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By
www.thegateacademy.com
Syllabus for Electromagnetic Theory
Coulomb's Law, Electric Field Intensity, Electric Flux Density, Gauss's Law, Divergence, Electric Field and Potential due to Point, Line, Plane and Spherical Charge Distributions, Effect of Dielectric Medium, Capacitance of Simple Configurations, Biot‐Savart’s Law, Ampere’s Law, Curl, Faraday’s Law, Lorentz Force, Inductance, Magnetomotive Force, Reluctance, Magnetic Circuits, Self and Mutual Inductance of Simple Configurations.
Analysis of GATE Papers
Year Percentage of Marks Overall Percentage
2015
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2.627% 20143.60
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20060.67
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Chapters
Page No.
#1. Electromagnetic Field
1 – 46
Introduction
1
Operators
2 – 7
Material and Physical Constants
7 – 8
Electromagnetic (EM Field)
8 – 9
Electric Field Intensity
9 – 12
Electric Dipole
12 – 17
Divergence of Current Density and Relaxation
18
Boundary Conditions
19 – 21
The Magnetic Vector Potential
21 – 25
Faraday’s Law
25 – 27
Maxwell’s Equation’s
27 – 28
Magnetic Field
28 – 32
Solved Examples
32 – 38
Assignment 1
39 – 41
Assignment 2
41 – 42
Answer Keys & Explanations
43 – 46
Module Test
47 – 50
Test Questions
47 – 48
Answer Keys & Explanations
49 – 50
alone will contribute immeasurably to success." …Harry Fosdick
Electromagnetic
Field
Learning Objectives
After reading this chapter, you will know: 1. Elements of Vector Calculus
2. Operators, Curl, Divergence
3. Electromagnetic Coulombs’ law, Electric Field Intensity, Electric Dipole, Electric Flux Density 4. Gauss's Law, Electric Potential
5. Divergence of Current Density and Relaxation 6. Boundary Conditions
7. Biot-Savart’s Law, Ampere Circuit Law, Continuity Equation
8. Magnetic Vector Potential, Energy Density of Electric & Magnetic Fields, Stored Energy in Inductance
9. Faraday’s Law, Motional EMF, Induced EMF Approach 10. Maxwell’s Equations
Introduction
Cartesian coordinates (x, y, z), −∞ < x < ∞, −∞ < y < ∞, −∞ < z < ∞ Cylindrical coordinates (ρ , ϕ, z), 0 ≤ ρ < ∞, 0 ≤ ϕ < 2π, −∞ < z < ∞ Spherical coordinates (r, θ, ϕ ) , 0 ≤ r < ∞, 0 ≤ θ ≤ π, 0 ≤ ϕ < 2π Other valid alternative range of θ and ϕ are---
(i) 0 ≤ θ < 2π, 0 ≤ ϕ ≤ π (ii) −π ≤ θ ≤ π, 0 ≤ ϕ ≤ π (iii) −π2≤ θ ≤ π 2⁄ , 0 ≤ ϕ < 2π (iv) 0 < θ ≤ π, −π ≤ ϕ < π Vector Calculus Formula
SL. No Cartesian Coordinates Cylindrical Coordinates Spherical Coordinates (a) Differential Displacement
dl = dx ax + dy ay + dz az dl = dρaρ + ρdϕaϕ +dzaz dl = draθdϕaϕ r + rdθaθ + r sin (b) Differential Area dS = dy dz ax = dx dz ay = dx dy az dS = ρ dϕ dz aρ = d ρ dz aϕ = ρdρd ϕ az ds = r2sin θ dθ dϕ ar = r sin θ dr dϕ aθ = r dr dθ aϕ (c) Differential Volume dv = dx dy dz dv = ρ dρ dϕ dz dv = r 2sin θ dθ dϕ dr
C
HA
PT
ER
1
Operators
1) ∇ V – Gradient, of a Scalar V 2) ∇ V – Divergence, of a Vector V 3) ∇ × V – Curl, of a Vector V 4) ∇2 V – Laplacian, of a Scalar V DEL Operator: ∇ = ∂ ∂xax+ ∂ ∂yay+ ∂ ∂z az(Cartesian) = ∂ ∂ρaρ+ 1 ρ ∂ ∂ϕaϕ+ ∂ ∂z az(Cylindrical) = ∂ ∂rar+ 1 r ∂ ∂θaθ+ 1 rsi n θ ∂ ∂ϕ aϕ(Spherical) Gradient of a Scalar fieldV is a vector that represents both the magnitude and the direction of maximum space rate of increase of V. ∇V =∂V ∂xax+ ∂V ∂yay+ ∂V
∂z az For Cartisian Coordinates =∂V ∂ρaρ+ 1 ρ ∂V ∂ϕaϕ+ ∂V
∂z az For Spherical Coordinates =∂V ∂rar+ 1 r ∂V ∂θaθ+ 1 rsi n θ ∂V
∂ϕ aϕ For Cylindrical Coordinates
The following are the fundamental properties of the gradient of a scalar field V 1. The magnitude of ∇V equals the maximum rate of change in V per unit distance. 2. ∇V points in the direction of the maximum rate of change in V.
3. ∇V at any point is perpendicular to the constant V surface that passes through that point. 4. If A = ∇V, V is said to be the scalar potential of A.
5. The projection of ∇V in the direction of a unit vector a is ∇V. a and is called the directional derivative of V along a. This is the rate of change of V in direction of a.
Example: Find the Gradient of the following scalar fields: (a) V = e−z sin 2x cosh y
(b) U = ρ2z cos 2ϕ (c) W = 10r sin2θ cos ϕ Solution: (a) ∇V =∂V ∂xax+ ∂V ∂yay+ ∂V ∂zaz = 2e−z cos 2x cosh y a
x + e−z sin 2x sinh y ay− e−z sin 2x cosh y az
(b) ∇U =∂U ∂ρaρ+ 1 ρ ∂U ∂ϕaϕ+ ∂U ∂zaz
= 2ρz cos 2ϕ aρ− 2ρz sin 2ϕ aϕ+ ρ2 cos 2 ϕ a z (c) ∇W =∂W ∂r ar+ 1 r ∂W ∂θaθ+ 1 r sin θ ∂W ∂ϕaϕ = 10 sin2θ cos ϕ a
Divergence of a Vector
Statement: Divergence of A at a given point P is the outward flux per unit volume as the volume shrinks about P. Hence, DivA = ∇. A = lim ∆v→0 ∮ A . dsS ∆v … … … (1)
Where, ∆v is the volume enclosed by the closed surface S in which P is located. Physically, we may regard the divergence of the vector field A at a given point as a measure of how much the field diverges or emanates from that point.
∇. A =∂Ax ∂x + ∂Ay ∂y ∂Az ∂z Cartisian System =1 ρ ∂ ∂ρ(ρAρ) + 1 ρ ∂Aϕ ∂ϕ + ∂Az ∂z Cylindrical System = 1 r2 ∂ ∂r(r2Ar) + 1 r sin θ ∂ ∂θ(Aθsin θ) + 1 r sin θ ∂Aϕ ∂ϕ Sphearical System From equation (1), ∮ A . dS S = ∫ ∇ . A dv V
This is called divergence theorem which states that the total outward flux of the vector field A through a closed surface S is same as the volume integral of the divergence of A.
Example: Determine the divergence of these vector field (a) P = x2yza
x+ xzaz
(b) Q = ρ sin ϕ aρ+ ρ2za
ϕ+ z cos ϕ az
(c) T = 1
r2cos θ ar+ r sin θ cos ϕ aθ+ cos θ aϕ
Solution: (a) ∇. P = ∂ ∂xPx+ ∂ ∂yPy+ ∂ ∂zPz = ∂ ∂x(x2yz) + ∂ ∂y(0) + ∂ ∂z(xz) = 2xyz + x (b) ∇ . Q =1 ρ ∂ ∂ρ(ρQρ) + 1 ρ ∂ ∂ϕQϕ+ ∂ ∂zQz =1 ρ ∂ ∂ρ(ρ2sin ϕ) + 1 ρ ∂ ∂ϕ(ρ2z) + ∂ ∂z (z cos ϕ) = 2 sin ϕ + cos ϕ (c) ∇. T =r12∂r∂ (r2T
r) +r sin θ1 ∂θ∂ (Tθsin θ) +r sin θ1 ∂ϕ∂ (Tϕ)
= 1 r2 ∂ ∂r(cos θ) + 1 r sin θ ∂ ∂θ(r sin 2θ cos ϕ) + 1 r sin θ ∂ ∂ϕ(cos θ) = 0 + 1
r sin θ2r sin θ cos θ cos ϕ + 0 = 2 cos θ cos ϕ
Curl of a Vector
Curl of a Vector field provides the maximum value of the circulation of the field per unit area and indicates the direction along which this maximum value occurs.
