \; A$.
Thus,$i_d = \frac{1}{4} (2 \times 10^{-6} e^{-1}) = 0.5 \times 10^{-6} e^{-1} \; A$.
Using the Ampere-Maxwell law,$\oint B \cdot dl = \mu_0 i_d \implies B(2 \pi r) = \mu_0 i_d$.
$B = \frac{\mu_0 i_d}{2 \pi (0.5)} = \frac{\mu_0 i_d}{\pi} = \frac{4 \pi \times 10^{-7} \times 0.5 \times 10^{-6} \times e^{-1}}{\pi} = 2 \times 10^{-13} \times e^{-1} \; T$.
Using $e^{-1} \approx 0.3678$,$B \approx 2 \times 10^{-13} \times 0.3678 \approx 0.7356 \times 10^{-13} \; T \approx 0.74 \times 10^{-13} \; T$.

(N/A) Radius of each circular plate,$r = 12 \; cm = 0.12 \; m$
Distance between the plates,$d = 5 \; cm = 0.05 \; m$
Charging current,$I = 0.15 \; A$
Permittivity of free space,$\varepsilon_{0} = 8.85 \times 10^{-12} \; C^{2} \; N^{-1} \; m^{-2}$
$(a)$ Capacitance $C$ is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A = \pi r^{2}$.
$C = \frac{8.85 \times 10^{-12} \times \pi \times (0.12)^{2}}{0.05} \approx 8.0032 \times 10^{-12} \; F = 80.032 \; pF$.
Since $q = CV$,differentiating with respect to time gives $\frac{dq}{dt} = C \frac{dV}{dt}$.
Given $\frac{dq}{dt} = I$,we have $\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{80.032 \times 10^{-12}} \approx 1.87 \times 10^{9} \; V/s$.
$(b)$ The displacement current $i_{d}$ between the plates is equal to the conduction current $I$ in the connecting wires. Thus,$i_{d} = 0.15 \; A$.
$(c)$ Yes,Kirchhoff's first rule is valid at each plate of the capacitor if we consider the total current as the sum of conduction current and displacement current.

(N/A) Radius of each circular plate,$R = 6.0 \; cm = 0.06 \; m$
Capacitance of a parallel plate capacitor,$C = 100 \; pF = 100 \times 10^{-12} \; F$
Supply voltage,$V = 230 \; V$
Angular frequency,$\omega = 300 \; rad \; s^{-1}$
$(a)$ The rms value of conduction current is given by $I = \frac{V}{X_C}$,where $X_C = \frac{1}{\omega C}$ is the capacitive reactance.
Therefore,$I = V \omega C = 230 \times 300 \times 100 \times 10^{-12} \; A = 6.9 \times 10^{-6} \; A = 6.9 \; \mu A$.
$(b)$ Yes,the conduction current is equal to the displacement current in the circuit.
$(c)$ The magnetic field $B$ at a distance $r$ from the axis between the plates is given by $B = \frac{\mu_0 r}{2 \pi R^2} I_0$,where $I_0 = \sqrt{2} I$ is the peak current.
Given $r = 3.0 \; cm = 0.03 \; m$ and $\mu_0 = 4 \pi \times 10^{-7} \; T \; m \; A^{-1}$.
$B = \frac{4 \pi \times 10^{-7} \times 0.03 \times \sqrt{2} \times 6.9 \times 10^{-6}}{2 \pi \times (0.06)^2} \approx 1.63 \times 10^{-11} \; T$.

(B) Ampere's circuital law,in its original form,states that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c$,where $I_c$ is the conduction current.
Maxwell observed that this law was inconsistent for non-steady currents,such as in a charging capacitor,where the current is not continuous through the dielectric.
To resolve this discrepancy,Maxwell proposed that a time-varying electric field acts as a source of magnetic field,similar to how a conduction current does.
He introduced a new term called 'displacement current' $(I_d = \epsilon_0 \frac{d\Phi_E}{dt})$.
Thus,the modified Ampere-Maxwell law is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d)$.

(N/A) Maxwell's equations are a set of four fundamental equations that describe how electric and magnetic fields are generated by charges,currents,and by each other. These equations are as follows:
$(1)$ Gauss's Law for Electricity: $\oint \overrightarrow{E} \cdot d\overrightarrow{A} = \frac{Q}{\epsilon_{0}}$
$(2)$ Gauss's Law for Magnetism: $\oint \overrightarrow{B} \cdot d\overrightarrow{A} = 0$
$(3)$ Faraday's Law of Induction: $\oint \overrightarrow{E} \cdot d\overrightarrow{l} = -\frac{d\Phi_{B}}{dt}$
$(4)$ Ampere-Maxwell Law: $\oint \overrightarrow{B} \cdot d\overrightarrow{l} = \mu_{0} i_{C} + \mu_{0} \epsilon_{0} \frac{d\Phi_{E}}{dt}$
By using these equations along with the Lorentz force law,all basic principles of electromagnetism can be mathematically represented.

(N/A) The most significant inference from Maxwell's equations is the existence of electromagnetic waves,which represent the coupled propagation of time-varying electric and magnetic fields in space.
Due to the propagation of these time-varying fields,electromagnetic waves are generated that travel through space at the speed of light,which is $c = 3 \times 10^{8} \text{ m/s}$.
Thus,Maxwell established a fundamental relationship between electric,magnetic,and optical phenomena.
In $1887$,the scientist Heinrich Hertz experimentally produced electromagnetic waves in the laboratory,confirming Maxwell's theoretical predictions. The subsequent application of this technology revolutionized modern communication systems.

