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

(A) James Clerk Maxwell proposed the concept of displacement current in $1861$. He realized that a time-varying electric field acts as a source of a magnetic field,just as a conduction current does. This modification to Ampere's law,known as the Ampere-Maxwell law,is expressed as $\oint B \cdot dl = \mu_0 (I_c + I_d)$,where $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ is the displacement current. Therefore,the correct answer is $A$.

(N/A) When a capacitor is being charged, the current flowing inside the capacitor is known as the $displacement \ current$ $(I_d)$.
Explanation:
$1$. In the conducting wires connected to the capacitor plates, the current is the $conduction \ current$ $(I_c)$, which is due to the actual flow of electrons.
$2$. Between the plates of the capacitor, there is no physical medium for electrons to flow, so the $conduction \ current$ is zero.
$3$. However, as the capacitor charges, the electric field $(E)$ between the plates changes with time.
$4$. According to Maxwell's modification of Ampere's Law, this time-varying electric field produces a magnetic field, which is equivalent to a current.
$5$. This current is defined as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, where $\epsilon_0$ is the permittivity of free space and $\frac{d\Phi_E}{dt}$ is the rate of change of electric flux.
$6$. Thus, the $displacement \ current$ ensures the continuity of the total current in the circuit.

(B) Ampere's circuital law states that $\oint B \cdot dl = \mu_0 I_{enclosed}$.
Consider a parallel plate capacitor being charged. If we choose a flat circular surface between the plates,the current passing through it is zero,implying $B = 0$.
However,if we choose a balloon-shaped surface with the same boundary,the conduction current $I_c$ passes through it,implying $B \neq 0$.
This contradiction arises because Ampere's law does not account for the time-varying electric field between the plates,which acts as a displacement current $I_d = \epsilon_0 \frac{d\phi_E}{dt}$.
Thus,the law is incomplete for non-steady currents.

(N/A) The Ampere-Maxwell law is a generalization of Ampere's circuital law,which accounts for the displacement current. It is expressed as:
$\oint B \cdot dl = \mu_0 (I_c + I_d)$
Where:
$1$. $\oint B \cdot dl$ is the line integral of the magnetic field around a closed loop.
$2$. $\mu_0$ is the permeability of free space.
$3$. $I_c$ is the conduction current.
$4$. $I_d$ is the displacement current,given by $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\epsilon_0$ is the permittivity of free space and $\Phi_E$ is the electric flux.
Thus,the complete equation is: $\oint B \cdot dl = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.

(A) The expression ${\epsilon _0}\left( {\frac{{d{\Phi _E}}}{{dt}}} \right)$ represents the displacement current,denoted by ${I_d}$.
According to Maxwell's modification of Ampere's circuital law,the displacement current is defined as ${I_d} = {\epsilon _0}\frac{{d{\Phi _E}}}{{dt}}$.
Since ${I_d}$ is a current,its $SI$ unit is the same as that of electric current.
The $SI$ unit of electric current is the Ampere $(A)$.

(A) Ampere's circuital law states that the line integral of the magnetic field $\vec{B}$ around any closed path is equal to $\mu_0$ times the total current $I$ passing through the surface enclosed by the path.
Maxwell noticed that this law was inconsistent for time-varying electric fields.
He introduced the concept of displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$ to maintain the continuity of current.
The modified Ampere's circuital law is given by $\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d) = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.
Therefore, the missing term that Maxwell added to the original law is $\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.

(B) Displacement current is defined as the current that arises in a region where the electric field is changing with time.
It was introduced by James Clerk Maxwell to maintain the consistency of Ampere's circuital law.
The mathematical expression for displacement current is given by $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\epsilon_0$ is the permittivity of free space and $\frac{d\Phi_E}{dt}$ is the rate of change of electric flux.
Unlike conduction current,it does not involve the actual flow of charge carriers but is associated with the time-varying electric field.

(N/A) The equation $i = i_c + i_d$ represents the total current in a circuit,which is the sum of the conduction current $(i_c)$ and the displacement current $(i_d)$.
$1$. Conduction current $(i_c)$: This is the current due to the actual flow of charges (electrons) through a conductor.
$2$. Displacement current $(i_d)$: This is the current that arises due to the time-varying electric field in a region,such as between the plates of a capacitor.
This equation was proposed by James Clerk Maxwell to generalize Ampere's Circuital Law,ensuring that the total current is continuous even in regions where there is no physical charge flow,such as across a capacitor gap.