That is, Curl A = ∇ × A = lim ΔS→0( ∮ A . dlL ∆S )max an… … … . . (2) ∇ × A = || ax ay az ∂ ∂x ∂ ∂y ∂ ∂z Ax Ay Az || = 1 ρ || aρ ρaϕ az ∂ ∂ρ ∂ ∂ϕ ∂ ∂z Aρ ρAϕ Az || = 1 r2sin θ || aρ raθ r sin θ aϕ ∂ ∂r ∂ ∂θ ∂ ∂ϕ Ar rAθ r sin θ Aϕ ||
From equation (2) we may expect that ∮ A dl = ∫(∇ × A
S
) . ds
L
This is called stoke’s theorem, which states that the circulation of a vector field A around a (closed) path L is equal to the surface integral of the curl of A over the open surface S bounded by L, Provided A and Δ × A are continuous no s.
Example: Determine the curl of each of the vector fields. (a) P = x2yz a
x+ xzaz
(b) Q = ρ sin ϕaρ+ ρ2 zaϕ+ z cos ϕaz
(c) T = 1
r2cos θ ar+ r sinθ cos ϕaθ+ cos ϕ aϕ
Solution: (a) ∇ × P = (∂Pz ∂y − ∂Py ∂z) ax+ ( ∂Px ∂z − ∂Pz ∂x) ay+ ( ∂Py ∂x − ∂Px ∂y) az = (0 − 0)ax+ (x2y − z)a y+ (0 − x2z)az = (x2y − z)a y− x2zaz (b) ∇ × Q = [1 ρ ∂Qz ∂ϕ − ∂Qϕ ∂z ] aρ+ [ ∂Qρ ∂z − ∂Qz ∂ρ] aϕ+ 1 ρ [ ∂ ∂ρ(ρQϕ) − ∂Qρ ∂ϕ] az = (−z ρ sin ϕ − ρ2) aρ+ (0 − 0)aϕ+ 1 ρ(3ρ2z − ρ cos ϕ)az = −1
(c) ∇ × T = 1 r sin θ[ ∂ ∂θ(Tϕsin θ) − ∂ ∂ϕTθ] ar +1 r[ 1 sin θ ∂ ∂ϕTr− ∂ ∂r(rTϕ)] aθ+ 1 r [ ∂ ∂r(rTθ) − ∂ ∂θTr] aϕ = 1 r sin θ[ ∂ ∂θ(cos θ sin θ) − ∂ ∂ϕ(r sin θ cos ϕ)] ar +1 r [ 1 sin θ ∂ ∂ϕ (cos θ) r2 − ∂ ∂r(r cos θ)] aθ +1 r[ ∂ ∂r(r2sin θ cos ϕ) − ∂ ∂θ (cos θ) r2 ] aϕ = 1
r sin θ(cos 2θ + r sin θ sin ϕ)ar+ 1 r(0 − cos θ)aθ +1 r(2r sin θ cos ϕ + sin θ r2 ) aϕ = (cos 2θ r sin θ+ sin ϕ) ar− cos θ r aθ+ (2 cos ϕ + 1 r3) sin θ aϕ Laplacian
(a) Laplacian of a scalar field V, is the divergence of the gradient of V and is written as ∇2V.
∇2V =∂2V
∂x2+
∂2V
∂y2+
∂2V
∂z2 → For Cartisian Coordinates
∇2V =1 ρ ∂ ∂ρ(ρ ∂V ∂ρ) + 1 ρ2 ∂2V ∂ϕ2+ ∂2V
∂z2 → For Cylindrical Coordinates
= 1 r2 ∂ ∂r(r2 ∂V ∂r) + 1 r2sin θ ∂ ∂θ(sin θ ∂V ∂θ) + 1 r2sin θ ∂2V
∂ϕ2→ For Spherical Coordinates
If ∇2V = 0, V is said to be harmonic in the region.
A vector field is solenoid if ∇.A = 0; it is irrotational or conservative if ∇ × A = 0 ∇. (∇ × A) = 0
∇ × (∇V) = 0
(b) Laplacian of Vector A̅
∇2A⃗⃗ = ⋯ is always a vector quantity
∇2A⃗⃗ = (∇2A
x)âx + (∇2Ay)ây + (∇2Az)âz
∇2A x → Scalar quantity ∇2A y→ Scalar quantity ∇2A z→ Scalar quantity ∇2V =−p ϵ...Poission’s E.q. ∇2V = 0 ...Laplace E.q. ∇2E = μσ ∂E⃗⃗ ∂t+ μE ∂2E ∂t2. . . wave E. q.
Example: The potential (scalar) distribution in free space is given as V = 10y4+ 20x3.
If ε0: permittivity of free space what is the charge density ρ at the point (2,0)? 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: Poission’s Equation ∇2 V = −ρ ε (∂2 ∂x2+ ∂2 ∂y2+ ∂2 ∂z2) (10 y4+ 20x3) = −ρ ε0 ∵ ε = εr ε0[ε = ε0 as εr = 1] 20 × 3 × 2x + 10 × 4 × 3y2 =−ρ ε0 At pt(2, 10) ⇒ 20 × 3 × 2 × 2 =−ρ ε0 ρ = −240ε0 Example: Find the Laplacian of the following scalar fields
(a) V = e−z sin 2x cosh y
(b) U = ρ2z cos 2ϕ
(c) W = 10r sin2 θ cos ϕ
Solution: The Laplacian in the Cartesian system can be found by taking the first derivative and later the second derivative.
(a) ∇2V =∂2V ∂x2+ ∂2V ∂y2+ ∂2V ∂z2 = ∂
∂x(2e−zcos 2x cosh y) + ∂
∂y(e−zsin 2x sinh y) + ∂
∂z(−e−zsin 2x cosh y) = −4e−zsin 2x cosh y + e−zsin 2x cosh y + e−zsin 2x cosh y
= −2e−zsin 2x cosh y
(b) ∇2U =1 ρ ∂ ∂ρ(ρ ∂U ∂ρ) + 1 ρ2 ∂2U ∂ϕ2+ ∂2U ∂z2 =1 ρ ∂ ∂ρ(2ρ2z cos 2ϕ) − 1 ρ24ρ2z cos 2ϕ + 0 = 4z cos 2ϕ − 4z cos 2ϕ = 0 (c) ∇2W = 1 r2 ∂ ∂r (r2 ∂W ∂r) + 1 r2sin θ ∂ ∂θ(sin θ ∂W ∂θ) + 1 r2 sin2θ ∂2W ∂ϕ2 = 1 r2 ∂ ∂r(10 r2 sin2θ cos ϕ) + 1 r2 sinθ ∂
∂θ(10r sin 2θ sin θ cos ϕ) −
10r sin2θ cos ϕ
r2 sin2θ
=20 sin2 θ cos ϕ
r +
20r cos 2θ sin θ cos ϕ r2sin θ +
10r sin 2θ cos θ cos ϕ r2sin θ −
10 cos ϕ r =10 cos ϕ
r (2 sin2 θ + 2 cos 2θ + 2 cos2 θ − 1) =10 cos ϕ
r (1 + 2 cos 2θ)
Stoke’s Theorem
Statement: Closed line integral of any vector A⃗⃗ integrated over any closed curve C is always equal to the surface integral of curl of vector A⃗⃗ integrated over the surface area ‘s’ which is enclosed by the closed curve ‘c’.
