Lectures in Electromagnetism (4th Year, Course 1)

Lectures in Electromagnetism (4th Year, Course 1)

Assistant Professor Dr. Hamad Rahman Jappor Department of Physics University of Babylon October, 2014

‫محاضرات في الكهرومغناطيسية‬ ‫المرحلة الرابعة‪( ،‬الكورس األول)‬

‫اعداد‬ ‫األستاذ مساعد الدكتور حمد رحمن جبر‬ ‫قسم الفيزياء‬ ‫جامعة بابل‬ ‫تشرين األول‪2014 ،‬‬

Contents 1 Vector Analysis 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17

Scalars and vectors Vectors represented in rectangular coordinate systems Vector addition and subtraction Vector multiplication Cylindrical coordinate Systems Spherical coordinate system Differential length, area and volume in Cartesian coordinates Differential elements in cylindrical coordinate systems Differential elements in spherical coordinate systems Line, surface, and volume integrals Divergence of a vector The vector operator (DEL Operator) Divergence theorem Gradient of a scalar field Curl of a vector Stokes’ theorem Laplacian of a scalar

2 Electrostatic Fields 2.1 Coulomb's Law and Field Intensity 2.2 Electric field intensity 2.3 Electric fields due to continuous charge distributions 2.4 Electric Flux Density 2.5 Gauss's Law-Maxwell's Equation 2.6 Applications of Gauss's Law 2.7 Electric Potential 2.8 Capacitance 2.9 Relationship between E and V-Maxwell 's Equation 2.10 An Electric Dipole and Flux Lines 2.11 Energy Density in Electrostatic Fields 3 Electric Fields in Material Space 3.1 Properties of Materials 3.2 Convection and Conduction Currents 3.3 Conductors

3.4 3.5 3.6 3.7

Continuity Equation Boundary conditions Poisson’s and Laplace’s equation Uniqueness Theorem

4

Magnetostatic Fields 4.1 Biot-Savart's Law 4.2 Ampere's Circuit Law-Maxwell's Equation 4.3 Applications of Ampere's Law 4.4 Magnetic Flux Density-Maxwell's Equation 4.5 Maxwell's Equations for Static EM Fields 4.6 Magnetic Scalar and Vector Potentials 4.7 Derivation of Biot-Savart's Law and Ampere's Law

5

Magnetic Forces, Materials, and Devices 5.1 Forces due to Magnetic Fields 5.2 Magnetic Torque and Moment 5.3 A Magnetic Dipole 5.4 Magnetization in Materials 5.5 Classification of Magnetic Materials 5.6 Magnetic Boundary Conditions 5.7 Inductors and Inductances 5.8 Magnetic Energy

6

Maxwell's Equations 6.1 Faraday's Law 6.2 Transformer and Motional EMFs 6.3 Displacement Current 6.4 Maxwell's Equations in Final Forms 6.5 Time-Varying Potentials 6.6 Time-Harmonic Fields 7

Electromagnetic Wave Propagation 7.1 7.2 7.3 7.4 7.5 7.6 7.7

Waves in General Wave Propagation in Lossy Dielectrics Plane Waves in Lossless Dielectrics Plane Waves in Free Space Plane Waves in Good Conductors Power and the Poynting Vector Monochromatic plane waves

7.8 7.9 7.10 7.11 7.12

Linear and circular polarizations Energy and momentum of electromagnetic waves Electromagnetic waves in matter Reflection and transmission of EM waves at normal incidence Reflection and Transmission at Oblique Incidence

Text Books: Matthew N. O. Sadiku, Jerry Sagliocca, and Oladega Soriyan, Elements of Electromagnetics, 3rd ed., Oxford University Press, (2000).

References: 1. David J. Griffiths, Introduction to Electrodynamics, 3rd ed., Prentice Hall, 1999; PHI Learning Private Limited, (2009). 2. William H. Hayt, Jr., and John A. Buck, Engineering Electromagnetics Electromagnetics, 8th ed., McGraw-Hill, (2012). 3. John R. Reitz, Frederick. J. Milford and Robert W. Christy, Foundations of Electromagnetic Theory, 4th ed., Addison Wesley, (1993).

Chapter 1: Vector Analysis Scalars and vectors  A scalar is a quantity that has only magnitude.  Time  Distance  Temperature  Speed  A vector is a quantity that has both magnitude and direction.  Force  Displacement  Velocity

Unit vector  A vector ⃗A has both magnitude and direction. ⃗ = |𝐴 ⃗|=𝐴 Magnitude of 𝐴 ⃗ is defined as a vector whose magnitude is unity and whose  A unit vector along A direction is along vector ⃗A. It is denoted by 𝑎̂A ⃗ 𝐴 𝑎̂𝐴 = ⃗| |𝐴

,

⃗ = 𝐴𝑎 𝐴 ̂𝐴

 Vector ⃗A is completely specified in terms of its magnitude A and direction 𝑎̂A

Vectors represented in rectangular coordinate systems  Any vector in space can be uniquely expressed in terms of x, y and z coordinates using a rectangular coordinate system.

1

Assist. Prof. Dr. Hamad Rahman Jappor

⃗ = 𝐴𝑥 𝑎 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧 𝐴𝑥 , 𝐴𝑦 , 𝐴𝑧 ⇒ Components of A in the direction of 𝑥, 𝑦, 𝑧 ̂𝑥 , 𝑎 𝑎 ̂𝑦 , 𝑎 ̂𝑧 ⇒ Unit vectors specifying the direction of 𝑥, 𝑦, 𝑧 axes

Position vector of a point in space  A point P in Cartesian coordinate system may be expressed as its x, y, z coordinates. The position vector of a point P is the directed distance from the origin O to the point P.  A point P (3,4,5) has the position vector ⃗𝑟𝑝 = 3𝑎 ̂𝑥 + 4𝑎 ̂𝑦 + 5𝑎 ̂𝑧

Vector addition and subtraction If

⃗ = 𝐴𝑥 𝑎 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧

and

⃗𝐵 ⃗ = 𝐵𝑥 𝑎 ̂𝑥 + 𝐵𝑦 𝑎 ̂𝑦 + 𝐵𝑧 𝑎 ̂𝑧

⃗ =𝐴 ⃗ + ⃗𝐵 ⃗ = (𝐴𝑥 + 𝐵𝑥 )𝑎 𝐶 ̂𝑥 + (𝐴𝑦 + 𝐵𝑦 )𝑎 ̂𝑦 + (𝐴𝑧 + 𝐵𝑧 )𝑎 ̂𝑧 ⃗ − ⃗𝐵 ⃗𝐷 ⃗ =𝐴 ⃗ = (𝐴𝑥 − 𝐵𝑥 )𝑎 ̂𝑥 + (𝐴𝑦 − 𝐵𝑦 )𝑎 ̂𝑦 + (𝐴𝑧 − 𝐵𝑧 )𝑎 ̂𝑧

Distance vector  Distance vector is the displacement from one point to another.  If two points A (Ax, Ay, Az) and B (Bx, By, Bz) are given, the distance vector from A to B is given by ⃗𝑟𝐴𝐵 = (𝐵𝑥 − 𝐴𝑥 )𝑎 ̂𝑥 + (𝐵𝑦 − 𝐴𝑦 )𝑎 ̂𝑦 + (𝐵𝑧 − 𝐴𝑧 )𝑎 ̂𝑧

Unit vector in the direction of given vector ⃗ be a vector in space given by 𝐴 ⃗ = 𝐴𝑥 𝑎  Let 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧 ⃗ is given by  A unit vector in the direction of 𝐴 ⃗ 𝐴𝑥 𝑎 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧 𝐴 ̂𝐴 = 𝑎 = 2 + 𝐴2 ⃗| |𝐴 √𝐴𝑥2 + 𝐴𝑦 𝑧 2


Example 1: ⃗ = 10𝑎 ⃗ = 2𝑎 If 𝐴 ̂𝑥 − 4𝑎 ̂𝑦 + 6𝑎 ̂𝑧 and ⃗𝐵 ̂𝑥 + 𝑎 ̂𝑦 find ⃗ along 𝑎 (i) Component of 𝐴 ̂𝑦 ⃗ −𝐵 ⃗⃗ (ii) Magnitude of 3𝐴 ⃗ + 2𝐵 ⃗⃗ (iii) A unit vector along 𝐴 Answer: (i) -4 ⃗ −𝐵 ⃗⃗ = (30𝑎 (ii) 3𝐴 ̂ 𝑥 − 12𝑎 ̂ 𝑦 + 18𝑎 ̂ 𝑧 ) − (2𝑎 ̂𝑥 + 𝑎 ̂𝑦 ) = 28𝑎 ̂𝑥 − 13𝑎 ̂𝑦 + 18𝑎 ̂𝑧 |3𝐴 ⃗ −𝐵 ⃗⃗ | = √282 + 132 + 182 =35.74 ⃗ + 2𝐵 ⃗⃗ = 14𝑎 (iii) 𝐴 ̂𝑥 − 2𝑎 ̂𝑦 + 6𝑎 ̂𝑧 ⃗ + 2𝐵 ⃗⃗ = A unit vector 𝑐̂ along 𝐴 =

⃗ + 2𝐵 ⃗⃗ 𝐴 ⃗ + 2𝐵 ⃗⃗ | |𝐴 14𝑎 ̂𝑥 − 2𝑎 ̂𝑦 + 6𝑎 ̂𝑧

√142 + 22 + 62 𝑐̂ = 0.9113𝑎 ̂𝑥 − 0.1302𝑎 ̂𝑦 + 0.3906𝑎 ̂𝑧

Vector multiplication-dot product  Dot product or scalar product is defined as the product of magnitudes of the two vectors and the cosine of the smaller angle between them. ⃗ ||𝐵 ⃗ = |𝐴 ⃗⃗ |cos𝜃𝐴𝐵 𝐴∙𝐵

𝜃𝐴𝐵 is the smaller angle between them

 Properties: )i) Dot product obeys the commutative law: 𝐴 ∙ 𝐵⃗ = 𝐵⃗ ∙ 𝐴 (ii) When two vectors are perpendicular the angle between them is θ =90o, cos90 = 0 ⃗ = 𝐴𝐵cos90 = 0 𝐴∙𝐵 If the dot product of two vectors is zero, they are perpendicular. (iii) Since 𝑎̂𝑥 , 𝑎̂𝑦 , 𝑎̂𝑧 are mutually perpendicular

𝑎̂𝑥 ∙ 𝑎̂𝑦 = 𝑎̂𝑦 ∙ 𝑎̂𝑧 = 𝑎̂𝑧 ∙ 𝑎̂𝑥 = 0 (iv) When two vectors are parallel the angle between them is either 0 or 180 ⃗ = 𝐴𝐵cos0 = 𝐴𝐵 𝐴∙𝐵

⃗ = 𝐴𝐵cos180 = −𝐴𝐵 𝐴∙𝐵

or

3


(v) The square of a vector is the square of its magnitude 𝐴 ∙ 𝐴 = 𝐴𝐴cos0 = 𝐴2

⟹

𝐴2 = 𝐴2

and any unit vector dotted with itself is unity,

𝑎̂𝐴 ∙ 𝑎̂𝐴 = 1 (vi) Dot product is equal to the sum of products of their corresponding components. ⃗ = 𝐴𝑥 𝑎 ⃗𝐵 ⃗ = 𝐵𝑥 𝑎 If 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧 and ̂𝑥 + 𝐵𝑦 𝑎 ̂𝑦 + 𝐵𝑧 𝑎 ̂𝑧 ⃗ . ⃗𝐵 ⃗ = (𝐴𝑥 𝑎 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧 ) ∙ (𝐵𝑥 𝑎 ̂𝑥 + 𝐵𝑦 𝑎 ̂𝑦 + 𝐵𝑧 𝑎 ̂𝑧 ) = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧

Cross product or vector product ⃗ is denoted  Cross product or vector Product: Cross product of two vectors ⃗A and B ⃗ and is defined as as ⃗A × B ⃗ ||𝐵 ⃗ =𝑎 ⃗⃗ |sin𝜃𝐴𝐵 𝐴×𝐵 ̂ 𝑛 |𝐴 ⃗ and such that 𝐴 , 𝐵 ⃗ and 𝑎  Where 𝑎 ̂𝑛 is a unit vector perpendicular to 𝐴 and 𝐵 ̂𝑛 forms a right handed system.  Geometrically, the cross product can be defined as a vector whose magnitude is ⃗ and whose direction is in equal to the area of the parallelogram formed by 𝐴 and 𝐵 ⃗ through the the direction of advance of a right handed screw as 𝐴 is turned in to 𝐵 smaller angle.

 Properties: ⃗ = −𝐵 ⃗ ×𝐴 (i) Anti-commutative: 𝐴 × 𝐵 ⃗ + 𝐶 ) = (𝐴 × 𝐵 ⃗ ) + (𝐴 × 𝐶 ) (ii) Distributive: 𝐴 × (𝐵 ⃗ ×B ⃗ × ⃗C) ≠ (A ⃗ ) × ⃗C (iii) Not associative: ⃗A × (B (iv) Cross product of two parallel vectors is zero. 4


⃗ ||𝐵 ⃗ ||𝐵 ⃗ = |𝐴 ⃗⃗ |sin𝜃𝐴𝐵 𝑎 ⃗⃗ |sin0 𝑎 𝐴×𝐵 ̂𝑛 = |𝐴 ̂𝑛 = 0 ⃗ ||A ⃗ |sin0 𝑎̂𝑛 = 0 (v) ⃗A × ⃗A = |A (vi) 𝑎̂𝑥 × 𝑎̂𝑥 = 𝑎̂𝑦 × 𝑎̂𝑦 = 𝑎̂𝑧 × 𝑎̂𝑧 = 0 (vii) 𝑎̂𝑥 × 𝑎̂𝑦 = 𝑎̂𝑧

,

𝑎̂𝑦 × 𝑎̂𝑥 = −𝑎̂𝑧 ,

𝑎̂𝑦 × 𝑎̂𝑧 = 𝑎̂𝑥

and

𝑎̂𝑧 × 𝑎̂𝑥 = 𝑎̂𝑦

𝑎̂𝑧 × 𝑎̂𝑦 = −𝑎̂𝑥

and

𝑎̂𝑥 × 𝑎̂𝑧 = −𝑎̂𝑦

⃗ = 𝐴𝑥 𝑎 (viii) If 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧 𝑎𝑥 ̂ ⃗ × ⃗𝐵 ⃗ = [𝐴𝑥 𝐴 𝐵𝑥

𝑎𝑦 ̂ 𝐴𝑦 𝐵𝑦

and

⃗𝐵 ⃗ = 𝐵𝑥 𝑎 ̂𝑥 + 𝐵𝑦 𝑎 ̂𝑦 + 𝐵𝑧 𝑎 ̂𝑧

𝑎𝑧 ̂ 𝐴𝑧 ] 𝐵𝑧

⃗ along ⃗𝐵 ⃗ on ⃗𝐵 ⃗ is called projection of 𝐴 ⃗ and is given by  Scalar component of 𝐴 𝐴𝐵 = 𝐴cos𝜃𝐴𝐵 ⃗ ||𝑎̂𝐵 |cos𝜃𝐴𝐵 = |𝐴 ⃗ ∙ 𝑎̂𝐵 =𝐴

5


⃗ along ⃗𝐵 ⃗ is the scalar component multiplied by a unit vector  Vector component of 𝐴 ⃗ along ⃗𝐵 ⃗ ∙ 𝑎̂𝐵 ) 𝑎̂𝐵 𝐴𝐵 = 𝐴𝐵 𝑎̂𝐵 = (𝐴

Scalar triple product ⃗,𝐵 ⃗ and 𝐶 the scalar triple product is defined  Given three vectors as 𝐴 ⃗ × 𝐶) = 𝐵 ⃗ ∙ (𝐶 × 𝐴) = 𝐶 ∙ (𝐴 × 𝐵 ⃗) 𝐴 ∙ (𝐵 ⃗ 𝐶] and is represented as [𝐴 𝐵  Geometrically the scalar triple product is equal to the volume of parallelepiped ⃗,𝐵 ⃗ and 𝐶 as sides having 𝐴  Properties: ⃗ 𝐶]= [𝐵 ⃗ 𝐶 𝐴]=[𝐶 𝐴 𝐵 ⃗ ] (i) [ 𝐴 𝐵 ⃗ × 𝐶) = 𝐵 ⃗ ∙ (𝐶 × 𝐴) = 𝐶 ∙ (𝐴 × 𝐵 ⃗) i.e. 𝐴 ∙ (𝐵 (ii) A change in the cyclic order of vectors changes the sign of scalar triple product. ⃗ 𝐶 ] = −[ 𝐵 ⃗ 𝐴 𝐶] [𝐴 𝐵 ⃗ = 𝐴𝑥 𝑎 (iii) If 𝐴 ̂𝑥 + 𝐴𝑦 𝑎 ̂𝑦 + 𝐴𝑧 𝑎 ̂𝑧

and

⃗𝐵 ⃗ = 𝐵𝑥 𝑎 ̂𝑥 + 𝐵𝑦 𝑎 ̂𝑦 + 𝐵𝑧 𝑎 ̂𝑧 ⃗ = 𝐶𝑥 𝑎 𝐶 ̂𝑥 + 𝐶𝑦 𝑎 ̂𝑦 + 𝐶𝑧 𝑎 ̂𝑧 𝐴𝑥

𝐴𝑦

𝐴𝑧

⃗ 𝐶 ] = [ 𝐵𝑥 [𝐴 𝐵

𝐵𝑦

𝐵𝑧 ]

𝐶𝑥

𝐶𝑦

𝐶𝑧

Vector triple product ⃗, 𝐶 For any three vectors 𝐴, 𝐵 ⃗ × 𝐶) = 𝐵 ⃗ (𝐴 ∙ 𝐶 ) − 𝐶 (𝐴 ∙ 𝐵 ⃗) 𝐴 × (𝐵

Cylindrical Coordinate Systems  Any point in space is considered to be at the intersection of three mutually perpendicular surfaces: 6


 A circular cylinder (ρ=constant)  A vertical plane (Φ=constant)  A horizontal plane (z=constant)  Any point in space is represented by three coordinates P(ρ, Φ, z)  ρ denotes the radius of an imaginary cylinder passing through P, or the radial distance from z axis to the point P.  Φ denotes azimuthal angle, measured from x axis to a vertical intersecting plane passing through P.  z denotes distance from xy-plane to a horizontal intersecting plane passing through P. It is the same as in rectangular coordinate system.

 A vector in cylindrical coordinate system may be specified using three mutually perpendicular unit vectors 𝑎 ̂𝜌 , 𝑎 ̂Φ , 𝑎 ̂𝑧  𝑎 ̂𝜌 , 𝑎 ̂Φ , 𝑎 ̂𝑧 form a right handed system because a right handed screw when rotated from 𝑎 ̂𝜌 to 𝑎 ̂Φ moves towards 𝑎 ̂𝑧 7


 These unit vectors specify directions along ρ, Φ and z axes.  Using these unit vectors any vector A may be expressed as ⃗ = 𝐴𝜌 𝑎 𝐴 ̂𝜌 + 𝐴Φ 𝑎 ̂Φ + 𝐴𝑧 𝑎 ̂𝑧  The magnitude of the vector is given by ⃗ | = √𝐴𝜌2 + 𝐴2Φ + 𝐴𝑧2 |𝐴

𝑎̂𝜌 ∙ 𝑎̂𝜌 = 𝑎̂Φ ∙ 𝑎̂Φ = 𝑎̂𝑧 ∙ 𝑎̂𝑧 = 1 𝑎̂𝜌 ∙ 𝑎̂Φ = 𝑎̂Φ ∙ 𝑎̂z = 𝑎̂𝑧 ∙ 𝑎̂𝜌 = 0 𝑎̂𝜌 × 𝑎̂Φ = 𝑎̂𝑧

,

𝑎̂Φ × 𝑎̂𝑧 = 𝑎̂𝜌

,

𝑎̂𝑧 × 𝑎̂𝜌 = 𝑎̂Φ Spherical coordinate system

 Any point in space is represented as the intersection of three surfaces:  A sphere of radius r from the origin (r=constant)  A cone centred around the z axis (θ=constant)  A vertical plane (Φ=constant)  Any point in spherical coordinate system is considered to be at the intersection of the above three planes.