(N/A) Stationary charges or charges moving with constant velocity (steady current) cannot be the source of electromagnetic waves. This is because a stationary charge produces only an electric field,and charges moving with constant velocity produce both electric and magnetic fields,but these fields do not change with time.
According to Maxwell,accelerated charges produce electromagnetic waves.
An oscillating charge is an example of accelerated motion. When a charge oscillates with a certain frequency,it produces electric and magnetic fields that oscillate in space. As this phenomenon repeats,the oscillating electric and magnetic fields propagate through space,which are called electromagnetic waves.
These waves propagate in a direction perpendicular to both the electric and magnetic fields. The electric and magnetic fields oscillate perpendicular to each other.
The frequency of the electromagnetic waves is equal to the frequency of the oscillating charge. The energy of the accelerated charge is imparted to the propagating electromagnetic waves.
It is easy to imagine that light consists of electromagnetic waves,but it is difficult to test this in a laboratory because modern electronic circuits can typically produce frequencies up to $10^{11} \text{ Hz}$,whereas the frequency of yellow light in the visible spectrum is about $6 \times 10^{14} \text{ Hz}$. Hence,to provide experimental proof of electromagnetic waves,Hertz's experiment was performed in the lower range of radio waves.
After the success of Hertz's experiment,Jagdish Chandra Bose,working in Kolkata,succeeded in producing and observing electromagnetic waves of much shorter wavelengths ($25 \text{ mm}$ to $5 \text{ mm}$). His experiment was confined to the laboratory. During this time,the Italian scientist Marconi succeeded in sending electromagnetic waves over several miles.

(C) According to Maxwell's theory of electromagnetism,an electric charge at rest produces only an electric field.
When a charge moves with a constant velocity,it produces both an electric field and a magnetic field,but it does not radiate energy.
However,when a charge is accelerated,it produces a time-varying electric field and a time-varying magnetic field,which propagate through space as electromagnetic waves.
Therefore,acceleration is the necessary condition for the emission of electromagnetic waves by charges.

(B) In $1887$,the German physicist $Heinrich \text{ } Hertz$ successfully produced and detected electromagnetic waves in the laboratory for the first time. This experimental verification confirmed the theoretical predictions made by $James \text{ } Clerk \text{ } Maxwell$ regarding the existence of electromagnetic waves. Therefore,the correct option is $B$.

(N/A) The primary difficulty in establishing the wave theory of light was that,since waves traditionally required a medium for propagation,it was unclear how light waves could travel through a vacuum.
James Clerk Maxwell resolved this by developing a set of equations describing electricity and magnetism. From these equations,he derived a wave equation that predicted the existence of electromagnetic waves.
Maxwell calculated the speed of these electromagnetic waves in free space and found that the theoretical value matched the measured speed of light. Consequently,he proposed that light is an electromagnetic wave.
According to Maxwell,light waves consist of oscillating electric and magnetic fields. $A$ changing electric field generates a time-varying magnetic field,and a changing magnetic field generates a time-varying electric field. This self-sustaining process allows electromagnetic waves to propagate through a vacuum without a material medium.
While ray optics (based on rectilinear propagation) explains phenomena like reflection and refraction,wave optics explains phenomena like diffraction and interference,where light bends around obstacles comparable in size to its wavelength (approximately $0.5 \ \mu m$).

(A) The existence of electromagnetic waves was theoretically predicted by James Clerk Maxwell in $1865$. However,it was Heinrich Hertz who experimentally produced and detected electromagnetic waves for the first time in $1887$ using a spark-gap transmitter. This experiment confirmed Maxwell's theory.

(A) In the early experiments of radio communication,specifically those conducted by Jagadish Chandra Bose and Guglielmo Marconi,a $Coherer$ was used as a detector for radio waves. $A$ $Coherer$ consists of a tube filled with metal filings that change their electrical resistance when exposed to electromagnetic waves,allowing the detection of the signal.

(B) The speed of electromagnetic waves in a vacuum is given by the relation derived from Maxwell's equations.
According to the theory of electromagnetism,the speed of light $c$ in free space is related to the permeability of free space ${\mu _0}$ and the permittivity of free space ${\epsilon _0}$ by the formula:
$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$
This relationship shows that the speed of light is determined by the fundamental electromagnetic constants of the vacuum.

(A) Radio waves are produced by the rapid acceleration and deceleration of electrons in conducting wires or antennas. When an alternating current $(AC)$ flows through an antenna,the electrons oscillate back and forth. This oscillating motion of charges creates a time-varying electric field,which in turn generates a time-varying magnetic field. These mutually perpendicular,time-varying electric and magnetic fields propagate through space as electromagnetic waves,specifically radio waves.

