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

(D) The Ampere-Maxwell Law is a modification of Ampere's Law that accounts for the displacement current.
The mathematical expression is given by:
$\oint \vec{B} \cdot d\vec{l} = \mu_{0} i_{c} + \mu_{0} \varepsilon_{0} \frac{d\phi_{E}}{dt}$
where:
- $\oint \vec{B} \cdot d\vec{l}$ is the line integral of the magnetic field.
- $i_{c}$ is the conduction current.
- $\varepsilon_{0} \frac{d\phi_{E}}{dt}$ is the displacement current $(i_{d})$.
- $\mu_{0}$ is the permeability of free space.
- $\varepsilon_{0}$ is the permittivity of free space.
Therefore,option $D$ is the correct representation.

(B) According to the principles of electromagnetism,electromagnetic waves are produced by accelerated charges.
An electric charge at rest produces only a static electric field.
$A$ charge moving with a constant velocity produces a steady electric current,which creates a constant magnetic field,but does not radiate energy as electromagnetic waves.
$A$ charge moving in a circular orbit is undergoing centripetal acceleration,which is a form of accelerated motion.
Therefore,a charge moving in a circular orbit acts as a source of electromagnetic waves.

(C) According to the principles of electromagnetism,a charge at rest produces only an electric field. $A$ charge moving with uniform velocity produces both an electric field and a magnetic field,but it does not radiate energy. An accelerating charge produces a time-varying electric field,which in turn produces a time-varying magnetic field. This continuous process results in the propagation of electromagnetic waves through space.

(B) Maxwell's equations are as follows:
$1$. Gauss's Law for electricity: $\oint \vec{E} \cdot d \vec{A} = \frac{Q}{\varepsilon_0}$
$2$. Gauss's Law for magnetism: $\oint \vec{B} \cdot d \vec{A} = 0$
$3$. Faraday's Law of induction: $\oint \vec{E} \cdot d \vec{\ell} = -\frac{d \phi_B}{dt}$
$4$. Ampere-Maxwell Law: $\oint \vec{B} \cdot d \vec{\ell} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d \phi_E}{dt}$
Comparing these with the given options,the equation $\oint \vec{B} \cdot d \vec{A} = \frac{Q}{\varepsilon_0}$ is incorrect because the magnetic flux through a closed surface is always zero.

(C) The modified Ampere-Maxwell law is given by the expression: $\oint \vec{B} \cdot d\vec{l} = \mu_0 (i_c + i_d)$.
Here,$i_d$ is the displacement current,which is defined as $i_d = \epsilon_0 \frac{d\phi_E}{dt}$.
Substituting this into the equation,we get: $\oint \vec{B} \cdot d\vec{l} = \mu_0 \left( i_c + \epsilon_0 \frac{d\phi_E}{dt} \right)$.
From this expression,it is evident that a time-varying electric field (represented by $\frac{d\phi_E}{dt}$) produces a magnetic field. This is the essence of the modified Ampere's law.

(C) According to Maxwell's theory,a charge at rest produces only an electric field.
$A$ charge moving with uniform velocity produces both an electric and a magnetic field,but these fields do not vary with time in a way that generates electromagnetic waves.
When a charge moves with acceleration or deceleration,it produces a time-varying electric field and a time-varying magnetic field.
These time-varying fields propagate through space as electromagnetic waves.
Therefore,electromagnetic waves are produced by accelerated or decelerated charges only.

(C) According to the classical theory of electromagnetism,an electric charge at rest produces only an electric field.
An electric charge moving with a constant velocity produces both an electric field and a magnetic field,but it does not radiate energy.
However,an accelerated electric charge produces a time-varying electric field,which in turn produces a time-varying magnetic field.
These time-varying fields propagate through space as electromagnetic waves.
Therefore,an accelerated electric charge is a source of electromagnetic waves.

