Physics Unit 6: Introduction to Electromagnetism PDF

# Unit 6 - Physics ## Introduction to Electromagnetism ### Divergence - When the Del operator operates scalarly on a vector function, the result is a scalar quantity called the divergence of the vector function. - It is the dot product of the del operator with a vector. - Thus, $\nabla \cdot A...

# Unit 6 - Physics ## Introduction to Electromagnetism ### Divergence - When the Del operator operates scalarly on a vector function, the result is a scalar quantity called the divergence of the vector function. - It is the dot product of the del operator with a vector. - Thus, $\nabla \cdot A$ denotes the divergence of A. - Divergence of A = $\nabla \cdot A$ = $(\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) \cdot [A_x i + A_y j + A_z k]$ - $\nabla \cdot A = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$ - The divergence of A at a given point P is the net outward flow of flux per unit volume over a closed surface - $div A = \nabla \cdot A = lim_{\Delta V \to 0} \frac{\oint_S A.ds}{\Delta V}$ - Where $\Delta V$ is the volume enclosed by the closed surface S in which point P is located. #### Physical interpretation of divergence - Divergence of a vector field A at a given point is a measure of how much field diverges or converges from that point. - A **positive divergence** of vector A indicates a source of that vector A at that point. - Similarly, a **negative divergence** of vector A indicates a "sink." - [Image of a circle with arrows radiating outward] (+ive divergence) - [Image of a circle with arrows pointing inward] (-)ve div. - [Image of a circle with arrows pointing vertically] zero div. - If a fluid is incompressible, there will not be any gain or loss in volume of the element. In that case, $\nabla \cdot V = 0$. - If the flux entering any element of space is the same as leaving it (i.e. div V = 0 everywhere), then such a function is called a solenoidal vector function. ### Curl - When the Del operator ($\nabla$) operates vectorally on a vector function, the result is a vector quantity known as curl of the vector, i.e. it is the vector multiplication (cross product) of the del operator and any one vector. - Thus, if A is a vector, then - Curl A = $\nabla \times A$ = $(\frac{\partial}{\partial x} i + \frac{\partial}{\partial y} j + \frac{\partial}{\partial z} k) \times [A_x i + A_y j + A_z k]$ - = $\begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ A_x & A_y & A_z \end{vmatrix}$ - $= ( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z} )i + (\frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} )j + ( \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y})k$ - The curl of vector quantity A is an axial vector (or rotational vector): - Whose magnitude is the maximum circulation of vector A per unit area as the area tends to zero. - And whose direction is the normal direction of the area when the area is oriented to make the circulation max. ve. - Curl A = $\nabla \times A$ = $lim_{\Delta S \to 0} \frac{\oint_L A \cdot dl}{\Delta S} max \hat{n}$ - Here, $\Delta S$ is bound by the curve L, and $\hat{n}$ is the unit vector normal to the surface. #### Physical interpretation of curl: - The curl provides the maximum value of the circulation on the field per unit area (or circulation density) and indicates the direction along which this max. value occurs. - The curl of a vector field A at a point P may be regarded as a measure of how much the field curls around P. - So if a vector has no curl, it is irrotational. - [Image of