(B) The displacement current $I_d$ in a capacitor is equal to the conduction current $I_c$ flowing through the connecting wires,which is defined as the rate of change of charge on the plates.
$I_d = I_c = \frac{dq}{dt}$
Given the charge $q = q_0 \cos(2\pi \nu t)$,we differentiate with respect to time $t$:
$I_d = \frac{d}{dt} [q_0 \cos(2\pi \nu t)]$
Using the chain rule,the derivative of $\cos(kx)$ is $-k \sin(kx)$:
$I_d = q_0 \cdot [-\sin(2\pi \nu t)] \cdot (2\pi \nu)$
$I_d = -2\pi \nu q_0 \sin(2\pi \nu t)$

(B) The capacitive reactance $X_{c}$ is given by $X_{c} = \frac{1}{2 \pi f C}$.
The conduction current $I_{c}$ in the circuit is given by $I_{c} = \frac{V}{X_{c}} = V(2 \pi f C)$.
According to Maxwell's modification of Ampere's law,the displacement current $I_{d}$ is equal to the conduction current $I_{c}$ in the circuit,so $I_{d} = I_{c} = 2 \pi f C V$.
Since $I_{d} \propto f$,if the frequency $f$ decreases,the displacement current $I_{d}$ will also decrease.

(N/A) Consider a circular loop of radius $r$ between the plates of a parallel-plate capacitor. According to the Ampere-Maxwell law,the line integral of the magnetic field around this loop is given by:
$\oint B \cdot dl = \mu_0 I_d$
$B(2\pi r) = \mu_0 \epsilon_0 \frac{d\phi_E}{dt}$
Since the electric field $E$ is uniform between the plates,the electric flux $\phi_E$ through the circular loop of radius $r$ is $\phi_E = E \cdot A = E(\pi r^2)$.
Substituting this into the equation:
$B(2\pi r) = \mu_0 \epsilon_0 \frac{d}{dt}(E \pi r^2)$
$B(2\pi r) = \mu_0 \epsilon_0 \pi r^2 \frac{dE}{dt}$
Dividing both sides by $2\pi r$,we get:
$B = \frac{\mu_0 \epsilon_0 r}{2} \cdot \frac{dE}{dt}$

(N/A) The displacement current $I_d$ in a capacitor is given by the formula $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,which is equivalent to $I_d = C \frac{dV}{dt}$.
Given:
Capacitance $C = 2 \,\mu F = 2 \times 10^{-6} \, F$
Displacement current $I_d = 1 \, mA = 10^{-3} \, A$
We need to find the rate of change of potential difference $\frac{dV}{dt}$.
Using the relation $I_d = C \frac{dV}{dt}$,we get:
$\frac{dV}{dt} = \frac{I_d}{C} = \frac{10^{-3}}{2 \times 10^{-6}} = 0.5 \times 10^3 = 500 \, V/s$.
Therefore,by changing the potential difference across the plates at a rate of $500 \, V/s$,an instantaneous displacement current of $1 \, mA$ can be established.

(N/A) The electric field $\vec{E}$ at a radial distance $a$ from an infinitely long charged wire with linear charge density $\lambda$ is given by Gauss's Law as $\vec{E} = \frac{\lambda}{2\pi\epsilon_0 a} \hat{r}$,where $\hat{r}$ is the radial unit vector in the $xy-$plane.
The moving charged wire constitutes a current $I = \lambda v$. The magnetic field $\vec{B}$ at a distance $a$ from the wire is given by Ampere's Law as $\vec{B} = \frac{\mu_0 I}{2\pi a} \hat{\phi} = \frac{\mu_0 \lambda v}{2\pi a} \hat{\phi}$,where $\hat{\phi}$ is the azimuthal unit vector.
The Poynting vector is defined as $\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$.
Substituting the expressions for $\vec{E}$ and $\vec{B}$:
$\vec{S} = \frac{1}{\mu_0} \left( \frac{\lambda}{2\pi\epsilon_0 a} \hat{r} \times \frac{\mu_0 \lambda v}{2\pi a} \hat{\phi} \right)$
$\vec{S} = \frac{\lambda^2 v}{4\pi^2 \epsilon_0 a^2} (\hat{r} \times \hat{\phi})$
Since $\hat{r} \times \hat{\phi} = \hat{k}$ (the unit vector along the $z-$axis),
$\vec{S} = \frac{\lambda^2 v}{4\pi^2 \epsilon_0 a^2} \hat{k}$.