(C) $A \rightarrow (iii)$: According to Gauss's law for electricity,the total electric flux through a closed surface is $\oint E \cdot dA = \frac{Q}{\varepsilon_0}$.
$B \rightarrow (i)$: According to Gauss's law for magnetism,the net magnetic flux through any closed surface is zero,i.e.,$\oint B \cdot dA = 0$.
$C \rightarrow (ii)$: According to Faraday's law of electromagnetic induction,the induced electromotive force $(EMF)$ is the line integral of the electric field,$\oint E \cdot dl = -\frac{d\phi_B}{dt}$.
$D \rightarrow (iv)$: According to the Ampere-Maxwell law,the line integral of the magnetic field is $\oint B \cdot dl = \mu_0(i_c + i_d)$,where $i_c$ is conduction current and $i_d$ is displacement current.

(C) Given:
Radius of plates, $r = 2 \,cm = 2 \times 10^{-2} \,m$
Distance between plates, $d = 0.1 \,mm = 10^{-4} \,m$
Rate of change of potential difference, $\frac{dV}{dt} = 5 \times 10^6 \,Vs^{-1}$
Permittivity of free space, $\varepsilon_0 = 8.85 \times 10^{-12} \,F/m$
Area of the plates, $A = \pi r^2 = \pi \times (2 \times 10^{-2})^2 = 4\pi \times 10^{-4} \,m^2$
The displacement current $I_d$ is given by:
$I_d = \varepsilon_0 \frac{d\phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt}$
Since $E = \frac{V}{d}$, we have $\frac{dE}{dt} = \frac{1}{d} \frac{dV}{dt}$
Therefore, $I_d = \varepsilon_0 \frac{A}{d} \frac{dV}{dt}$
Substituting the values:
$I_d = (8.85 \times 10^{-12}) \times \frac{4\pi \times 10^{-4}}{10^{-4}} \times (5 \times 10^6)$
$I_d = 8.85 \times 10^{-12} \times 4\pi \times 5 \times 10^6$
$I_d = 8.85 \times 20\pi \times 10^{-6} \,A$
$I_d \approx 8.85 \times 62.83 \times 10^{-6} \,A$
$I_d \approx 556 \times 10^{-6} \,A = 0.556 \,mA$
Thus, the correct option is $C$.

(D) The conduction current is given by $I_c = \frac{V}{R} = V \sigma A / d$,where $\sigma = 1/\rho$ is the conductivity. The amplitude is $I_{c,0} = \frac{V_0}{\rho} \frac{A}{d}$.
The displacement current is given by $I_d = \epsilon_0 \epsilon_r A \frac{dE}{dt} = \epsilon_0 \epsilon_r A \frac{d}{dt} (V/d) = \frac{\epsilon_0 \epsilon_r A}{d} \omega V_0 \cos(\omega t)$. The amplitude is $I_{d,0} = \frac{\epsilon_0 \epsilon_r A \omega V_0}{d}$.
The ratio of the amplitudes is $\frac{I_{d,0}}{I_{c,0}} = \frac{\epsilon_0 \epsilon_r A \omega V_0 / d}{V_0 A / (\rho d)} = \epsilon_0 \epsilon_r \omega \rho$.
Given $\epsilon_r = 80$,$\rho = 0.25 \, \Omega m$,$f = 0.4 \times 10^9 \, Hz$,and $\omega = 2 \pi f = 2 \pi (0.4 \times 10^9) = 0.8 \pi \times 10^9 \, rad/s$.
Using $\epsilon_0 = \frac{1}{36 \pi \times 10^9} \, F/m$,the ratio is $\frac{1}{36 \pi \times 10^9} \times 80 \times (0.8 \pi \times 10^9) \times 0.25 = \frac{80 \times 0.8 \times 0.25}{36} = \frac{16}{36} = \frac{4}{9}$.