a circle with an arrow pointing up and an arrow pointing out of the page] (a) curl at P point out of the page - [Image of a circle with an arrow pointing vertically] (b) Curl at P is zero. - The curl of a vector A gives the measure of the angular velocity at every point of a vector field. - Curl V = $\frac{1}{2} \omega$. - Means, the angular velocity at any point is equal to half the Curl of the linear velocity at that point. ### Gauss's Divergence theorem - Curl - Stoke's theorem: - $\oint_V (\nabla . A)dV = \oint_S A.ds$ - $\int_L A.dl = \oint_S (\nabla \times A) ds$ ### Divergence of the electrostatic field - When positive charge is placed inside the closed surface, according to Gauss, the normal component of the electric field (i.e., the net outward electric flux) over any closed surface of any shape is equal to $\frac{1}{\epsilon_0}$ times the net charge enclosed by the surface, i.e. - $\oint_S E.ds=\frac{Q}{\epsilon_0}$ ① - Let P be the charge density at a point within an infinitesimal volume element $dv$, then the charge $Pdv$ may be considered as a point charge. - Thus, if the surface S encloses the volume V, then - $\oint_S E.ds=\frac{1}{\epsilon_0}\int_V Pdv$ ② - By using Gauss Divergence theorem - $\oint_S E.ds = \int_V (\nabla. E)dv$ ③ - From equations ② & ③ - $\int_V (\nabla . E)dv=\frac{1}{\epsilon_0}\int_V P dv$ - or $\int_V (div E - \frac{P}{\epsilon_0} dv=0$ - $\implies div E - \frac{P}{\epsilon_0} = 0$ - $\implies div E = \frac{P}{\epsilon_0}$ ### Curl of the electrostatic field - [Image of a circle with a point charge at the center, arrows pointing outward] - Potential difference between two points is independent of the different path followed from one point to another. - Let us consider a point charge Q at the origin, so - $V_{AB} = - \int_{BEA} E.dl = -\int_{BDA} E.dl$ ① - So by rearranging, we get - $\int_{BEA} E.dl + \int_{ADB} E.dl = 0 \implies \oint_{BEADB} E.dl = 0$ - or - $\oint_{BEADB} E.dl = 0$ ② - Multiply both sides with a test charge q, - $\oint_{BEADB} qE.dl =0$ ③ - According to this equation, the work done in moving a test charge around a closed path in a static electric field is zero. - Now curl of vector E is given by - $\nabla \times E = lim_{\Delta S \to 0}\frac{\oint_L E.dl}{\Delta S} \hat{n} $ ④ - where $\Delta S$ is the area bounded by the closed bath. - n is the unit vector normal to the area, with direction s.t. $\oint_L E. dl$ is maximum. - From eq" (2) for static fields $\oint E.dl = 0$ for any path, so - $\nabla \times E=0$ ### Poisson's and Laplace's Equations for Electrostatic Potential #### (i) Poisson's Equation: - The differential form of Gauss's law may be written as $\nabla . E = \frac{P}{\epsilon_0}$ ① - Where E is the electric field intensity, and P is the volume density of the charge. - By definition, the electric field at a point may be written as the negative gradient of electric potential, $\phi$, i.e. - $ E = -grad \phi = - \nabla \phi$ ② - So eq" (1) may be written as - $div E= \nabla.E = \nabla. (-\nabla \phi)$ - = $-\nabla^2 \phi = \frac{P}{\epsilon_0}$ - or $\nabla^2 \phi = -\frac{P}{\epsilon_0}$ - or $(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}) \phi = -\frac{P}{\epsilon_0}$, known as Poisson's equation. #### (ii) Laplace's Equation: - $\nabla.P=0$ means zero volume charge density. - But assuming (i.e., the point of observation is in empty space) then P = 0 and we have $\nabla^2 \phi = 0$, which is Laplace's equation. - Here the $\nabla$ operator is known as laplace's operator. - or $\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} =0$ #### Applications/ Significance of Laplace's and Poisson's equations: - Using Laplace and Poisson's equations, we can obtain potential at any point in between two surfaces when potential at two surfaces are given. - We can also obtain Capacitance between these two surfaces. ### Biot-Savart Law - [Image of a wire carrying a current I, with a point P a distance R from the current element I*dl] - When current flows through a conductor, magnetic field is produced. - *Statement* - According to Biot-Savart law, the magnetic field intensity produced at point P due to a differential current element I * dl is proportional to the product I * dl and the sine of the angle *alpha* between the element and the line joining point P to the element, and is inversely proportional to the square of the distance R between point P and the element, i.e. - (i) $dH \propto Idl$ - (ii) $dH \propto sin \alpha$ - (iii) $ dH \propto \frac{1}{R^2}$ - So we can write - $dH \propto \frac{Idl sin \alpha} {R^2}$ - Or $dH = \frac{K Idl sin \alpha}{R^2}$ - $dH = \frac{Idl sin \alpha}{4 \pi R^2}$ (∵ K= $\frac{\mu_0}{4 \pi}$) ① - If $\hat{n}$ be the unit vector in the direction from dl to P, then $dl \times \hat{n} = dl sin \alpha$ - $\implies d\vec{H} = \frac{Idl \times \hat{n}}{4\pi R^2} Amp/m.$ ② - Eq" ② gives the differential magnetic field intensity at point P. - Total magnetic field intensity can be obtained by integrating eq" ② (i.e., field intensity can be obtained by integrating eq" ②, i.e. $\vec{H} = \int d\vec{H}$) - $ \vec{H} = \int \frac{Idl \times \hat{n}}{4\pi R^2}$ - $ \vec{H} = \int \frac{Idl \times \vec{R}}{4\pi R^2 |R|}$ - $ \vec{H} = \int \frac{Idl \times \vec{R}}{4\pi R^3} $ - $ \vec{H} = \frac{\mu_0}{4\pi} \int \frac{Idl \times \vec{R}}{R^3} $ ③ - Eq" ③ gives the integral form of Biot-Savart law. ### Divergence of static magnetic field - According to Gauss's law for magnetic field, net outgoing magnetic flux from any surface is always zero. This is due to the fact that magnetic field lines always form closed loops. - i.e. - $\phi_B = 0$ - Or $\int_S B.ds=0$ ① - Using Gauss's divergence theorem - $\int_S B.ds = \int_V (\nabla. B)dv$ ② - So from Eq" ① & ② - $\nabla .B = divB = 0$. - This equation proves that magnetic monopole doesn't exist in nature. ### Curl of static magnetic field - Lets consider a current $\Delta I$ flowing through the area $\Delta S$. The current $\Delta I$ is in the perpendicular direction to the area $\Delta S$. Now, according to Ampere's circuital law, we have: - $\oint_L \vec{H}.\overrightarrow{dl} = \Delta I$ ① - Now, dividing both sides of this equation by $\Delta S$, we get: - $\oint_L \frac{\vec{H}.\overrightarrow{dl}}{\Delta S} = \frac{\Delta I}{\Delta S}$ - Now, taking the limit as $\Delta S$ approaches to zero, we get - $lim_{\Delta S \to 0} \oint_L \frac{\vec{H}.\overrightarrow{dl}}{\Delta S} = lim_{\Delta S \to 0} \frac{\Delta I}{\Delta S}$ ② - By the definition of curl, L.H.S. of eq"(2) is equal to curl of H - So $curl \ \vec{H} = lim_{\Delta S \to 0} \frac{\Delta I}{\Delta S} = \vec{J}$ ③ - So $curl \ \vec{H} = \nabla \times \vec{H} = \vec{J}$ - In cartesian coordinate system, - $curl \ \vec{H}= \begin{vmatrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ H_x & H_y & H_z \end{vmatrix}$ ### Faraday's Law: - [Image of a loop of wire with a magnetic field going through it ] - Faraday proposed that if a current could produce a magnetic field, then the reverse effect should be possible, that is, a changing magnetic field should also produce a current in the loop of