Physics Vidyapith ✍️

Physical Significance: The physical significance of Maxwell's equations obtained from integral form are given below: Maxwell's First Equation: 1. The total electric displacement through the surface enclosing a volume is equal to the total charge within the volume. 2. It represents Gauss Law. 3. This law is independent of time. Charge acts as source or sink for the lines of electric force. Maxwell's Second Equation: 1. The total magnetic flux emitting through any closed surface is zero. An isolated magnet do not exist monopoles. 2. There is no source or sink for lines of magnetic force. 3. This is time independent equation. Maxwell's Third Equation: 1. The electromotive force around the closed path is equal to the time derivative of the magnetic displacement through any surface bounded by the path. 2. This gives relation between electric field $E$ and magnetic induction $B$. 3. This expression is time varying i.e. $E$ is generate

Electromagnetic Wave: An electromagnetic wave is the combined effect of an electric field and magnetic field which carry energy from one place to another. When an electric field and the magnetic field are applied perpendicular to each other then a wave propagates perpendicular to both the electric field and the magnetic field. This wave is called the electromagnetic wave. Characteristics of Electromagnetic Wave: 1.) Electric and Magnetic Fields: Electromagnetic waves are produced through the mutually perpendicular interaction of electric and magnetic fields. The propagation of the wave is also perpendicular to both the electric field and the magnetic field. 2.) Wave Nature of electromagnetic waves: Electromagnetic waves are characterized by their wave-like behavior, so they exhibit the properties such as wavelength, frequency, amplitude, and velocity. This wave-like behavior of electromagnetic waves can undergo phenomena like interference, diffraction, and polarization. 3.

In electromagnetic waves, the electric field vector and magnetic field vector are mutually perpendicular to each other (Proof) The general solution of the wave equation for the electric field vector and magnetic field vector are respectively given below $\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$ $\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$ Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant. Now $\overrightarrow{\nabla} \times \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $ $\overrightarrow{\nabla} \times \overrightarrow{E}

Electromagnetic waves are transverse in nature: (Proof) The general solution of the wave equation for the electric field and magnetic field are respectively given below $\overrightarrow{E}= E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(1)$ $\overrightarrow{B}= B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(2)$ Here $E_{\circ}$ and $B_{\circ}$ are the complex amplitude of electric field vector $\overrightarrow{E}$ and magnetic field vector $\overrightarrow{B}$ respectively and $\overrightarrow{k}$ is the propagation constant. Now $\overrightarrow{\nabla}. \overrightarrow{E}= \left( \hat{i} \frac{\partial}{\partial x} + \hat{i} \frac{\partial}{\partial x} +\hat{i} \frac{\partial}{\partial x} \right). \left( \hat{i}E_{x} + \hat{j}E_{y} + \hat{k}E_{z} \right) $ $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\partial}{\partial x} \left(E_{x} \right)+ \frac{\partial}{\partial y} \left(E_{y} \rig

The electromagnetic wave equations in conducting media: For electric field vector: $ \nabla^{2}.\overrightarrow{E}-\mu \epsilon\frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} - \sigma \mu \frac{\partial \overrightarrow{E}}{\partial t}=0 \qquad(1)$ For magnetic field vector: $\nabla^{2}.\overrightarrow{B} - \mu \epsilon \frac{\partial^{2} B}{\partial t^{2}}-\sigma \mu \frac{\partial \overrightarrow{B}}{\partial t}=0 \qquad(2)$ The wave equation of electric field vector: $\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$ The wave equation of magnetic field vector: $\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$ Now the solution of electromagnetic wave for electric field vector. Differentiate with respect to $t$ of equation $(3)$ $\frac{\partial \overrightarrow{E}}{\parti

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}+\mu \epsilon \frac{\partial \overrightarrow{E}}{\partial t}$ For Conducting Media: Cur