∮ A⃗⃗ . dL⃗ = ∫ ∫(∇ × A⃗⃗ )
S dS⃗
The theorem is valid irrespective of (i) Shape of closed curve ‘C’ (ii) Type of vector ‘A’
(iii) Type of co-ordinate system
Divergence Theorem
∯ A⃗⃗
S
dS⃗ = ∭ V⃗⃗ . A⃗⃗ dv
V
Statement: Closed surface integral of any vector A⃗⃗ integrated over any closed surface area. S is always equal to the volume integral of the divergence of vector A⃗⃗ integrated over the volume V which is enclosed by the closed surface are ‘S’ the theorem holds good, irrespective
(i) Shape of closed surface (ii) Type of coordinate system (iii) Type of vector A⃗⃗
Material & Physical Constants
(a) Material Constants
Material Conductivity (σ ) S/m Relative Permittivity (εr)
Air 0 1.0006 Aluminum 3.186 × 107 1.0 Bakelite 10−14 5 Brass 2.564 × 107 1 Carbon 3 × 104 − Copper 5.8 × 107 1 Glass 10−13 6 Graphite 105 − Mica 10−15 6 Paper − 3 Paraffin 10−15 2.1 S V S C
Plexiglas − 3.4 Polystyrene 10−16 2.7 PVC − 2.7 Porcelain − 5 Quartz 10−17 5 Rubber 10−13 5 Rutile − 100 Soil(Clay) (Sandy) 5 × 10−3 2 × 10−3 14 10 Urban ground 2 × 10−4 4 Vaseline − 2.2 Terflon 10−15 2.1 Water (Distilled) (Fresh) (Sea) 10−4 10−2 to 10−3 4 to 5 80 80 80 Wood − 2 Transformer oil − 2 to 3 Ebonite − 2.6 Epoxy − 4 (b) Physical Constants
Permittivity of free space, ∊0 = 8.854×10-12 F/m = (1/36 π) 10-9 F/m
Permeability of free space, μ0 = 4π×10-7 H/m
Impedance of free space, η0 = 120 π Ohms = 377 Ohms
Velocity of free space, c = 3 ×108 m/sec = 3 ×1010 cm/sec
Charge of an Electron, q = 1.602 × 10-19 C
Mass of electron, m = 9.107 × 10 -31 kg
Boltzman’s constant, k = 1.38×10-23 J/0K
Planck’s constant, h = 1.054× 10-34 J-s
Base of natural logarithm, e = 2.718
Electromagnetic (EM Field)
In general, electromagnetic field is regarded as interplay between time varying electric and magnetic fields. The study of electromagnetic can be accomplished with study of electrostatics, magneto statics and time varying electric and magnetic fields.
Electrostatics deals with field related to stationary charge(s). The charge can be positive or negative. The unit of charge is called a coulomb. The charge of an electron is
e = −1.6019 × 10−19 Coulombs q = |e|
Charge may be distributed in space or may be concentrated in a small volume at a point.
A charge that occupies a volume in space may be considered to be a point charge for analysis purposes if this volume is small compared to the surrounding dimensions.
A charge density defines charge distribution on a line (or) over a surface (or) throughout a volume.
Coulomb’s Law
Statement: “The force between any two point charges Q1 and Q2 is proportional to the product of the
two charges, inversely proportional to square of the distance between the two charges, and directed along the line connecting the two charges”.
The Mathematical expression of Coulomb’s law is F = Q1Q2
4π ε R2 R̂
Where (1/4π ε) is proportionality factor R is the distance between the two charges. R
̂ is the unit vector pointing form Q1 to Q2 (or) Q2 to Q1.
The proportionality factor depends on the material in which the charges are located. ‘∊’ is a material constant and is called the permittivity of the material and its units are “Farad/meter”.
Force is measured in “Newton”.
ε = ε0 εr, where εris relative permittivity and ε0 is permittivity of free space.
Electric Field Intensity
Electric Field Intensity is defined as the force per unit charge, when placed in an electric field and its unit is Newton/Coulomb (or) Volt/meter.
The electric field can be viewed as starting at a positive charge and ending at a negative charge. In the electric field E of a charge (say Q1), if we introduce another charge (say Q2), there will be a force
acting on this charge Q2.
i.e., F = Q2 E
E⃗⃗ = Q
4πε0R2âr. . . due to pt. chrage
E⃗⃗ = ∫ δLdl 4πε0R2âr C . . . (Line Charge) E⃗⃗ = ∫ ∫ psds 4πεR2âr S . . . (Surface Charge) E⃗⃗ = ∫ ∫ ∫ ρvdv 4πε0R2âr V . . . (Volume Charge)
General Form of Coulomb’s Law R ⃗⃗ = r − r⃗⃗ ′ R = |R⃗⃗ | = |r − r⃗⃗ | ′ R ̂ = R |R⃗⃗ | ⃗⃗⃗⃗⃗ = r − r⃗⃗ ′ |r − r⃗⃗ |′ Q r âr
Electrical field E⃗⃗ = Q 4πε0R2× âR = Q 4πε0R3R⃗⃗ E⃗⃗ = Q(r − r⃗⃗ )′ 4πε0|r − r⃗⃗ |′ 3 Equipotential surface
An equipotential surface has its every point at equal potential. Properties:
The movement of charge over such a surface would require no work.
Tangential to such surface is zero electric field.
Electric field is always perpendicular to an equipotential surface for static fields, a conductor surface is always an equipotential surface.
Electric Field Intensity of a Finite Line Charge
Consider a line charge with uniform charge density ρL extending from A to B along the z – axis as shown in
For a finite line charge E = ρL
4πε0ρ[−(sin α2− sin α1)ar+ (cos α2− cosα1)az]
Where ρL is line charge density and r is the perpendicular distance from the line to point of interest. As a special case, for an infinite line charge α1=+π
2 and α2 = −π
2 And E = ρL
2πε0ρar
The E Field Due to a Line Charge
z r B A x P (x, y, z) Y 0 α2 α1 Q ̂R R ⃗⃗ = r − r⃗⃗ ′ r′ ⃗⃗ r Q x y O
For an infinite sheet of charge, E = ρs
2ε0an
Where ρs is surface charge density of sheet and an is a unit vector normal to sheet.
Example: Point charges 1 mC and −2 mC are located at (3, 2, −1) and (−1, −1, 4), respectively. Calculate the electric force on a 10 nC charge located at (0, 3, 1) and the electric field intensity at that point.
Solution: F = ∑ QQk 4πε0R2aR = ∑ QQk(r − rk) 4πε0|r − rk|3 k=1,2 k=1,2 = Q 4πε0{ 10−3[(0, 3, 1) − (3, 2, −1)] |(0, 3, 1) − (3, 2, −1)|3 − 2 × 10−3[(0, 3, 1) − (−1, −1, 4)] |(0, 3, 1) − (−1, −1, 4)|3 } =10−3× 10 × 10−9 4π ×1036π−9 [ (−3, 1, 2) (9 + 1 + 4)3/2− 2(1, 4, −3) (1 + 16 + 9)3/2] = 9 × 10−2[(−3, 1, 2) 14√14 + (−2, −8, 6) 26√26 ] F = −6.507 ax− 3.817 ay+ 7.506az mN At that point, E = F Q = (−6.507 − 3.817, 7.506) × 10−3 10 × 10−9 E = −650 × 7ax− 381 × 7ay+ 750 × 6az kV/m
Example: The finite sheet 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 on the z = 0 plane has a charge density ρs= xy(x2+ y2+ 25)3 2⁄ n C/m2. Find
(a) The total charge on the sheet (b) The electric field at (0, 0, 5)
(c) The force experienced by a −1 mC charge located at (0, 0, 5) Solution:
(a) Q = ∫ ρS SdS= ∫ ∫ xy(x01 01 2+ y2+ 25)3/2 dx dy nC
Since x dx = 1/2 d(x2), we now integrate with respect to x2 (or change variable:
x2= u so that x dx = du 2⁄ . Q =1 2 ∫ y 1 0 ∫(x2+ y2+ 25)3 2⁄ d(x2)dy nC 1 0 =1 2∫ y 2 5 1 0 (x2+ y2+ 25)5 2⁄ | 0 1 dy =1 5∫ 1 2 1 0 [(y2+ 26)5 2⁄ − (y2+ 25)5 2⁄ ]d(y2)
= 1 10× 2 7[(y2+ 26)7 2⁄ − (y2+ 25)7/2]|0 1 = 1 35[(27)7 2⁄ + (25) 7 2− 2(26)7 2⁄ ] Q = 33.15 nC (b) E = ∫ρS dS aR 4πε0r2 S = ∫ρS dS (r − r′) 4πε0 |r − r′|3 S
Where r − r′= (0, 0, 5) − (x, y, 0) = (−x, −y, 5). Hence,
E = ∫ ∫10 −9xy(x2+ y2+ 25)3 2⁄ (−xa x− yay+ 5az)dx dy 4π ×1036π (x−9 2+ y2+ 25)3 2⁄ 1 0 1 0 = 9 [− ∫ x2 dx 1 0 ∫ y dy ax 1 0 − ∫ x dx 1 0 ∫ y2 dy a y+ 5 1 0 ∫ x dx 1 0 ∫ y dy az 1 0 ] = 9 (−1 6 , −1 6 , 5 4) = (−1.5, −1.5,11.25) V/m (c) F = qE = (1.5, 1.5, −11.25) mN
Electric Dipole
Two equal and opposite electric charges, separated by a very short distance is called electric dipole and is shown below.