wire. - Faraday's law states that: "The electromotive force induced in a closed path is equal to the negative of the time rate of change of the magnetic flux enclosed by the path." - E.m.f = -N $\frac{d \phi}{dt}$ ① - Where N is the number of turns of the coil, and the negative sign is according to Lenz's Law. - Or, E.m.f. = -N $\frac{\int_S B.ds}{dt}$ = -$\frac{d\int_S B.ds}{dt}$ ### Displacement current and magnetic field arising from time-dependent electric field: - [Image of a capacitor in a circuit] - According to Maxwell, in empty free space, the conduction current is zero, but a magnetic field is present due to displacement current. - In this circuit diagram, current flowing through resistance is given by $I_R = \frac{V}{R}$ ① - This current is due to the actual motion of charge, hence it is called conduction current, and conduction current per unit area is referred to as conduction current density ($J_c$). - The current flowing through the capacitor is given by - $I_C= \frac{dQ}{dt}$ - But $Q=CV$ - $\implies$ $I_C = C \frac{dV}{dt}$ ② - The current flows through the capacitor only when the voltage is changing. This current is referred to as displacement current $I_d$ and the displacement current per unit area is called displacement current density ($J_d$). - $I_d = \frac{I_d}{A} = \frac{dQ}{dt}$ ③ - Here - $C = capacitance$ - $C=\epsilon A$ - $A = common \ area \ of \ plates$ - $\epsilon= permittivity \ of \ medium$ - $d= spacing \ between \ the \ plates$ - $E= electric \ field \ intensity \ across \ the \ capacitor, E=\frac{V}{d}$ - So - $I_d = C\frac{dV}{dt}$ - = $\epsilon A \frac{d}{dt}[Ed]$ - = $\epsilon A d \frac{dE}{dt}$ - $I_d = \epsilon A d \frac{dE}{dt}$ - $= \epsilon A \frac{d}{dt} \frac{D}{\epsilon}$ - = $A\frac{dD}{dt}$ - But $D=\epsilon E$ or $E = \frac{D}{\epsilon}$ - So, - $I_d = \frac{D D}{A} = displacement \ current \ density$ - Or - $J_d = \frac{dD}{dt}$ ④ - Now, the total current density is given by - $\vec{J} = \vec{J_c} + \vec{J_d}$ ⑤ - So we reconsider Ampere's circuital law for time varying electric fields as - $\nabla \times \vec{H} = \vec{J} = \vec{J_c} + \vec{J_d} = \vec{J_c} + \frac{\partial \vec{D}}{\partial t}$ ⑥ - So we define the displacement current as - $I_d = \int_S \vec{J_d} ds = \int_S \frac{\partial \vec{D}}{\partial t}.ds$ - Using Ampere's circuital law, we know - $\oint_L \vec{H}.\overrightarrow{dl} = I_{enclosed} = \oint_S \vec{J}ds = \oint_S (\vec{J_c}+\vec{J_d}).ds$ - If there is no conduction current (i.e. $J_c = 0$ then - $\oint_L \vec{H}.\overrightarrow{dl} = \oint_S \vec{J_d}.ds$ - So, from eq" ④, we get - $\oint_L \vec{H}.\overrightarrow{dl} = \int_S \frac{\partial \vec{D}}{\partial t}ds = \frac{d}{dt} \int_S \vec{D}.ds$ - $\oint_L \vec{H}.\overrightarrow{dl} = \frac{dQ}{dt} = I$ (∵ $\int B.ds = \int_V \vec{P}.dv = Q$) - Hence magnetic field arises through the time varying electric field due to the generation of displacement current $I_d$. ### Maxwell's Equations - There are four fundamental equations of electromagnetism known as Maxwell's equations, which define the relation between electric and magnetic fields. #### Maxwell's first equation: - Maxwell's first equation is based on Gauss's law for E. According to this law, total electric flux passing through a closed surface S in vacuum is equal to the product of total charge (Q) contained inside and $\epsilon_0$. - i.e., - $\phi_E = \oint_S E. ds = \frac{Q}{\epsilon_0}$ ① - For a medium, - $\phi_E = \oint_S E. ds = \frac{Q}{\epsilon_o \epsilon_r}$ ② - If charge inside the closed surface is continuously