(4/9) Let the distance between the two plates of the capacitor be $d$ and the voltage applied be $V(t) = V_0 \sin(2\pi vt)$.
The electric field is given by $E = \frac{V_0}{d} \sin(2\pi vt)$.
By Ohm's law,the conduction current density is $J_c = \sigma E = \frac{E}{\rho} = \frac{V_0}{\rho d} \sin(2\pi vt)$.
Let $J_0^c = \frac{V_0}{\rho d}$,so $J_c = J_0^c \sin(2\pi vt)$.
The displacement current density is $J_d = \epsilon \frac{\partial E}{\partial t} = \epsilon \frac{\partial}{\partial t} \left[ \frac{V_0}{d} \sin(2\pi vt) \right] = \frac{\epsilon (2\pi v) V_0}{d} \cos(2\pi vt)$.
Let $J_0^d = \frac{2\pi v \epsilon V_0}{d}$,so $J_d = J_0^d \cos(2\pi vt)$.
The ratio of the amplitudes is $\frac{J_0^d}{J_0^c} = \frac{2\pi v \epsilon V_0}{d} \times \frac{\rho d}{V_0} = 2\pi v \epsilon \rho$.
Substituting the given values: $\epsilon = 80\epsilon_0$,$\rho = 0.25\,\Omega m$,and $v = 4 \times 10^8\, Hz$.
$\frac{J_0^d}{J_0^c} = 2\pi v (80\epsilon_0) \rho = 160\pi \epsilon_0 v \rho$.
Using $4\pi\epsilon_0 = \frac{1}{9 \times 10^9}$,we have $160\pi\epsilon_0 = 40 \times (4\pi\epsilon_0) = \frac{40}{9 \times 10^9}$.
$\frac{J_0^d}{J_0^c} = \frac{40}{9 \times 10^9} \times (4 \times 10^8) \times 0.25 = \frac{40 \times 4 \times 10^8 \times 0.25}{9 \times 10^9} = \frac{40}{90} = \frac{4}{9}$.

(N/A) $(i)$ The displacement current density is given by $\vec{J}_d = \epsilon_0 \frac{\partial \vec{E}}{\partial t}$.
Given $\vec{E}(s,t) = \mu_0 I_0 \nu \cos(2\pi \nu t) \ln(s/a) \hat{k}$.
$\vec{J}_d = \epsilon_0 \frac{\partial}{\partial t} [\mu_0 I_0 \nu \cos(2\pi \nu t) \ln(s/a) \hat{k}] = \epsilon_0 \mu_0 I_0 \nu \ln(s/a) \hat{k} \cdot \frac{d}{dt} [\cos(2\pi \nu t)]$.
Since $\epsilon_0 \mu_0 = 1/c^2$,we have $\vec{J}_d = \frac{1}{c^2} I_0 \nu \ln(s/a) \hat{k} \cdot (-2\pi \nu \sin(2\pi \nu t)) = -\frac{2\pi \nu^2 I_0}{c^2} \ln(s/a) \sin(2\pi \nu t) \hat{k}$.
Using $\lambda = c/\nu$,$\vec{J}_d = \frac{2\pi I_0}{\lambda^2} \ln(a/s) \sin(2\pi \nu t) \hat{k}$.
$(ii)$ The total displacement current $I_d = \int \vec{J}_d \cdot d\vec{A} = \int_0^a J_d (2\pi s ds)$.
$I_d = \int_0^a \frac{2\pi I_0}{\lambda^2} \ln(a/s) \sin(2\pi \nu t) (2\pi s ds) = \frac{4\pi^2 I_0 \sin(2\pi \nu t)}{\lambda^2} \int_0^a s \ln(a/s) ds$.
Let $x = s/a$,then $ds = a dx$. The integral becomes $a^2 \int_0^1 x \ln(1/x) dx = a^2 \int_0^1 -x \ln x dx = a^2 [1/4] = a^2/4$.
Thus,$I_d = \frac{4\pi^2 I_0 \sin(2\pi \nu t)}{\lambda^2} \cdot \frac{a^2}{4} = I_0 \sin(2\pi \nu t) (\frac{\pi a}{\lambda})^2$.
$(iii)$ Comparing $I_d$ with $I(t) = I_0 \sin(2\pi \nu t)$,we find $I_d = I(t) (\frac{\pi a}{\lambda})^2$. Since $a << \lambda$,the displacement current is much smaller than the conduction current.

(A) In $1887$,while performing experiments to produce electromagnetic waves,Heinrich Hertz observed that the production of sparks across the gap of the receiver was enhanced when the metal surface of the receiver was illuminated by ultraviolet light from an arc lamp. This phenomenon was later identified as the photoelectric effect.