Electric Dipole
The electric dipole moment, p = Q d
The dipole moment is directed from –Q to + Q
The electric field intensity of a dipole varies as 1/R3, where as the electric field intensity of a
point charge varies as 1/R2
The electric field due to the dipole is given by E = p 4πε0r3(2 cos θ ar+ sin θ aθ) R > > d +Q Q R P d θ
Electric Flux Density (D) D = ε0 E
This vector has the same direction as E, but it is independent of ‘∊’ and therefore of material properties.
The unit of ‘D’ is Coulomb/meter2.
Electric Flux Ψ in terms of D is defined as Ψ = ∫ D. ds
S
Gauss’s Law
Gauss’s law states that the total electric flux Ψ through any closed surface is equal to the total charge enclosed by that surface. Thus
Ψ = Qenc
Ψ = Qenc i. e. , Ψ = ∮ dΨ
S
= ∮ D ∙ ds
Total charge enclosed Q = ∫ ρv dv
v or Q = ∮ D ∙ dS S = ∫ ρv V dv
By applying divergence theorem, ∮ D ∙ dS = ∫ ∇
v
∙ D dv So, ρv= ∇ ∙ D
Which is one of the four Maxwell’s equation and it states that the volume charge density is the same as the divergence of the electric flux density.
When, at any point if charge density is zero, then divergence of electric flux density and divergence of electric field intensity is zero.
Curl of static electric field intensity is zero. Mathematically, ∇ × E = 0
Thus, electrostatic field is Irrotational (curl free) and Non-Solenoidal (non zero divergence).
Gauss’s law is an alternate form of Coulomb‘s law.
Gauss’s law may be used either to calculate the equivalent charges from known electric fields or electric fields due to known charges.
Example: Find the Total flux in a cylinder of radius r and length L placed in a uniform electric field E parallel to the axis of cylinder.
dA B
dA⃗⃗
A L C dA
Solution:
The total flux ϕE is the sum of flux coming out of surface A, B and C ϕE= ∫ E dA cos θ A + ∫ E dA cos θ B + ∫ E dA cos θ C θA = 0° θB= 90° θC = 180° dAA = πr2 dAB= 2πrl dAC= πr2 ϕE= E. A + 0 − E. A = 0
As ϕE = 0, charge enclosed Qenclosed = 0 and hence D and E are 0 at all point inside the plane of uniformly charged circular ring.
Example: Find the electric flux inside and outside a symmetrically charged sphere of radius ‘A’
Solution: By Guss’s Law ∫ E. dA surface = 1 ε0Qenclosed ∫ E dA = E (4πR2) = 1 ε0Q E = 1 4πε0 Q R2
Where Q is the charge in the sphere of radius A and enclosed by sphere of radius R. Suppose there is a point charge Q at the origin O. Then the electric field at the distance R will be
E = 1 4πε0
Q R2
E.g.: Find the field for a long st. wire of charge.
∫ EdA =λh
ε0, E(2πrh) =
λh ε0
E = λ
2πε0× r, Where, λ = Line Change / Unit Length
h E E E Gaussian Surface R A 0 r
Example: If A = ρ sin ϕ μp+ ρ2 μp, and L is the contour of fig. given below, the circulation Solution: ∮ A. dL C = ( ∫ + ab ∫ + bc ∫ + cd ∫ da ) A. dL Along ab, dϕ = 0, ϕ = 0 A. dL = 0 ∫ A. dL b a = 0 Along bc, dρ = 0 A.dL = ρ3 dϕ ∫ A. dL c b = ∫ ρ3dϕ π 0 = 23π = 8π Along cd, dϕ = 0, ϕ = π, A.dL = 0 ∫ A. dL = 0 d C
Along da, dρ = 0, A.dL = ρ3dϕ
∫ A. dL = ρ3∫ dϕ = (1)3(−π) = −π 0 𝜋 a d ∫ A. dL a d = 0 + 8π + 0 − π = 7π
Example: Let J =800 sin θ
r2+4 × Ur A/m2 the total current flowing in a portion of spherical surface having
radius r = 0.8 bounded by 0.1 π < θ < 0.3π, 0 < ϕ < 2π will be Solution: I = ∫ ∫ J. n d s S ∫ ∫ 800 sin θ (0.8)2+ 4 3π 0.1π × (0.8)2sin θ 2π 0 dθ dϕ = 154.8 A ∮A. dL c d 2 y L b a 1 2 −1 −2 1 C x
Example: In a certain region where the relative permitivity is 2.4, D = 2ux− 4uy+ 5uz nc m⁄ 2. Polarization = ? Solution: D = ε0E + P ⇒ where D = ε0εrE ε0E = D/εr p = D −D εr= D εr(εr− 1) p = (2ux − 4uy+ 5uz) 2.4 − 1 2.4 p = 1.2ux − 2.3uy+ 2.9uz nc m⁄ 2
Electric Potential
The scalar electric potential is defined using fundamental ideas of force and work related to the electric field. When a charge is allowed to move due to force in the electric field, work is said to be done as expressed below.
Work, W = (Force) (Displacement) dW = −F. dI
= (QE) (Displacement) = QE. dl
The total work done in moving a point charge Q from A to B is,
W = −Q ∫ E. dl
B A
Where negative sign indicates that the work is being done by an external agent.
Work done per unit charge is potential difference. Potential difference between two points is difference of absolute potentials at the two points.
Absolute Potential
The potential at any point is the work per unit charge required to bring a unit charge form infinity to the point.
In simple, it is potential difference between any point and a reference point at infinity. Due to a point charge, Q
Absolute potential at a point r = a is Va = Q/4π ε a
Absolute potential at a point r = b is Vb = Q / 4π ε b
Potential difference between ‘a’ and ‘b’ is Vab = Va – Vb
= Q 4πε( 1 a− 1 b)
The potential, only depends on the distance between points ‘a’ and ‘b’ and the point charge, regardless of path between ‘a’ and ‘b’.
In moving a charge along a closed path in electrostatic field, total work done is zero. Thus, ∮ E ∙ dl
L
= 0
By applying stoke’s theorem ∮ E ∙ dl
L
= ∫(∇ × E). ds = 0
S
∇ × E = 0
Any surface on which the potential is same throughout is known an equipotential surface.
The direction of E is everywhere normal to the equipotential surface.
No work is done in moving a charge from one point to other point along equipotential surface. The lines of constant potential are always perpendicular to the electric field intensity. Electric scalar potential is scalar.
Electric field intensity and electric potential are related as, E = − ∇ V. The negative sign shows that the direction of E is opposite to the direction in which V increases; E is directed from higher to lower level of V.
An electric flux line is an imaginary path or line drawn in such a way that its direction at any point is the direction of the electric field at that point.
Example: Two point charges −4 μC and 5 μC are located at (2, −1, 3) and (0, 4, −2), respectively. Find the potential at (1, 0, 1), assuming zero potential at infinity.
Solution: Let, Q1= −4 μC, Q2 = 5 μC V(r) = Q1 4πε0|r − r1| + Q2 4πε0|r − r2| + C0 If V(∞) = 0, C0= 0, |r − r1| = |(1, 0, 1) – (2, − 1, 3)| = |(−1, 1, −2)| = √6 |r − r2| = |(1, 0, 1) – (0, 4, −2,)| = |(1, −4, 3)| = √26 Hence, V(1, 0, 1) = 10 −6 4π × 1036 π−9 [−4 √6+ 5 √26] = 9 × 103 (−1.633 + 0.9806) = − 5.872 kV
In electrostatics, charges are considered to be stationary. This certainly does not mean that charges cannot move. A conductor is a material that allows free movement of charge within its volume. In other words, if a charge is introduced into a conductor, it can move freely until something prevents it from moving. This something may be an electric field or the surface of the conductor. The movement of charges is merely a mechanism to reach the steady state. After charges have reached their final state, the conductor has no effect on the charges.
Conductors in electrostatic field are said to be perfect conductors. A perfect conductor (σ = ∞) can not contain an electrostatic field within it. E = 0, ρv= 0, Vab= 0 inside a conductor.
Unlike conductors, dielectrics are materials in which charges are not free to move.
A perfect dielectric is a material which has bound charges but no free charges.
A material is linear if a particular property like permittivity does not change when the fields are changed.
A homogeneous material is a material whose physical properties do not vary from point to point in space.
