Electromagnetic Wave and Radiating System CT - 1

Electromagnetic Wave: Electromagnetic Waves or EM waves are waves that are created as a result of vibrations an electric field and a magnetic field.

Microwave: Microwave are defined as electromagnetic radiations with a frequency ranging between 300MHz to 300 GHz. Example : Satellite communication signal.

Del Operator / Nabla Operator

Also known as vector differential operator.

• collection of partial derivative operators

• a metatheatrical operator commonly used in vector calculus

• “Operator” is similar to a function

• It’s defined as a vector, but it doesn’t have magnitude. So, it isn’t a “true” vector.

$\nabla = \frac{\delta }{\delta x} i + \frac{\delta }{\delta y} j + \frac{\delta }{\delta z} k$

Uses of Del operator

• Gradient : The gradient of a scalar field is a vector that points in the direction in which the field is most rapidly increasing, with the scalar part equal to the rate of change.

  $G radient = \vec\nabla\vec V$

• Divergence: A divergence shows how the field behaves towards away from a point. The divergence of a vector field is scalar field.

  $Divergence = \vec\nabla.\vec V$

• Curl: A curl is used to measure the rotational extent of the field about a particular point. The curl of a vector field is a vector field.

  $Curl = \vec \nabla \times \vec V$

  

The Divergence Theorem :

The divergence theorem states that the surface integral of the normal component of a vector point function $F$ over a closed surface $S$ is equal to the volume integral of the divergence of $\vec F$

$\int_x\int_y\int_z \vec \nabla . \vec F \space d\vec v = \int_v \vec \nabla . \vec F \space d\vec v = \oint_s \vec F \space . \space d\vec s$

Stokes Theorem

Statement: The surface integral of the curl of a function over a surface bounded by a closed surface is equal to the line integral of the particular vector function around the surface.

$\oint_C \vec F \vec dr = \int\int_s \vec F \times \vec\nabla \space \vec ds$

Electric Field

• a vector field

• the force per unit charge exerted/applied on a positive test charge at rest at a specific point

$E = \frac{F}{q}$

Electric Displacement Vector

• is the charge per unit area that would be displaced across a layer of conductor placed across an electric field

• denoted by $D$

$D = \epsilon_0E + P$

$\text{Polarization density, P } = \frac{\Delta p}{\Delta V}$

$\Delta p = \text{dipole moment } \\ \Delta V = \text{volume element}$

Isotropic medium

• all properties are uniform and independent of direction

• does not depend on the direction

• the velocity of light is the same in all direction

• Example : Glass

Anisotropic medium

• the properties are different in all direction

• direction-dependent

• the velocity of light differs according to direction

Ampere’s Circuit Law: The line integral of Magnetic field intensity along a closed path is equal to the current distribution passing through that loop.

$\oint H \cdot dL = I_{enc} = \int\int J \cdot ds$

$J = \text{current density}$

Maxwell Four Equation

Maxwell first equation: $\vec \nabla . \vec D = \rho_v$

Statement : The total electric displacement through the surface enclosing a volume is equal to the total charge with the volume. The more charge density the more Electric field.

$\oint \vec D \cdot d \vec s = Q_{enc} \space ...(1)$

$\oint\oint \vec D . d \vec s = \int\int\int \nabla .\vec D d\vec v \space ... (2) [\text{from Divergence theorem}]$

$\int\int\int \nabla .\vec D d\vec v \space = Q_{enc} \space .. (3) [\text{from equation 1 and 2}]$

The volume charge density,

$\rho_v = \frac{dQ}{dv} \\ \text{or, } dQ = \rho_v dv \\ \text{or, } Q = \int\int\int \rho_v dv$

$\int\int\int \vec \nabla . \vec Dd\vec v = \int\int\int \rho_v d \\ \nabla . D = \rho_v$

Maxwell Second equation $\vec \nabla \cdot \vec B = 0$

Statement: The total outward flux of magnetic induction $(B)$ through any closed surface is equal to zero.