(D) The conduction current density is $J_{c} = \sigma E = \frac{E}{\rho} = \frac{V(t)}{\rho d}$.
The displacement current density is $J_{d} = \varepsilon \frac{\partial E}{\partial t} = \varepsilon \frac{1}{d} \frac{dV}{dt} = \frac{\varepsilon}{d} V_{0} (2 \pi f) \cos(2 \pi ft)$.
Given $J_{c} = 10^{x} J_{d}$,we have $\frac{V_{0} \sin(2 \pi ft)}{\rho d} = 10^{x} \frac{\varepsilon}{d} V_{0} (2 \pi f) \cos(2 \pi ft)$.
This simplifies to $\tan(2 \pi ft) = 10^{x} \cdot \rho \cdot \varepsilon \cdot 2 \pi f$.
Given $f = 900 \, Hz$,$t = \frac{1}{800} \, s$,$\rho = 0.25 \, \Omega m$,and $\varepsilon = 80 \varepsilon_{0}$.
$2 \pi ft = 2 \pi \times 900 \times \frac{1}{800} = \frac{9 \pi}{4} = 2 \pi + \frac{\pi}{4}$.
So,$\tan(2 \pi ft) = \tan(\pi/4) = 1$.
Now,$1 = 10^{x} \times 0.25 \times 80 \varepsilon_{0} \times 2 \pi \times 900$.
Using $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9}$,we have $\varepsilon_{0} = \frac{1}{36 \pi \times 10^{9}}$.
$1 = 10^{x} \times 20 \times \frac{1}{36 \pi \times 10^{9}} \times 1800 \pi = 10^{x} \times \frac{20 \times 1800}{36 \times 10^{9}} = 10^{x} \times \frac{36000}{36 \times 10^{9}} = 10^{x} \times 10^{-6}$.
$10^{x} = 10^{6}$,which gives $x = 6$.

(A) The charge on the capacitor is given by $q = CV$.
Substituting the given voltage $V = V_{0} \sin \omega t$,we get $q = C V_{0} \sin \omega t$.
The displacement current $I_{d}$ is equal to the rate of change of charge on the capacitor plates,which is $I_{d} = \frac{dq}{dt}$.
$I_{d} = \frac{d}{dt} (C V_{0} \sin \omega t)$.
Since $C$ and $V_{0}$ are constants,$I_{d} = C V_{0} \frac{d}{dt} (\sin \omega t)$.
Using the derivative of $\sin \omega t$,which is $\omega \cos \omega t$,we get $I_{d} = C V_{0} \omega \cos \omega t$.
Therefore,$I_{d} = V_{0} \omega C \cos \omega t$.

(D) The capacitance of the parallel plate capacitor is given by $C = \frac{\varepsilon_0 A}{d}$.
Substituting the given values: $C = \frac{8.85 \times 10^{-12} \times 1}{2 \times 10^{-3}} = 4.425 \times 10^{-9} \, F$.
The angular frequency is $\omega = 2 \pi f = 2 \times \pi \times 50 = 100 \pi \, rad/s$.
The amplitude of the current $I_0$ is given by $I_0 = V_0 \omega C$.
Substituting the values: $I_0 = 20 \times (100 \pi) \times (4.425 \times 10^{-9})$.
$I_0 = 2000 \times 3.14159 \times 4.425 \times 10^{-9} \approx 27.79 \times 10^{-6} \, A$.
Therefore,the amplitude of the displacement current is $27.79 \, \mu A$.

(C) The displacement current $I_{d}$ in a parallel plate capacitor is given by the formula: $I_{d} = \varepsilon_{0} A \frac{dE}{dt} = \varepsilon_{0} A \frac{d}{dt} (\frac{V}{d}) = \frac{\varepsilon_{0} A}{d} \frac{dV}{dt}$.
Given values:
$I_{d} = 4.425 \times 10^{-6} \,A$
$\frac{dV}{dt} = 10^{6} \,V s^{-1}$
$A = 40 \,cm^{2} = 40 \times 10^{-4} \,m^{2} = 4 \times 10^{-3} \,m^{2}$
$\varepsilon_{0} = 8.85 \times 10^{-12} \,C^{2} N^{-1} m^{-2}$
Rearranging the formula to solve for $d$:
$d = \frac{\varepsilon_{0} A}{I_{d}} \frac{dV}{dt}$
Substituting the values:
$d = \frac{8.85 \times 10^{-12} \times 4 \times 10^{-3} \times 10^{6}}{4.425 \times 10^{-6}}$
$d = \frac{8.85 \times 4 \times 10^{-9}}{4.425 \times 10^{-6}}$
$d = 2 \times 4 \times 10^{-3} \,m = 8 \times 10^{-3} \,m$
Comparing this with $x \times 10^{-3} \,m$,we get $x = 8$.

(C-D) According to the modified Maxwell-Ampere law for a world with magnetic charges,the magnetic $emf$ is given by $\oint B \cdot dl = \mu_{0} \epsilon_{0} \frac{d\phi_{E}}{dt}$.
The electric flux $\phi_{E}$ through the circuit is $\phi_{E} = E \cdot A = E \cdot (l \cdot x)$,where $x$ is the position of the rod.
The rate of change of electric flux is $\frac{d\phi_{E}}{dt} = E \cdot l \cdot \frac{dx}{dt} = E \cdot l \cdot v$.
Substituting this into the $emf$ equation,we get $emf = \mu_{0} \epsilon_{0} E l v$.
Since $c^{2} = \frac{1}{\mu_{0} \epsilon_{0}}$,we have $emf = \frac{E v l}{c^{2}}$.
The direction of the induced magnetic current depends on the direction of the electric field $E$. If $E$ is directed outward,the current is clockwise; if $E$ is directed inward,the current is counter-clockwise. Thus,both options $(c)$ and $(d)$ are physically possible depending on the orientation of $E$.