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 Three unit vectors of the spherical coordinate system are shown in the figure.  Unit vector 𝑎̂𝑟 lies along the radially outward direction to the spherical surface. It lies on the cone θ =constant and the plane Φ=constant  The unit vector 𝑎̂θ is normal to the conical surface and lies in Φ=constant plane and is tangential to the spherical surface.  Unit vector 𝑎̂Φ is the same as in cylindrical coordinate system. It is normal to Φ=constant plane and is tangential to both the cone and the sphere.  The unit vectors are mutually perpendicular and forms a right handed set. A right hand screw when rotated from 𝑎̂𝑟 to 𝑎̂θ will move it towards 𝑎̂Φ direction.  A vector 𝐴 in spherical coordinate system may be expressed as ⃗ = 𝐴𝑟 𝑎 𝐴 ̂𝑟 + 𝐴θ 𝑎 ̂θ + 𝐴Φ 𝑎 ̂Φ ̂𝑟 , 𝑎 𝑎 ̂θ , 𝑎 ̂Φ are unit vectors along r, θ, Φ directions  Magnitude of the vector is given by ⃗ | = √𝐴𝑟2 + 𝐴2θ + 𝐴2Φ |𝐴  The unit vectors 𝑎̂𝑟 , 𝑎̂θ , 𝑎̂Φ are mutually orthogonal. Thus

𝑎̂𝑟 . 𝑎̂𝑟 = 𝑎̂θ . 𝑎̂θ = 𝑎̂Φ . 𝑎̂Φ = 1 𝑎̂𝑟 . 𝑎̂θ = 𝑎̂θ . 𝑎̂Φ = 𝑎̂Φ . 𝑎̂𝑟 = 0 𝑎̂𝑟 × 𝑎̂θ = 𝑎̂Φ 𝑎̂θ × 𝑎̂Φ = 𝑎̂𝑟 𝑎̂Φ × 𝑎̂𝑟 = 𝑎̂θ

9


Differential length, area and volume in Cartesian coordinates

 Differential displacement is given by 𝑑𝑙 = 𝑑𝑥𝑎 ̂ 𝑥 + 𝑑𝑦𝑎 ̂ y + 𝑑𝑧𝑎 ̂z  Differential normal area is given by 𝑑𝑆 = 𝑑𝑦𝑑𝑧𝑎̂ 𝑥 = 𝑑𝑥𝑑𝑧𝑎̂ 𝑦 = 𝑑𝑥𝑑𝑦𝑎̂ 𝑧  Differential volume is given by 𝑑v = 𝑑𝑥𝑑𝑦𝑑𝑧  𝑑𝑙 and 𝑑𝑆 are vectors whereas 𝑑v is a scalar.  If we move from P to Q, 𝑑𝑙 = 𝑑𝑦𝑎̂y  If we move from Q to S, 𝑑𝑙 = 𝑑𝑦𝑎̂y + 𝑑𝑧𝑎̂z  If we move from D to Q, 𝑑𝑙 = 𝑑𝑥𝑎̂𝑥 + 𝑑𝑦𝑎̂y + 𝑑𝑧𝑎̂z  In general, the differential surface area is defined as 𝑑𝑆 = 𝑑𝑠𝑎̂𝑛 where dS is the area of the surface element and 𝑎̂𝑛 is a unit vector normal to the surface dS.

 The different surfaces in figure (1) is described as

ABCD ⇒ 𝑑𝑆 = 𝑑𝑦𝑑𝑧𝑎̂𝑥

ADSP ⇒ 𝑑𝑆 = −𝑑𝑥𝑑𝑧𝑎̂𝑦

PQRS ⇒ 𝑑𝑆 = −𝑑𝑦𝑑𝑧𝑎̂𝑥

ABQP ⇒ 𝑑𝑆 = 𝑑𝑥𝑑𝑦𝑎̂𝑧

BCRQ ⇒ 𝑑𝑆 = 𝑑𝑥𝑑𝑧𝑎̂𝑦

DCRS ⇒ 𝑑𝑆 = −𝑑𝑥𝑑𝑦𝑎̂𝑧

10


Differential normal areas in cylindrical coordinate systems  Differential displacement is given by 𝑑𝑙 = 𝑑𝜌𝑎 ̂𝜌 + 𝜌𝑑Φ𝑎 ̂ Φ + 𝑑𝑧𝑎 ̂z  Differential normal area is given by 𝑑𝑆 = 𝜌𝑑Φ𝑑𝑧𝑎 ̂𝜌 = 𝑑𝜌𝑑𝑧𝑎̂Φ = 𝜌𝑑Φ𝑑𝜌𝑎̂𝑧  Differential volume is given by 𝑑v = 𝜌𝑑𝜌𝑑Φ𝑑𝑧

Differential normal areas in spherical coordinate systems  Differential displacement is given by 𝑑𝑙 = 𝑑𝑟𝑎 ̂ 𝑟 + 𝑟𝑑θ𝑎 ̂ θ + 𝑟sinθ𝑑Φ𝑎 ̂Φ  Differential normal area is given by 𝑑𝑆 = 𝑟 2 sinθ𝑑θ𝑑Φ𝑎̂𝑟 = 𝑟sinθ𝑑𝑟𝑑Φ𝑎̂θ = 𝑟𝑑𝑟𝑑θ𝑎̂Φ  Differential volume is given by 𝑑v = 𝑟 2 sinθ𝑑𝑟𝑑θ𝑑Φ 11


For easy reference, the differential length, surface, and volume elements for the three coordinate systems are summarized in the following table:

Differential elements

Coordinate system Rectangular (Cartesian)

Length 𝑑𝑙

𝑑𝑥𝑎 ̂𝑥 +𝑑𝑦𝑎 ̂y +𝑑𝑧𝑎 ̂z

Surface 𝑑𝑆

𝑑𝑦𝑑𝑧𝑎̂ 𝑥 +𝑑𝑥𝑑𝑧𝑎̂ 𝑦 +𝑑𝑥𝑑𝑦𝑎̂ 𝑧

Volume 𝑑v

𝑑𝑥𝑑𝑦𝑑𝑧

Cylindrical 𝑑𝜌𝑎 ̂𝜌 +𝜌𝑑Φ𝑎 ̂Φ +𝑑𝑧𝑎 ̂z

Spherical 𝑑𝑟𝑎 ̂𝑟 +𝑟𝑑θ𝑎 ̂θ +𝑟sinθ𝑑Φ𝑎 ̂Φ

𝜌𝑑Φ𝑑𝑧𝑎 ̂𝜌 +𝑑𝜌𝑑𝑧𝑎̂Φ +𝜌𝑑Φ𝑑𝜌𝑎̂𝑧

𝑟 2 sinθ𝑑θ𝑑Φ𝑎̂𝑟 + 𝑟sinθ𝑑𝑟𝑑Φ𝑎̂θ + 𝑟𝑑𝑟𝑑θ𝑎̂Φ

𝜌𝑑𝜌𝑑Φ𝑑𝑧

𝑟 2 sinθ𝑑𝑟𝑑θ𝑑Φ

12


Line integrals  Line integral is defined as any integral that is to be evaluated along a line. A line indicates a path along a curve in space. 𝑏

 ∫ 𝐴⃗. 𝑑𝑙⃗ represents a line integral where each element of length 𝑑𝑙⃗ on the curve is a

multiplied according to scalar dot product rule by the local value of 𝐴⃗ and then these products are summed to get the value of the integral.  Let 𝐴⃗ be a vector field in space and ab a curve from point a to point b. Let the curve ab is subdivided in to infinitesimally small vector elements 𝑑𝑙⃗1 , 𝑑𝑙⃗2 , 𝑑𝑙⃗3 , . . . . . . , 𝑑𝑙⃗𝑟  Let the dot products 𝐴⃗1 . 𝑑𝑙⃗1 , 𝐴⃗2 . 𝑑𝑙⃗2 , 𝐴⃗3 . 𝑑𝑙⃗3 , . . . . . . , 𝐴⃗𝑟 . 𝑑𝑙⃗𝑟 are taken where 𝐴⃗1 , 𝐴⃗2 , 𝐴⃗3 , . . . . . . , 𝐴⃗𝑟 are the value of the vector field at the junction points of the vector element 𝑑𝑙⃗1 , 𝑑𝑙⃗2 , 𝑑𝑙⃗3 , . . . . . . , 𝑑𝑙⃗𝑟  Then the sum of these products ∑𝑏𝑎 𝐴⃗𝑟 . 𝑑𝑙⃗𝑟 along the entire length of the curve is known as the line integral of 𝐴⃗ along the curve ab.

 As an example if 𝐹⃗ represents the force acting on a moving particle along a curve ab, then the line integral of 𝐹⃗ over the path described by the particle represents the work done by the force in moving the particle from a to b.  The line integral around a closed curve is called closed line integral

13

Surface integrals  Consider a vector field 𝐴⃗ continuous in a region of space containing a smooth surface S.  The surface integral of 𝐴⃗ through S can be defined as 𝜓 = ∫ 𝐴⃗ . 𝑑𝑆⃗. 𝑆  Consider a small incremental surface area on the surface S denoted by dS. Let 𝑎̂𝑛 be a unit normal to the surface dS.  The magnitude of flux crossing the unit surface normally is given by |𝐴⃗| cos𝜃 𝑑𝑆 = 𝐴⃗ . 𝑎̂𝑛 𝑑𝑆 = 𝐴⃗ . 𝑑𝑆𝑎̂𝑛 = 𝐴⃗ . 𝑑𝑆⃗  Where 𝑑𝑆⃗ denote the vector area having magnitude equal to dS and whose direction is in the direction of the unit normal. 𝑑𝑆⃗ = 𝑑𝑆𝑎̂𝑛  The total flux crossing the surface is obtained by integrating 𝐴⃗ . 𝑑𝑆⃗ over the surface of interest. 𝜓 = ∫𝑆𝐴⃗ . 𝑑𝑆⃗  For a closed surface defining a volume the surface integral becomes closed surface integral and is denoted by 𝜓 = ∮ 𝐴⃗ . 𝑑𝑆⃗ 𝑆

 It represents the net outward flow of flux from surface S

Volume integrals  Let V be a volume bounded by the surface S. Let 𝜑(𝑥, 𝑦, 𝑧) be a function of position defined over V. If the volume V is subdivided in to n elements of volumes dv1, dv2 , dv3,......, dvn.  In each part let us choose an arbitrary point 𝜑(𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 ).  Then the limit of the sum ∑ 𝜑(𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 )𝑑v𝑖 as 𝑛 → ∞ and 𝑑v𝑖 → 0 is called the volume integral of 𝜑(𝑥, 𝑦, 𝑧) over V and is denoted by ∫v 𝜑 𝑑v.

Scalar and vector fields  Let ℛ be a region of space at each point of which a scalar Φ = Φ(x,y,z) is given, then Φ is called a scalar point function and ℛ is called a scalar field. 14

Examples:  Temperature distribution in a medium  Distribution of atmospheric pressure in space.  Let ℛ be a region of space at each point of which a vector v ⃗⃗ = v ⃗⃗(𝑥, 𝑦, 𝑧) is given, then ⃗⃗ is called a vector point function and ℛ is called a scalar field. v Examples:  The velocity of a moving fluid at any instant  Gravitational force in a region.

Divergence of a vector ⃗⃗ is the outflow of flux from a small closed  The divergence of the vector flux density A surface per unit volume as the volume shrinks to zero. ⃗⃗ . 𝑑𝑆⃗ ∮A 𝑆 ⃗ ⃗ ⃗ ⃗ Divergence of A = div A = lim ∆v→0 ∆v ⃗⃗ . 𝑑𝑆⃗ is the net outflow of flux of a vector field A ⃗⃗ from a closed surface S ∮ A 𝑆

Divergence of a vector in Cartesian coordinates  To evaluate the divergence of a vector field ⃗A⃗ at point P(x0, y0, z0) first construct a ⃗⃗ is obtained as differential volume around point P, The closed surface integral of A ⃗⃗ . 𝑑𝑆⃗ = (∫ ∮A 𝑆

+∫

front

back

+∫ +∫ left

+∫ +∫

right

top

) ⃗A⃗ . 𝑑𝑆⃗

bottom

The divergence of 𝐴⃗ at point P(x0, y0, z0) in a Cartesian coordinates is given by div ⃗A⃗ =

𝜕A𝑥 𝜕Ay 𝜕𝐴𝑧 + + 𝜕𝑥 𝜕𝑦 𝜕𝑧 15

In cylindrical coordinates: ⃗⃗ = div A

1 𝜕 1 𝜕AΦ 𝜕𝐴𝑧 (𝜌A𝜌 ) + + 𝜌 𝜕𝜌 𝜌 𝜕Φ 𝜕𝑧

In spherical coordinates: div ⃗A⃗ =

1 𝜕 2 1 𝜕 1 𝜕𝐴Φ (𝑟 ) (𝐴 𝐴 + sin θ) + 𝑟 𝜃 𝑟 2 𝜕𝑟 𝑟sin θ 𝜕θ 𝑟sin θ 𝜕Φ

Examples:  Divergence of the velocity of water in a container after the outlet has opened

is zero because water is an incompressible fluid. Volume of water entering and leaving different regions of the closed surface is equal.  When the valve on a steam boiler is opened, there is a net outward flow of steam for each elemental volume. In this case, the divergence has a positive value. It indicates a source of vector quantity at that point.  When an evacuated glass bulb is broken, there is a sudden inrush of air and there is a net inward flow of air for each elemental volume. In this case, the divergence has a negative value. It indicates a sink of that vector quantity.

⃗⃗ (del operator) The vector operator ∇ ⃗⃗.  The del operator is the vector differential operator and is denoted by ∇  The vector differential operator is not a vector in itself, but when it operates on a scalar function, the result is a vector.  This operation is useful in defining: ⃗⃗𝑉  The gradient of a scalar ∇ 16

⃗⃗. 𝐴⃗  The divergence of a vector ∇ ⃗⃗ × 𝐴⃗  The curl of a vector ∇  The Laplacian of a scalar ∇2 𝑉 In Cartesian coordinates the del operator is defining as, ⃗⃗= ∇

𝜕 𝜕 𝜕 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧

Warning! The del operator is defined only in Cartesian coordinates, i.e., the ∇ operator ⃗⃗. 𝐴⃗ is does not have a specific form in other coordinate systems. When ∇ written for the divergence of 𝐴⃗ in other coordinate systems, it does not mean that a del operator can be defined for these systems. For example, the divergence in cylindrical coordinates will be written as div ⃗A⃗ =

1 𝜕 1 𝜕AΦ 𝜕𝐴𝑧 + (𝜌A𝜌 ) + 𝜌 𝜕𝜌 𝜌 𝜕Φ 𝜕𝑧

This does not imply that ⃗⃗= ∇

1 𝜕 1 𝜕( ) 𝜕( ) (𝜌 )𝑎̂𝜌 + 𝑎̂Φ + 𝑎̂ 𝜌 𝜕𝜌 𝜌 𝜕Φ 𝜕𝑧 𝑧

Divergence theorem  The integral of the normal component of any vector field over a closed surface is equal to the integral of the divergence of this vector field throughout the volume enclosed by the closed surface.  The total outward flux of a vector field ⃗A⃗ through the closed surface S is the same as the volume integral of the divergence of ⃗A⃗

⃗⃗ . 𝑑𝑆⃗⃗ = ∫∇ ⃗⃗. ⃗A⃗ 𝑑v ∮A 𝑆

v

 Volume integrals are easier to evaluate than surface integrals. Using Divergence theorem, we can convert surface integral to a volume integral and then easily evaluate it.

Divergence Theorem-Explanation  Let the volume v bounded by the surface S is subdivided in to a number of elementary volumes Δv.  The flux diverging from each such cell enters or converges on the adjacent cells unless the cell contains a portion of the outer surface. 17

 As a result, the divergence of the flux density throughout the volume leads to the same result as determining the net flux crossing the enclosing surface.

Example 2: Determine the divergence of the following vector fields ⃗⃗⃗ = 𝑥 2 𝑦𝑧𝑎 (a) 𝑃 ̂𝑥 + 𝑥𝑧𝑎 ̂𝑧 ⃗⃗⃗ = 𝜌sin Φ 𝑎 (b) 𝑄 ̂𝜌 + 𝜌2 𝑧𝑎 ̂Φ + 𝑧 cos Φ 𝑎 ̂𝑧 ⃗⃗ = (c) ⃗𝑇

1

cos θ 𝑎 ̂𝑟 + 𝑟sinθ cos Φ 𝑎 ̂𝜃 + cos θ 𝑎 ̂Φ

𝑟2

Solution: ⃗⃗. P ⃗⃗ = (𝜕P𝑥 + 𝜕Py + 𝜕𝑃𝑧) (a) ∇ 𝜕𝑥

⃗⃗. ⃗P⃗ = ∇

𝜕 𝜕𝑥

(𝑥 2 𝑦𝑧) +

⃗⃗. ⃗Q⃗ = (1 (b) ∇

𝜕

𝜌 𝜕𝜌

⃗⃗. ⃗Q⃗ = ∇

1 𝜕 𝜌 𝜕𝜌

𝜕𝑦

𝜕𝑧 𝜕

(0) +

𝜕𝑦

(𝜌Q 𝜌 ) +

(𝜌2 sin Φ) +

1 𝜕QΦ 𝜌 𝜕Φ 1 𝜕 𝜌 𝜕Φ

𝜕 𝜕𝑧

+

(𝑥𝑧) = 2𝑥𝑦𝑧 + 𝑥

𝜕𝑄𝑧 𝜕𝑧

)

(𝜌2 𝑧) +

𝜕 𝜕𝑧

(𝑧 cos Φ)

⃗⃗. ⃗Q⃗ = 2sin Φ + cos Φ ∇ ⃗⃗. ⃗T⃗ = ( 12 𝜕 (𝑟 2 𝑇𝑟 ) + 1 𝜕 (𝑇𝜃 sin θ) + 1 𝜕𝑇Φ ) (c) ∇ 𝑟 𝜕𝑟 𝑟sin θ 𝜕θ 𝑟sin θ 𝜕Φ ⃗⃗. ⃗T⃗ = ∇

1 𝜕 𝑟2

𝜕𝑟

(cos θ) +

1

𝜕

𝑟sin θ 𝜕θ

(𝑟sin2 θ cos Φ) +

1

𝜕

𝑟sin θ 𝜕Φ

(cos θ)

= 2 cos θcos Φ

Example 3: ⃗⃗ = 𝜌2 cos 2 𝜙 â𝜌 + 𝑧sin𝜙 âΦ over the surface of the cylinder 0 ≤ Determine the flux of ⃗D 𝑧 ≤ 1, 𝜌 = 4 . Verify divergence theorem. 18

Solution If 𝜓 is the flux through the given surface 𝜓 = 𝜓 𝑇 + 𝜓𝐵 + 𝜓𝑆 For 𝜓 𝑇 , 𝑧 = 1, 𝑑𝑆⃗ = 𝜌𝑑𝜌𝑑𝜙𝑎̂𝑧 4

⃗⃗ . 𝑑𝑆⃗ = ∫ 𝜓𝑇 = ∫ D

2𝜋

(𝜌2 cos 2 𝜙 â𝜌 + 𝑧sin𝜙 âΦ ) ∙ 𝜌𝑑𝜌𝑑𝜙𝑎̂𝑧 = 0

∫

𝜌=0 𝜙=0

For 𝜓𝐵 , 𝑧 = 0, 𝑑𝑆⃗ = 𝜌𝑑𝜌𝑑𝜙(−𝑎̂𝑧 ) 4

𝜓𝐵 = ∫ ⃗D⃗ . 𝑑𝑆⃗ = ∫

2𝜋

(𝜌2 cos 2 𝜙 â𝜌 + 𝑧sin𝜙 âΦ ) ∙ 𝜌𝑑𝜌𝑑𝜙(−𝑎̂𝑧 ) = 0

∫

𝜌=0 𝜙=0

For 𝜓𝑆 , 𝜌 = 4, 𝑑𝑆⃗ = 𝜌𝑑𝜙𝑑𝑧𝑎̂𝜌 1

𝜓𝑆 = ∫ ⃗D⃗ . 𝑑𝑆⃗ = ∫

2𝜋

∫

(𝜌2 cos 2 𝜙 â𝜌 + 𝑧sin𝜙 âΦ ) ∙ 𝜌𝑑𝜙𝑑𝑧𝑎̂𝜌

𝑧=0 𝜙=0 1 2𝜋

⃗⃗ . 𝑑𝑆⃗ = ∫ 𝜓𝑆 = ∫ D

∫

2𝜋 2

64cos 𝜙𝑑𝜙𝑑𝑧 = 64 ∫

𝑧=0 𝜙=0

𝜙=0

2𝜋

=

cos 2 𝜙𝑑𝜙

2𝜋

64 ∫ (1 + cos2𝜙)𝑑𝜙 = 32 ∫ 1𝑑𝜙 = 64𝜋 2 𝜙=0 𝜙=0

⃗⃗⃗ . 𝑑𝑆⃗ = ∫∇ ⃗⃗⃗⃗. ⃗D ⃗⃗ 𝑑v Applying divergence theorem ∮D 𝑆

v

1 𝜕 1 𝜕 𝜕 1 𝜕 3 2 1 𝜕 (𝜌 cos 𝜙) + 𝐷Φ + 𝐷𝑧 = 𝑧sin𝜙 (𝜌𝐷𝜌 ) + 𝜌 𝜕𝜌 𝜌 𝜕𝜙 𝜕𝑧 𝜌 𝜕𝜌 𝜌 𝜕𝜙 1 ⃗⃗ ∙ ⃗D ⃗⃗ = 3𝜌cos 2 𝜙 + 𝑧cos𝜙 ∇ 𝜌 4 2𝜋 1 1 ⃗⃗⃗⃗ ⃗ ⃗⃗ ∫∇ . D 𝑑v = ∫ ∫ ∫ (3𝜌cos 2 𝜙 + 𝑧cos𝜙) 𝜌𝑑𝜌𝑑𝜙𝑑𝑧 𝜌 v 𝜌=0 𝜙=0 𝑧=0 ⃗⃗ ∙ ⃗D ⃗⃗ = ∇