(C) According to the Ampere-Maxwell law,the magnetic field $B$ at a distance $r$ from the axis between the plates of a charging capacitor is given by the line integral of the magnetic field around a circular loop of radius $r$.
For $r > R$,the displacement current enclosed by the loop is equal to the total displacement current $I_D$ passing through the capacitor.
Applying Ampere-Maxwell law: $\oint B \cdot dl = \mu_0 I_{enclosed}$.
Since the current is $I_D$,we have $B(2 \pi r) = \mu_0 I_D$.
Therefore,the magnetic field is $B = \frac{\mu_0 I_D}{2 \pi r}$.

(A) The displacement current $I_d$ is given by the formula $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$,where $\Phi_E$ is the electric flux through the surface bounded by the loop.
Since the electric field $E$ exists only between the plates (within the radius $R = 10 \ cm$),the flux through the loop of radius $r = 20 \ cm$ is $\Phi_E = E \cdot A_{plate} = E \cdot \pi R^2$.
Thus,$I_d = \varepsilon_0 \frac{d}{dt}(E \cdot \pi R^2) = \varepsilon_0 \pi R^2 \frac{dE}{dt}$.
Given: $R = 10 \ cm = 0.1 \ m$,$\frac{dE}{dt} = 3.6 \times 10^{12} \ V/(m \cdot s)$,and $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2 \Rightarrow \varepsilon_0 = \frac{1}{36 \pi \times 10^9} \ F/m$.
Substituting these values:
$I_d = \left( \frac{1}{36 \pi \times 10^9} \right) \cdot \pi \cdot (0.1)^2 \cdot (3.6 \times 10^{12})$
$I_d = \frac{1}{36 \times 10^9} \cdot 0.01 \cdot 3.6 \times 10^{12}$
$I_d = \frac{3.6 \times 10^{10}}{36 \times 10^9} = \frac{36 \times 10^9}{36 \times 10^9} = 1 \ A$.

(B) Given: Displacement current,$i_{d} = 150 \times 10^{-6} \ A$. Capacitance,$C = 30 \times 10^{-6} \ F$.
The displacement current $i_{d}$ in a capacitor is related to the rate of change of potential difference across its plates by the formula:
$i_{d} = C \frac{dv}{dt}$
Rearranging the formula to solve for the rate of change of potential $\frac{dv}{dt}$:
$\frac{dv}{dt} = \frac{i_{d}}{C}$
Substituting the given values:
$\frac{dv}{dt} = \frac{150 \times 10^{-6} \ A}{30 \times 10^{-6} \ F} = 5 \ Vs^{-1}$
Therefore,the capacitor is charged at a rate of $5 \ Vs^{-1}$.

(C) The scientist Maxwell further investigated Ampere's law and identified that it was incomplete for time-varying electric fields.
He introduced the concept of displacement current,$I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,to modify Ampere's law.
This modification,known as the Ampere-Maxwell law,resolves the inconsistency in Ampere's original circuital law regarding the continuity of current in circuits containing capacitors.

(B) The displacement current $I_d$ is given by $I_d = \varepsilon_0 \frac{d\phi_E}{dt}$.
Using the Ampere-Maxwell law for a circular path of radius $r$ inside the capacitor: $\oint B \cdot dl = \mu_0 I_d$.
Since the electric field is uniform,the flux through a circular area of radius $r$ is $\phi_r = \phi_E \left( \frac{\pi r^2}{A} \right)$,where $A$ is the total area of the plate.
Thus,$B(2\pi r) = \mu_0 \varepsilon_0 \frac{d}{dt} \left( \phi_E \frac{\pi r^2}{A} \right) = \mu_0 \varepsilon_0 \frac{r^2}{A} \frac{d\phi_E}{dt} \cdot \pi$.
$B = \frac{\mu_0 \varepsilon_0 r}{2A} \frac{d\phi_E}{dt}$.
Given $\frac{d\phi_E}{dt} = 7 \times 10^{14}$,$r = 0.1 \text{ m}$,$A = 1 \text{ m}^2$,$\varepsilon_0 = 8.85 \times 10^{-12}$,$\mu_0 = 4\pi \times 10^{-7}$.
$B = \frac{(4\pi \times 10^{-7})(8.85 \times 10^{-12})(0.1)}{2(1)} (7 \times 10^{14}) \approx 7.79 \times 10^{-6} \text{ T} = 0.779 \times 10^{-5} \text{ T}$.