An isotropic material is one whose properties are independent of direction in space.
Divergence of Current Density and Relaxation
Charges in conductors under static conditions are distributed on the surface of the conductor in such a way that the potential energy in the system is minimum.
If charges are placed in the interior of a conductor, they will move to the surface. This motion of charges constitutes a current. Since the current ceases, once the charges are static on the surface, the current is a transient current.
Example: A block of silicon made in the form of a sphere of radius 100 mm is given. The conductivity of silicon is 4 × 10-4 S/m, its relative permittivity is 12 and both are constant.
Suppose that by some means a volume charge density ρv = 10-6 C/m3 is placed in the
interior of the sphere at t = 0; calculate
(a) The current produced by the charge as they move to the surface (b) The time constant of the charge decay in the silicon
(c) The divergence of the current density during the transient
Solution: At time t = 0, the charges start moving toward the surface. The charge density must satisfy the continuity equation at all times, ∇. J = −∂𝛅𝐯
∂t
In addition, the current in the conducting material must satisfy Ohm’s law, J = σE and Gauss’s law ∇. E =ρv ℰ ∴ ∆. σE =σρv ε = − ∂ρv ∂t ⇒∂δv ∂t + σ ερv= 0
This is a homogenous linear differential equation. By separately variables, ∂ρv ρv = −σ ε ∂t and integrating ln ρv= − σt ε + ln ρvo ρv = ρ0 e−t/τ where τ = ε/σ
This equation shows that the introduction of charge at some interior point of material results in a decay of volume charge density 𝛒𝐯.
At a radius ‘R’, the total current crossing the surface defined by the sphere of radius ‘R’ is I(R, t) = − d(QR) dt = − 4πR3 3 d dt[ρv(t)] = 15.8 R3 exp [ −t 2.65 × 10−7]
The current depends on the location and increases with the radius. Therefore, it is not constant in space or time.
The time constant of the charge decay is ε/σ. This time constant depends on material alone and is called the relaxation time.
A long time constant (poor conductors) means charges take longer to ‘relax’ or to reach the surface.
A short time constant (good conductor) means the charges quickly reach their static state (at the surface). In the present case,
τ = ε/σ = 2.65 × 10−7 sec
the divergence of the current density is ∇ . J = − ∂ρv ∂t = − − ∂ ∂t [ρ0e −t (ϵ σ⁄ )] = 3.77 exp (−t 2.65 × 10⁄ −7)
Boundary Conditions
If the field exists in a region consisting of two different media, the condition that the field must satisfy at the interface separating the media are called “Boundary Conditions”.
(a) Dielectric – Dielectric Boundary Conditions: The tangential component Et of field undergoes no change on the boundary and it is said to be continuous across the boundary,
E1t = E2t ⇒D1t
ε1 =
D2t
ε2
Hence Dt undergoes some change across the boundary and said to be discontinuous. For normal component, D1n− D2n = ρS
Where ρS is the free charge density placed at the boundary. If ρS= 0 then
D1n = D2n ε1E1n = ε2E2n
So the normal component of E is discontinuous across the interface. Law of refraction,tanθ1
tan θ2 = εr1 εr2
Where θ1 and θ2 are angles which E1 and E2 makes with the normal at the interface.
(b) Conductor – Dielectric Boundary Condition: No electric field, E may not exist inside conductor that is
E = 0, ρV = 0 Et= 0 = Dt
Dn= ρS ε2En = ρS
Thus, an electric field must be external to the conductor and must be normal to the surface. (c) Conductor – Free Space Boundary Condition:
Dt= εo Et= 0 Dn= εo En = ρS Biot–Savart’s Law
The Biot – Savart’s Law is used to compute the magnetic field generated by a steady current, i.e., a continuous flow of charges, for example through a wire, which is constant in time and in which charge is neither building up nor depleting at any point. The equation is as follows:
B = ∫μ0Idl × r̂
4π r2 or (equivalently) B = ∫
μ0Idl × r
4π r3 (in SI units)
Magnetic Field dB at P Due to Current Element I dl
dl I α r P dB (Inside)
Where,
I is the current,
dl is a vector, whose magnitude is the length of the differential element of the wire, and whose direction is the direction of conventional current,
B is the net magnetic field, μ0 is the magnetic constant,
r̂ is the displacement unit vector in the direction pointing from the wire element towards the point at which the field is being computed,
r = rr̂ is the full displacement vector from the wire element to the point at which the field is being computed, the symbols in boldface denote vector quantities.
Magnetic field B, at point P due to a straight line conductor, B = μoI
4πρ(cos α2− cos α1)aϕ
Where I is current of conductor and ρ is perpendicular distance.
Field at P Due to a Straight Filamentary Conductor
When the conductor is semi infinite, α1= 90 and α2= 0
B = μoI 4πρaϕ
For a infinite conductor α1= 180 and α2 = 0 B = μoI
2πρaϕ
Ampere's Circuit Law
In classical electromagnetism, Ampère's circuit law, discovered by André-Marie Ampère in 1826, relates the integrated magnetic field around a closed loop to the electric current passing through the loop.
It states that the line integral of B around a closed path is the same as the net current Ienc enclosed by the path multiplied by μ permeability.
In SI units (the version in cgs units is in a later section), the "integral form" of the original Ampère's circuital law is:
∮ B. dl = μ0 Ienc C
By applying stoke’s Theorem to the right side of equation, B I A P ρ α1 α2
∮ B. dl = ∫(∇ × B). ds = μ0Ienc S C But Ienc = ∫ J. ds S ∇ × B = μoJ Where
∮ C is the closed line integral around the closed curve C.
B is the magnetic field in tesla.
". " is the vector dot product.
dl is an infinitesimal element (differential) of the curve C (i.e., a vector with magnitude equal to the length of the infinitesimal line element, and direction given by the tangent to the curve C, see below),
∬ denotes an integral over the surface S enclosed by the curve C (see below). The double integral sign is meant simply to denote that the integral is two-dimensional in nature.
μ0 is the magnetic constant also called the absolute permeability of free space.
Jf is the free current density through the surface S enclosed by the curve C
dS is the vector area of an infinitesimal element of surface S (that is, a vector with magnitude equal to the area of the infinitesimal surface element, and direction normal to surface S. The direction of the normal must correspond with the orientation of C by the right hand rule, see below for further discussion),
Ienc is the net free current that penetrates through the surface S.
∇ × H = J ≠ 0, that is a magnetostatic field is not conservative.
This is one of the four Maxwell’s equation. An isolated magnetic charge does not exist.
Thus the total flux through a closed surface in a magnetic field must be zero; that is, ∮ B. ds= 0
By applying divergence theorem, ∮ B. ds
S
= ∫ ∇. B dv = 0
V
or ∇. B = 0
This is the fourth Maxwell’s equation.
The Magnetic Vector Potential (A)
The magnetic vector potential is defined based on the divergence free condition of ‘B’, the magnetic flux density.
The definition of ‘A’ is based entirely on the mathematical properties of the vector ‘B’, not on its physical characteristics.
∴‘A’ is viewed as an auxiliary function rather than fundamental field quantity. Since the magnetic vector potential is a vector function, both its curl and divergence must be specified.
The magnetic vector potential does not have a simple physical meaning in the sense that it is not a measurable physical quantity like B or H.
Because the magnetic vector potential relates to the magnetic flux density through ‘A’, it is at right angles to the magnetic flux density ‘B’.
The magnetic vector potential is always in the direction of the current (and perpendicular to B). It is a sort of current distributed in space. It is sometimes even called a “fuzzy current” since it is spread around the current, as shown in below figure
∇ . B = 0, ∴ B = ∇ × A
∇. A = 0 and is called the coulomb guage for static fields. Ampere’s law is ∇ × H = J ∴ ∇ × B = μJ Since, B = ∇ × A, ∇ × ∇ × A = μ J ∇(∇. A). ∇2A = μ J Taking ∇. A = 0 −∇2A = μ J i.e., ∇2A = − μ J
The above equation is a vector Poisson’s equation.
Energy Density of Electric and Magnetic Fields
Electric and magnetic fields store energy. In a vacuum, the (volumetric) energy density (in SI units) is given by U =ε0 2 E2+ 1 2μ0B 2
The electric (Potential) energy in an electrostatic field is given by WE=
1
2∫ D. E dv = 1
2∫ εE2dv
The energy in an magnetostatic field in given by WM =
1
2∫ B. H dv = 1
2∫ μH2 dv
Capacitance of capacitor (Parallel Plate) is defined as the ratio of the magnitude of the charge on one of the plate to the potential difference between them.