19

2𝜋

1

2𝜋

1

4

3𝜌3 =∫ ∫ ( cos 2 𝜙 + 𝑧𝜌cos𝜙) 𝑑𝜙𝑑𝑧 3 𝜙=0 𝑧=0 0 ∫ (64cos 2 𝜙 + 4𝑧cos𝜙)𝑑𝜙𝑑𝑧

=∫

𝜙=0 𝑧=0 1

2𝜋

4𝑧 2 2 = ∫ (64𝑧cos 𝜙 + cos𝜙) 𝑑𝜙 2 𝜙=0 0 2𝜋

2𝜋

(64cos2

=∫

2𝜋 2

𝜙 + 2cos𝜙)𝑑𝜙 = ∫

𝜙=0

64cos 𝜙𝑑𝜙 + ∫

𝜙=0 2𝜋

2cos𝜙𝑑𝜙

𝜙=0

2𝜋 64 2𝜋 = ∫ (1 + cos2𝜙)𝑑𝜙 = 32 ∫ 𝑑𝜙 + ∫ cos2𝜙𝑑𝜙 2 𝜙=0 𝜙=0 𝜙=0 2𝜋

= 32 ∫

𝑑𝜙 = 64𝜋

𝜙=0

Example 4: ⃗⃗ = 𝑥 2 𝑎̂𝑥 + 𝑥𝑦𝑎̂𝑦 + 𝑦𝑧𝑎̂𝑧 for the volume of Verify divergence theorem for the flux of ⃗D cube with 1unit for each side. The cube is situated in the first octant of the coordinate system with one corner on the origin

⃗⃗⃗ . 𝑑𝑆⃗⃗ = ∫∇ ⃗⃗. D ⃗⃗⃗ 𝑑v ∮D 𝑆

v

⃗⃗⃗ . 𝑑𝑆⃗ = (∫ ∮D 𝑆

∫ front

front 1

+∫ back

+∫ +∫ left

right

+∫ +∫ top

)

bottom

1

⃗⃗⃗ . 𝑑𝑆⃗ = ∫ ∫(𝑥 2 𝑎̂𝑥 + 𝑥𝑦𝑎̂𝑦 + 𝑦𝑧𝑎̂𝑧 ) ∙ 𝑑𝑦𝑑𝑧𝑎̂𝑥 D y=0 𝑧=0

20

and 𝑥 = 1

1 1

= ∫ ∫ 𝑑y𝑑𝑧 = 1 0 0 1

1

⃗⃗ . 𝑑𝑆⃗⃗ = ∫ ∫(𝑥2 𝑎̂ 𝑥 + 𝑥𝑦𝑎̂ 𝑦 + 𝑦𝑧𝑎̂ 𝑧 ) ∙ 𝑑𝑦𝑑𝑧(−𝑎̂ 𝑥 ) ∫ ⃗D back

and 𝑥 = 0

y=0 𝑧=0

=0 1

1

⃗⃗ . 𝑑𝑆⃗⃗ = ∫ ∫(𝑥2 𝑎̂ 𝑥 + 𝑥𝑦𝑎̂ 𝑦 + 𝑦𝑧𝑎̂ 𝑧 ) ∙ 𝑑𝑥𝑑𝑧(−𝑎̂ 𝑦 ) ∫ ⃗D left

and 𝑦 = 0

z=0 𝑥=0

=0 1

∫

1

⃗D ⃗⃗ . 𝑑𝑆⃗⃗ = ∫ ∫(𝑥2 𝑎̂ 𝑥 + 𝑥𝑦𝑎̂ 𝑦 + 𝑦𝑧𝑎̂ 𝑧 ) ∙ 𝑑𝑥𝑑𝑧𝑎̂ 𝑦

right

z=0 𝑥=0 1 1

= ∫ ∫ 𝑥𝑑𝑥𝑑𝑧 = 1

0 0

1

1 2

⃗⃗ . 𝑑𝑆⃗⃗ = ∫ ∫(𝑥2 𝑎̂ 𝑥 + 𝑥𝑦𝑎̂ 𝑦 + 𝑦𝑧𝑎̂ 𝑧 ) ∙ 𝑑𝑥𝑑𝑦𝑎̂ 𝑧 ∫ ⃗D top

and 𝑧 = 1

y=0 𝑥=0 1 1

= ∫ ∫ 𝑦𝑑𝑥𝑑𝑦 = 0 0 1

∫

and 𝑦 = 1

1 2

1

⃗D ⃗⃗ . 𝑑𝑆⃗⃗ = ∫ ∫(𝑥2 𝑎̂ 𝑥 + 𝑥𝑦𝑎̂ 𝑦 + 𝑦𝑧𝑎̂ 𝑧 ) ∙ 𝑑𝑥𝑑𝑦(−𝑎̂ 𝑧 )

bottom

and 𝑧 = 0

y=0 𝑥=0

=0 1 1 + =2 2 2 𝑆 𝜕D𝑥 𝜕Dy 𝜕𝐷𝑧 𝜕 2 𝜕 𝜕 ⃗⃗. D ⃗⃗⃗ = (𝑥 ) + (𝑥𝑦) + (𝑦𝑧) = 3𝑥 + 𝑦 ∇ + + = 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 ⃗⃗⃗ . 𝑑𝑆⃗ = 1 + ∮D

1

1

1

⃗⃗. ⃗D ⃗⃗ 𝑑v = ∫ ∫ ∫ (3𝑥 + 𝑦)𝑑𝑥𝑑𝑦𝑑𝑧 = 2 ∫∇ 0

v

0

⃗⃗⃗ . 𝑑𝑆⃗⃗ = ∫∇ ⃗⃗. ⃗D ⃗⃗ 𝑑v ∮D 𝑆

0

Thus divergence theorem is verified

v

21

Gradient of a scalar field  The gradient of a scalar field V is a vector that represents the magnitude and direction of the maximum space rate of increase of V.  Let V be a scalar field and let V1, V2 and V3 be contours on which V is constant.  Consider the difference in the field dV between points P1 and P2

𝜕𝑉 𝜕𝑉 𝜕𝑉 𝑑𝑥 + 𝑑𝑦 + 𝑑𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑉 𝜕𝑉 𝜕𝑉 = ( 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂ ) . (𝑑𝑥𝑎̂𝑥 + 𝑑𝑦𝑎̂𝑦 + 𝑑𝑧𝑎̂𝑧 ) 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑧

𝑑V =

For convenience Let

(

𝜕𝑉 𝜕𝑉 𝜕𝑉 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂ ) = 𝐺⃗ 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑧

Then 𝑑V = 𝐺⃗ . ⃗⃗⃗⃗ 𝑑𝑙 = 𝐺cos𝜃 𝑑𝑙 , Then 𝑑V is maximum when 𝜃 = 0𝑜 𝑑𝑙 𝑑V =𝐺 | 𝑑𝑙 𝑀𝐴𝑋

𝑑V = 𝐺cos𝜃 𝑑𝑙

𝑖. 𝑒. , when 𝑑𝑙⃗ is in the direction of 𝐺⃗

Magnitude of 𝐺⃗ is equal to the maximum space rate of change of V Direction of 𝐺⃗ is along the maximum space rate of change of V 𝐺⃗ is defined as the gradient of V, and is denoted by grad V, which may be written in rectangular coordinates as 𝜕𝑉 𝜕𝑉 𝜕𝑉 ⃗⃗V = grad V = ∇ 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂ 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑧 22

For cylindrical coordinates, 𝜕𝑉 1 𝜕𝑉 𝜕𝑉 ⃗⃗V = grad V = ∇ 𝑎̂𝜌 + 𝑎̂Φ + 𝑎̂ 𝜕𝜌 𝜌 𝜕Φ 𝜕𝑧 𝑧 For spherical coordinates, 𝜕𝑉 1 𝜕𝑉 1 𝜕𝑉 ⃗⃗V = grad V = ∇ 𝑎̂𝑟 + 𝑎̂𝜃 + 𝑎̂ 𝜕𝑟 𝑟 𝜕θ 𝑟sin θ 𝜕Φ Φ Gradient of scalar Field-Important Relations ∎

∇(V + U) = ∇V + ∇U

∎

∇(𝑉𝑈 ) = 𝑉∇𝑈 + 𝑈∇𝑉

∎

𝑉 𝑈∇𝑉 − 𝑉∇𝑈 ∇[ ] = 𝑈 𝑈2

∎

𝑛

𝑛−1

∇(𝑉 ) = 𝑛𝑉

∇𝑉

Gradient of scalar Field-Important Points ⃗⃗V is equal to the maximum space rate of change of V  Magnitude of ∇ ⃗⃗V is along the maximum space rate of change of V  Direction of ∇ ⃗⃗V at any point is perpendicular to the constant V surface that passes through  ∇ that point. ⃗⃗ = ∇ ⃗⃗ ⃗⃗V ,V is called the scalar potential of A  If A ⃗⃗V in the direction of a given unit vector â is ∇ ⃗⃗V. â  The projection of ∇  It is called the directional derivative of V along â  It indicates the rate of change of V in the direction of â

Example 5: Find grad 𝛷 when 𝛷 = 3𝑥 2 𝑦 − 𝑦 3 𝑧 2 . Find the directional derivative in the direction of 3𝑎̂𝑥 + 4𝑎̂𝑦 + 12𝑎̂𝑧 at (2, -1, 0) Solution: 𝜕𝛷 𝜕𝛷 𝜕𝛷 ⃗⃗𝛷 = grad 𝛷 = ∇ 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂ 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑧 ⃗⃗𝛷 = grad 𝛷 = ∇

𝜕(3𝑥 2 𝑦 − 𝑦 3 𝑧 2 ) 𝜕(3𝑥 2 𝑦 − 𝑦 3 𝑧 2 ) 𝜕(3𝑥 2 𝑦 − 𝑦 3 𝑧 2 ) 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧

⃗⃗𝛷 = (6𝑥𝑦)𝑎̂𝑥 + (3𝑥 2 − 3𝑦 2 𝑧 2 )𝑎̂𝑦 + (−2𝑦 3 𝑧)𝑎̂𝑧 grad 𝛷 = ∇ ⃗⃗𝛷 = (−12)𝑎̂𝑥 + (12)𝑎̂𝑦 At (2, -1, 0) ∇ 23

⃗⃗𝛷. 𝑎̂ = (−12𝑎̂𝑥 + 12𝑎̂𝑦 ) ∙ Directional derivative ∇

(3𝑎̂𝑥 + 4𝑎̂𝑦 + 12𝑎̂𝑧 ) √9 + 16 + 144

= (−12𝑎̂𝑥 + 12𝑎̂𝑦 ) ∙ (0.23𝑎̂𝑥 + 0.31𝑎̂𝑦 + 0.92𝑎̂𝑧 ) = 0.96

Example 6: Find grad V when V = 10𝑟sin2 θcos Φ Solution: 𝜕𝑉 1 𝜕𝑉 1 𝜕𝑉 ⃗⃗V = grad V = ∇ 𝑎̂𝑟 + 𝑎̂𝜃 + 𝑎̂ 𝜕𝑟 𝑟 𝜕θ 𝑟sin θ 𝜕Φ Φ ⃗⃗V = 10sin2 θcos Φ 𝑎̂𝑟 + 10sin2 θcos Φ 𝑎̂𝜃 − 10sin θcos Φ 𝑎̂Φ ∇

Curl of a vector ⃗⃗ is an axial or rotational vector whose magnitude is the maximum  The curl of a vector A circulation (closed line integral) of ⃗A⃗ per unit area as the area tends to zero and whose direction is the normal direction of the area when the area is oriented so as to make the circulation maximum. ⃗⃗ around a closed path 𝑙 is the integral ∮A ⃗⃗ . 𝑑𝑙⃗  The circulation of a vector field A 𝑙

⃗⃗ × ⃗A⃗ = lim (Curl ⃗A⃗)𝑁 = ∇

∆SN →0

⃗⃗ . 𝑑𝑙⃗ ∮A 𝑙

∆SN

where ∆SN is the planar area enclosed by the closed line integral, The N subscript indicates that the component of the curl is that component which is normal to the surface enclosed by the closed path. It may represent any component in any coordinate system.  Consider the differential area in the yz plane.

24

 Closed line integral of ⃗A vector around abcd is obtained as below: ⃗ . 𝑑𝑙 = (∫ + ∫ + ∫ + ∫ ∮A 𝑙

𝑎𝑏

𝑏𝑐

𝑐𝑑

⃗ . 𝑑𝑙 )A

𝑑𝑎

The resultant curl will be the sum of component curls about x, y, z axes 𝜕𝐴 𝜕𝐴𝑦 𝜕𝐴𝑥 𝜕𝐴 𝜕𝐴𝑥 𝜕𝐴𝑧 ⃗ = ( 𝑍 − 𝑦) 𝑎 ⃗ ×𝐴 )𝑎 ∇ ̂𝑥 + ( − ̂𝑦 + ( − ̂𝑧 )𝑎 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝑎 ̂𝑥

𝑎 ̂𝑦

𝑎 ̂𝑧

|𝜕 ⃗ = ⃗ ×𝐴 ∇ 𝜕𝑥 |

𝜕 𝜕𝑦

𝜕| 𝜕𝑧 |

𝐴𝑥

𝐴𝑦

𝐴𝑧

The expressions for curl in cylindrical and spherical coordinates are

⃗ ×𝐴 = ∇

⃗ ×𝐴 = ∇

𝑎̂𝜌

𝜌𝑎̂Φ

𝑎̂𝑧

1 |𝜕 𝜌 |𝜕𝜌

𝜕 𝜕Φ

𝜕| 𝜕𝑧|

𝐴𝜌

𝜌𝐴Φ

𝐴𝑧

𝑎̂𝑟

𝑟𝑎̂θ

𝑟sin𝜃𝑎̂Φ

1 |𝜕 𝑟 2 sin𝜃 |𝜕𝑟

𝜕 𝜕θ

𝜕 𝜕Φ

𝐴𝑟

𝑟𝐴θ

𝑟sin𝜃𝐴Φ

| |

Properties of curl:  The curl of a vector field is another vector field ⃗ × V does not exist  The curl of a scalar field V, ∇ ⃗ − (A ⃗ ·∇ ⃗ × (𝐴 × 𝐵 ⃗ ) = 𝐴(∇ ⃗ ∙𝐵 ⃗)−𝐵 ⃗ (∇ ⃗ ∙ 𝐴) + (B ⃗ ·∇ ⃗ )A ⃗ )B ⃗  ∇ ⃗ × (V𝐴) = V∇ ⃗ ×𝐴+∇ ⃗ V×𝐴  ∇ ⃗ · (𝐴 × 𝐵 ⃗)=𝐵 ⃗ ·∇ ⃗ ×𝐴−𝐴·∇ ⃗ ×𝐵 ⃗  ∇ ⃗ . (∇ ⃗ V) = ∇2 V  div grad V= ∇ ⃗ ∙ (∇ ⃗ × 𝐴) = 0  div curl 𝐴 = ∇ ⃗ ×∇ ⃗V=0  curl grad V, ∇ 25


Example 7: Determine the curl of the following vector fields: ⃗⃗ = 𝑥 2 𝑦𝑧𝑎 (a) 𝑃 ̂𝑥 + 𝑥𝑧𝑎 ̂𝑧 ⃗⃗ = 𝜌sin Φ 𝑎 (b) 𝑄 ̂𝜌 + 𝜌2 𝑧𝑎 ̂Φ + 𝑧 cos Φ 𝑎 ̂𝑧 ⃗ = 12 cos 𝜃 𝑎 (c) ⃗𝑇 ̂𝑟 + 𝑟sin𝜃 cos Φ 𝑎 ̂𝜃 + cos 𝜃 𝑎 ̂Φ 𝑟

Solution: 𝑎 ̂𝑥

𝑎 ̂𝑦

𝑎 ̂𝑧

𝜕 ⃗ ×𝑃 ⃗⃗ = | (a) ∇ | 𝜕𝑥

𝜕 𝜕𝑦

𝜕| 𝜕𝑃𝑦 𝜕𝑃𝑥 𝜕𝑃𝑍 𝜕𝑃𝑦 𝜕𝑃𝑥 𝜕𝑃𝑧 =( − ̂𝑥 + ( − ̂𝑦 + ( − ̂𝑧 )𝑎 )𝑎 )𝑎 𝜕𝑧 | 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑥 𝜕𝑦

𝑃𝑥

𝑃𝑦

𝑃𝑧 = (0 − 0)𝑎 ̂𝑥 + (𝑥 2 𝑦 − 𝑧)𝑎 ̂𝑦 + (0 − 𝑥 2 𝑧)𝑎 ̂𝑧 = (𝑥 2 𝑦 − 𝑧)𝑎 ̂𝑦 − 𝑥 2 𝑧𝑎 ̂𝑧

𝑎̂𝜌 1 𝜕 ⃗ = || (b) ⃗∇ × 𝑄 𝜌 𝜕𝜌 𝑄𝜌

𝜌𝑎̂Φ 𝜕 𝜕Φ 𝜌𝑄Φ

𝑎̂𝑧 𝜕 | 𝜕𝑧| 𝑄𝑧

⃗ = [1 𝜕𝑄𝑍 − 𝜕𝑄Φ] 𝑎̂𝜌 + [𝜕𝑄𝜌 − 𝜕𝑄𝑧] 𝑎̂Φ + 1 [ 𝜕 (𝜌Q Φ ) − 𝜕𝑄𝜌] 𝑎 ⃗ ×𝑄 ∇ ̂𝑧 𝜌 𝜕Φ

𝜕𝑧

𝜕𝑧

𝜕𝜌

26

𝜌 𝜕𝜌

𝜕Φ


−𝑧

=(

𝜌

−1

=

⃗ = (c) ⃗∇ × 𝑇

⃗ ×𝑇 ⃗ = ∇

𝜌

1

sin Φ − 𝜌2 ) 𝑎̂𝜌 + (0 − 0)𝑎̂Φ + (3𝜌2 𝑧 − 𝜌cos Φ)𝑎 ̂𝑧 𝜌

(𝑧sin Φ + 𝜌3 )𝑎̂𝜌 + (3𝜌𝑧 − cos Φ)𝑎 ̂𝑧

1 𝑟 2 sin𝜃

1

[

𝑎̂𝑟

𝑟𝑎̂𝜃

𝑟sin𝜃𝑎̂Φ

𝜕

| 𝜕𝑟

𝜕

𝜕

𝜕𝜃

𝜕Φ

𝑇𝑟

𝑟𝑇θ

𝑟sin𝜃𝑇Φ

𝜕

𝑟sin𝜃 𝜕𝜃

(𝑇Φ sin𝜃) −

1

1 𝜕 𝑇 𝑟 sin𝜃 𝜕Φ r

+ [ =

1

𝜕

[

𝑟sin𝜃 𝜕θ 1

+ [

1

=

𝜕 (cos 𝜃)

sin𝜃 𝜕Φ

1

𝜕

𝑟 𝜕𝑟

1

𝜕 𝜕𝑟

𝜕

𝑇 ] 𝑎̂𝑟 𝜕Φ 𝜃 1

(𝑟𝑇Φ )] 𝑎̂𝜃 + [

𝑟2

−

𝜕 𝜕Φ 𝜕 𝜕𝑟

(𝑟𝑇𝜃 ) −

𝜕 𝑇 ]𝑎 ̂Φ 𝜕𝜃 r

(𝑟sin𝜃 cos Φ)] 𝑎̂𝑟

(𝑟cos𝜃)] 𝑎̂𝜃

(𝑟 2 sin𝜃 cos Φ) −

𝜕 (cos 𝜃) 𝜕θ

𝑟2

]𝑎 ̂Φ

1

𝑟sin𝜃

(cos2𝜃 + 𝑟sin𝜃 sin Φ)𝑎̂𝑟 + (0 − cos𝜃)𝑎̂𝜃 𝑟

1

+ (2𝑟sin𝜃 cos Φ + 𝑟 cos2𝜃

=(

𝜕

𝑟 𝜕𝑟

(cos𝜃 sin𝜃) −

𝑟

+ [

−

|

𝑟sin𝜃

+ sin Φ) 𝑎̂𝑟 −

sin 𝜃 𝑟2

cos𝜃 𝑟

̂Φ )𝑎

𝑎̂𝜃 + (2 cos Φ +

1 𝑟3

̂Φ ) sin𝜃𝑎

Stokes’ theorem  Stokes’ theorem states that the circulation of a vector field ⃗A around a closed path L is ⃗ over the open surface S bounded by L equal to the surface integral of the curl A ⃗ × 𝐴 are continuous on S provided that ⃗A and ∇

⃗ . 𝑑𝑙 = ∫(∇ ⃗ ) ∙ 𝑑𝑆 ⃗ ×A ∮A 𝐿

𝑆

which is Stokes’ theorem

27


The surface S is subdivided in to a large number of cells.  There is cancellation on every interior path. Therefore, the sum of line integrals around Lk’s is the same as the line integrals around the bounding curve L.