(A) The displacement current $I_d$ is given by the formula $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$,where $\epsilon_0$ is the permittivity of free space and $\frac{d\Phi_E}{dt}$ is the rate of change of electric flux.
Given $\epsilon_0 = 8.854 \times 10^{-12} \text{ F/m}$ and $\frac{d\Phi_E}{dt} = 9 \pi \times 10^3 \text{ Vm s}^{-1}$.
Substituting the values: $I_d = (8.854 \times 10^{-12}) \times (9 \times 3.14159 \times 10^3)$.
$I_d \approx 8.854 \times 10^{-12} \times 28.27 \times 10^3 \approx 250.3 \times 10^{-9} \text{ A} = 0.25 \mu \text{A}$.

(D) The displacement current $I$ in a parallel plate capacitor is given by the formula $I = \varepsilon_0 \frac{d\phi_E}{dt}$,where $\phi_E$ is the electric flux.
Since the electric flux $\phi_E$ through the area $A$ of the plates is given by $\phi_E = E_{field} \cdot A$,where $E_{field}$ is the electric field between the plates.
Substituting this into the displacement current formula,we get $I = \varepsilon_0 \frac{d}{dt}(E_{field} \cdot A)$.
Since the area $A$ is constant,we have $I = \varepsilon_0 A \frac{dE_{field}}{dt}$.
Given that the rate of change of the electric field is $E$,we substitute $\frac{dE_{field}}{dt} = E$.
Thus,$I = \varepsilon_0 A E$.
Solving for the area $A$,we get $A = \frac{I}{\varepsilon_0 E}$.

(A) The displacement current is given by the formula:
$I_d = \varepsilon_0 \frac{d\phi_E}{dt} = \varepsilon_0 \frac{\Delta \phi_E}{\Delta t}$
Since the electric flux $\phi_E = E \cdot A$ and the electric field $E = \frac{V}{d}$, we have:
$I_d = \varepsilon_0 \frac{A}{d} \frac{\Delta V}{\Delta t}$
Given values are:
$\Delta V = 200 \, V$
$\Delta t = 1 \, \mu s = 10^{-6} \, s$
$d = 1 \, mm = 10^{-3} \, m$
$A = 20 \, cm^2 = 20 \times 10^{-4} \, m^2$
$\varepsilon_0 = 8.85 \times 10^{-12} \, F/m$
Substituting these values:
$I_d = \frac{8.85 \times 10^{-12} \times 20 \times 10^{-4} \times 200}{10^{-3} \times 10^{-6}}$
$I_d = \frac{8.85 \times 20 \times 200 \times 10^{-16}}{10^{-9}}$
$I_d = 35400 \times 10^{-7} \, A = 3.54 \times 10^{-3} \, A \approx 3.5 \, mA$

(B) Maxwell's four equations are the fundamental equations of electromagnetism, which include:
$1$. Gauss' law for electricity $( \nabla \cdot E = \rho / \epsilon_0)$.
$2$. Gauss' law for magnetism $( \nabla \cdot B = 0)$.
$3$. Faraday's law of induction $( \nabla \times E = -\partial B / \partial t)$.
$4$. Ampere-Maxwell law $( \nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \partial E / \partial t)$.
Le-Chatelier's law of equilibrium is a principle in chemistry related to chemical reactions, not electromagnetism. Therefore, it is not described by Maxwell's equations.