Description of Displacement Current: The concept of displacement current was first introduced by Maxwell purely on the theoretical ground. Maxwell postulates that "It is not only current in a conductor that produces a magnetic field but a changing electric field (or time varying electric field) in vacuum or in dielectric also produces the magnetic field. It means that a changing electric field is equivalent to a current which flows as long as the electric field is changing. This equivalent current in a vacuum or dielectric produces the same magnetic effect as an ordinary or conductor current in a conductor. This equivalent current is known as displacement current ". According to the Maxwell modified ampere's law. $\oint \overrightarrow{B}. \overrightarrow{dl}= \mu_{\circ}i+\mu_{\circ}i_{d}$ Where $i_{d}$ = Displacement Current

What is the energy density in the electromagnetic wave in free space? The total energy stored in electromagnetic waves per unit volume due to the electric field and the magnetic field is called energy density in the electromagnetic wave in free space. $U=\epsilon_{0} E^{2}=\frac{B^{2}}{\mu_{0}}$ Derivation of Energy density in electromagnetic waves in free space: The energy per unit volume due to the electric field is $U_{E}= \frac{1}{2} \overrightarrow{E}.\overrightarrow{D} \qquad(1)$ The energy per unit volume due to the magnetic field is $U_{B}= \frac{1}{2} \overrightarrow{B}.\overrightarrow{H} \qquad(2)$ The total energy density of electromagnetic waves is $U=U_{E}+U_{B} \qquad(3)$ Now substitute the value of $U_{E}$ and $U_{B}$ in equation$(3)$ then we get $U=\frac{1}{2} \left( \overrightarrow{E}.\overrightarrow{D}+\overrightarrow{B}.\overrightarrow{H} \right)$ $U=\frac{1}{2} \left( \overrightarrow{E}.\epsilon_{0}\overrightarrow{E}+\overri

Poynting Vector: The rate of flow of energy per unit area in plane electromagnetic wave is known as Poynting vector. It is represented by $\overrightarrow{S}$. It is a vector quantity. $\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H}$ $\overrightarrow{S}=\frac{1} {\mu_{0}} (\overrightarrow{E} \times \overrightarrow{B})$ Poynting Theorem (Work energy theorem): The most important aspect of electrodynamics is: Energy density stored with an electromagnetic wave Energy Flux associated with an electromagnetic wave To derive the energy density and energy flux. We consider the conservation of energy in small volume elements in space. The work done per unit volume by an electromagnetic wave: $W=\overrightarrow{J}.\overrightarrow{E} \qquad(1)$ This work done also consider as energy dissipation per unit volume. This energy dissipation must be connected with the net decrease in energy density and energy flow out of the volume.

We know that the electromagnetic wave propagates perpendicular to both electric field and magnetic field which can describe as $\overrightarrow{k} \times \overrightarrow{E}= \omega \overrightarrow{B} \qquad(1)$ If $\hat{n}$ is a unit vector in the direction of the propagation then $\overrightarrow{k}=k \hat{n}$ Substitute these values in equation$(1)$ then we get $k(\hat{n} \times \overrightarrow{E})= \omega \overrightarrow{B}$ $\overrightarrow{B}= \frac{k}{\omega}(\hat{n} \times \overrightarrow{E}) \qquad(2)$ But the value of $k$ and $\omega$ is $k=\frac{2\pi}{\lambda}$ $\omega=2 \pi \nu$ Then value of $\frac{k}{\omega}=\frac{1}{c}$ Now substitute the value of $\frac{k}{\omega}$ in equation$(2)$ then we get $\overrightarrow{B}= \frac{1}{c}(\hat{n} \times \overrightarrow{E})$ The magnitude form of the above equation can be written as $B=\frac{E}{c}$ $\frac{E}{B}=c$ $\frac{E}{\mu_{0}H}=c \qquad (\because B=\mu_{0} H)$ $\frac{