Example 8: ⃗ . 𝑑𝑙 around the path shown below. Verify Stokes’ If ρ cos Φ âρ + sinΦâΦ , evaluate ∮ A Theorem

Solution: 𝑏

𝑐

𝑑

𝑎

⃗ . 𝑑𝑙 = [∫ + ∫ + ∫ + ∫ ∮A 𝐿

𝑎

Along 𝑎𝑏 𝜌 = 2

𝑏

𝑐

] ⃗A . 𝑑𝑙

𝑑

𝑑𝑙 = 𝜌 𝑑Φ𝑎̂Φ 28


𝑏

30

∫ ⃗A . 𝑑𝑙 = ∫ 𝑎

𝜌 sinΦ 𝑑Φ = [−2 cos Φ]

Φ=60

30 60

= −(√3 − 1)

Along 𝑏𝑐 Φ = 30 𝑑𝑙 = 𝑑𝜌𝑎̂𝜌 𝑐

5

5

𝜌2 21√3 ⃗ ∫ A . 𝑑𝑙 = ∫ 𝜌 cosΦ 𝑑𝜌 = cos 30 [ ] = 2 2 4 𝑏 ρ=2 Along cd 𝜌 = 5 𝑑

𝑑𝑙 = 𝜌 𝑑Φ𝑎̂Φ

60

∫ ⃗A . 𝑑𝑙 = ∫ 𝑐

𝜌 sinΦ 𝑑Φ = [−5 cos Φ]

Φ=30

Along 𝑑𝑎 Φ = 60 𝑎

5 = (√3 − 1) 30 2 60

𝑑𝑙 = 𝑑𝜌𝑎̂𝜌 2

2

𝜌2 21 ⃗ . 𝑑𝑙 = ∫ 𝜌 cosΦ 𝑑𝜌 = cos 60 [ ] = − ∫ A 2 5 4 𝑎 ρ=5 ⃗ . 𝑑𝑙 = −(√3 − 1) + ∮A 𝐿

21√3 5 21 + (√3 − 1) − = 4.941 4 2 4

⃗⃗⃗ × ⃗A) ∙ 𝑑𝑆 Using Stokes’ theorem ∮ ⃗A . 𝑑𝑙 = ∫ (∇ 𝐿

𝑆

𝜕𝐴𝜌 𝜕𝐴𝑧 𝜕𝐴𝜌 1 𝜕𝐴𝑍 𝜕𝐴Φ 1 𝜕 ⃗ ×𝐴 =[ ∇ − − ̂ ] 𝑎̂𝜌 + [ ] 𝑎̂Φ + [ (𝜌𝐴Φ ) − ]𝑎 𝜌 𝜕Φ 𝜕𝑧 𝜕𝑧 𝜕𝜌 𝜌 𝜕𝜌 𝜕Φ 𝑧 𝑑𝑆 = 𝜌𝑑𝛷𝑑𝜌𝑎̂𝑧 1 1 ⃗ × 𝐴 = (0 − 0)𝑎̂𝜌 + (0 − 0)𝑎̂Φ + (1 + 𝜌) sinΦ 𝑎 ∇ ̂ 𝑧 = (1 + 𝜌) sinΦ 𝑎 ̂𝑧 𝜌 𝜌 60

⃗ × ⃗A) ∙ 𝑑𝑆 = ∫ ∫(∇ 𝑆

5

∫

Φ=30

1

𝜌=2 𝜌

(1 + 𝜌) sinΦ𝜌𝑑𝜌𝑑𝛷

60

5

= ∫ sinΦ𝑑𝛷 ∫ (1 + 𝜌) 𝑑𝜌 30

2

60 𝜌2 5 = −[cosΦ] [ρ + ] = 4.941 30 2 2 ⃗ . 𝑑𝑙 ⃗ × ⃗A) ∙ 𝑑𝑆 = 4.941 = ∮A ∫(∇ 𝑆

Stokes’ Theorem is thus verified

𝐿

Laplacian of a scalar 29


 The Laplacian of a scalar field V, written as ∇2 V is the divergence of the gradient of V. It is another scalar field ⃗ ∙∇ ⃗ V = ∇2 V  In Cartesian coordinates, Laplacian is V = ∇ =(

𝜕 𝜕 𝜕 𝜕V 𝜕V 𝜕V 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂𝑧 ) ∙ ( 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂ ) 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑧 𝜕2V 𝜕2V 𝜕2V ∇ V= 2+ 2+ 2 𝜕𝑥 𝜕𝑦 𝜕𝑧 2

 A scalar field V is said to be harmonic in a given region if its Laplacian vanishes in that region. ∇2 V = 0 ⇒ Laplace's Equation

 In cylindrical coordinates, 1 𝜕 𝜕V ∇ V= (𝜌 ) 𝜌 𝜕𝜌 𝜕𝜌 2

1 𝜕2V 𝜌2 𝜕Φ2

𝜕2V 𝜕𝑧 2

 In spherical coordinates,

1 𝜕 𝜕V ∇ V = 2 (𝑟 2 ) 𝑟 𝜕𝑟 𝜕𝑟 2

1 𝜕 𝜕V 1 𝜕2V (sin𝜃 ) + 2 2 𝑟 2 sin𝜃 𝜕𝜃 𝜕𝜃 𝑟 sin 𝜃 𝜕Φ2

 Laplacian of a vector ⃗A denoted as ∇2 ⃗A is defined as

⃗ · ⃗A) − (∇ ⃗ × ⃗∇ × ⃗A) ∇2 ⃗A = ⃗∇(∇ Example 9 Find the Laplacian of the following scalar fields (a) V = 𝑒 −𝑧 sin2𝑥 cosh𝑦 (b) U = 𝜌2 zcos2Φ

Solution: 𝜕2V 𝜕2V 𝜕2V (a) ∇ V = 2 + 2 + 2 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕 2 −𝑧 𝜕 2 −𝑧 𝜕 2 −𝑧 = 2 (𝑒 sin2𝑥 cosh𝑦) + 2 (𝑒 sin2𝑥 cosh𝑦) + 2 (𝑒 sin2𝑥 cosh𝑦) 𝜕𝑥 𝜕𝑦 𝜕𝑧 2

30


=

𝜕 𝜕 −𝑧 𝜕 (2𝑒 −𝑧 cos2𝑥 cosh𝑦) + (𝑒 sin2𝑥 sinh𝑦) + (−𝑒 −𝑧 sin2𝑥 cosh𝑦) 𝜕𝑥 𝜕𝑦 𝜕𝑧

= −2𝑒 −𝑧 sin2𝑥 cosh𝑦 1 𝜕 𝜕V 1 𝜕2V 𝜕2V (b) ∇ V = + (𝜌 ) + 2 𝜌 𝜕𝜌 𝜕𝜌 𝜌 𝜕Φ2 𝜕𝑧 2 2

1 𝜕 𝜕 2 1 𝜕2 𝜕2 2 2 = (𝜌 zcos2Φ) + 2 (𝜌 zcos2Φ) (𝜌 (𝜌 zcos2Φ)) + 2 𝜌 𝜕𝜌 𝜕𝜌 𝜌 𝜕Φ2 𝜕𝑧 =

1 𝜕 𝜌 𝜕𝜌

(2𝜌2 zcos2Φ) +

1 𝜕 𝜌2 𝜕Φ

(−2𝜌2 zsin2Φ) = 4zcos2Φ − 4zcos2Φ = 0

31


Chapter 2: Electrostatic Fields Coulombs Law and field intensity

 Coulombs law states that the force F between two point charges Q1and Q2 separated in a vacuum or free space by a distance which is large when compared to their size is:  Along the line joining Q1 and Q2  Directly proportional to the product Q1 Q2 of the charges.  Inversely proportional to the square of the distance between them 𝐹=𝑘

Q1 Q 2 𝑅2

Q1, Q2 ⇒ Quantity of positive or negative charges R ⇒ Separation between charges Q1 and Q2 k ⇒ Proportionality constant When Q is in coulombs and R is in meters the constant k is found to be 1 𝑘= 4𝜋𝜀0 𝜀0 ⇒ the permittivity of free space which is 8.854 × 10−12 F/m 𝑘=

1 = 9 × 109 m/F 4𝜋𝜀0

Q1Q2 2 4𝜋𝜀0𝑅  The force acts along the line joining Q1 and Q2. In order to incorporate this information we may write Q1 Q 2 𝐹= 𝑎̂ 4𝜋𝜀0 𝑅2 𝑅 ∎

Incorporating these values 𝐹 =

Where 𝑎̂𝑅 is a unit vector in the direction of the force The force F12 on Q2 due to Q1 is given by 𝐹12 =

Q1 Q 2 𝑎̂ 4𝜋𝜀0 𝑅2 𝑅12 32


Where 𝑎̂𝑅12 is a unit vector directed from Q1 to Q2 If 𝑟1 and 𝑟2 are the position vectors of the points where Q1 and Q2 is situated ⃗ 12 = 𝑟2 − 𝑟1 Vector joining Q1 and Q2 is R

𝑅 = |𝑅⃗12 |

Now

𝑎̂𝑅12 =

𝐹12 =

𝐹12 =

⃗R12 ⃗R12 = ⃗ 12 | R |R

⃗ 12 Q1 Q 2 R Q1 Q 2 ⃗R = ( ) 4𝜋𝜀0 𝑅2 R 4𝜋𝜀0 𝑅 3 12

Q1 Q 2 ⃗ R 4𝜋𝜀0 𝑅3 12

𝐹12 =

Q1 Q 2 (𝑟2 − 𝑟1 ) 4𝜋𝜀0 ȁ𝑟2 − 𝑟1 ȁ3

The force F21 on Q1 due to Q2 is given by 𝐹21 = |𝐹12 |𝑎̂𝑅21 = |𝐹12 |(−𝑎̂𝑅12 ) 𝐹21 =

Q1 Q 2 Q1 Q 2 𝑎 ̂ = − 𝑎̂ 𝑅 4𝜋𝜀0 𝑅2 21 4𝜋𝜀0 𝑅2 𝑅12

Example 1 A 2mC positive charge is located at P1(3,-2,-4) in vacuum and a 5μC negative charge is located at P2(1,-4,2). Find the force on the negative charge. Solution: 𝑄1 = +2 × 10−3 𝐶

𝑄2 = −5 × 10−6 𝐶

⃗𝑟1 = 3𝑎 ̂𝑥 − 2𝑎 ̂𝑦 − 4𝑎 ̂𝑧 ⃗𝑟2 = 𝑎 ̂𝑥 − 4𝑎 ̂𝑦 + 2𝑎 ̂𝑧 ⃗R12 = 𝑟2 − 𝑟1 = −2𝑎 ̂𝑥 − 2𝑎 ̂𝑦 + 6𝑎 ̂𝑧 𝑅 = |𝑅⃗12 | = √4 + 4 + 36 = √44 𝐹12 =

Q1 Q 2 4𝜋𝜀0 𝑅12 2

𝑎̂𝑅12 33


𝑎̂𝑅12 = 𝐹12

⃗ 12 −2𝑎 ̂𝑥 − 2𝑎 ̂ 𝑦 + 6𝑎 ̂𝑧 R = = −0.3𝑎 ̂𝑥 − 0.3𝑎 ̂𝑦 + 0.9𝑎 ̂𝑧 ⃗ 12 | |R √44

(2 × 10−3 )(−5 × 10−6 ) = ̂𝑥 − 0.3𝑎 ̂𝑦 + 0.9𝑎 ̂𝑧 ) (−0.3𝑎 4𝜋 × 8.854 × 10−12 × 44 = −2.043(−0.3𝑎 ̂𝑥 − 0.3𝑎 ̂𝑦 + 0.9𝑎 ̂ 𝑧 ) = 0.61𝑎 ̂𝑥 + 0.61𝑎 ̂𝑦 − 1.84𝑎 ̂𝑧

Electric field intensity ⃗ at a point is defined as the force on a unit positive test  Electric field intensity E charge placed at that point. ⃗F ⃗E = Q  The direction of electric field intensity is the same as that of the force and is measured in newtons/coulomb  The electric field intensity at a point with position vector r due to a point charge at a point with position vector r is obtained as 𝐸⃗ =

Q Q(𝑟 − 𝑟1 ) 𝑎 ̂ = 𝑅 4𝜋𝜀0 𝑅2 4𝜋𝜀0 ȁ𝑟 − 𝑟1 ȁ3

 If there are N point charges Q1,Q2,……Qn located at points with position vectors 𝑟1 , 𝑟2 , . . . . . . 𝑟𝑛 , the electric field intensity at point 𝐸⃗ with position vector r is the vector sum of the electric field intensities produced by charges Q1, Q2,……Qn 𝐸⃗ =

Q1 (𝑟 − 𝑟1 ) Q 2 (𝑟 − 𝑟2 ) Q 𝑛 (𝑟 − 𝑟𝑛 ) + + ∙ ∙ ∙ ∙ ∙ ∙ + 4𝜋𝜀0 ȁ𝑟 − 𝑟1 ȁ3 4𝜋𝜀0 ȁ𝑟 − 𝑟2 ȁ3 4𝜋𝜀0 ȁ𝑟 − 𝑟𝑛 ȁ3 𝑛

1 Q 𝑚 (𝑟 − 𝑟𝑚 ) 𝐸⃗ = ෍ ȁ𝑟 − 𝑟𝑚 ȁ3 4𝜋𝜀0 𝑚=1

Example 2: Find the electric field intensity at that point P(1, 1, 1) caused by four identical 3nC charges located at P1(1, 1, 0), P2(−1, 1, 0), P3(−1,−1, 0), and P4 (1,−1, 0). 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧: We find that 𝑟 ⃗ =𝑎 ̂𝑥 + 𝑎 ̂𝑦 + 𝑎 ̂𝑧 , 𝑟 ⃗1 = 𝑎 ̂𝑥 + 𝑎 ̂𝑦 , 𝑟 ⃗ −𝑟 ⃗1 = 𝑎 ̂𝑧 ⃗𝑟 − ⃗𝑟2 = 2𝑎 ̂𝑥 + 𝑎 ̂𝑧 , ⃗𝑟 − ⃗𝑟3 = 2𝑎 ̂𝑥 + 2𝑎 ̂𝑦 + 𝑎 ̂𝑧 , and ⃗𝑟 − ⃗𝑟4 = 2𝑎 ̂𝑦 + 𝑎 ̂𝑧 The magnitudes are ȁ𝑟 ⃗ −𝑟 ⃗ 1ȁ = 1 ,

ȁ𝑟 ⃗ −𝑟 ⃗ 2 ȁ = √5 , ȁ𝑟 ⃗ −𝑟 ⃗ 3 ȁ = 3 , and

ȁ𝑟 ⃗ −𝑟 ⃗ 4 ȁ = √5 because 𝑄⁄4𝜋𝜀0 = 3 × 10−9 ⁄4𝜋 × 8.854 × 10−12 = 26.96 V 34


𝑛

1 Q 𝑚 (𝑟 − 𝑟𝑚 ) 𝑄 (𝑟 − 𝑟1 ) (𝑟 − 𝑟2 ) (𝑟 − 𝑟3 ) (𝑟 − 𝑟4 ) ⃗ = E = + + + ෍ ⌈ ⌉ ȁ𝑟 − 𝑟𝑚 ȁ3 4𝜋𝜀0 4𝜋𝜀0 ȁ𝑟 − 𝑟1 ȁ3 ȁ𝑟 − 𝑟2 ȁ3 ȁ𝑟 − 𝑟3 ȁ3 ȁ𝑟 − 𝑟4 ȁ3 𝑚=1

= 26.96 ⌈

̂ 𝑥 + 2𝑎 ̂𝑦 + 𝑎 ̂𝑧 2𝑎 ̂𝑦 + 𝑎 ̂𝑧 𝑎 ̂𝑧 2𝑎 ̂𝑥 + 𝑎 ̂ 𝑧 2𝑎 + + + ⌉ (1)3 (3)3 (5)3/2 (5)3/2

⃗E = 6.82𝑎 ̂𝑥 + 6.82𝑎 ̂𝑦 + 32.8𝑎 ̂𝑧

V/m

Electric fields due to continuous charge distributions

Point charge

Line charge

Surface charge

Volume charge

 A charge may be located on a point, along a line, on a surface or in a volume. Accordingly we have four types of charge distributions:  Point charges  Line charges  Surface charges  Volume charges.  Point Charge: A charge that is located on a body whose dimensions are much smaller than other relevant dimensions is called point charge. A collection of charges on a pinhead may be considered as a point charge. Total Charge 𝑄 = ෍ 𝑄𝑖

Charge distributions  Line Charge: A charge that is distributed along a fine line, as in the case of a sharp electron beam in a cathode ray tube, is considered as a line charge distribution.  It is convenient to associate a line charge density 𝜌𝐿 with a line charge distribution. Charge element 𝑑𝑄 = 𝜌𝐿 𝑑𝑙 Total charge 𝑄 = ∫𝐿 𝜌𝐿 𝑑𝑙 35


 Surface Charge: A charge that is distributed over a surface is considered as a surface charge distribution.  It is convenient to associate a surface charge density 𝜌𝑆 with a surface charge distribution. Charge element 𝑑𝑄 = 𝜌𝑆 𝑑𝑆 Total charge 𝑄 = ∫𝑆 𝜌𝑆 𝑑𝑆  Volume Charge: A charge that is distributed throughout a specified volume is considered as a volume charge distribution.  It is convenient to associate a volume charge density 𝜌v with a volume charge distribution. Charge element 𝑑𝑄 = 𝜌v 𝑑v Total charge 𝑄 = ∫v 𝜌v 𝑑v

Electric field intensity of Charge distributions  Electric field intensity of a point charge is given by Q 𝐸⃗ = 𝑎̂ 4𝜋𝜀0 𝑅2 𝑅  By replacing the charge Q by charge elements and integrating we get the electric field intensity of various charge distributions. 𝜌𝐿 𝑑𝑙 For line charge distributions, 𝐸⃗ = ∫ 𝑎̂ 4𝜋𝜀0 𝑅2 𝑅 𝜌𝑆 𝑑𝑆 For surface charge distributions, 𝐸⃗ = ∫ 𝑎̂ 4𝜋𝜀0 𝑅2 𝑅 𝜌v 𝑑v For volume charge distributions, 𝐸⃗ = ∫ 𝑎̂ 4𝜋𝜀0 𝑅2 𝑅

Electric field intensity of a finite line charge  Align the line charge along the Z axis symmetrically with respect to the origin.  Take two elemental lengths 𝑑𝑧 of the line charge that are symmetrical about the origin.  The charge associated with each elemental length 𝑑𝑧 is 𝑑𝑄 = 𝜌𝐿 𝑑𝑧 and can be treated as a point charge.  The vertical components of dE1 and dE2 gets cancelled leaving only the 𝑎̂𝜌 radial component along  The radial components of dE1 and dE2 get added.  There is no variation of the field along 𝛷 direction. 36