(D) The displacement current $I_d$ is defined by the formula $I_d = \varepsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E$ is the electric flux.
Since the electric flux $\phi_E = E \cdot A = \frac{V}{d} \cdot A$ (where $V$ is the potential difference and $d$ is the distance between plates),the displacement current is proportional to the rate of change of the potential difference: $I_d = \frac{\varepsilon_0 A}{d} \frac{dV}{dt}$.
Therefore,for a displacement current to exist,the potential difference $V$ across the plates must be changing with time (i.e.,$\frac{dV}{dt} \neq 0$).

(A) The total displacement current $I_d$ between the plates is equal to the conduction current $i$ in the wires,so $I_d = i = \varepsilon_0 A \frac{dE}{dt} = \varepsilon_0 (\pi R^2) \frac{dE}{dt}$.
The displacement current density $J_d$ is uniform across the plates,given by $J_d = \frac{I_d}{A} = \frac{i}{\pi R^2}$.
The displacement current $I_d'$ in the annular region between $r = \frac{R}{2}$ and $r = R$ is the integral of the current density over that area:
$I_d' = J_d \times A' = \frac{i}{\pi R^2} \times [\pi R^2 - \pi (\frac{R}{2})^2]$
$I_d' = \frac{i}{\pi R^2} \times [\pi R^2 - \frac{\pi R^2}{4}]$
$I_d' = \frac{i}{\pi R^2} \times [\frac{3}{4} \pi R^2] = \frac{3}{4} i$.

(C) The Hertz experiment was the first experimental demonstration of the existence of electromagnetic waves.
In this experiment,Heinrich Hertz used an induction coil to create a high-voltage spark across a spark gap,which acted as a transmitter to produce electromagnetic waves.
He also used a separate loop with a small spark gap,which acted as a receiver to detect these waves.
Therefore,the Hertz experiment is used for both the production and detection of electromagnetic waves.
Thus,the correct option is $(c)$.

(B) According to the principle of continuity of current,the conduction current in the connecting wires is equal to the displacement current between the plates of the capacitor.
Since the conduction current in the circuit is given as $i$,the displacement current $i_d$ between the capacitor plates must also be equal to $i$.
Therefore,$i_d = i$.

(A) The displacement current $I_d$ in a parallel plate capacitor is equal to the conduction current $I$ flowing through the circuit.
The formula for the current in a capacitor is given by $I = C \frac{dV}{dt}$.
Given that the capacitance $C = \frac{1}{2} \text{ F}$ and the displacement current $I_d = I$,we substitute these values into the equation:
$I = \frac{1}{2} \cdot \frac{dV}{dt}$.
Rearranging the equation to solve for $\frac{dV}{dt}$:
$\frac{dV}{dt} = 2I$.
Therefore,the correct option is $A$.

(A) The four Maxwell's equations are as follows:
$1$. Gauss's Law in Electrostatics: $\oint \vec{E} \cdot d \vec{s} = \frac{q}{\epsilon_0}$ $(A-IV)$
$2$. Faraday's Law of Induction: $\oint \vec{E} \cdot d \vec{l} = -\frac{d \phi_B}{d t}$ $(B-I)$
$3$. Gauss's Law in Magnetism: $\oint \vec{B} \cdot d \vec{A} = 0$ $(C-II)$
$4$. Ampere-Maxwell Law: $\oint \vec{B} \cdot d \vec{l} = \mu_0 i_C + \mu_0 \epsilon_0 \frac{d \phi_E}{d t}$ $(D-III)$
Therefore,the correct matching is $A-IV, B-I, C-II, D-III$.

(C) Maxwell's equations describe how electric and magnetic fields are generated and altered by each other and by charges and currents.
Faraday's Law of Induction,given by $\oint \vec{E} \cdot d\vec{l} = -\frac{\partial \phi_B}{\partial t}$,states that a time-varying magnetic flux induces an electromotive force $(EMF)$.
In static conditions,the magnetic flux $\phi_B$ is constant with respect to time,so $\frac{\partial \phi_B}{\partial t} = 0$,which reduces the equation to $\oint \vec{E} \cdot d\vec{l} = 0$.
While the equation holds true in static conditions,it is specifically defined to describe the phenomenon of electromagnetic induction that occurs only under time-varying conditions.
However,the question asks for an equation that is valid for time-varying conditions but not valid for static conditions in its general form. Actually,all Maxwell's equations are valid for both,but the term involving time variation is only non-zero when conditions are time-varying. Among the choices,the Faraday's law equation explicitly represents the dynamic coupling.