C =Q V Wcharging = ∫ q Cdq = 1 2 Q2 C = 1 2CV2 Q 0 = Wstored = 1 2QV I A A
Capacitance of Simple Systems
Type Capacitance Resistance Comment Parallel-Plate Capacitor εA/d d
σA A: Area d: Distance Coaxial Cable / Cylindrical Capacitor 2πεl ln (a2/a1) ln a 1 a 1 ⁄ 2πσl a1 : Inner Radius a2 : Outer Radius l : Length Concentric Spheres 4πεa1a2
a2− a1 1 a −1b 4πσ a1 : Inner Radius a2 : Outer Radius Sphere 4πεa 1 4πσa a : Radius Circular Disc 8εa a : Radius
Example: The point charges −1 nC, 4 nC, and 3 nC are located at (0, 0, 0), (0, 0, 1), and (1, 0, 0), respectively. Find the energy in the system
Solution: W= W1+ W2+ W3 = 0 + Q2V21+ Q3(V31+ V32) = Q2 Q1 4πε0 |(0, 0, 1) − (0, 0, 0)|+ Q3 4πε0[ Q1 |(1, 0, 0) − (0, 0, 0)|+ Q2 |(1, 0, 0) − (0, 0, 1)|] = 1 4πε0(Q1Q2+ Q1Q3+ Q2Q3 √2 ) = 1 4π1036π−9 (−4 − 3 + 12 √2) 10 −18= 9 (12 √2− 7) nJ = 13.37 nJ Alternatively, W = 1 2 ∑ QkVk= 1 2 (Q1V1+ Q2V2+ Q3V3) 3 k=1 = Q1 2 [ Q2 4πε0(1)+ Q3 4πε0(1)] + Q2 2 [ Q1 4πε0(1)+ Q3 4πε0(√2)] + Q3 2 [ Q1 4πε0(1)+ Q2 4πε0(√2)] = 1 4πε0(Q1Q2+ Q1Q3+ Q2Q3 √2 ) = 9 ( 12 √2− 7) nJ = 13.37 nJ As obtained in the first solution.
Stored Energy in Inductance
The energy (measured in joules, in SI) stored by an inductor is equal to the amount of work required to establish the current through the inductor, and therefore the magnetic field. This is given by Estored =1
2LI2
The quantitative definition of the self inductance of a wire loop in SI units (weber per ampere known as henries) is
L = Nϕ I
Where ϕ denotes the magnetic flux through the area spanned by the loop, and N is the number of wire turns. The flux linkage thus is
Inductance of a Solenoid
A solenoid is a long, thin coil, i.e., a coil whose length is much greater than the diameter. Under these conditions, and without any magnetic material used, the magnetic flux density B within the coil is practically constant and is given by
B = μ0NI/ l
Where μ0 is the magnetic constant, N the number of turns, I the current and l the length of the coil.
Ignoring end effects the total magnetic flux through the coil is obtained by multiplying the flux density B by the cross-section area A and the number of turns N
ϕ = μ0N2IA/l
From which it follows that the inductance of a solenoid is given by L = μ0N2A/l
Inductance of Simple Systems
Coaxial Cable, High Frequency l 2πln a1 a μ a1 : Outer radius a : Inner radius l : Length Toroidal Core (Circular
cross-section) L = μ0 μr
N2r2
D
L = Inductance (H)
μ0 = Permeability of free space = 4π × 10−7H/m
μr = Relative permeability of core material
N = Number of turns
r = Radius of coil winding (m) D = Overall diameter of toroid (m)
The Continuity Equation
Consider figure below which shows an isolated volume, charged with a charge density ρv. No charge
leaves or enters the volume and, therefore, the charge is conserved. This is a trivial example of conversation of charge.
Now, consider figure below where we connect the volume through wire and allow the charge to flow through the wire to some other body (not shown)
The rate of decrease of charge in volume v is the current out of the volume. Because charges flow there is a current I in the wire. At the same time, the charge that flows out of the volume over a time dt (i.e., the charge that flows through the wire) is dQ. The time rate of decrease of charge in the volume ‘v’ is = dQ/dt.
∴ The rate of decrease of the charge in volume ‘v’ must equal the current out of the volume. I = −dQ/dt
I
ρv
ds
This is the basic definition of current. The current is considered to be positive because it flows through the surface of the volume ‘v’ in the direction of ds.
The total charge in volume ‘v’ is Q =∫ ρv vdv
∴ I = − d dt⁄ ∫ ρv
v dv = ∫
dρv dt dv
v
The current I flowing out of the volume ‘v’ in terms of the current density J that flows through the surface enclosing the volume ‘v’ is
I = ∮ J. ds
S
∴ ∮ J. ds = ∫ − dρv⁄ dvdt
v
The surface ‘s’ is a closed surface (encloses a volume). Thus, we can apply the divergence theorem to the LHS of the above equation to convert the closed surface integral to a volume integral.
∫ ∇ ∙ J dv
v
= ∫ (− dρv⁄ ) dv dt
v
Since both integrals are taken over the same volume, we get ∇. J = − dρv⁄ dt
This is the general form of the continuity equation.
This expression holds at any point in space and is not limited to conductors.
In the particular case of steady currents, the charge density ρv does not vary with time. The rate of
change of charge with time is zero and the charge decreases from volume ‘v’ must be replenished constantly to maintain a steady current.
∇. J = 0, If ρv= Constant
The above equation means that a steady current must flow in closed circuits, it cannot end in a point because the divergence at that point would not be zero, invalidating the requirement of steady current. This also means that the total current entering any volume must equal the total leaving this volume.
The steady current density is conservative, since ∮ E . dI = 0 i. e. , ∮ J σ⁄ . dI = 0
∴ ∇ × J = 0
The steady current density is solenoidal, Since, ∮ J . ds = 0 i. e. , ∇ . J = 0
∴ The current density is an irrotational, solenoidal field.
Faraday’s Law
Faraday found that the electromotive force (EMF) produced around a closed path is proportional to the rate of change of the magnetic flux through any surface bounded by that path.
In practice, this means that an electrical current will be induced in any closed circuit when the magnetic flux through a surface bounded by the conductor changes. This applies whether the field itself changes in strength or the conductor is moved through it.
Electromagnetic induction underlies the operation of generators, all electric motors, transformers, induction motors, synchronous motors, solenoids, and most other electrical machines.
Faraday's law of electromagnetic induction states that: Vemf= −dΨ
dt, Thus
Vemf is the electromotive force (emf) in Volts
ΨB is the magnetic flux in webers
Thus a static magnetic field produces no current flow, but a time varying field, produces an induced voltage in a closed circuit and causes a flow of current.
For the common but special case of a coil of wire, composed of N loops with the same area, Faraday's law of electromagnetic induction states that
Vemf= −N
dΨ dt Where
Vemf is the electromotive force (emf) in Volts N is the number of turns of wire
ΨB is the magnetic flux in Webers through a single loop
An emf - produced field is non-conservative
Motional EMF
When a conductor moves through a magnetic field an emf is produced in the conductor.
The charges in the conductor are carried along with the moving conductor and thus experience a magnetic force acting upon them which causes them to move inside the conductor.
As the conduction charges pile up at the end of conductor creating an electric field in the conductor.
The conduction electrons will stop piling up when the electric force on the interior conduction charges is equal to the magnetic force on those same charges so that the net force on the conduction charges is zero.
Motion in a Perpendicular Field
At Equilibrium, when the net force is zero. FE= FM
qE = vB sin(θ)q E = vB sin(90o)
E = vB
The charges that pile up create a voltage or emf across the length of the rod that is constant. E = V
L, Vemf = EL = vBL
v = Velocity of Conductor
Conduction Electrons Built up of Conductor B-Field Force Equilibrium v L E q v FB FM
Conduction Charge Magnetic Force on Electric Force on Conduction Charge Fr − − − − + + + + − −
Induced EMF Approach
As the rod moves, it sweeps out area A. The change in magnetic flux can be found using Faraday's Law, Vemf= − d dtϕM= − d dt∫ Bcos(θ′)dA =− Bd dt ∫ cos(180°)dA = Bd dt A = B d dtLx = BL dx dt= BLv
The results are the same, a conductor moving in a perpendicular magnetic field produces an emf across its length
Vemf = vLB
If the ends of the conductor are connected to an external circuit them the emf can act like a battery. The end where the positive charges would build up would act like the positive pole of a battery. Alternately, the flow of current is in the same direction as the force that magnetic field exerts on positive conduction charges due to the conductors motion.