 So we need to calculate only the radial component along 𝑎̂𝜌

𝑑𝐸 = Radial components of (𝑑𝐸1 + 𝑑𝐸2 ) = 2 × Put cos𝛼 =

𝜌 𝑅

Put 𝑅 = √𝜌2 + 𝑧 2

𝑑Q cos𝛼 𝑎̂𝜌 4𝜋𝜀0 𝑅2

𝑑Q 𝜌 𝑎̂ 4𝜋𝜀0 𝑅3 𝜌 𝜌𝐿 𝑑𝑧 𝜌 𝑑𝐸⃗ = 2 × 𝑎̂ 4𝜋𝜀0 (𝜌2 + 𝑧 2 )3/2 𝜌

𝑑𝐸⃗ = 2 ×

 The electric field intensity at P due to the entire line charge is 𝐿/2 𝜌𝐿 𝜌 1 ⃗𝐸 = 2 ∫ 𝑑𝑧 𝑎̂𝜌 2 2 3/2 𝑧=0 4𝜋𝜀0 (𝜌 + 𝑧 )  Using the standard integral 1 𝑥 ∫ 2 𝑑𝑥 = (𝑎 + 𝑥 2 )3/2 𝑎2 √𝑎2 + 𝑥 2 𝐿/2

𝜌𝐿 𝜌 𝑧 𝜌𝐿 𝜌 𝐿 𝐸⃗ = [ ] 𝑎̂𝜌 = [ ] 𝑎̂ 2𝜋𝜀0 𝜌2 √𝜌2 + 𝑧 2 2𝜋𝜀0 𝜌2 √4𝜌2 + 𝐿2 𝜌 0 𝐸⃗ =

𝜌𝐿 𝐿 [ ] 𝑎̂ 2𝜋𝜀0 𝜌 √4𝜌2 + 𝐿2 𝜌

For a finite line charge 𝐸⃗ =



𝜌𝐿 𝐿 [ ] 𝑎̂ 2𝜋𝜀0 𝜌 √4𝜌2 + 𝐿2 𝜌

For an infinite line charge, L→∞ and 37


𝐸⃗ = lim

L→∞

𝜌𝐿 1 𝜌𝐿 𝑎̂ [ ] 𝑎̂𝜌 = 2𝜋𝜀0 𝜌 √4𝜌2 /𝐿2 + 1 2𝜋𝜀0 𝜌 𝜌

For an infinite line charge 𝐸⃗ =

𝜌𝐿 𝑎̂ 2𝜋𝜀0 𝜌 𝜌

Electric field intensity of an infinite surface charge  Assume that the infinite sheet charge is located in the x - z plane.  Assume that the infinite sheet charge is composed of line charge distributions with density 𝜌𝐿 = 𝜌𝑆 𝑑𝑧 C/m  The electric field intensity of an infinite line charge is radially directed away from the line charge and its magnitude is 𝜌𝐿 𝜌𝑆 𝑑𝑧 𝑑𝐸 = = 2𝜋𝜀0 𝑅 2𝜋𝜀0 𝑅  Consider one more line charge 𝜌́ 𝐿 symmetrically located with respect to the origin  The vertical components of dE1 and dE2 gets cancelled leaving only the radial component along 𝑎̂𝑦

𝑑𝐸 = y components of (𝑑𝐸1 + 𝑑𝐸2 ) = 2 × Put cos𝛼 =

𝑑 𝑅

𝑑𝐸⃗ =

𝜌𝑆 𝑑z cos𝛼 𝑎̂𝑦 2𝜋𝜀0 𝑅

𝜌𝑆 𝑑 𝑑z𝑎̂𝑦 𝜋𝜀0 𝑅2

 The electric field intensity at P due to the entire sheet charge is 38


∞

𝜌𝑆 𝑑 𝑑𝑧 𝑎̂𝑦 2 𝑧=0 𝜋𝜀0 𝑅

𝐸⃗ = ∫

Put 𝑅 =

√𝑑 2

+

𝑧2

Using the standard integral 𝐸⃗ =

𝜌𝑆 𝑑 ∞ 1 𝐸⃗ = ∫ 𝑑𝑧 𝑎̂𝑦 𝜋𝜀0 𝑧=0 (𝑑 2 + 𝑧 2 ) 1 1 𝑧 −1 ∫ 2 𝑑𝑥 = tan ( ) (𝑎 + 𝑥 2 ) 𝑎 𝑎

𝜌𝑆 𝑑 1 𝑧 ∞ 𝜌𝑆 𝜋 𝜌𝑆 −1 ∙ 𝑎̂𝑦 = 𝑎̂ [ tan ] 𝑎̂𝑦 = 2𝜋𝜀0 𝑑 𝑑 0 𝜋𝜀0 2 2𝜀0 𝑦

𝐸⃗ =

𝜌𝑆 𝑎̂ 2𝜀0 𝑦

𝑦>0

𝐸⃗ = −

𝜌𝑆 𝑎̂ 2𝜀0 𝑦

𝑦>d we can make the following assumptions 𝑟2 − 𝑟1 ⋍ 𝑑cos𝜃

and

𝑟1 𝑟2 ⋍ 𝑟 2 49


Q 𝑑cos𝜃 4𝜋𝜀0 𝑟 2 𝑑cos𝜃 = 𝑑⃗ ∙ 𝑎̂𝑟 where 𝑑⃗ = 𝑑𝑎̂𝑧 𝑄𝑑⃗ ∙ 𝑎̂𝑟 V= 4𝜋𝜀0 𝑟 2 ⃗⃗ , dipole moment By defining p ⃗⃗ = Qd Then V =

V=

𝑎̂𝑟 is a unit vector in r direction

⃗p⃗ ∙ 𝑎̂𝑟 4𝜋𝜀0 𝑟 2

The magnitude of dipole moment is equal to the product of charge and distance and its direction is from −Q to +Q  The electric field due to the dipole with centre at the origin is given by 𝜕V 1 𝜕V ⃗E⃗ = −∇ ⃗⃗𝑉 = − [ 𝑎̂𝑟 + 𝑎̂ ] There is no field variation along ϕ direction 𝜕𝑟 𝑟 𝜕θ 𝜃 Q𝑑cos𝜃 Q𝑑sin𝜃 Q 𝑑cos𝜃 = 𝑎 ̂ + 𝑎 ̂ where V = 𝑟 𝜃 2𝜋𝜀0 𝑟 3 4𝜋𝜀0 𝑟 3 4𝜋𝜀0 𝑟 2

⃗⃗ = E

𝑝 ሺ2cos𝜃𝑎̂𝑟 + sin𝜃𝑎̂𝜃 ሻ 4𝜋𝜀0 𝑟 3

When the dipole centre is at the origin,

Where 𝑝 = |p ⃗⃗| = Q𝑑

V=

⃗p⃗ ∙ 𝑎̂𝑟 4𝜋𝜀0 𝑟 2

When the dipole centre is not at the origin, but at 𝑟⃗́

V=

⃗⃗ ∙ (𝑟⃗ − 𝑟⃗́) p 3 4𝜋𝜀0 |𝑟⃗ − 𝑟⃗́|

 A point charge is a monopole and its field varies inversely as r 2 and its potential varies inversely as r. 50


 The electric field due to a dipole varies inversely as r3 and its potential varies inversely as r2.  The electric field due to successive higher order multipoles varies inversely as r 4, r5, r6, ...... while their potential varies inversely as r3, r4 , r5,......

Flux lines and equipotential surfaces  An electric flux line is an imaginary line or path drawn in such a way that its direction at any point is the direction of the electric field at that point.  A surface on which the potential is the same throughout is called equipotential surface.  The intersection of an equipotential surface and a plane results in a path or line called equipotential line.  No work is done in moving a charge from one point to another along an equipotential line or surface ∫ 𝐸⃗⃗ ∙ 𝑑𝑙⃗ = 0  This implies flux lines (direction of E) are always normal to equipotential surfaces.

Energy density in electrostatic fields  Consider a region free of electric fields. Let there be three point charges Q 1,Q2,Q3 at infinity.  To determine the energy present in the assembly of charges, we have to determine the amount of work necessary to assemble them.  No work is required to transfer Q1 from infinity to P1 because the space is initially charge free.  The work done in transferring Q2 from infinity to P2 is equal to the product of Q2 and the potential V21 at P2 due to Q1.  The work done in positioning Q3 at P3 is equal to Q3(V32+V31). 51


 The total work in positioning the three charges is WE=W1+W2+W3=0+Q2V21+ Q3(V31+V32ሻ …... ሺ1ሻ  If the charges were positioned in the reverse order WE=W3+W2+W1=0+Q2V23+ Q1(V12+V13ሻ …... ሺ2ሻ  Adding (1) and (2) 2𝑊𝐸 = 𝑄1 ሺ𝑉12 + 𝑉13 ሻ + 𝑄2 ሺ𝑉21 + 𝑉23 ሻ + 𝑄3 ሺ𝑉31 + 𝑉32 ሻ = 𝑄1 𝑉1 + 𝑄2 𝑉2 + 𝑄3 𝑉3 1 𝑊𝐸 = ሺQ1 V1 + Q 2 V2 + Q 3 V3 ሻ 2 𝑉1 , 𝑉2 , 𝑉3 ⟹ Total potentials at 𝑃1 , 𝑃2 , and 𝑃3 𝑚=𝑛

1 If there are n point charges 𝑊𝐸 = ∑ Q 𝑚 V𝑚 2 𝑚=1

 In order to obtain an expression for the energy stored in a region of continuous charge distribution, each charge is replaced by 𝜌v 𝑑v, and the summation becomes an integral, 1 𝑊𝐸 = ∫𝜌v V𝑑v , 2 v 1 ⃗⃗. 𝐷 ⃗⃗)V𝑑v 𝑊𝐸 = ∫(∇ 2 v

but

⃗⃗. 𝐷 ⃗⃗ሻ = 𝜌v ሺ∇

⃗⃗. 𝐷 ⃗⃗)V = ∇ ⃗⃗. V𝐷 ⃗⃗ − 𝐷 ⃗⃗. ∇ ⃗⃗V Using the identity (∇ 1 1 ⃗⃗ ∙ V𝐷 ⃗⃗)𝑑v − ∫(𝐷 ⃗⃗ ∙ ∇ ⃗⃗V)𝑑v 𝑊𝐸 = ∫(∇ 2 v 2 v Applying divergence theorem to the first term on the RHS, of this equation, 𝑊𝐸 =

1 1 ⃗⃗ ∙ 𝑑𝑆⃗ − ∫(𝐷 ⃗⃗ ∙ ∇ ⃗⃗V)𝑑v ∮V𝐷 2 𝑆 2 v

If we allow the volume v to increase to include all space, the surface area goes to infinity. 52


⃗⃗ as 1/r2 so that VD varies as 1/r3. The surface area (dS) V varies as 1/r and 𝐷 ⃗⃗. 𝑑𝑆⃗ decreases effectively as 1/r. Hence, the surface increases as r2. The quantity V𝐷 1

⃗⃗ ∙ 𝑑𝑆⃗) goes to zero as r → ∞ and S→ ∞ integral ( ∮ V𝐷 2

𝑆

1 ⃗⃗ ∙ ∇ ⃗⃗V)𝑑v ⃗⃗V 𝑊𝐸 = − ∫(𝐷 Putting 𝐸⃗⃗ = −∇ 2 v 1 1 1 ⃗⃗ ∙ 𝐸⃗⃗ )𝑑v = ∫(𝜀0 𝐸⃗⃗ ∙ 𝐸⃗⃗ )𝑑v = ∫𝜀0 𝐸 2 𝑑v 𝑊𝐸 = ∫(𝐷 2 v 2 v 2 v 1 1 ⃗⃗ ∙ 𝐸⃗⃗ )𝑑v = 𝜀0 E 2 𝑑v 𝑑𝑊𝐸 = (𝐷 2 2 𝑑𝑊𝐸 1 1 ⃗⃗ ∙ ⃗E⃗) = 𝜀0 E 2 = (𝐷 𝑑v 2 2 𝑑𝑊𝐸 = 𝑤𝐸 is defined as the energy density in electrostatic field 𝑑v

Energy density is a quantity which when integrated on overall space yields the total energy. 'All space' is the volume containing the entire field. 1 1 𝐷2 2 ⃗ ⃗ ⃗⃗ 𝑤𝐸 = (𝐷 ∙ 𝐸 ) = 𝜀0 𝐸 = 2 2 2𝜀0 𝑊𝐸 = ∫ 𝑤𝐸 𝑑v

Example 5: Three point charges -1nC, 4nC,and 3nC are located at (0,0,0),(0,0,1), and (1,0,0), respectively. Find the energy in the system.

Solution: 3

1 1 𝑊𝐸 = ∑ Q 𝑘 V𝑘 = ሺQ1 V1 + Q 2 V2 + Q 3 V3 ሻ 2 2 𝑘=1

53


Q1 Q2 Q3 Q2 Q1 Q3 + + [ ]+ [ ] 2 4𝜋𝜀0 ሺ1ሻ 4𝜋𝜀0 ሺ1ሻ 2 4𝜋𝜀0 ሺ1ሻ 4𝜋𝜀0 ሺ√2ሻ

𝑊𝐸 =

+ 𝑊𝐸 =

Q3 Q1 Q2 + [ ] 2 4𝜋𝜀0 ሺ1ሻ 4𝜋𝜀0 ሺ√2ሻ

1 Q2 Q3 1 12 (Q1 Q 2 + Q1 Q 3 + )= (−4 − 3 + ) nJ = 13.37nJ 4𝜋𝜀0 4𝜋𝜀0 √2 √2

Example 6: A charge distribution with spherical symmetry is given by 𝜌𝑣 = {

𝜌0

0≤𝑟≤𝑅

0 Determine the energy stored in the region r < R

𝑟>𝑅

Solution: The electric field of spherical charge distribution is For 𝑟 ≤ 𝑅

𝐸⃗⃗ =

𝜌𝑉 r 𝑎̂ 3𝜀0 𝑟

1 1 ⃗⃗. 𝐸⃗⃗ )𝑑v = ∫𝜀0 𝐸 2 𝑑v 𝑊𝐸 = ∫(𝐷 2 v 2 v 𝜌0 r 𝜌0 r but 𝐸⃗⃗ = 𝑎̂𝑟 ⇒ 𝐸 = ⇒ 3𝜀0 3𝜀0 1 𝜌0 2 r 2 𝜌0 2 𝑊𝐸 = ∫ 𝜀0 𝑑v = ∫ r 2 𝑑v 2 2 v 9𝜀0 18𝜀0 v 𝜌0 2 = ∫ r 2 r 2 sin𝜃 𝑑𝑟𝑑𝜃𝑑𝜙 18𝜀0 v

𝜌0 2 r 2 𝐸 = 9𝜀0 2 2

𝑅 𝜋 𝑅 𝜌0 2 2𝜋 𝜋 𝜌0 2 2𝜋 4 = ∫ ∫ ∫ r sin𝜃𝑑𝜃𝑑𝑟𝑑𝜙 = ∫ 𝑑𝜙 ∫ sin𝜃𝑑𝜃 ∫ r 4 𝑑𝑟 18𝜀0 𝜙=0 𝜃=0 𝑟=0 18𝜀0 𝜙=0 𝜃=0 𝑟=0 𝑅

𝜌0 2 r5 2𝜋𝜌0 2 R5 𝑊𝐸 = 4𝜋 [ ] = 18𝜀0 5 0 45𝜀0 54


Chapter 3: Electric Fields in Material Space Conduction and convection currents  The current through a given area is the electric charge passing through the area per unit time. Its unit is Amperes. 𝑑𝑄 𝐼= 𝑑𝑡  One ampere current is produced if charge is transferred at the rate of one coulomb per second.  The concept of current density is useful in defining the events occurring at a point.  If current ΔI flows through a surface ΔS, then current density is ΔI Jn = Δ𝑆 And ΔI = Jn ΔS Assuming the current density is perpendicular to the surface.  If the current density is not normal to the surface ΔI = J cos𝜃 Δ𝑆 = ⃗J ∙ ∆𝑆⃗  The total current flowing through the surface is I = ∫J⃗ ∙ 𝑑𝑆⃗ 𝑆

Convection currents  Convection current is produced when current flows through an insulating medium such as liquid, rarefied gas, or a vacuum.  It does not involve conductors and hence does not satisfy Ohm’s law.  A beam of electrons in a vacuum tube is an example of convection current.

55


 Consider a beam of electrons with volume charge density 𝜌v flowing in the y direction with a velocity 𝑢𝑦 . 𝑢 ⃗⃗ = 𝑢𝑦 𝑎̂𝑦  The current through the filament is ∆𝑄 ሺ𝜌𝑣 ∆𝑣ሻ ሺ𝜌𝑣 ∆𝑆∆𝑙ሻ ∆𝑙 ∆𝐼 = = = = 𝜌v ∆𝑆 ( ) = 𝜌v ∆𝑆𝑢𝑦 ∆𝑡 ∆𝑡 ∆𝑡 ∆𝑡  The y directed current density Jy is given by ∆𝐼 J𝑦 = = 𝜌v 𝑢𝑦 ∆𝑆  In general ⃗J = 𝜌v 𝑢 ⃗⃗ , ⃗J is the convection current density (A/m2).

Conduction currents  Conduction current requires a conductor as a medium.  A conductor contains a large number of free electrons that provide conduction current due to an impressed electric field.  When an electric field 𝐸⃗⃗ is applied the force on an electron with charge –e is 𝐹⃗ = −𝑒𝐸⃗⃗  Since the electron is not in free space, it will not be accelerated under the influence of the electric field.  It suffers constant collisions with the atomic lattice and drifts from one atom to another.  If the electron with mass m is moving in an electric field 𝐸⃗⃗ with an average drift velocity 𝑢 ⃗⃗ the average change in momentum of the free electron must match the applied force.  According to Newton’s law, 𝑚𝑢 ⃗⃗ 𝑒𝜏 = −𝑒𝐸⃗⃗ 𝐨𝐫 𝑢 ⃗⃗ = − 𝐸⃗⃗ 𝜏 ⇒ Average time interval between collisions 𝜏 𝑚  This indicates that the drift velocity is directly proportional to the applied field.  If there are n electrons per unit volume, the electronic charge density is given by 𝜌v = −𝑛𝑒 Thus conduction current density is 𝑛𝑒 2 𝜏 𝑛𝑒 2 𝜏 𝐽⃗ = 𝜌𝑣 𝑢 ⃗⃗ = 𝐸⃗⃗ = 𝜎𝐸⃗⃗ , 𝜎= 𝑚 𝑚 𝐽⃗ = 𝜎𝐸⃗⃗ Point form of Ohm’s law 56

is the conductivity


Conductors  A conductor contains free electrons, which accounts for its conductivity.  In an isolated conductor, when an external electric field is applied, the positive charges moves in the same direction as the applied field.  The negative charges moves in the opposite direction.  These free charges accumulate on the surface of the conductor and form an induced surface charge.  The induced charges set up an internal induced field 𝐸⃗⃗𝑖 which cancels the externally applied field 𝐸⃗⃗𝑒 Thus, a perfect conductor cannot contain an electrostatic field within it.

A perfect conductor ⃗⃗V = 0 which implies V=0  Inside a conductor 𝐸⃗⃗ = 0 and − ∇  Thus, a conductor is an equipotential body. Also, in a conductor, 𝜎 ⟶ ∞ and as per the equation, 𝐽⃗ = 𝜎𝐸⃗⃗ the electric field intensity 𝐸⃗⃗ ⟶ 0  According to Gauss’s law, 𝜀 ∮ 𝐸⃗⃗ ∙ 𝑑𝑆⃗ = ∫v 𝜌v 𝑑v 𝑆

 If 𝐸⃗⃗ = 0, the charge density 𝜌v = 0 and which implies  When the two ends of the conductor are maintained at a potential difference V, the electric field is not zero inside the conductor,  In this case, there is no static equilibrium, since the applied voltage prevents the establishment of such equilibrium.  An electric field must exist inside the conductor to sustain the flow of current.  The opposition to the flow is called resistance. Resistance of a conductor having non-uniform cross section is 𝑅=

𝑉 ∫ 𝐸⃗⃗ ∙ 𝑑𝑙⃗ = 𝐼 ∫ 𝜎𝐸⃗⃗ ∙ 𝑑𝑆⃗ 57


Example 1: 1 ሺ2 cos 𝜃𝑎̂𝑟 + sin 𝜃𝑎̂Φ ሻ 𝐴/𝑚2 calculate the current passing through 𝑟3 (a) A hemispherical shell of radius 20 cm (b) A spherical shell of radius 10 cm

If ⃗𝐽⃗ =

Solution: ሺaሻ 𝐼 = ∫ 𝐽⃗ ∙ 𝑑𝑆⃗

where 𝑑𝑆⃗ = r 2 sin𝜃 𝑑𝜙𝑑𝜃 𝑎̂𝑟

1 1 2 ሺ2 ሻ cos 𝜃𝑎 ̂ + sin 𝜃𝑎 ̂ ∙ r sin𝜃 𝑑𝜙𝑑𝜃 𝑎 ̂ = ∫2 cos 𝜃 sin𝜃 𝑑𝜃𝑑𝜙 𝑟 Φ 𝑟 3 𝑟 𝑆 𝑆𝑟

𝐼=∫

1 𝜋/2 2𝜋 = ∫ ∫ 2 cos 𝜃 sin 𝜃𝑑𝜃𝑑𝜙| 𝑟 𝜃=0 𝜙=0 2

=

𝜋/2

4𝜋 sin 𝜃 | 0.2 2 0

𝑟=0.2

𝜋/2 2 = 2𝜋 ∫ sin 𝜃 cos 𝜃 𝑑𝜃| 𝑟 𝜃=0 𝑟=0.2

= 10𝜋 = 31.4 A

(b) In the second case, 0 ≤ 𝜃 ≤ 𝜋 and r = 0.1. Hence, 𝜋 4𝜋 sin2 𝜃 𝐼= | =0 0.2 2 0

Continuity equation  Electric charges can be neither created nor destroyed according to law of conservation of electric charges.  Consider an arbitrary volume v bounded by surfaces S as shown in figure.  A net charge Qin exists within this region.  If a net current I flows across this surface the charge in the volume must decrease at a rate that equals the current.  If a net current flows across the surface in to the volume, the charge in the volume must increase at the rate equal to the current.