(A) The original Ampere's circuital law is given by $\oint B \cdot dl = \mu_0 I_c$,where $I_c$ is the conduction current.
Maxwell observed that this law was inconsistent for non-steady currents,such as in a charging capacitor.
He proposed that a changing electric field creates a current,which he called the displacement current,denoted by $I_d = \epsilon_0 \frac{d\phi_E}{dt}$.
Thus,the modified Maxwell-Ampere law is $\oint B \cdot dl = \mu_0 (I_c + I_d)$.
Therefore,the missing term identified by Maxwell is the displacement current.

(A) Maxwell's equations are a set of four fundamental equations that describe how electric and magnetic fields are generated by charges,currents,and by each other. These are:
$1$. Gauss's Law for Electricity: $\oint E \cdot dA = \frac{q_{enclosed}}{\epsilon_0}$
$2$. Gauss's Law for Magnetism: $\oint B \cdot dA = 0$
$3$. Faraday's Law of Induction: $\oint E \cdot dl = -\frac{d\Phi_B}{dt}$
$4$. Ampere-Maxwell Law: $\oint B \cdot dl = \mu_0 I_{enclosed} + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$
Therefore,the correct option is $A$.

(B) In $1887$,the German physicist $Heinrich \text{ } Hertz$ successfully generated and detected electromagnetic waves in the laboratory for the first time. This experiment confirmed the theoretical predictions made by $James \text{ } Clerk \text{ } Maxwell$ regarding the existence of electromagnetic waves and their propagation at the speed of light.

(N/A) Maxwell logically demonstrated that a changing electric field induces a magnetic field. This phenomenon is of fundamental importance in physics.
Historically, the connection between electricity and magnetism was first observed by Oersted, who showed that a current-carrying wire produces a magnetic field.
This phenomenon is crucial for explaining the existence and propagation of electromagnetic waves, including radio waves, gamma rays, and visible light.
Specifically, the production of a magnetic field by a changing electric field is mathematically described by the displacement current term added to Ampere's circuital law, which is essential for understanding the charging of a capacitor.

(N/A) When applying Ampere's circuital law to a loop enclosing a charging capacitor, we encounter a contradiction.
Consider a loop of radius $r$ outside the capacitor plates as shown in figure $(a)$. Applying Ampere's circuital law gives $\oint \vec{B} \cdot d\vec{l} = \mu_0 i(t)$, which yields a non-zero magnetic field $B = \frac{\mu_0 i(t)}{2\pi r}$.
Now, consider a surface bounded by the same loop but passing through the region between the capacitor plates, as shown in figures $(b)$ and $(c)$. Since there is no conduction current flowing between the plates, the current enclosed by this surface is zero $(\sum I = 0)$.
Applying Ampere's circuital law to this surface gives $\oint \vec{B} \cdot d\vec{l} = \mu_0 (0) = 0$, which implies $B = 0$.
This is a contradiction because the magnetic field at the same point $P$ cannot be both non-zero and zero. This inconsistency led Maxwell to propose the existence of displacement current $i_d = \epsilon_0 \frac{d\Phi_E}{dt}$ to complete the law.

(N/A) Ampere's circuital law states that $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$. However,this law is inconsistent for time-varying electric fields,such as inside a charging capacitor.
Consider a parallel plate capacitor with plate area $A$ and charge $Q$. The electric field $E$ between the plates is given by $E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A \epsilon_0}$.
The electric flux $\Phi_E$ through a surface $A$ inside the capacitor is $\Phi_E = E A = \left( \frac{Q}{A \epsilon_0} \right) A = \frac{Q}{\epsilon_0}$.
Therefore,$Q = \epsilon_0 \Phi_E$.
Differentiating with respect to time $t$,we get $\frac{dQ}{dt} = \epsilon_0 \frac{d\Phi_E}{dt}$.
Since the conduction current $I_c = \frac{dQ}{dt}$,we define the displacement current $I_d$ as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$.
Definition: Displacement current is the current that arises due to the time-varying electric field in a region,even in the absence of actual charge carriers.
$SI$ Unit: The $SI$ unit of displacement current is the Ampere $(A)$.

(N/A) The displacement current allows the laws of electric and magnetic fields to be represented more symmetrically.
Faraday's law of electromagnetic induction states that a time-varying magnetic field produces an electric field.
Maxwell's modification to Ampere's law (the Ampere-Maxwell law) states that a time-varying electric field (represented by displacement current) produces a magnetic field.
Thus,electric and magnetic fields changing with time produce each other.
This symmetry in the laws of electromagnetism leads to the prediction of the existence of electromagnetic waves,which propagate through space as self-sustaining oscillations of electric and magnetic fields.

(N/A) $(1)$ Gauss's law for electricity: $\oint \overrightarrow{E} \cdot d \overrightarrow{A} = \frac{Q}{\epsilon_{0}}$
$(2)$ Gauss's law for magnetism: $\oint \overrightarrow{B} \cdot d \overrightarrow{A} = 0$
$(3)$ Faraday's law: $\oint \overrightarrow{E} \cdot d \vec{l} = -\frac{d \Phi_{B}}{d t}$
$(4)$ Ampere-Maxwell law: $\oint \overrightarrow{B} \cdot d \vec{l} = \mu_{0} i_{c} + \mu_{0} \epsilon_{0} \frac{d \Phi_{E}}{d t}$
Where:
$\overrightarrow{E} = \text{electric field}$
$\overrightarrow{B} = \text{magnetic field}$
$d \overrightarrow{A} = \text{area element}$
$d \vec{l} = \text{line element}$
$\frac{d \Phi_{B}}{d t} = \text{rate of change of magnetic flux}$
$\frac{d \Phi_{E}}{d t} = \text{rate of change of electric flux}$
$i_{c} = \text{conduction current}$
$i_{d} = \epsilon_{0} \frac{d \Phi_{E}}{d t} = \text{displacement current}$
$\mu_{0} = \text{permeability of vacuum}$