From Gauss’s Law, $\oint\oint \vec B \cdot d\vec A = 0$

$\oint\oint \vec B \cdot d\vec A = \int_v \vec \nabla .\vec B d\vec v = \vec \nabla . \vec B = 0$

Maxwell Third equation: $\vec \nabla \times \vec E = - \frac{d\vec B}{dt}$

Statement: The electromotive force around a closed path is equal to negative rate of change of magnetic flux linked with the path. That means change of magnetic flux will create the electric field.

Faraday’s Law, $\epsilon_{emf} = - \frac{d \phi}{dt}$

Gauss Law, $\epsilon_{emf} = \oint_c \vec E.dl$

$\oint_c \vec E.dl = -\frac{d \phi}{dt} = \frac{d\oint_S - \vec B . da}{dt} = -\oint_S \frac{d\vec B}{dt} da$

or, $\oint_S \vec \nabla \times \vec E \space da = -\oint_S \frac{d\vec B}{dt} da$

or, $\oint_S (\vec \nabla \times \vec E + - \frac{d\vec B}{dt} )\space da = 0$

or, $\vec \nabla \times \vec E = - \frac{d\vec B}{dt}$

Maxwell Fourth Equation or Maxwell-Ampere’s Law: $\vec \nabla \times \vec H = \vec J + \frac{\delta \vec D}{\delta t}$

Statement: The magnetomotive force around a closed path is equal to the summation of conductor current and displacement current through any surface bounded by the path. Changing electric field creates magnetic field.

Ampere’s circuit law, $\oint_c \vec B . \vec {dl} = \mu_0i$

By Stroke’s theorem, $\oint_c \vec B . \vec dl = \oint_S (\vec \nabla \times \vec B) \vec dS$

$\oint_S (\vec \nabla \times \vec B) \vec dS = \mu_0i = \mu_0\oint_S \vec J . \vec dS$

or, $\oint (\vec \nabla \times \vec B - \mu_0 \vec J) \vec dS = 0$

or, $\vec \nabla \times \vec B = \mu_0 \vec J$

or, $\vec \nabla \times \vec H = \vec J$

Modified Ampere’s circuit law,

$\oint_c \vec B .\vec dl = \mu_0i + i_d$ [$i_d \text{ : displacement current}$]

$\vec \nabla \times \vec H = \vec J + \vec J_d = \vec J + \epsilon_0 \frac{\delta \vec E}{\delta t} = \vec J + \frac{\delta \vec D}{\delta t} \space \space [E = \frac{D}{\epsilon_0}]$

Derive equation of continuity by Maxwell’s 4th equation

$\vec \nabla \times \vec H = \vec J + \frac{\delta D}{\delta t} \\$or, $0 = \vec \nabla . (\vec J + \frac{\delta D}{\delta t})$

or, $\vec \nabla . \vec J = - \frac{\delta \rho}{\delta t}$

Why Ampere’s Law is not correct ?

Ampere’s Law is valid only for steady current or when the electric field does not change with time.

$\vec \nabla.\vec J = - \frac{\delta \rho}{\delta t}$, equation of continuity

From Maxwell fourth equation, $\vec \nabla \times \vec H = \vec J$

or, $\vec \nabla . (\vec \nabla \times H) = \vec \nabla . \vec J$

or, $\vec \nabla . \vec J = 0$, [$\vec \nabla (\vec \nabla \times H) = 0$] it doesn’t match with continuity equation.

Drawback of Ampere’s Law

• applicable for constant current or when the electric field does not vary the time

• valid only for the points where there is only conduction current and no displacement current

• doesn’t match with continuity equation

Problems:

• Type 1: Electron and proton number is given and you have to calculate the electric flux.

  Find the electric flux through the surface of a sphere containing $m$ protons and $n$ electrons.