58


 The current leaving the volume is the total outward flux of the current density vector through the surface S. 𝑑𝑄𝑖𝑛 𝐼𝑜𝑢𝑡 = ∮ ⃗J ∙ 𝑑𝑆⃗ = − ⋯ ⋯ ሺ1ሻ 𝑑𝑡 𝑆 Using divergence theorem,

⃗⃗. ⃗J) 𝑑v ⋯ ⋯ ሺ2ሻ ∮J⃗ . 𝑑𝑆⃗ = ∫(∇ 𝑆

𝑑𝑄𝑖𝑛 𝑑 𝜕𝜌v − = − ∫𝜌v 𝑑v = − ∫ 𝑑v 𝑑𝑡 𝑑𝑡 v v 𝜕𝑡

v

⋯ ⋯ ሺ3ሻ

 Substituting (2) and (3) in (1) 𝜕𝜌v 𝜕𝜌v ⃗ ⃗ ⃗ ∇ . J = − ⃗⃗. ⃗J) 𝑑v = − ∫ ∫(∇ 𝑑v or 𝜕𝑡 𝜕𝑡 v v  This equation is called continuity of current equation or continuity equation.  It states that there can be no accumulation of charge at any point.  For steady currents

𝜕𝜌v 𝜕𝑡

⃗⃗. ⃗J = 0 = 0 and hence ∇

 The total current leaving a volume is the same as the total current entering it.

Boundary conditions  When the field exists in a medium consisting of two different media, the conditions the field must satisfy are called boundary conditions.  For the electrostatic field the following boundary conditions are important.  Dielectric – dielectric interface.  Conductor – dielectric.

Dielectric-dielectric Boundary conditions  Consider the boundary between two dielectrics with permittivities 𝜀1 = 𝜀0 𝜀r1

and

𝜀2 = 𝜀0 𝜀r2

59


 The fields in the two media can be expressed as 𝐸⃗⃗1 = 𝐸⃗⃗1𝑡 + 𝐸⃗⃗1𝑛

𝐸⃗⃗2 = 𝐸⃗⃗2𝑡 + 𝐸⃗⃗2𝑛

 Apply the equation ∮𝐸⃗⃗ ∙ 𝑑𝑙⃗ = 0 to the path abcda in the figure. 𝑙

𝐸⃗⃗ ∙ 𝑑𝑙⃗ = 𝐸1𝑡 ∆𝑤 − 𝐸1𝑛

∮ 𝑎𝑏𝑐𝑑𝑎

∆ℎ ∆ℎ ∆ℎ ∆ℎ − 𝐸2𝑛 − 𝐸2𝑡 ∆𝑤 + 𝐸2𝑛 + 𝐸1𝑛 =0 2 2 2 2

𝐸1𝑡 ∆𝑤 − 𝐸2𝑡 ∆𝑤 = 0 𝐸1𝑡 = 𝐸2𝑡  Tangential components of 𝐸⃗⃗ are equal at the boundary.  𝐸⃗⃗𝑡 undergoes no change on the boundary and it is continuous across the boundary. 𝐷1𝑡 𝐷2𝑡 𝐸1𝑡 = 𝐸2𝑡 = 𝜀2 𝐷1𝑡 = 𝜀1 𝐷2𝑡 𝜀1 𝜀2 ⃗⃗ undergoes some change across the boundary.  The tangential component of 𝐷 ⃗⃗ is said to be discontinuous across the boundary.  So 𝐷  The boundary conditions for the normal components are obtained by applying Gauss’s law on a small pillbox shaped volume as in the figure. Assuming Δh→0 ⃗⃗ ∙ 𝑑𝑆⃗ = 𝑄 ∮𝐷 𝑆

𝐷1𝑛 ∆𝑆 − 𝐷2𝑛 ∆𝑆 = ∆𝑄 = 𝜌S ∆𝑆 𝐷1𝑛 − 𝐷2𝑛 = 𝜌S

 If no free charge exists at the boundary, 𝜌S = 0 𝐷1𝑛 = 𝐷2𝑛 60


⃗⃗ are equal at the boundary.  Normal components of 𝐷 ⃗⃗𝑛 undergoes no change on the boundary and it is continuous across the boundary.  𝐷 𝐸1𝑛 𝜀2 𝐷1𝑛 = 𝐷2𝑛 𝜀1 𝐸1𝑛 = 𝜀2 𝐸2𝑛 = 𝐸2𝑛 𝜀1  The normal component of 𝐸⃗⃗ undergoes some change across the boundary.  So 𝐸⃗⃗𝑛 is said to be discontinuous across the boundary.

Conductor-Dielectric boundary conditions

 The interface between a perfect conductor and a dielectric shown in figure  for the tangential component, the boundary condition is, 𝐷𝑡 = 𝜀0 𝜀𝑟 𝐸𝑡 = 0 because 𝐸⃗⃗ = 0 inside the conductor.  The boundary condition for the normal components is, 𝐷𝑛 = 𝜀0 𝜀𝑟 𝐸𝑛 = 𝜌S 61


Poisson’s and Laplace’s equation  Gauss’s law in point form is given by ⃗ ∙𝐷 ⃗ =∇ ⃗ ∙ ε𝐸⃗ = 𝜌v ∇ ⃗V  We know that ⃗E = −∇ ⃗ ∙ (−ε∇ ⃗ V) = 𝜌v ∇ 𝜌 ⃗ ∙∇ ⃗V=− v ∇ 𝜀

∇2 V = −

𝜌v 𝜀

 This equation is known as Poisson’s equation. ⃗ ∙∇ ⃗V=( ∇

𝜕 𝜕 𝜕 𝜕V 𝜕V 𝜕V 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂𝑧 ) ∙ ( 𝑎̂𝑥 + 𝑎̂𝑦 + 𝑎̂ ) 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝑧

𝜕 𝜕V 𝜕 𝜕V 𝜕 𝜕V 𝜕2V 𝜕2V 𝜕2V = ( )+ ( )+ ( )= 2 + 2 + 2 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 𝜕𝑧 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕2V 𝜕2V 𝜕2V 𝜌v ∇ V= 2+ 2+ 2 =− or 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜀

𝜕2V 𝜕2V 𝜕2V 𝜌v + + = − 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 𝜀

2

 In the case of a charge free region, Poisson’s equation reduces to Laplace’s equation as given below. 𝜕2V 𝜕2V 𝜕2V + + =0 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2  In cylindrical and spherical systems,

∇2 V = 0

1 𝜕 𝜕V ∇ V= (𝜌 ) 𝜌 𝜕𝜌 𝜕𝜌 2

1 𝜕 𝜕V ∇ V = 2 (𝑟 2 ) 𝑟 𝜕𝑟 𝜕𝑟 2

1 𝜕2V 𝜌2 𝜕Φ2

𝜕2V =0 𝜕𝑧 2

1 𝜕 𝜕V 1 𝜕2V =0 (sin𝜃 ) + 2 2 𝑟 2 sin𝜃 𝜕𝜃 𝜕𝜃 𝑟 sin 𝜃 𝜕Φ2

 Laplace’s equation is very useful in finding out the potential V of a set of conductors maintained at different potentials as in the case of capacitor plates. ⃗V  Electric field ⃗E can be obtained once we obtain potential V from 𝐸⃗ = −∇

Uniqueness Theorem If a solution to Laplace’s Equation can be found that satisfies the boundary conditions, then the solution is unique. 62


Proof: (by contradiction)  Assume that there are two solutions V1 and V2 of Laplace’s equations both of which satisfies the given boundary conditions. ∇2 V1 = 0 ∇2 V2 = 0 , V1 =V2 ∇2 (V2 − V1 ) = 0

That is

∇2 Vd = 0

Letting (V2 − V1 ) = Vd

 V1 and V2 must reduce to the same potential along the boundary Letting (V2 − V1 ) = Vd = 0 ⃗ . ⃗A 𝑑v = ∮ ⃗A . 𝑑𝑆  From divergence theorem ∫∇ 𝑆

v

 The above equation is true for any vector, so let ⃗A = Vd ⃗⃗⃗ ∇ Vd ⃗ . (Vd ⃗∇Vd )𝑑v = ∮(Vd ⃗∇Vd ). 𝑑𝑆 ⋯ (1) ∫∇ v

𝑆

⃗ ) = ψ∇ ⃗ ∙ ⃗A + ⃗A ∙ ∇ ⃗ψ  Using the vector identity ⃗∇ ∙ (ψA ⃗ ∙ (Vd ⃗∇Vd ) = Vd ⃗∇ ∙ (∇ ⃗ Vd ) + (∇ ⃗ Vd ) ∙ ∇ ⃗ Vd ∇ 2

2 ⃗ Vd ) = Vd ∇ Vd + (∇

∇2 Vd = 0

Putting

2

⃗ ∙ (Vd ⃗∇Vd ) = (∇ ⃗ Vd ) ⋯ (2) ∇  Substituting equation (2) into equation (1) gives 2

⃗ Vd ) 𝑑v = ∮(Vd ∇ ⃗ Vd ). 𝑑𝑆 ∫ (∇ 𝑉

𝑆

Putting Vd = 0

2

⃗ Vd ) 𝑑v = 0 ∫ (∇ 𝑉

⃗ Vd = 0 ∇

2

⃗ Vd | 𝑑v = 0 ∫ |∇ 𝑉

Vd = V2 − V1 = constant everywhere in v

At the boundary we have seen that V2 –V1 = 0 or V1 =V2 everywhere. So V1 and V2 cannot be different solutions of the same problem. Uniqueness theorem is proved.

General Procedure for solving Poisson’s and Laplace’s equations  Solve Laplace’s (if ρv = 0) or Poisson’s equation ( if ρv ≠ 0) using either (a) Direct integration if V is a function of one variable, or (b) Separation of variables if V is a function of more than one variable. The solution at this point is not unique but expressed in terms of unknown integration constants. 63 Assist. Prof. Dr. Hamad Rahman Jappor

 Apply boundary conditions to determine a unique solution for V. ⃗ V and D ⃗⃗ from 𝐷 ⃗ = ε𝐸⃗  After obtaining V, find 𝐸⃗ using 𝐸⃗ = −∇  Find the charge induced on a conductor using 𝑄 = ∫ 𝜌S 𝑑𝑆 where 𝜌S = 𝐷𝑛 and ⃗ normal to the conductor. 𝐷𝑛 is the component of 𝐷  If required, find the capacitance between the conductors using C=Q/V

Example 2: Using Laplace’s theorem obtains the potential distribution between two spherical conductors separated by a single dielectric. The inner spherical conductor of radius 𝑎 is at a potential V0 and the outer conductor of radius b is at potential zero. Also evaluate the electric field.

Solution: Laplace’s equation in spherical coordinates is 1 𝜕 2 𝜕V 1 𝜕 𝜕V 1 𝜕2V 2 ∇ V = 2 (𝑟 =0 )+ 2 (sin𝜃 ) + 2 2 𝑟 𝜕𝑟 𝜕𝑟 𝑟 sin𝜃 𝜕𝜃 𝜕𝜃 𝑟 sin 𝜃 𝜕Φ2 Since V is a function of r only 1 𝜕 𝜕V ∇2 V = 2 (𝑟 2 ) = 0 𝑟 𝜕𝑟 𝜕𝑟 1 𝜕 2 𝜕V Since 2 ≠ 0 we get (𝑟 )=0 𝑟 𝜕𝑟 𝜕𝑟 𝜕V Integrating with respect to 𝑟 𝑟 2 = 𝐾1 𝜕𝑟 𝐾1 + 𝐾2 𝑟 Constants K1 and K2 are found out by applying boundary conditions. Again integrating with respect to r V = −

(i) 𝑟 = 𝑎, (ii) 𝑟 = 𝑏,

V = V0 V=0

𝐾1 + 𝐾2 𝑎 } ⋯ (1) 𝐾1 0 = − + 𝐾2 𝑏 64 Assist. Prof. Dr. Hamad Rahman Jappor

V0 = −

Solving equations 1 𝐾1 = −V0

𝑎𝑏 (𝑏 − 𝑎)

𝐾2 =

−V0 𝑎 (𝑏 − 𝑎)

V0 𝑎𝑏 V0 𝑎 − (𝑏 − 𝑎)𝑟 (𝑏 − 𝑎)

𝜕V 1 𝜕V 1 𝜕V 𝐸⃗ = − [ 𝑎̂𝑟 + 𝑎̂𝜃 + 𝑎̂ ] 𝜕𝑟 𝑟 𝜕θ 𝑟sin θ 𝜕Φ Φ

⃗ V, 𝐸⃗ = −∇

Since 𝑉 is a function of 𝑟 only 𝐸⃗ = − 𝐸⃗ = −

Then V =

𝜕V 𝑎̂ 𝜕𝑟 𝑟

𝜕 V0 𝑎𝑏 V0 𝑎 V0 𝑎𝑏 − 𝑎̂ V/m [ ] 𝑎̂𝑟 = (𝑏 − 𝑎)𝑟 2 𝑟 𝜕𝑟 (𝑏 − 𝑎)𝑟 (𝑏 − 𝑎)

Example 3: Find the potential at any point between the plates of a parallel plate capacitor and electric field. Solution: Laplace’s equation in rectangular coordinates is 𝜕2V 𝜕2V 𝜕2V + + =0 𝜕𝑥 2 𝜕𝑦 2 𝜕𝑧 2 Since 𝑉 is a function of 𝑦 only Integrating with respect to 𝑦

𝜕2V =0 𝜕𝑦 2 𝜕V = 𝐾1 𝜕𝑦

Again integrating with respect to 𝑦

𝑉 = 𝐾1 𝑦 + 𝐾2

Constants K1 and K2 are found out by applying boundary conditions. (i) y = 0,

V = V0

(ii) 𝑦 = 𝑑,

Substituting in V = 𝐾1 𝑦 + 𝐾2 𝑉=−

V=0

𝐾2 = V0

𝐾1 = −

V0 𝑑

V0 𝑦 + V0 d

⃗V=− 𝐸⃗ = −∇

𝜕V 𝜕 V0 V0 𝑎̂𝑦 = − (− 𝑦 + 𝑉0 ) 𝑎̂𝑦 = 𝑎̂𝑦 𝜕𝑦 𝜕𝑦 𝑑 𝑑 65


Example 4: Find the potential distribution between the conductors.

Solution: Laplace’s equation in cylindrical coordinates is 1 𝜕 𝜕V 1 𝜕2V 𝜕2V ∇ V= (𝜌 ) + 2 2 + 2 = 0 𝜌 𝜕𝜌 𝜕𝜌 𝜌 𝜕𝜙 𝜕𝑧 1 𝜕 𝜕V Since V is a function of 𝜌 only (𝜌 ) = 0 𝜌 𝜕𝜌 𝜕𝜌 2

Since

1 ≠0 𝜌

we get

𝜕 𝜕V (𝜌 ) = 0 𝜕𝜌 𝜕𝜌

𝜕V = 𝐾1 𝜕𝜌 Again integrating with respect to 𝜌 V = 𝐾1 ln𝜌 + 𝐾2 Constants K1 and K2 are found out by applying boundary conditions. Integrating with respect to 𝜌

(i) 𝜌 = 𝑎,

V = V0

𝜌

V0 = 𝐾1 ln𝑎 + 𝐾2 }

(ii) 𝜌 = 𝑏,

V=0

⋯ (1)

0 = 𝐾1 ln𝑏 + 𝐾2

Solving equations 1 𝐾1 = −

V0 ln(𝑏/𝑎)

Then V = −

𝐾2 =

V0 ln𝑏 ln(𝑏/𝑎)

V0 V0 ln(𝜌/𝑏) ln𝜌 + ln𝑏 = V0 Volt ln(𝑏/𝑎) ln(𝑏/𝑎) ln(𝑎/𝑏)

⃗V=− 𝐸⃗ = −∇

𝜕V 𝜕 ln(𝜌/𝑏) V0 𝑎̂𝜌 = − [𝑉0 𝑎̂ V/m ] 𝑎̂𝜌 = 𝜕𝜌 𝜕𝜌 ln(𝑎/𝑏) 𝜌ln(𝑏/𝑎) 𝜌 66


Chapter 4: Magnetostatic Fields Biot-Savart’s Law  Direct currents flow only in closed loops.  We can find out the contributions to the magnetic field due to differential lengths of such current carrying conductors.  A current element is the current I flowing through a differential vector length 𝑑𝑙 of a filamentary conductor.  A filamentary conductor is the limiting case of a cylindrical conductor of circular cross section as the radius approaches zero.

Biot-Savart’s law states that the magnetic field intensity dH produced at a point P by a differential current element Idl is proportional to the product Idl and the sine of the angle α between the element and the line joining P to the element and is inversely proportional to the square of the distance R between P and the element and its direction can be obtained by right handed screw rule. 𝑑𝐻 ∝

𝐼𝑑𝑙 sin 𝛼 𝑅2

𝑑𝐻 = 𝑘

𝐼𝑑𝑙 sin 𝛼 𝑅2

k is a proportionality constant whose value is 1/4𝜋 𝐼𝑑𝑙 sin 𝛼 𝑑𝐻 = 4𝜋𝑅2 This equation can be modified by incorporating the direction of the magnetic field intensity

⃗ = 𝑑𝐻

𝐼𝑑𝑙 × 𝑎̂𝑅 𝐼𝑑𝑙 × 𝑅⃗ = 4𝜋𝑅2 4𝜋𝑅3

𝑅 = |𝑅⃗|

and

𝑎̂𝑅 =

𝑅⃗ 𝑅

dH ⇒ Magnetic field intensity at P Idl ⇒ Current element R⇒ Distance from current element to point P 𝑎̂𝑅 ⇒ Unit vector directed from current element to P The units of the magnetic field intensity H are evidently amperes per meter (A/m). 67


 When current flows through a sheet of vanishingly small thickness we cannot measure current density 𝐽 in amperes per square meter as it becomes infinite.  In this case surface current density is measured in amperes per meter width and is ⃗ designated as 𝐾

 If the surface current density is uniform, the total current in any width b is I=Kb  where the width b is measured perpendicular to the direction of current flow.  For a nonuniform surface current density, integration is necessary: 𝐼 = ∫ K𝑑𝑁  Where 𝑑𝑁 is a differential element of the path across which the current is flowing.  Thus the differential current element 𝐼𝑑𝑙 , where 𝑑𝑙 is in the direction of the current, ⃗ or current density 𝐽, may be expressed in terms of surface current density 𝐾  Let ΔS be the cross sectional area of the wire. Then ⃗ 𝑑𝑆 = 𝐽𝑑v 𝐼𝑑𝑙 = 𝐾  In terms of distributed current sources, the Biot-Savart’s law may be expressed in the following ways:

𝐼𝑑𝑙 × 𝑎̂𝑅 2 𝐿 4𝜋𝑅

⃗ =∫ 𝐻

for line current

⃗ × 𝑎̂𝑅 𝑑𝑆 𝐾 for surface current 2 4𝜋𝑅 S

⃗ =∫ 𝐻

𝐽 × 𝑎̂𝑅 𝑑v for volume current 2 v 4𝜋𝑅

⃗ =∫ 𝐻

68


Magnetic field of a linear conductor Let dl ⇒ Small current element in the conductor AB P⇒ Point where the magnetic field is required.

ρ ⇒ Perpendicular distance between the conductor and the point P α1 ,α2 ⇒Angles subtended by the lower and upper ends of AB.