$(A)$ time-varying magnetic field is generated by a time-varying electric current.
$1$. A permanent magnet produces a static magnetic field.
$2$. An electric field changing linearly with time produces a constant displacement current $(I_d = \epsilon_0 \frac{d\Phi_E}{dt})$, which results in a constant magnetic field, not a time-varying one.
$3$. Direct current produces a constant magnetic field.
$4$. A decelerating charged particle is an accelerated charge, which produces time-varying electric and magnetic fields (electromagnetic waves).
$5$. An antenna fed with a digital signal involves rapidly changing currents, which produce time-varying magnetic fields.
Therefore, both $(D)$ and $(E)$ are sources of time-varying magnetic fields. However, based on the provided options, the most appropriate choice is $(D)$ and $(E)$ (Note: If the provided options are restricted, $(D)$ is the primary physical source mentioned).

(A) The Maxwell's equations are as follows:
$1$. Ampere-Maxwell law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d\phi_E}{dt}$ (Matches $A-IV$)
$2$. Faraday's law of induction: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$ (Matches $B-III$)
$3$. Gauss' law for electricity: $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0}$ (Matches $C-I$)
$4$. Gauss' law for magnetism: $\oint \vec{B} \cdot d\vec{A} = 0$ (Matches $D-II$)
Therefore,the correct matching is $A-IV, B-III, C-I, D-II$.

(D) The Maxwell's equations in integral form are as follows:
$A$. Gauss's law of magnetostatics states that the net magnetic flux through any closed surface is zero: $\oint \vec{B} \cdot d\vec{a} = 0$ $(A-II)$.
$B$. Faraday's law of electromagnetic induction states that the induced electromotive force in a closed loop is equal to the negative rate of change of magnetic flux through the loop: $\oint \vec{E} \cdot d\vec{l} = -\frac{d}{dt} \int \vec{B} \cdot d\vec{a}$ $(B-III)$.
$C$. Ampere's law (in its original form) relates the line integral of the magnetic field around a closed loop to the current passing through the surface enclosed by the loop: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I$ $(C-IV)$.
$D$. Gauss's law of electrostatics states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of free space: $\oint \vec{E} \cdot d\vec{a} = \frac{1}{\epsilon_0} \int \rho dV$ $(D-I)$.
Thus,the correct matching is $A-II, B-III, C-IV, D-I$.

(D) The displacement current $(I_d)$ in a capacitor is equal to the conduction current $(I_c)$ flowing through the circuit.
The capacitive reactance $(X_C)$ is given by the formula:
$X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C}$
Given:
$V_{max} = 40 \, V$
$f = 4 \, kHz = 4 \times 10^3 \, Hz$
$C = 12 \, \mu F = 12 \times 10^{-6} \, F$
Calculating $X_C$:
$X_C = \frac{1}{2 \times 3.1416 \times 4 \times 10^3 \times 12 \times 10^{-6}}$
$X_C = \frac{1}{8 \pi \times 12 \times 10^{-3}} = \frac{1}{0.3016} \approx 3.317 \, \Omega$
The maximum current $(I_{max})$ is:
$I_{max} = \frac{V_{max}}{X_C} = \frac{40}{3.317} \approx 12.06 \, A$
Thus, the maximum displacement current is nearly $12 \, A$.

(A) According to the modified Ampere's circuital law,the total current is the sum of conduction current $(I_C)$ and displacement current $(I_D)$: $\oint B \cdot dl = \mu_0(I_C + I_D)$.
In the wire,there is only conduction current $I_C = I$,and the displacement current $I_D = 0$.
In the gap between the plates,there is no conduction current $(I_C = 0)$,but the changing electric field produces a displacement current $I_D$.
By the principle of continuity of current,the displacement current in the gap must be equal to the conduction current in the wire,i.e.,$I_D = I_C = I$.
This displacement current flows in the same direction as the conduction current to maintain the continuity of the total current in the circuit.

(C) The displacement current density $J_d$ is uniform across the cross-section of the capacitor plates.
$J_d = \frac{I}{A} = \frac{6 \ A}{16 \ cm^2}$.
The displacement current $I_d$ through a smaller area $A_0$ is given by $I_d = J_d \times A_0$.
$I_d = \left( \frac{6}{16} \right) \times 3.2 \ A$.
$I_d = 0.375 \times 3.2 \ A = 1.2 \ A$.
Since $1 \ A = 1000 \ mA$,the displacement current is $1.2 \times 1000 \ mA = 1200 \ mA$.