When an electromagnetic wave strikes a surface then its momentum changes. the rate of change of momentum is equal to the applied force. this force acting on the unit area of the surface exerts a pressure called radiation pressure$(P_{rad})$. Let us consider a plane electromagnetic wave incident normally on a perfectly absorbing surface of area $A$ for a time $t$. If energy $U$ is absorbed during this time then momentum $P$ delivered to the surface is given according to Maxwell's prediction by $P=\frac{U}{C} \qquad(1)$ If $S$ is the energy flow per unit area per unit time i.e. Poynting vector then the energy density $U=SAt \qquad(2)$ From equation $(1)$ and equation $(2)$ $P=\frac{SAt}{c}$ $P=UAt \qquad (\because U=\frac{S}{c})$ $\frac{P}{t}=UA \qquad (3)$ If average force $(F)$ acting on the surface, is equal to the average rate of change of momentum $(P)$, is delivered to the surface then $F=\frac{P}{t} \qquad(4)$ Now from equation$(3)$ and equation$(4

Derivation of energy flow in the electromagnetic wave in free space: The Poynting vector is given by $\overrightarrow{S}=\overrightarrow{E} \times \overrightarrow{H} \qquad(1)$ $\overrightarrow{S}=\frac{1}{\mu_{0}} ( \overrightarrow{E} \times \overrightarrow{B} ) \qquad(2) \qquad (\because \overrightarrow{B}= \mu_{0} \overrightarrow{H})$ We know that the characteristic impedance equation i.e. $\overrightarrow{B}=\frac{1}{\mu_{0}c}(\hat{n} \times \overrightarrow{E}) \qquad(3)$ Now substitute the value of $\overrightarrow{B}$ in equation$(2)$ $\overrightarrow{S}=\frac{1}{\mu_{0}c} [\overrightarrow{E} \times (\hat{n} \times \overrightarrow{E})]$ $\overrightarrow{S}=\frac{1}{\mu_{0}c} [(\overrightarrow{E}.\overrightarrow{E}) \hat{n}- (\overrightarrow{E}.\hat{n}) \overrightarrow{E})] \qquad(4)$ As $\overrightarrow{E}$ is perpendicular to $\hat{n}$ so $\overrightarrow{E} . \hat{n}=0$ then we get for above equation$(4)$ $\overrightarrow{S}=\frac{1}{\mu_{0}c} E^{2} \hat{n}

Derivation of momentum of electromagnetic wave: Maxwell's had also predicted that electromagnetic waves transport linear momentum in the direction of propagation. Let a particle which has mass $m$ moving with velocity then the momentum of a particle, $\overrightarrow{P}=m\overrightarrow{v} \qquad(1)$ According to mass-energy relation $U=mc^{2}$ Here $U$ - Total energy of the particle $m=\frac{U}{c^{2}} \qquad(2)$ From equation $(1)$ and equation $(2)$ $\overrightarrow{P}=\frac{U}{c^{2}} \overrightarrow{v} \qquad(3)$ If the electromagnetic wave is propagating along the x-axis then $\overrightarrow{v}=c \hat{i}$ Put this value in the above equation $(3)$ $\overrightarrow{P}=\frac{U}{c} \hat{i} \qquad(4)$ We know that the equation of energy flow in electromagnetic wave $\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{n}$ Here wave is propagating along x-axis i.e  $\hat{n}=\hat{i}$ $\overrightarrow{S}= \frac{1}{\mu_{0} c} E^{2} \hat{i} \qquad(5)$

The electromagnetic wave equations in free space: For electric field vector: $\nabla^{2} \overrightarrow{E}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$ For magnetic field vector: $\nabla^{2} \overrightarrow{B}=\frac{1}{c^{2}} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} \qquad(2)$ The wave equation of electric field vector: $\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$ The wave equation of magnetic field vector: $\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$ Now the solution of electromagnetic wave for electric field vector. Differentiate with respect to $t$ of equation $(3)$ $\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$ Again differentiate with respect to $t$ of the above equation:

The electromagnetic wave equations in non-conducting media : For electric field vector: $\nabla^{2} \overrightarrow{E}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{E}}{\partial t^{2}} \qquad(1)$ For magnetic field vector: $\nabla^{2} \overrightarrow{B}=\frac{1}{v^{2}} \frac{\partial^{2} \overrightarrow{B}}{\partial t^{2}} \qquad(2)$ The wave equation of electric field vector: $\overrightarrow{E}(\overrightarrow{r},t)=E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(3)$ The wave equation of magnetic field vector: $\overrightarrow{B}(\overrightarrow{r},t)=B_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)} \qquad(4)$ Now the solution of electromagnetic wave for electric field vector. Differentiate with respect to $t$ of equation $(3)$ $\frac{\partial \overrightarrow{E}}{\partial t}=i \omega E_{\circ} e^{i(\overrightarrow{k}. \overrightarrow{r} - \omega t)}$ Again diff