⃗ at P due to an element 𝑑𝑙 at (0, 0, z) is By Biot - Savart’s law the contribution 𝑑𝐻 ⃗ = 𝑑𝐻

𝐼𝑑𝑙 × 𝑅⃗ 4𝜋𝑅3

𝐵𝑢𝑡 𝑑𝑙 = 𝑑𝑧 𝑎̂𝑧 and 𝑅⃗ = 𝜌𝑎̂𝜌 − 𝑧𝑎̂𝑧 ,

so 𝑑𝑙 × 𝑅⃗ = 𝜌 𝑑𝑧 𝑎̂𝜙

𝐼𝜌𝑑𝑧 𝑎̂ 4𝜋𝑅3 𝜙 𝐼𝜌𝑑𝑧 𝐼𝜌𝑑𝑧 ⃗ =∫ 𝐻 𝑎 ̂ = ∫ 𝑎̂ 𝜙 4𝜋𝑅3 4𝜋(𝜌2 + 𝑧 2 )3/2 𝜙 ⃗ = 𝑑𝐻

𝐵𝑢𝑡 𝑧 = 𝜌 cot 𝛼

𝑑𝑧 = −𝜌cosec 2 𝛼𝑑𝛼

⃗ =− 𝐻

𝐼 𝛼2 𝜌2 cosec 2 𝛼𝑑𝛼 𝐼 𝛼2 𝜌2 cosec 2 𝛼𝑑𝛼 ∫ 𝑎̂ = − ∫ 𝑎̂ 4𝜋 𝛼1 (𝜌2 + 𝜌2 cot2 𝛼)3/2 𝜙 4𝜋 𝛼1 [𝜌2 (1 + cot 2 𝛼)]3/2 𝜙

⃗ =− 𝐻

𝐼 𝛼2 𝜌2 cosec 2 𝛼𝑑𝛼 𝐼 𝛼2 𝜌2 cosec 2 𝛼𝑑𝛼 ∫ 𝑎̂ = − ∫ 𝑎̂ 4𝜋 𝛼1 [𝜌2 cosec 2 𝛼]3/2 𝜙 4𝜋 𝛼1 𝜌3 cosec 3 𝛼 𝜙 69


𝛼2 𝐼 𝐼 𝛼 ⃗ =− 𝐻 𝑎̂𝜙 ∫ sin𝛼𝑑𝛼 = 𝑎̂𝜙 [cos 𝛼]𝛼21 4𝜋𝜌 4𝜋𝜌 𝛼1

⃗ = 𝐻

𝐼 (cos 𝛼2 − cos 𝛼1 )𝑎̂𝜙 4𝜋𝜌

𝑎̂𝜙 = 𝑎̂𝑙 × 𝑎̂𝜌

where 𝑎̂𝑙 is a unit vector along the line current and 𝑎̂𝜌 is a unit vector along the perpendicular line from the line current to the field point.  When the conductor is semi-infinite, so that point A is now at O(0,0,0) while B is at (0,0,∞) and α1 = 90o , α2 = 0o ⃗ = 𝐻

𝐼 𝐼 (cos 0 − cos 90)𝑎̂𝜙 = 𝑎̂ 4𝜋𝜌 4𝜋𝜌 𝜙

 When the conductor is of infinite length, point A is at (0,0,-∞) while B is at (0,0,∞) and α1 = 180o , α2 = 0o ⃗ = 𝐻

𝐼 𝐼 (cos 0 − cos 180)𝑎̂𝜙 = 𝑎̂ 4𝜋𝜌 2𝜋𝜌 𝜙

Magnetic field at the centre of a circular current loop ⃗ = ∮ 𝑑𝐻 ⃗ where 𝑑𝐻 ⃗ is the field  The magnetic field intensity at O is given by 𝐻 intensity at O due to any current element 𝐼𝑑𝑙  The direction of 𝐼𝑑𝑙 at any point P on the circular wire is given by the tangent at P in the direction of current flow.  The unit vector at P directed towards O is along the radius PO so that α = 90o. ⃗ = 𝑑𝐻

𝐼𝑑𝑙 × 𝑎̂𝑅 𝐼𝑑𝑙 sin 𝛼 𝐼𝑑𝑙 sin 90 𝐼𝑑𝑙 = 𝑎 ̂ = 𝑎 ̂ = 𝑎̂ 𝑛 𝑛 4𝜋𝑅2 4𝜋𝑅2 4𝜋𝑎2 4𝜋𝑎2 𝑧

⃗  Total field intensity at the centre of the circular wire is obtained by integrating 𝑑𝐻 around the circular path.

⃗ =∮ 𝐻

𝐼 𝐼 𝑑𝑙 𝑎 ̂ = ∮ 𝑑𝑙𝑎̂𝑧 𝑧 4𝜋𝑎2 4𝜋𝑎2

𝐼 2𝜋𝑎𝑎̂𝑧 4𝜋𝑎2 𝐼 ⃗ = 𝐻 𝑎̂ 2𝑎 𝑧 =

70


Magnetic field at a line through the centre of a circular current loop  Let P be a point at a distance h from the centre of a circular current loop.  Consider two diametrically opposite elements of the loop dl and 𝑑𝑙̀.  The field intensity at P distant R from the current element is given by ⃗ = 𝑑𝐻

𝐼𝑑𝑙 × 𝑎̂𝑅 𝐼𝑑𝑙 sin 𝛼 𝐼𝑑𝑙 = 𝑎 ̂ = 𝑎̂ 𝑛 4𝜋𝑅2 4𝜋𝑅2 4𝜋𝑎2 𝑧

 Since 𝑑𝑙 and 𝑎̂𝑅 are perpendicular 𝐼𝑑𝑙 ⃗ = 𝑑𝐻 𝑎̂ 4𝜋𝑅2 𝑛 𝑎̂𝑛 ⇒ A unit vector perpendicular to the plane containing 𝑑𝑙 and 𝑎̂𝑅

 The resultant field intensity at the point P is given by: ⃗ = 𝐻

𝐼𝑎2 𝑎̂ 2(𝑎2 + ℎ2 )3/2 𝑍

Example 1: Find the magnetic field at (0, 0, 5) due to side OA of the triangular loop carrying a current of 10A and lying in the xy plane.

71


Solution: ⃗ due to OA To find ⃗H ⃗ = 𝐻

𝐼 (cos 𝛼2 − cos 𝛼1 )𝑎̂𝜙 4𝜋𝜌

𝑎̂𝜙 = 𝑎̂𝑙 × 𝑎̂𝜌 𝜌=5 cos 𝛼2 = ⃗ = 𝐻

cos 𝛼1 = cos90 = 0 2 √29

𝐼 = 10

10 2 ( ) 𝑎̂ = 0.0591𝑎̂𝜙 4𝜋 ∙ 5 √29 𝜙

𝑎̂𝜙 = 𝑎̂𝑥 × 𝑎̂𝑧 = −𝑎̂𝑦

𝑤ℎ𝑒𝑟𝑒 𝑎̂𝑙 = 𝑎̂𝑥 , 𝑎̂𝜌 = 𝑎̂𝑧

⃗ = −0.0591𝑎̂𝑦 𝐴/𝑚 𝐻 Example 2: A circular loop located on x2+y2 = 9 carries a direct current of 10A along 𝑎̂𝜙 . Determine magnetic field at (0, 0, 4) and (0, 0, -4) Solution: ⃗ = 𝐻

𝐼𝑎2 𝑎̂ 2(𝑎2 + ℎ2 )3/2 𝑍

𝑎=3

ℎ=4

𝐼 = 10 at (0,0,4)

10 × 32 90 ⃗ = 𝐻 𝑎 ̂ = 𝑎̂ = 0.36𝑎̂𝑍 𝐴/𝑚 𝑍 250 𝑍 2(32 + 42 )3/2 𝑎=3

ℎ = −4

𝐼 = 10 at (0,0, −4)

10 × 32 90 ⃗ = 𝐻 𝑎 ̂ = 𝑎̂ = 0.36𝑎̂𝑍 𝐴/𝑚 𝑍 250 𝑍 2(32 + (−4)2 )3/2

Ampere’s Current law ⃗  Ampere’s circuital law states that the line integral of the tangential component of 𝐻 around a closed path is the same as the net current enclosed by the path. ⃗ ∙ 𝑑𝑙 = 𝐼𝑒𝑛𝑐 ∮𝐻 𝑙

72


⃗ around a closed path is called magneto motive  Line integral of the magnetic field 𝐻 force.  So Ampere’s law can be stated as: The magneto motive force around a closed path is equal to the current enclosed by that path.

But 𝐼𝑒𝑛𝑐 = ∫𝐽 ∙ 𝑑𝑆 𝑆

⃗ ∙ 𝑑𝑙 = ∫𝐽 ∙ 𝑑𝑆 ∮𝐻 𝑙

𝑆

 Applying Stokes theorem to the LHS of the above equation ⃗ ×𝐻 ⃗ ) ∙ 𝑑𝑆 = ∫𝐽 ∙ 𝑑𝑆 ∫(∇ 𝑆

𝑆

⃗ ×𝐻 ⃗ =𝐽 ∇  This is the Ampere’s law in differential form or point form and is the third of the Maxwell’s equations.

Magnetic field of an infinite line current  Let an infinitely long filamentary current element be placed along the z axis.  To determine H at a point P we select a closed Amperian path that passes through P.  Since this path encloses the whole current, ⃗ ∙ 𝑑𝑙 𝐼𝑒𝑛𝑐 = ∮𝐻 𝑙

𝐼 = ∮𝐻𝜙 𝑎̂𝜙 ∙ 𝑑𝑙𝑎̂𝜙 𝑙

= 𝐻𝜙 ∮𝑑𝑙 = 𝐻𝜙 2𝜋𝜌 𝑙

𝐼 2𝜋𝜌 𝐼 ⃗ = 𝐻 𝑎̂ 2𝜋𝜌 𝜙 𝐻𝜙 =

73


Magnetic field by a solenoid  When the coils of the solenoid are closely spaced, each turn can be regarded as a circular loop, and the net magnetic field is the vector sum of the magnetic field for each loop.  This produces a magnetic field that is approximately constant inside the solenoid, and nearly zero outside the solenoid.  The ideal solenoid is approached when the coils are very close together and the length of the solenoid is much greater than its radius. Then we can approximate the magnetic field as constant inside and zero outside the solenoid.

Apply Ampere’s law for the closed path 1234 ⃗ ∙ 𝑑𝑙 = 𝐼𝑒𝑛𝑐 ∮𝐻 𝐿

⃗ ∙ 𝑑𝑙 = ∫ 𝐻 ⃗ ∙ 𝑑𝑙 + ∫ 𝐻 ⃗ ∙ 𝑑𝑙 + ∫ 𝐻 ⃗ ∙ 𝑑𝑙 + ∫ 𝐻 ⃗ ∙ 𝑑𝑙 ∮𝐻 𝐿

12

23

34

41

= 𝐻𝐿 + 0 + 0 + 0 = 𝐻𝐿 Current enclosed by the path is = NI 𝑁𝐼 𝐻𝐿 = 𝑁𝐼 ⇒ 𝐻= 𝐿 𝑁𝐼 ⃗ = 𝐻 𝑎ො 𝐿 𝑥 ⃗ = 𝐵

𝜇𝑁𝐼 𝑎ො 𝐿 𝑥

,

𝐵=

𝜇𝑁𝐼 𝐿

⃗ = 𝜇𝐼 𝑁 = 𝜇𝐼 time number of turns per unit length 𝐵 𝐿

Magnetic flux density ⃗ = 𝜀0 𝐸⃗  Electric flux density and field intensity is relates as 𝐷 ⃗ = 𝜇0 𝐻 ⃗  Similarly, we can define magnetic flux density by the relation 𝐵 74

𝜇0 ⇒ Permiability of free space = 4𝜋 × 10−7 𝐻/m ⃗ ⇒ Magneic flux density 𝐵 ⃗ ⇒ Magneic field intensity 𝐻  Magnetic flux through a surface is obtained by integrating flux density throughout the surface ⃗ ∙ 𝑑𝑆 𝜓 = ∫𝐵 𝑆

⃗ is tangential at every point.  Magnetic flux lines is a path to which 𝐵  Each flux line is closed and has no beginning or end.  In an electrostatic field the flux crossing a closed surface is the same as the charge enclosed ⃗ ∙ 𝑑𝑆 = 𝑄 𝜓 = ∮𝐷  Therefore, it is possible to have an isolated electric charge and the flux lines produced by it need not be closed.  It is not possible to have an isolated magnetic pole.  Magnetic flux lines always close upon themselves in contrast to electric flux lines.  Electric flux lines need not be closed. Therefore, it is possible to have an isolated electric charge.

It is not possible to isolate the north and south poles of a magnet.

Magnetic flux lines are always closed. An isolated magnetic pole cannot exist 75

Gauss’s Law for magnetostatic fields  The total flux through a closed surface in a magnetic field is zero. ⃗ ∙ 𝑑𝑆 = 0 ∮𝐵  By applying divergence theorem, ⃗ . 𝑑𝑆 = ∫ ∇ ⃗ .B ⃗ 𝑑v = 0 ∮B

⃗ .B ⃗ =0 ∇

v

 This is the Maxwell’s fourth equation. It states that magnetostatic fields have no sources or sinks.

Maxwell’s Equations for static EM fields Differential form

Integral form

Remarks

⃗ ∙𝐷 ⃗ = 𝜌v ∇

⃗ ∙ 𝑑𝑆 = ∫ 𝜌v 𝑑v ∮𝐷 v

Gauss’ s Law

⃗ ∙𝐵 ⃗ =0 ∇

⃗ ∙ 𝑑𝑆 = 0 ∮𝐵

Nonexistence of magnetic monopole

⃗ × 𝐸⃗ = 0 ∇

∮ 𝐸⃗ ∙ 𝑑𝑙 = 0

Conservativeness of electrostatic field

⃗ ×𝐻 ⃗ =𝐽 ∇

⃗⃗⃗⃗ ⃗ ∙ 𝑑𝑙 = ∫ 𝐽 ∙ 𝑑𝑆 ∮𝐻

Ampere’s Law

𝑆

𝑆

𝐿

𝐿

𝑆

Magnetic scalar and vector potentials ⃗V  In electrostatics electric field intensity and potential are related as 𝐸⃗ = −∇  Similar to this we can relate magnetic field intensity with two magnetic potentials: 1. Magnetic scalar potential 2. Magnetic vector potential ⃗  Magnetic scalar potential Vm (in amperes) is related to by the relation 𝐻 ⃗ = −∇ ⃗⃗⃗ Vm 𝐻

𝑖𝑓

𝐽 = 0 ⋯ (1)

⃗ ×𝐻 ⃗ = 𝐽 ⋯ (2)  According to Maxwell’s equation ∇ ⃗ × (∇ ⃗ V) = 0 ⋯ (3)  For any scalar ∇ ⃗ ×𝐻 ⃗ = ∇ ⃗ × (−∇ ⃗ Vm ) = 0 ⋯ (4) ∇

⃗ = −∇ ⃗ Vm 𝑖𝑓 𝐽 = 0 𝐻

Equation (4) is true only if ⃗𝐽 = 0 76

 The definition of Magnetic scalar potential according to equation (1) must satisfy equations (2) and (4).  For this, the condition 𝐽 = 0 must be satisfied.  Vm satisfies Laplace’s equation also ∇2 Vm = 0

𝑖𝑓

𝐽=0

 For a magnetostatic field ⃗ .𝐵 ⃗ = 0 ⋯ (5) ∇  For any vector ⃗ . (∇ ⃗ × 𝐴) = 0 ⋯ (6) ∇  We can define another potential called magnetic vector potential 𝐴 (in Wb/m) that satisfies equations (5) and (6) simultaneously ⃗ =∇ ⃗ ×𝐴 𝐵

𝐴 is called magnetic vector potential

⃗ from it.  In many EM problems, it is more convenient to first find 𝐴 and then find 𝐵  We can derive Magnetic vector potentials from Biot-Savart’s law. To do this, we write Biot-Savart’s law as 𝜇0 𝐼𝑑𝑙 × 𝑅⃗ ∫ 4𝜋 𝐿 𝑅3  Let the source 𝑑𝑙̀ be located at (𝑥́ , 𝑦́ , 𝑧́ ) and let (x, y, z) be the point where we want to find the magnetic potential. 𝑅 = |𝑟 − 𝑟́ | = [(𝑥 − 𝑥 ́ )2 + (𝑦 − 𝑦 ́ )2 + (𝑧 − 𝑧 ́ )2 ]1/2 ⃗ = 𝐵

(𝑥 − 𝑥 ́ )𝑎 ̂𝑥 + (𝑦 − 𝑦 ́ )𝑎 ̂𝑦 + (𝑧 − 𝑧 ́ )𝑎 ̂𝑧 1 𝑅⃗ ∇( ) = − = − [(𝑥 − 𝑥 ́ )2 + (𝑦 − 𝑦 ́ )2 + (𝑧 − 𝑧 ́ )2 ]3/2 𝑅 𝑅3 𝑅⃗ 1 𝑎ො𝑅 ⃗ ( ) − 3=∇ (= 2 ) 𝑅 𝑅 𝑅 𝜇 1 ⃗ = − 0 ∫𝐼𝑑𝑙́ × ∇ ⃗ ( ) ⋯ (1) 𝐵 4𝜋 𝐿 𝑅 Using the vector identity ⃗ × (𝜙𝐹 ) = 𝜙∇ ⃗ × 𝐹 + (∇ ⃗ 𝜙) × 𝐹 ∇ or

⃗ 𝜙) = 𝜙∇ ⃗ ×𝐹−∇ ⃗ × (𝜙𝐹 ) 𝐹 × (∇

77

1 ́ where 𝜙 a scalar is field and 𝐹 is a vector field. Taking 𝜙 = and 𝐹 = 𝑑𝑙 , we have 𝑅 ́ 1 1 𝑑𝑙 ́ ́ ⃗ ( )= ∇ ⃗ × 𝑑𝑙 − ∇ ⃗ ×( ) 𝑑𝑙 × ∇ 𝑅 𝑅 𝑅 ⃗ operates with respect to (x, y, z) where R is a function of (𝑥́ , 𝑦́ , 𝑧́ ) and (x, y, z). Since ∇ ́ ́ ⃗ × 𝑑𝑙 while 𝑑𝑙 is a function of (𝑥́ , 𝑦́ , 𝑧́ ), So ∇ = 0. Hence, ́ 1 𝑑𝑙 ́ ⃗ ( ) = −∇ ⃗ × ( ) ⋯ (2) 𝑑𝑙 × ∇ 𝑅 𝑅

Substituting in equation (1) ́ ́ 𝜇0 𝐼 𝑑𝑙 𝜇0 𝐼𝑑𝑙 ⃗ = ⃗ × ( ) =∇ ⃗ ×∫ 𝐵 ∫∇ 4𝜋 𝐿 𝑅 𝐿 4𝜋𝑅 Comparing with

⃗ =∇ ⃗ ×𝐴 𝐵

́ 𝜇0 𝐼𝑑𝑙 𝐴=∫ 𝐿 4𝜋𝑅

For line current

Similarly ⃗ 𝑑𝑆 𝜇0 𝐾 𝑆 4𝜋𝑅

For surface current

𝜇0 𝐽𝑑v 𝑉 4𝜋𝑅

For volume current

𝐴=∫

𝐴=∫

Magnetic flux from vector potential ⃗ . 𝑑𝑆 = ∫(∇ ⃗ × 𝐴) ∙ 𝑑𝑆 𝜓 = ∫B 𝑆

𝑆

By Stokes’s Theorem

⃗ . 𝑑𝑙 ⃗ × 𝐴) ∙ 𝑑𝑆 = ∮A ∫(∇ 𝑆

𝐿

⃗ . 𝑑𝑙 ⃗ . 𝑑𝑆 = ∮A 𝜓 = ∫B 𝑆

𝐿

⃗ . 𝑑𝑙 𝜓 = ∮A 𝐿

78

Chapter 5: Magnetic Forces, Materials, and Devices Magnetic force on a charged particle

 A magnetic field can exert a force only on a moving charge.  The magnetic force 𝐹𝑚 experienced by a charge Q moving with a velocity 𝑢 ⃗ in a ⃗ is given by magnetic field B ⃗ 𝐹𝑚 = 𝑄𝑢 ⃗ ×B  An electric field can exert a force on a stationary charge and the force is given by 𝐹𝑒 = 𝑄𝐸⃗  For a moving charge Q in the presence of both electric and magnetic fields the total force on the charge is given by ⃗) 𝐹 = 𝐹𝑚 + 𝐹𝑒 = 𝑄(𝐸⃗ + 𝑢 ⃗ ×B  This is called Lorentz force equation  When the mass of the charged particle moving in an electromagnetic field with velocity 𝑢 ⃗ is m, 𝑑𝑢 ⃗ ⃗) 𝐹=𝑚 = 𝑄(𝐸⃗ + 𝑢 ⃗ ×𝐵 𝑑𝑡  The above equation relates mechanical force to electrical force.