distributed and P is the charge volume density, then - $Q = \int_V Pdv$ ② - From eq" (1) + (2) - $\oint_S E.ds = \frac{1}{\epsilon_0}\int_V Pdv$ or - $\oint_S E.ds = \int_V Pdv$ ③ - This is the integral form of Maxwell's first equation. - Using Gauss's divergence theorem, from eq" ③: - $\oint_S E.ds = \int_V (\nabla. E)dv = \frac{1}{\epsilon_0} \int_V Pdv$ - or - $div E = \nabla . E = \frac{P}{\epsilon_0}$ ④ - This is the differential form of Maxwell's first equation. - *If medium is homogeneous and isotropic, then* - $\nabla . E = \frac{P}{\epsilon_0 \epsilon_r} \implies \nabla. (\epsilon E) = P$ - $\nabla . D = P \implies \nabla. D = P$ - *Here, $D = \epsilon E$ is called displacement vector* - *For free space (P=J=0) - so Maxwell's first equation becomes $\oint_S E.ds = 0$ or $\nabla.E =0$ or $\nabla.D =0$ * #### Maxwell's Second Equation: - Maxwell's second equation is based on Gauss's Law for a magnetic field. According to this law, net outgoing magnetic flux from any surface is always zero. This is due to the fact that magnetic field lines always form closed loops. - i.e., - $\phi_B = 0$ - or $\oint_S \vec{B}. d\vec{s} = 0$ - *This is the integral form of Maxwell's second equation.* - Using Gauss's divergence theorem in eq" ①: - $\nabla . B = div \vec{B} = 0$ - *This is the differential form of Maxwell's second equation.* #### Maxwell's Third Equation: - Maxwell's third equation is based on Faraday's law of electromagnetic induction. According to this law, whenever magnetic flux linked with a circuit changes, an e.m.f. is induced in the circut, and is directly proportional to the rate of change of the magnetic flux linked with the circuit. - According to Lenz's law, its direction is always opposite to the changes responsible for its production. - i.e., - $e = - \frac{d \phi_B}{dt}$ ① - Since, the line integral of electric field is also equal to the e.m.f. - i.e., $e = \oint_C \vec{E}.d\vec{l}$ - and $\phi_B = \int_S \vec{B}.d\vec{s}$ - So from eq" ①, $\oint_C \vec{E}.d\vec{l} = -\frac{d}{dt}\int_S \vec{B}.ds$ (② - or $\oint_C \vec{E}. \overrightarrow{dl} = -\frac{d}{dt} \int_S \vec{B}.d\vec{s}$ - *This is the integral form of Maxwell's third equation.* - Using Stokes's curl theorem - $\oint_C \vec{E}. \overrightarrow{dl} = \oint_S (\nabla \times \vec{E}).d\vec{s}$ - So - $\oint_C \vec{E}. \overrightarrow{dl} = \oint_S (\nabla \times \vec{E}).d\vec{s} = -\frac{d}{dt}\oint_S \vec{B}.d\vec{s}$ - or $Curl \ \vec{E} = \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$ - *This is the differential form of Maxwell's third equation.* - *For static fields (w.r.t time) i.e. $\frac{\partial \vec{B}}{\partial t} = \frac{\partial \vec{E}}{\partial t} = 0$ * - So - $\nabla \times \vec{E} = 0$ - or $\oint_C \vec{E}.d\vec{l} = 0$ #### Maxwell's Fourth Equation: - Maxwell's fourth equation is based on Ampere's law. According to this law, the line integral of the magnetic field over a closed path is equal to μ times the net current passing through the closed path: - i.e. $\oint_L \vec{B}.\overrightarrow{dl} = \mu_o I$ - $ \mu_o I =\oint_S \vec{J}.d\vec{s}$ - If current density is J, then $I = \oint_S \vec{J}.d\vec{s}$ - So, $\oint_L \vec{B}.\overrightarrow{dl} = \mu_0 \oint_S \vec{J}.d\vec{s}$ ① - From Stokes's curl theorem - $\oint_L \vec{B}.\overrightarrow{dl} = \oint_S (\nabla \times \vec{B}).d\vec{s}$ ② - From eq" ① & ② - $\nabla\times \vec{B} = \mu_o \vec{J}$ ③ - But, there is an inconsistency in this law as taking div - $\nabla.(\nabla \times \vec{B}) = \mu_o (\nabla. \vec{J})$ - But, we know that divergence of the curl of a vector is always zero, i.e. $\nabla.