(D) The expression is $X = \varepsilon_0 L \frac{\Delta V}{\Delta t}$.
We know that the electric flux $\phi_E = E \cdot A$,where $E = \frac{V}{L}$ and $A = L^2$.
Thus,$\phi_E = \frac{V}{L} \cdot L^2 = V \cdot L$.
Substituting this into the expression for $X$,we get $X = \varepsilon_0 \frac{\Delta \phi_E}{\Delta t}$.
According to the definition of displacement current $i_d = \varepsilon_0 \frac{d\phi_E}{dt}$,the quantity $X$ represents the displacement current.
Therefore,the dimension of $X$ is the same as that of current.

(A) The conducting current density is given by $j_c = \sigma E$.
For an electromagnetic wave,$E = E_0 \sin(\omega t - kx)$,so the maximum conducting current density is $(j_c)_{\max} = \sigma E_0$.
The displacement current density is given by $j_d = \epsilon_0 \frac{\partial E}{\partial t}$.
Substituting $E$,we get $j_d = \epsilon_0 E_0 \omega \cos(\omega t - kx)$,so the maximum displacement current density is $(j_d)_{\max} = \epsilon_0 E_0 \omega$.
The ratio is $\frac{(j_c)_{\max}}{(j_d)_{\max}} = \frac{\sigma E_0}{\epsilon_0 \omega E_0} = \frac{\sigma}{\epsilon_0 \omega}$.
Given $\sigma = 10 \text{ mho/m}$,$f = 100 \times 10^6 \text{ Hz}$,and $\omega = 2 \pi f = 2 \pi \times 10^8 \text{ rad/s}$.
Also,$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9$,so $\frac{1}{\epsilon_0} = 4 \pi \times 9 \times 10^9 = 36 \pi \times 10^9$.
Ratio $= \frac{10 \times 36 \pi \times 10^9}{2 \pi \times 10^8} = \frac{360 \pi \times 10^9}{2 \pi \times 10^8} = 180 \times 10 = 1800$.

(B) The four Maxwell's equations are as follows:
$1$. Gauss's law for electricity: $\oint \overrightarrow{E} \cdot d\vec{a} = \frac{q_{enclosed}}{\epsilon_0} = \frac{1}{\epsilon_0} \int \rho dv$. This matches $C-IV$.
$2$. Gauss's law for magnetism: $\oint \overrightarrow{B} \cdot d\vec{a} = 0$.
$3$. Faraday's law of induction: $\oint \overrightarrow{E} \cdot d\vec{l} = -\frac{d}{dt} \oint \overrightarrow{B} \cdot d\vec{a}$. This matches $A-II$.
$4$. Ampere-Maxwell law: $\oint \vec{B} \cdot d\vec{l} = \mu_0(I + \epsilon_0 \frac{d\phi_E}{dt})$. This matches $B-III$.
$5$. Ampere's circuital law (original form for steady currents): $\oint \overrightarrow{B} \cdot d\vec{l} = \mu_0 I$. This matches $D-I$.
Therefore,the correct matching is $A-II, B-III, C-IV, D-I$.

(C) An oscillating charge produces electromagnetic waves of the same frequency as the frequency of oscillation of the charge.
Since the frequency of the oscillating particle is $8 \times 10^9 \text{ Hz}$,the frequency of the electromagnetic waves produced will also be $8 \times 10^9 \text{ Hz}$.

(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 values are $I_d = 4.0 \text{ A}$ and $C = 6 \text{ }\mu\text{F} = 6 \times 10^{-6} \text{ F}$.
Rearranging the formula to find the rate of change of potential difference: $\frac{dV}{dt} = \frac{I_d}{C}$.
Substituting the values: $\frac{dV}{dt} = \frac{4.0}{6 \times 10^{-6}} = \frac{4.0}{6} \times 10^6 \text{ V/s}$.
Calculating the value: $\frac{dV}{dt} = 0.666... \times 10^6 \text{ V/s}$.
Rounding to two decimal places,we get $\frac{dV}{dt} \approx 0.67 \times 10^6 \text{ V/s}$.
Comparing this with $\alpha \times 10^6 \text{ V/s}$,we find $\alpha = 0.67$.