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations for free space. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified Form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$

Maxwell's Equations: Maxwell's equation of the electromagnetic wave is a collection of four equations i.e. Gauss's law of electrostatic, Gauss's law of magnetism, Faraday's law of electromotive force, and Ampere's Circuital law. Maxwell converted the integral form of these equations into the differential form of the equations. The differential form of these equations is known as Maxwell's equations. $\overrightarrow{\nabla}. \overrightarrow{E}= \frac{\rho}{\epsilon_{0}}$ $\overrightarrow{\nabla}. \overrightarrow{B}=0$ $\overrightarrow{\nabla} \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}$ $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \overrightarrow{J}$ Modified Form: $\overrightarrow{\nabla} \times \overrightarrow{B}= \mu \left(\overrightarrow{J}+ \epsilon \frac{ \partial \overrightarrow{E}}{\partial t} \right)$ For free space

We know the equation of continuity is $\overrightarrow{\nabla}. \overrightarrow{J}+ \frac{\partial \rho}{\partial t}=0 \qquad(1)$ According to Maxwell's first differential equation $\overrightarrow{\nabla}. \overrightarrow{D}=\rho \qquad(2)$ From equation $(1)$ and equation$(2)$ $\overrightarrow{\nabla}. \overrightarrow{J}+ \frac{\partial }{\partial t}(\overrightarrow{\nabla}. \overrightarrow{D})=0$ $\overrightarrow{\nabla}. (\overrightarrow{J}+ \frac{\partial \overrightarrow{D} }{\partial t}) =0 $ Where the term → $(\overrightarrow{J}+ \frac{\partial \overrightarrow{D} }{\partial t})$ → solenoidal vector and it is also regarded as total current density for time varying electric field. $D$ → The displacement vector $\frac{\partial \overrightarrow{D} }{\partial t}$ → Displacement current density The above equation is known as the " Equation of continuity for current density ".

Definition: The mathematical representation of the law of conservation of charge in differential form is called the " continuity equation" . Mathematical representation of Equation of continuity: If $\overrightarrow{J}$ is the current density of a closed surface $\overrightarrow{S}$ then the current through a closed surface is $i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(1)$ Let $V$ be the volume enclosed by the surface $S$. So the total charge in this volume- $q=\oint_{V} \rho. dV \qquad(2)$ By the law of conservation of charge i.e. "Charge can neither be created nor destroyed". If some charge flows out from the volume per unit time giving rise to current density, the charge in the volume decreases at the same rate. So the current $i=-\frac{\partial q}{\partial t}$ $i=-\frac{\partial}{\partial t} (\oint_{V} \rho. dV) \qquad (from \: equation(2) \: )$ $i=-\oint_{V} \frac{\partial \rho}{\p

Maxwell's fourth equation is the differential form of Ampere's circuital law. $\overrightarrow{\nabla} \times \overrightarrow{H} = \overrightarrow{J} + \frac{\partial{\overrightarrow{D}}}{\partial{t}}$ Derivation: According to Ampere's circuital law $\oint_{l} \overrightarrow{B}. \overrightarrow{dl}=\mu_{0} i \qquad(1)$ According to Stroke's theorem- $\oint_{l} \overrightarrow{B}.\overrightarrow{dl}=\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} \qquad(2)$ From equation$(1)$ and equation$(2)$ $\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} i \qquad(3)$ where $i=\oint_{S} \overrightarrow{J}. \overrightarrow{dS} \qquad(4)$ So from equation$(3)$ and equation$(4)$ $\oint_{S} (\overrightarrow{\nabla} \times \overrightarrow{B}). \overrightarrow{dS} = \mu_{0} \oint_{S} \overrightarrow{J}. \overrightarrow{dS}$ $\oint_{S} (\over