(\nabla \times \vec{B}) = 0$, so - $\nabla.\vec{J} = 0$ ④ - But, from the continuity equation - $\nabla. \vec{J} + \frac{\partial \rho}{\partial t} = 0$ ⑤ - On comparing equations ④ & ⑤, we can say that Ampere's law is valid only for steady currents (i.e. for which $J = 0$ or constant, $\frac{\partial \rho}{\partial t}=0$) - Maxwell removed this inconsistency by giving a suggestion that total current density J can be considered as made of two types of currents: - free current density $J_f$ and - displacement current density $J_d$, i.e. - $\vec{J} = \vec{J_f} + \vec{J_d}$ - So eq" ③ now becomes - $\nabla \times \vec{B} = \mu_o (\vec{J_f} + \vec{J_d})$ - Again, taking divergence both sides, we get - $\nabla.(\nabla \times \vec{B}) = \mu_o(\nabla. \vec{J_f} + \nabla. \vec{J_d}) = 0$ - $\implies \nabla. \vec{J_f} + \nabla. \vec{J_d} = 0 \implies \nabla. \vec{J_f} = -\nabla. \vec{J_d}$ ⑥ - But, from the continuity equation, - $\nabla. \vec{J} + \frac{\partial \rho}{\partial t} = 0$ - or $\nabla . \vec{J_f} = -\frac{\partial \rho}{\partial t}$ ⑦ - From eq" ⑥ & ⑦ - $\nabla. \vec{J_d} = \frac{\partial \rho}{\partial t}$ - or $\nabla. \vec{J_d} = \frac{\partial}{\partial t}(\nabla. \vec{D})$ (From Maxwell's 1st eq") - $\implies \vec{J_d} = \frac{\partial \vec{D}}{\partial t}$ - Thus, after Maxwell's modification, equation ① becomes: - $\oint_L \vec{B}.\overrightarrow{dl} = \mu_o \int_S (\vec{J_f} + \vec{J_d}).d\vec{s}$ ⑧ - $\implies$ $\oint_L \vec{B}.\overrightarrow{dl} = \mu_o \int_S (\vec{J_f} + \frac{\partial \vec{D}}{\partial t}).d\vec{s}$ ⑧ - This is the integral form of Maxwell's fourth equation. - Now, using Stokes's curl theorem, - $\oint_L \vec{B}.\overrightarrow{dl} = \oint_S (\nabla \times \vec{B}).d\vec{s}$ - So eq"⑧ becomes - $\oint_S(\nabla \times \vec{B}).d\vec{s} = \mu_o \int_s (\vec{J_f} + \frac{\partial \vec{D}}{\partial t}).d\vec{s}$ - $\implies \nabla \times \vec{B} = \mu_o (\vec{J_f} + \frac{\partial \vec{D}}{\partial t}) = \mu_o \vec{J_f} + \mu_o \epsilon \frac{\partial \vec{E}}{\partial t}$ ⑨ - This is the differential form of Maxwell's fourth equation: - *For static fields (i.e. $\frac{\partial \vec{E}}{\partial t} = \frac{\partial \vec{B}}{\partial t} =0$) - $\nabla \times \vec{B} = \mu_o \vec{J_f}$ * ### Generalized forms of Maxwell's Equations | Maxwell equation | Differential form | Integral Form| Remarks | |---|---|---|---| | I | $\nabla.E = \frac{\rho}{\epsilon_0}$ or $\nabla.D = \rho$ | $\oint_S E.ds = \frac{1}{\epsilon_0}\int_V Pdv$ | Gauss's law for Electrostatics | | II | $\nabla. B = 0$| $\oint_S B.ds = 0$| Gauss's law for Magnetostatics | | III |$\nabla \times E = -\frac{\partial B}{\partial t} $ | $\oint_C E.dl = - \frac{d}{dt}\int_S B.ds$ | Faraday's Law | | IV | $\nabla \times B = \mu_o (\vec{J_f} + \frac{\partial \vec{D}}{\partial t})$ | $\oint_C B.dl = \mu_o \int_S (\vec{J_f} + \frac{\partial \vec{D}}{\partial t})ds $ | Ampere's Law | ### Flow of Energy and Poynting Vector - When electromagnetic waves travel in space, they carry energy. - The energy density is always associated with electric and magnetic fields. - The vector Poynting vector is defined as the rate of flow of energy per unit area per second. - i.e., Poynting Vector = Energy / Area x time - or Poynting vector = Power / Area = Power Density (watt/m^2) - Mathematically, it can be represented as - $P =\vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{\mu}$ - The direction of this energy flow is perpendicular to both electric and magnetic field intensity.