(D) The displacement current $I_d$ in a capacitor is given by the formula: $I_d = C \frac{dV}{dt}$.
Here,$C = 2.5 \mu \text{F} = 2.5 \times 10^{-6} \text{ F}$ and $I_d = 0.25 \text{ mA} = 0.25 \times 10^{-3} \text{ A}$.
Rearranging the formula to find the rate of change of potential difference $\frac{dV}{dt}$:
$\frac{dV}{dt} = \frac{I_d}{C}$.
Substituting the given values:
$\frac{dV}{dt} = \frac{0.25 \times 10^{-3}}{2.5 \times 10^{-6}}$.
$\frac{dV}{dt} = \frac{0.25}{2.5} \times 10^{3} = 0.1 \times 1000 = 100 \text{ V s}^{-1}$.
Thus,the magnitude of the rate of change of the potential difference is $100 \text{ V s}^{-1}$.

(D) According to Maxwell's modification of Ampere's Law,the displacement current $i_d$ is defined as:
$i_d = \varepsilon_0 \frac{d \phi_{E}}{dt}$
Since $i_d$ represents a current,its dimensions are the same as those of electric current.
Therefore,the dimensions of $\left(\varepsilon_0 \frac{d \phi_{E}}{dt}\right)$ are equivalent to the dimensions of electric current.

(C) The displacement current $I_d$ is given by $I_d = \epsilon_0 \frac{d\phi_e}{dt} = \epsilon_0 \frac{d}{dt}(EA \cos 0^{\circ})$.
Since $E = \frac{\sigma}{\epsilon_0}$,we have $I_d = \epsilon_0 A \frac{d}{dt}(\frac{\sigma}{\epsilon_0}) = A \frac{d\sigma}{dt}$.
Given that the surface charge density $\sigma$ increases at a constant rate,$\frac{d\sigma}{dt}$ is constant,so $I_d$ is constant.
Using Ampere-Maxwell law,the magnetic field $B$ at a distance $r$ from the axis (inside the plates,$r < R$) is $B(2\pi r) = \mu_0 I_{enclosed} = \mu_0 I_d (\frac{\pi r^2}{\pi R^2})$.
Thus,$B = \frac{\mu_0 I_d r}{2\pi R^2}$.
For $r > R$ (outside the plates),$B(2\pi r) = \mu_0 I_d$,so $B = \frac{\mu_0 I_d}{2\pi r}$.
The magnetic field is non-zero everywhere and reaches its maximum value at the boundary $r = R$ (the imaginary cylindrical surface connecting the peripheries of the plates).

(A) The displacement current $I_d$ in a capacitor is equal to the conduction current $I_c$ flowing through the connecting wires,which is given by the formula $I_d = I_c = C \frac{dV}{dt}$.
Given:
Capacitance $C = 1 \ \mu F = 1 \times 10^{-6} \ F$.
Rate of change of voltage $\frac{dV}{dt} = 4 \ V/s$.
Substituting these values into the formula:
$I_d = (1 \times 10^{-6} \ F) \times (4 \ V/s) = 4 \times 10^{-6} \ A = 4 \ \mu A$.
Therefore,the displacement current is $4 \ \mu A$.

(B) The displacement current $I_d$ is defined as $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the surface.
For a parallel plate capacitor,the electric field $E$ between the plates is uniform and given by $E = \frac{q}{\epsilon_0 A}$,where $q$ is the charge on the plates.
The electric flux $\Phi_E$ through an area $A'$ parallel to the plates is $\Phi_E = E \cdot A' = \frac{q}{\epsilon_0 A} \cdot A'$.
Substituting $A' = A/2$,we get $\Phi_E = \frac{q}{\epsilon_0 A} \cdot \frac{A}{2} = \frac{q}{2\epsilon_0}$.
The displacement current is $I_d = \epsilon_0 \frac{d}{dt} \left( \frac{q}{2\epsilon_0} \right) = \frac{1}{2} \frac{dq}{dt}$.
Since the charging current $I = \frac{dq}{dt}$,we have $I_d = \frac{I}{2}$.

(C) The existence of electromagnetic waves was first experimentally confirmed by the German physicist Heinrich Hertz in $1887$. He used a spark-gap transmitter to produce and detect electromagnetic waves,thereby validating Maxwell's theoretical predictions.

(A) The correct answer is $A$.
Displacement current $(i_{d})$ is defined as the current that arises due to a time-varying electric field.
The mathematical expression for displacement current is given by:
$i_{d} = \varepsilon_{0} \frac{d}{dt}(\phi_{E})$
where $\phi_{E}$ is the electric flux.
Since $\phi_{E} = E \cdot A$,for a constant area $A$,the expression becomes:
$i_{d} = \varepsilon_{0} A \frac{dE}{dt}$
Thus,a changing electric field is the source of displacement current.

(A) In Hertz's experiment,the two metal rods connected to the high-voltage induction coil act as the plates of a capacitor. When the potential difference between the rods becomes sufficiently high,the air between them breaks down,causing a spark to jump across the gap. This oscillation of charge creates electromagnetic waves.