Maxwell’s Equations
Static Time - Varying Integral Form Differential Form Differential Form Integral - Form ∮ H. dl = ∫ J. ds S L ∮ E. dl = 0 L ∮ D. dS = ∫ ρvdv V ∮ B. dS = 0 S ∇ × H = Jc ∇ × E = 0 ∇. D = ρv ∇. B = 0 ∇ × H = Jc+ JD ∇ × E = − ∂B ∂t⁄ ∇. D = ρv ∇. B = 0 ∮ H. dl = ∫ (JC+ ∂D ∂t) . ds S L ∮ E. dl =− ∂∂t L ∫ B. ds S ∮ D. ds = ∫ ρvdv V S ∮ B. ds = 0 ∇ × H = Jc+ JD is modified Ampere’s law
∇ × E = − ∂B ∂t⁄ is Faraday’s law
∇ × E = 0 conservative nature of electrostatic field ∇. D = ρv is Gauss’s law for electric field
∇. B = 0 is Gauss’s law for magnetic field (Non-existence of magnetic monopole) D = εE Electric flux density
B = μH Magnetic flux density
JC = σE Conduction current density, (this relation is referred to as Ohm’s law)
JD = ∂D ∂t⁄ Displacement current density
x L Area
σ is conductivity expressed in Siemens/ meter μ is permeability expressed in Henry / meter ε is permittivity expressed in Farad / meter
For a fields to “Qualify” as an electro magnetic field, it must satisfy all four Maxwell’s equations.
The ratio of conduction current density of displacement current density is referred to as loss tangent, i.e., loss tangent = (Jc / JD) = (σ / ωε).
If (σ / ωε) > > 1, the medium is referred to as high loss medium.
If (σ / ωε) < < 1, the medium is referred to as low loss medium.
If σ
ωε = 0, Loseless medium
The frequency at which (Jc/ JD) is equal to one is referred to as Transition Frequency, fq
i.e., When f = fq, Loss Tangent is one.
fq = σ/2πε is called Transition Frequency.
From continuity of current equation ∇. JC = − ∂ρv⁄ ∂t And Gauss’s law, ∇. D = ρv
And Ohms law, JC = σ E,
We can find, ρv = ρV0e−(σ ε⁄ )t
ρv = ρV0e−t τ
⁄ Where τ = ε/σ and is called relaxation time constant.
The above relaxation indicates that at an interior point of any conductor if we place a charge, it decreases to (1/e) times of initial charge in one relaxation time constant.
When operating frequency ‘f’, in a medium with σ, μ, ε is greater than fq, then that medium is
regarded as a Dielectric. f >> fq Dielectric
When operating frequency ‘f’, in a medium with σ, μ, ε is less than fq, then that medium is
regarded as a Conductor. f << fq Conductor
f = fq describes the transition point of medium behavior.
Magnetic Field
The magnetic field at a point is defined as being equal to the force acting on a unit magnetic pole placed at that point. [Unit of magnetic field is ampere per meter (A/m)]
Magnetomotive Force (mmf): Magnetomotive force is the flux producing ability of an electric current in a magnetic circuit. [It is something similar to electromotive force in an electric circuit].
[Unit of magnetomotive force is ampere (A)] - Note: Although some books use the term ampere-turns, it is strictly not correct as turns is not a dimension]
mmf ℑ = Σ I
Consider a coil having N turns as shown. It will link the flux path with each turn, so that total current linking with the flux would be
Σ I = N.I
Thus from Ampere’s Law, the mmf produced by a coil of N turns would be N I, and N I = H.I
Field Produced by a Long Straight Conductor: If a circular path of radius r is considered around the conductor carrying a current I, then the field Hr along this path would be constant by symmetry. ∴ By Ampere’s Law, 1.I = Hr .2π r
or Hr= I
2πr at a radial distance r from the conductor.
Field Produced Inside a Toroid
Consider a toroid (similar to a ring) wound uniformly with N turns. If the mean radius of the magnetic path of the toroid is a, then the magnetic path length would be 2π.a, and the total mmf produced would be N I. Thus from Ampere’s Law magnetic field H = NI
2π ainside the toroid.
[Variation of the magnetic field inside the cross section of the toroid is usually not necessary to be considered and is assumed uniform]
Reluctance of a Magnetic Path: A magnetic material presents a Reluctance S to the flow of magnetic flux when an mmf is applied to the magnetic circuit.
[This is similar to the resistance shown by an electric circuit when an emf is applied] Thus mmf = Reluctance × flux or ℑ = S. ϕ
For a uniform field, ℑ = N I = H.l, and ϕ = B. A = μ H. A ∴ H.l = S . μ H. A
So that the magnetic reluctance S = l
μ A
,
where l = length and A = cross-section[Unit of magnetic reluctance is henry−1 (H−1)]
Magnetic Permeance Λ is the inverse of the magnetic reluctance. Thus Λ = 1
S
=
μ Al
[Unit of magnetic permeance is henry (H)] Toroid
a I
N turn I
Self Inductance: While the reluctance is a property of the magnetic circuit, the corresponding quantity in the electrical circuit is the inductance.
Induced emf e +Ndϕ dt = L di dt, Nϕ = Li, L = Nϕ i
The self inductance L of a winding is the flux linkage produced in the same winding due to unit current flowing through it. For a coil of N turns, if the flux in the magnetic circuit is ϕ, the flux linkage with the coil would be N. ϕ.
also since NI = Sϕ, L =N
2
S = N2μA
l
Thus the inductance of a coil of N turn can be determined from the dimensions of the magnetic circuit.
Mutual Inductance: When two coils are present in the vicinity of each other’s magnetic circuit, mutual coupling can take place. One coil produces a flux which links with the second coil, and when a current in the first coil varies, an induced emf occurs in the second coil. Induced emf in coil 2 due to current in coil 1 e2= N2dϕ12 dt = M12 di1 dt , N2ϕ12= M12i1, M12= N2ϕ12 i1
The mutual inductance M12, of coil 2 due to a current in coil 1, is the flux linkage in the coil 2 due to unit current flowing in coil 1. Also since N1I1= Sϕ1 and a fraction k12 of the primary flux would link with the secondary, ϕ12= k12× ϕ1
∴ M12=
k12N1N2 S =
k12N1N2μ A l
k12 is know as the coefficient of coupling between the coils.
k12= k21 so that M12= M21. For good coupling, k12 is very nearly equal to unity.
Analysis of Electromagnetic Circuit
Electromagnetic circuit can be analyzed in a manner similar to the analysis of resistive circuit. Consider the following two winding transformer would in a three limb core.
Cross section areas of the core, and the effective length of magnetic path are indicated. It is assumed that the cross – section does not change at the corners. m. m. f. s
ℑ
1and ℑ
2 are produced in the twowindings and equal to N1I1 and N2I2.
relactances of each outer limb Sl= l1 μoμrAl,
relactances of each part of top and bottom yokes Sy=
ly μoμrAy, relactancesof middle limb Sm = ll
μoμrAm, [length of middle limb same as outer] I1 I2 Al N2 N1 Am ll ly Ay
if the fluxes flowing in the path are ϕl (outer limbs, yokes) and ϕm (centre limb), then an equivalent
circuit similar to the electrical equivalent circuit may be drawn as follows.
The fluxes can be calculated using laws similar to Ohm’s law and Kirchoff’s law as follows. ϕm= ϕl+ ϕl Similar to Kirchoff’s current law
ℑ1+ ℑ2 = Smϕm+ (2Sy+ Sl)ϕl Similar to Kirchoff’s voltage law and Ohm’s law
Only one loop was considered as both outer limbs are identical and must therefore have the same flux. If the limbs were different, then there would have been one additional flux term and one additional equation. The only unknowns are ϕm and ϕl which can be calculated. In the case of three
phase transformers, the winding current would have different phase angles, so that the corresponding mmfs too would have different phase angles. The analysis of this would be similar to the analysis of three phase problem, but no equivalent being there for inductance and capacitances in the corresponding equations. The above analysis are valid only in the linear region of the magnetization characteristic where the permeability can be assumed to be constant. However, when saturation occurs, the analysis is more complicated.