  $\text{Flux, } \phi_E = \frac{q_{enc}}{\epsilon_0}$,

  $q_{enc} = m\cdot q_p + n\cdot q_e$

  $q_p = 1.6\times 10^{-19}, q_e =- 1.6\times 10^{-19}, \epsilon_0 = 8.85 \times 10^{-12}$

• Type 2: Surface Charge density and $x,y$ is given

  A cube of side L contains a flat plate with variable surface charge density, $\rho = -3xy$ . If the plate extends from $x = \text{ 0 to L}$ and $y = \text{ 0 to M}$. What is the total electric flux through the wall of the cube ?

  $\phi_E = \frac{q_{enc}}{\epsilon_0}$

  $q_{enc} = \int_s \rho da = \int_{y=0}^M \int_{x=0}^{L}-3xy \space dx dy$ = … [Applicable for continuous charge]

• Type 3: Electric Flux density is given and find the charge density $\rho$

  The electric flux density is given as, $\vec D = x^3 \hat i + x^2 y \hat j .$ Find the charge density $\rho$ inside a cube of side $2m$ placed centered at the origin with its side along the co-ordinate axes.

  $\vec \nabla . \vec D = \rho$

  or, $\rho = (\frac{d}{dx} \hat i + \frac{d}{dy} \hat j + \frac{d}{dz} \hat k) (x^3 \hat i + x^2y\hat k) = 3x^2$

  Charge, $Q = \int_v \rho dv = \int_{-1}^1 \int_{-1}^1\int_{-1}^1 3x^2 dx dy dz =8C$

  

• Type 4: Sphere

  $dv = r^2sin\theta drd\theta d\phi$ [$0<r<2, \space 0<\theta<\pi, 0<\phi<2\pi$]

  $\int uv dx = u\int v dx - \int (u' \int vdx) dx$

  The volume charge density inside a hallow sphere is $\rho = 10e^{-30\pi} \text{cm}^{-2}$. Find the total charge enclosed within the sphere. Also find electric flux density on the surface of the sphere.

  $Q = \int_v \rho dv = \int_{\phi = 0}^{2\pi} \int_{\theta = 0}^{\pi} \int_{r=0}^{2} 10 e^{-20\pi} r^2 sin\theta dr d\theta d\phi = \frac{\pi}{100}$

  $D = \epsilon.E = \frac{1}{4\pi r^2} \cdot \frac{\pi}{100} = \frac{1}{400 r^2}$

• Type 5:

  In a conducting medium the magnetic field is given as $H = y^2x\hat i + 2(x+1)yz \hat j - (x+1)z^2 \hat k$. Find the conduction current density $(J)$ at point $(2,0,-1)$ ; Also find the current $y=1, 0<x<1, 0<z<1$ .

  $J = \vec \nabla \times \vec H = \begin{bmatrix} \hat i & \hat j & \hat k \\ \frac{\delta}{\delta x} & \frac{\delta}{\delta y} & \frac{\delta}{\delta z} \\ y^2 x & 2(x+1)y^3 & -(x+1)z \end{bmatrix} \\\space \\ = -(2xy+2y)\hat i +(z^2 + y^2) \hat j$

  At point $(2, 0,-1),$ $J = \hat j$

  At $y=1,$ $\vec J = (1-z^2) \hat j$

  $I = \int_S \vec J dA = \int_z \int_x J dx dy = \int_0^1 \int_0^1 \hat (1+z^2) dx dz = \frac{2}{3} A$

• Type 6: EMF

  Find the $emf$ induced in a square loop with sides of length lying in the $yz$ plane is a region in which magnetic field charge overtime, $B(t) = B_0 e^{-5t/t_0} \hat i,$ $\epsilon = \frac{-d}{dt} \int_s \vec B \space \hat \eta \space \vec da, \text{unit normal } \hat \eta = \hat i$.

  $\text{emf} = -\frac{d}{dt} \int_s \vec B \cdot \vec \eta \space d\vec a = -\frac{d}{dt} \int_s B_0 e^{-5t/t_0} \hat i . \hat i d\vec a \\= -\frac{d}{dt} B_0e^{-5t/t_0} \int_s da \\ = -\frac{d}{dt} B_0e^{-5t/t_0} a^2 \\ = \frac{5a^2B_0}{t_0} e^{-5t/t_0}$