Analysis in the Presence of a Non- Linear Magnetization Characteristic: Only a simple circuit having a non – linear magnetic characteristic and a series air gap will be considered to illustrate the method of analysis. It is assumed that there is no fringing of flux around that air gap so that the flux density will be the same in both the air gap as well as the magnetic core. Bm= Ba. The characteristic of the
magnetic core is alo known. The air has a linear characteristic with permeability μo
Let the cross section the core (and air gap) be A, the length of the magnetic path be lm in the magnetic material and la in the air gap. Let the number of turn in the coil be N and the current I the winding be I. Then from Ampere’s law
NI = Hmlm+ Hala I1 ℑ2= N2I2 Sm Sy Sy Sl Sl ℑ1= N1I1 Sy Sy ℑ2 Sy Sy Sl Sy Sy Sm ϕm ϕl ϕl ℑ 1 Sl ⇒
for the air gap, Ha= Ba μo = Bm μo ∴ NI = Hmlm+Bm μo la, or Bm= −a Hm+ b
Since this equation has been written in terms of the parameters of the magnetic material, intersection of this straight line with the magnetization characteristic would give the operating position.
Solved Examples
Example 1
Given that D̅ = r2a r
̅ + 2sinθa̅̅̅ in Spherical Coordinate System, where D is the electric flux θ density, find the charge density ρ?
Solution:
Given: ρv= 4 (r +1
rcos θ)
Electric Flux Density, D⃗⃗ = Dra⃗ r+ Dθa⃗ θ+ Dϕa⃗ ϕ In Spherical Co-ordinate System,
Where Dr= r2 , D
θ= 2 sin(θ) , Dϕ = 0
The relation between D⃗⃗ and volume charge density ρ, is given by ∇ . D⃗⃗ = ρv… … … (1)
Which is point form of Gauss law
∇ . D⃗⃗ in Spherical Co-ordinates is given by ∇ . D⃗⃗ = 1 r2 ∂ ∂r(r2Dr) + 1 r sinθ ∂ ∂θ(Dθsin θ) + 1 r sinθ ∂Dϕ ∂ϕ ∴ ρv = 1 r2 ∂ ∂r(r2r2) + 1 r sinθ ∂ ∂θ(2 sin2θ) + 0 = 1 r24r3+ 1
r sinθ 2 × 2 sinθ cos θ = 4r + 4 rcos θ ∴ ρv = 4 (r +1 rcos θ) Bm = −a Hm+ b Bm B Hm H
Example 2
Evaluate the integral, ∫ 𝑟 C⃗⃗ . d⃗ r, where C is the helical path described by, x = cos t, y = sin t, z = t, joining the points given by t = 0 and t = π/2
Solution:
Let I = ∫ r
c
. dr
Where C is the helical path described by x = cos (t), y = sin (t), z = t joining the point P at t = 0 and the point, Q at t = π/2.
Position vector, r in Cartesian co-ordinates is given by r = x a⃗ x+ y a⃗ y+ z a⃗ z and dr = dx a⃗ x+ dy a⃗ y+ dz a⃗ z ∴ I = ∫(x dx + y dy + z dz) =1 2[x2+ y2+ z2]PQ c With t = 0, P(x, y, z) = (1, 0, 0) With t = π/2, Q(x, y, z)=(0,1,π/2) I =1 2[x2|10+ y2|01+ z2|0π/2] = 1 2[−1 + 1 + π2 4] = π2 8 Example 3
Three electrostatic point charges are located in the xy-plane as given below +Q at (−a/2, 0), +Q at (a/2, 0) and −2Q at (0, a√3/2)
Calculate the co-ordinates of the point, P, on the y-axis, where the potential due to these charges is zero. Also, calculate the magnitude of the electric field strength at p. At the point, P, what is the angle between the equi-potential passing through p and the y-axis? Solution:
Three point charges + Q at A (−a
2, 0) , +Q at B ( a
2, 0) and − 2Q at C (0, √3
2 a) are located in the xy-plane as shown in figure below.
(i) Let P(0, y) be the point on y-axis, where the potential, V = V1+ V2+ V3 due to these charges is zero. V = V1+ V2+ V3= 0 … … … . . (1) (a 2, 0) OC = √3 2 a P(0, y) +Q +Q –2Q y O x A C (−a2, 0) B z
V1= Q 4πε0(AP), AP = √ a2 4 + y2… … … . . . (2) V2= Q 4πε0(BP), BP = √ a2 4 + y2= AP … … … … . . … (3) V3= −2Q 4πε0(CP), CP = (y − OC) = (y − √3 2 a) … … . (4) From equations (2), (3) and (4)
V = Q 4πε0[ 1 AP+ 1 BP− 2 CP] … … … . (5) V = 0 if 1 AP+ 1 BP= 2 CPor 2 AP= 2 CPor AP = CP √a2 4 + y2= y − √3 2 a or a2 4 + y2= y2+ 3 4a2− y√3a y√3a =3 4a2− a2 4 or y√3a = a2 4 or y = a 2√3 ∴ Coordinates of the point P on y-axis are (0,2√3a ) (ii) To find E⃗⃗ :
Find equation (5), V at(0,y) is given by
V = 2Q 4πε0[ 1 AP− 1 CP] = Q 2πε0 [ 1 √a2 4 + y2 − 1 y − √2 a]3 E⃗⃗ at P(0, y) is given by E⃗⃗ = −grad(V) = −∇V ⇒ − [∂ ∂xV a⃗ x+ ∂ ∂yV a⃗ y+ ∂ ∂zV a⃗ z] ∂ ∂xV = 0, ∂ ∂zV = 0 ∂ ∂yV = Q 2πε0[(− 1 2) ( a2 4 + y2) −3/2 2y + (y −√3 2 a) −2 ] E⃗⃗ = − Q 2πε0[−y ( a2 4 + y2) −3/2 + (y −√3 2 a) −2 ] a⃗ y At P(x, y) = P (0, a 2√3) E⃗⃗ = − Q 2πε0[− a 2√3( a2 4 + a2 12) −32 + ( a 2√3− √3 2 a) −2 ] a⃗ y = − Q 2πε0[ 3 2a2] a⃗ y= − 3Q 4πε0a2 a⃗ y ∴ E = Ey = 3Q 4πε0a2
(iii) The direction of E⃗⃗ at P is the direction of the normal to the equipotential surface (V = 0) at that point, in the direction of the decreasing values of V.
From equation (6) it follows that the direction of E⃗⃗ is in the –ve y-direction. ∴ The angle between the equipotential surface and the y-axis is zero.
Example 4
Given an irrotational vector field
F⃗ = (k1 xy + k2z3)a⃗ x + (3x2− k3z)a⃗ y+ (3xz2− y)a⃗ z
Find ∇. F⃗ at(1,1, −2) Solution:
Given: Irrotational vector field, F⃗ = Fxa⃗ x+ Fya⃗ y+ Fza⃗ z,
Where Fx = (k1x y + k2z3), F
y= (3 x2− k3z), Fz= (3 x z2− y)
In Cartesian coordinates, Div (F⃗ ) = ∇. F⃗ is given by ∇. F⃗ = ∂ ∂x(Fx) + ∂ ∂y(Fy) + ∂ ∂z(Fz) ∂ ∂xFx = k1y , ∂ ∂yFy= 0, ∂ ∂zFz = 6 x z At the point (x, y, z). ∇ . F⃗ = k1y + 6 x z At (1, 1, −2), ∇ . F⃗ = (k1− 12) Example 5
Given E = 10e−j(4x−kt)ŷ V/m in free space.
(A) Write all the four maxwell’s equations in free space (B) Find ∇ × E
(C) Find H
Solution: Maxwell’s equations are Electro magnetic equations relating the field variables, which are vectors: E⃗⃗ , H⃗⃗ , B⃗⃗ and D⃗⃗
Point or Differential Form Integral Form ∇ x H⃗⃗ = ∂ ∂t D⃗⃗ + J ∮ H⃗⃗ . dl⃗⃗⃗ = ∫ ( ∂ ∂t D + J ) s . (ds) (i) ∇ x E⃗⃗ = − ∂ ∂t B⃗⃗ ∮ E⃗⃗ . dl⃗⃗⃗ = ∫ − ∂ ∂t s B ⃗⃗ . ds⃗⃗⃗⃗ (ii) ∇ . D⃗⃗ = ρ ∮D. s d s = ∫ρdVv (iii) ∇ . B⃗⃗ = 0 ∮B. d s s = 0 (iv) Contained in the above is the equation osf continuity,
∇ . J = − ∂
∂t ρ ∮ J . ds⃗⃗⃗⃗ = ∫ − ∂ ∂tρ dv
Characteristics of the medium in which the fields exist given rise to the following Constitutive relation:
D
⃗⃗ = ϵ E⃗⃗ , B⃗⃗ = μ H⃗⃗ , J = σ E⃗⃗