Electric force 𝐹𝑒 is  Independent of velocity  It can perform work on the charge  It can change the kinetic energy of the charge

Magnetic force 𝐹𝑚 is  Dependent of velocity  Force is normal to velocity and magnetic field  It cannot perform work on the charge since the motion of the particle is at right angles to the force  It cannot change the kinetic energy of the charge

Magnetic force on a current element  The differential magnetic force that a differential element charge, dQ moving with ⃗ is a velocity 𝑢 ⃗ experiences in a magnetic field of flux density B ⃗ ⋯ (1) 𝑑𝐹 = 𝑑𝑄𝑢 ⃗ ×B

79

 The current in a conductor or in a beam of electrons or ions can be expressed as charge per unit time. Q 𝐼= t  Where t is the time taken for the charge to move a distance l. Q 𝑙 𝐼 ∙ 𝑙 = ∙ 𝑙 = Q ∙ = 𝑄𝑢 t t  Rewriting the equation for incremental charges and distances and using vector notation 𝐼𝑑𝑙 = 𝑑𝑄𝑢 ⃗ ⋯ (2)  Equation (2) shows that an elemental charge dQ moving with velocity 𝑢 ⃗ thereby producing convection current element 𝑑𝑄𝑢 ⃗ is equivalent to a conduction current element 𝐼𝑑𝑙  Substituting equation(2) in (1) ⃗ 𝑑𝐹 = 𝐼𝑑𝑙 × B  If the current I is through a closed path L, the force on the whole loop is ⃗ 𝐹 = ∮𝐼𝑑𝑙 × 𝐵 𝐿

 We have the following relations between current elements. ⃗ 𝑑𝑆 = 𝐽𝑑v 𝐼𝑑𝑙 = 𝐾  Using these relations ⃗ 𝑑𝑆 × 𝐵 ⃗ =𝐾 ⃗ ×𝐵 ⃗ 𝑑𝑆 𝑑𝐹 = 𝐾 ⃗ =𝐽×𝐵 ⃗ 𝑑v 𝑑𝐹 = 𝐽𝑑v × 𝐵  The total force in these cases is given by ⃗ ×𝐵 ⃗ 𝑑𝑆 𝐹 = ∫𝐾 𝑆

⃗ 𝑑v 𝐹 = ∫𝐽 × 𝐵 v

Magnetic force between two current elements  Consider two current loops C1 and C2 with currents I1 and I2.  The loops are divided in to very small vector line segments 𝑑𝑙  The current elements in the loops are denoted by 𝐼𝑑𝑙  Consider two current elements in the two loops and 80

 According to Biot-Savart’s law, current elements produce magnetic fields.  The magnetic field produced by at is 𝐼𝑑𝑙2 at 𝐼𝑑𝑙1 is ⃗2 = 𝑑𝐵

𝜇0 𝐼2 𝑑𝑙2 × 𝑎ො𝑅21 4𝜋𝑅21 2

⃗ 2 produced by the  Hence the force on current element 𝐼𝑑𝑙1 due to the field 𝑑𝐵 current element 𝐼𝑑𝑙2 is ⃗2 = 𝑑𝐹1 = 𝐼1 𝑑𝑙1 × 𝑑𝐵

𝜇0 𝐼1 𝑑𝑙1 ×(𝐼2 𝑑𝑙2 × 𝑎ො𝑅21 ) 4𝜋𝑅21 2

 The total force 𝐹1 on current loop 1 due to current loop 2 is 𝐹1 =

𝑑𝑙1 ×(𝑑𝑙2 × 𝑎ො𝑅21 ) 𝜇0 𝐼1 𝐼2 ∮ ∮ 4𝜋 𝑙1 𝑙2 𝑅21 2

⃗ 1 produced by loop 1  Similarly the total force 𝐹2 on loop 2 due to magnetic field 𝐵 is obtained by interchanging the subscripts 𝐹2 =

𝑑𝑙2 ×(𝑑𝑙1 × 𝑎ො𝑅12 ) 𝜇0 𝐼1 𝐼2 ∮ ∮ 4𝜋 𝑙1 𝑙2 𝑅12 2

Example 1 A charged particle of mass 2kg and charge 3C starts at point(1,-2,0) with velocity 4𝑎 ̂𝑥 + 3𝑎 ̂𝑧 m/s in an electric field 12𝑎 ̂𝑥 + 10𝑎 ̂𝑦 V/m. At time t=1s determine its acceleration, velocity, and kinetic energy. Solution 𝑄𝐸⃗ 3 𝐹 = 𝑚𝑎 = 𝑄𝐸⃗ ⟹ 𝑎 = = (12𝑎 ̂𝑥 + 10𝑎 ̂𝑦 ) = 18𝑎 ̂ 𝑥 + 15𝑎 ̂ 𝑦 𝑚/s 2 𝑚 2 𝑑𝑢 ⃗ 𝑑 𝑎= = (𝑢 , 𝑢 , 𝑢 ) = 18𝑎 ̂𝑥 + 15𝑎 ̂𝑦 𝑑𝑡 𝑑𝑡 𝑥 𝑦 𝑧 81

𝑑𝑢𝑥 = 18 ⇒ 𝑢𝑥 = 18𝑡 + 𝐶1 𝑑𝑡 𝑑𝑢𝑦 = 15 ⇒ 𝑢𝑦 = 15𝑡 + 𝐶2 𝑑𝑡 𝑑𝑢𝑧 = 0 ⇒ 𝑢𝑧 = 𝐶3 𝑑𝑡 At t = 0 𝑢 ⃗ = 4𝑎 ̂ 𝑥 + 3𝑎 ̂𝑧 𝑢𝑥 = 18𝑡 + 𝐶1

𝑢𝑦 = 15𝑡 + 𝐶2

𝑢𝑧 = 𝐶3

𝑢𝑥 (𝑡 = 0) = 4

⇒

4 = 0 + 𝐶1 ⇒ 𝐶1 = 4

𝑢𝑦 (𝑡 = 0) = 0

⇒

0 = 0 + 𝐶2 ⇒ 𝐶2 = 0

𝑢𝑧 (𝑡 = 0) = 3

⇒

𝐶3 = 3

⇒ 𝐶3 = 3

𝑢 ⃗ (𝑡) = (𝑢𝑥 , 𝑢𝑦 , 𝑢𝑧 ) = (18𝑡 + 4, 15𝑡, 3) 𝑢 ⃗ (𝑡 = 1𝑠) = 22𝑎 ̂ 𝑥 + 15𝑎 ̂ 𝑦 + 3𝑎 ̂𝑧 𝑚/s 1 1 Kinetic energy = 𝑚|𝑢 ⃗ |2 = (2)(222 + 152 + 32 ) = 718𝐽 2 2

Magnetic torque and moment  When a current loop is placed in a magnetic field, it experiences a force tending to rotate it, The rotating tendency of the loop is expressed in terms of torque.  The torque on the loop is the vector product of the moment arm and the force. ⃗ =𝑟×𝐹 𝑇

 Let a rectangular loop of length l and width w is placed in a uniform magnetic field ⃗ , 𝐼𝑑𝑙 is parallel to 𝐵 ⃗ along sides 1-2 and 3-4 of the loop. 𝐵 ⃗ =0 and so the force acting on these sides is zero.  So 𝐼𝑑𝑙 × 𝐵  The force acting on lengths 2-3 and 3-4 are 3

3

⃗ = 𝐼 ∫ 𝑑𝑧𝑎 ⃗ = 𝐼𝐵𝑙 𝐹0 = 𝐼 ∫ 𝑑𝑙 × B ̂𝑧 × B 2 1

2 1

⃗ = 𝐼 ∫ 𝑑𝑧(−𝑎 ⃗ = −𝐼𝐵𝑙 𝐹́0 = 𝐼 ∫ 𝑑𝑙 × B ̂𝑧 ) × B 4

4

𝐹 = 𝐹0 + 𝐹́0 = 𝐼𝐵𝑙 − 𝐼𝐵𝑙 = 0

82

 No force is exerted on the loop as a whole.  When the axis of the loop is pivoted at the centre axis, the forces are acting at different points with reference to the axis, and this creates a couple. ⃗ (force makes an angle  If the normal to the plane of the loop makes an angle α with B α with moment arm) the magnitude of the torque on the loop is ⃗ = |𝐹0 | sin 𝛼 𝑤 = 𝐵𝐼𝑙𝑤 sin 𝛼 = 𝐵𝐼𝑆 sin 𝛼 𝑇

𝑙𝑤 = 𝑆,

Area of the loop

= 𝐼𝑆 𝐵 sin 𝛼 Defining 𝑚 ⃗⃗ = 𝐼𝑆𝑎ො𝑛 as the dipole moment (in A.m2) of the loop, 𝑎ො𝑛 ⇒ Unit vector normal to the plane of the loop ⃗ = 𝐼𝑆 𝑎ො𝑛 × 𝐵 ⃗ =𝑚 ⃗ 𝑇 ⃗⃗ × 𝐵  The magnetic dipole moment is the product of current and area of the loop and its direction is normal to the loop.  Torque direction is the direction of the axis of rotation. In this case it is along the z direction. ⃗ =𝑚 ⃗ 𝑇 ⃗⃗ × 𝐵

𝑚 ⃗⃗ = 𝐼𝑆𝑎ො𝑛

 A bar magnet or a small filamentary current loop is a magnetic dipole.  While isolated electric charges exist, it is not possible to have an isolated magnetic pole.  The net magnetic dipole moment per unit volume is called magnetisation.  Magnetization at any point in the magnetized bar is obtained by 𝑚 ⃗⃗ amperes/meter ∆v→0 ∆v  The current produced by the bound charges is called a bound current or Amperian current. ⃗⃗ = lim 𝑀

83

 We may now define a volume current density 𝐽𝑏 throughout the body and a surface ⃗ 𝑏 on the surface of the body as following: current density 𝐾 ⃗𝑏 = M ⃗⃗⃗ × 𝑎ො𝑛 𝐾 ⃗ ×𝑀 ⃗⃗ 𝑗𝑏 = ∇

⃗ 𝑏 is the bound surface current density in 𝐾 amperes per meter) 𝑗𝑏 is the bound volume current density in amperes per meter square)

 We have seen that the magnetic properties may be derived by regarding the material as being made up of many small current loops.  In the interior of the material, incomplete cancellation of currents in adjacent loops results in a net volume current density.  At the surface of the material, the currents are not cancelled, giving a net equivalent surface current density.

Magnetization in materials        

Any material is composed of atoms. Each atom is composed of electrons orbiting about a central positive nucleus. The electrons spin about their own axes. Therefore, electrons orbiting around the nucleus and electrons spinning about their axes produce an internal magnetic field. Both of the electron motion produces internal magnetic field similar to the one produced by a current loop. The equivalent loop has a magnetic moment of 𝑚 ⃗⃗ = 𝐼𝑏 𝑆𝑎ො𝑛 Without an external field applied to the material the sum of the magnetic moments is zero due to the random orientation of the moments. When an external magnetic field is applied the magnetic moments of the electrons ⃗ so that the net magnetic moment is not zero. align themselves with 𝐵

 From Maxwell’s equation 84

⃗ ×𝐻 ⃗ =𝐽 ∇

⃗ 𝐵 ⃗∇ × ( ) = 𝐽 𝜇0

 In a material medium 𝐽 = 𝑗𝑓 + 𝑗𝑏 ⃗ 𝐵 ⃗∇ × ( ) = 𝑗𝑓 + 𝑗𝑏 𝜇0

𝑗𝑓 is the free volume current density

⃗ ×𝐻 ⃗ +∇ ⃗ ×𝑀 ⃗⃗ = ∇ ⃗ × (𝐻 ⃗ +𝑀 ⃗⃗ ) = ∇ ⃗ ×𝐵 ⃗ = 𝜇0 ∇ ⃗ × (𝐻 ⃗ +𝑀 ⃗⃗ ) ∇ ⃗ = 𝜇0 (𝐻 ⃗ +𝑀 ⃗⃗ ) 𝐵

⋯ (1)

⃗⃗ depends linearly on 𝐻 ⃗ for linear materials. So we can write 𝑀 ⃗⃗ = 𝜒𝑚 𝐻 ⃗ 𝑀

𝜒𝑚 ⇒ Magnetic Susceptibility of the medium

 Substituting in equation (1) ⃗ = 𝜇0 (𝐻 ⃗ + 𝜒𝑚 𝐻 ⃗ ) = 𝜇0 (1 + 𝜒𝑚 )𝐻 ⃗ = 𝜇𝐻 ⃗ = 𝜇0 𝜇𝑟 𝐻 ⃗ 𝐵

𝜇𝑟 = 1 + 𝜒𝑚 =

𝜇 𝜇0

𝜇𝑟 ⇒ Relative Permeability of the medium

⋯ (1)

Classification of magnetic materials  Magnetic materials may be classified in to three: 1. Diamagnetic 2. Paramagnetic 3. Ferromagnetic

 A material is said to be nonmagnetic if 𝜒𝑚 = 0 (or 𝜇𝑟 = 1); it is magnetic otherwise. Free space with 𝜒𝑚 = 0 (or 𝜇𝑟 ≈ 1) are regard as nonmagnetic.

85

Properties of diamagnetic materials  Diamagnetism occurs in materials where the magnetic fields due to orbital and spin motion of electrons completely cancel.  When placed inside a magnetic field, it gets feebly magnetized in direction opposite to that of the magnetic field and reduces the net field.  The magnetic susceptibility is small.  A diamagnetic material is feebly repelled by a magnet.  A rod of the material suspended in a magnetic field gets slowly set at right angles to the direction of the field. ⃗⃗ has a small negative value.  𝑀  𝜇𝑟 is just less than 1  Behaviour is not affected by temperature.  Does not obey Curies law, χm ∝ 1/ T

Properties of paramagnetic materials          

Paramagnetism occurs in materials where the magnetic fields produced by orbital and spin motion of electrons do not completely cancel. On application of external magnetic field it gets magnetised along the direction of the applied field and increases the net field. When subjected to non-uniform magnetic field it moves from weaker part of the magnetic field to stronger part. A rod of paramagnetic material suspended in a magnetic field slowly gets along the direction of external magnetic field. A paramagnetic material is feebly attracted by a magnet. ⃗⃗ has a small positive value. 𝑀 𝜇𝑟 is just less than 1 Susceptibility has a small positive value. It tends to lose its magnetism if subjected to temperature. Obeys Curies law , 𝜒𝑚 ∝ 1/ 𝑇

Properties of ferromagnetic materials 

Ferromagnetism occurs in materials whose atoms have relatively large permanent magnetic moment.  On application of external magnetic field it gets strongly magnetised along the direction of the applied field and retains its magnetism when removed from the field.  They are strongly attracted by a magnet. 86

 On being subjected to non-uniform magnetic field it moves from weaker part of the magnetic field to stronger part. ⃗⃗ has a large positive value.  𝑀    

𝜇𝑟 is very large. Value of flux density inside the material is very much greater than in vacuum. Susceptibility has a large positive value. Ferromagnetism decreases with temperature. At a critical temperature called Curie temperature, ferromagnetic properties disappear and the material becomes paramagnetic.

Magnetic boundary conditions  Consider the boundary between two isotropic homogeneous linear materials with permeabilities μ1 and μ2. ⃗ may be decomposed in to a  At the boundary, the magnetic flux density 𝐵 normal component Bn and a tangential component Bt.

 Applying Gauss’s law for magnetic field on a small pillbox shaped volume, ⃗ ∙ 𝑑S⃗ = 0 ∮B 𝑆

𝐵1𝑛 ∆𝑆 − 𝐵2𝑛 ∆𝑆 = 0 ⇒ 𝐵1𝑛 = 𝐵2𝑛 𝑖. 𝑒., 𝐵1𝑛 = 𝐵2𝑛

⃗2−𝐵 ⃗ 1 ). 𝑎̂𝑁 = 0 (𝐵

⃗2−𝐵 ⃗ 1 ). 𝑎̂𝑁 = 0 (𝐵

⃗ is continuous across the boundary between the two  Normal component of 𝐵 adjacent media. 87

𝐵1𝑛 = 𝐵2𝑛 𝜇1 𝐻1𝑛 = 𝜇2 𝐻2𝑛 𝐻1𝑛 𝜇2 = 𝐻2𝑛 𝜇1 𝐻2𝑛 =

or

𝜇1 𝐻 𝜇2 1𝑛

⃗ are not continuous across the boundary  The normal components of 𝐻  The normal components becomes continuous only when μ1 = μ2.  In that case there is no boundary between the two materials.

 Now construct a closed contour C across the boundary and apply Ampere’s circuital law for this closed path. ⃗ ∙ 𝑑𝑙 = 𝐼𝑒𝑛𝑐𝑙 ∮𝐻 𝐶

𝐻2𝑡 ∆𝑙 + 𝐻2𝑛

∆ℎ ∆ℎ ∆ℎ ∆ℎ + 𝐻1𝑛 − 𝐻1𝑡 ∆𝑙 − 𝐻1𝑛 − 𝐻2𝑛 = 𝐾𝑢 ∆𝑙 2 2 2 2

𝐻2𝑡 ∆𝑙 − 𝐻1𝑡 ∆𝑙 = 𝐾𝑢 ∆𝑙 𝐻2𝑡 − 𝐻1𝑡 = 𝐾𝑢

⇒

𝐻2𝑡 − 𝐻1𝑡 = 𝐾𝑢

⃗ normal to the plane of the 𝐾𝑢 ⇒ Component of 𝐾 closed path

⃗ is discontinuous across the boundary between the  Tangential component of 𝐻 two adjacent media. 88

 The above equation may be written in vector notation as ⃗2−𝐻 ⃗ 1 ) × 𝑎̂𝑁 = 𝐾 ⃗ (𝐻 ⃗ =0  If both the media across the boundary are not conductors, then 𝐾 𝐻2𝑡 − 𝐻1𝑡 = 0

𝐻2𝑡 = 𝐻1𝑡

⃗ is continuous across the boundary between the two  Tangential component of 𝐻 ⃗ =0 adjacent media if 𝐾

𝐵2𝑡 𝐵1𝑡 = 𝜇2 𝜇1 𝐵2𝑡 =

𝜇2 𝐵 𝜇1 1𝑡

⃗ is discontinuous across the boundary between the  Tangential component of B ⃗ =0 two adjacent media if 𝐾 𝐻2𝑡 = 𝐻1𝑡 𝐵1𝑡 𝐵2𝑡 = ⋯ (1) 𝜇1 𝜇2 𝐵1𝑛 = 𝐵2𝑛 ⋯ (2)

Dividing eq. (1) by eq. (2) gives 𝐵1𝑡 𝐵2𝑡 = ⋯ (3) 𝜇1 𝐵1𝑛 𝜇2 𝐵2𝑛 𝐵1 sin 𝜃1 𝐵2 sin 𝜃2 = 𝜇1 𝐵1 cos 𝜃1 𝜇2 𝐵2 cos 𝜃2

⇒

tan 𝜃1 𝜇1 = tan 𝜃2 𝜇2

89

tan 𝜃1 tan 𝜃2 = 𝜇1 𝜇2

Inductance  A closed loop carrying a current I produces a magnetic field which causes a flux 𝛹 according to ⃗ . 𝑑𝑆 𝜓 = ∫B 𝑆

 If the loop has N turns the total flux linkage λ is defined as 𝜆 = 𝑁𝜓  According to Biot-Savart’s law B is proportional to the current I and hence flux 𝜓 is proportional to I.  Flux linkage λ is thus proportional to current I. 𝜆∝𝐼 𝜆 = 𝐿𝐼  Here L is a proportionality constant called the inductance of the circuit .  Inductance L of an inductor is defined as the ratio of magnetic flux linkage λ to the current I through the inductor. Its unit is Henry. 𝜆 𝑁𝜓 𝜆 = 𝐿𝐼 ⟹ 𝐿= = 𝐼 𝐼  The inductance defined thus is the self-inductance of the coil since the flux linkages are produced by the inductor itself.  The magnetic energy stored in an inductor is given by 1 2𝑊𝑚 𝑊𝑚 = 𝐿𝐼 2 ⟹ 𝐿= 2 2 𝐼  So inductance may be considered as a measure of the magnetic energy that can be stored in an inductor.

Self-inductance of a solenoid  Find the self-inductance per unit length of an infinitely long solenoid.  The magnetic field inside an infinitely long solenoid is 𝑁𝐼 𝐵=𝜇 𝑙  If S is the cross sectional area of the solenoid, then the total flux through this solenoid is 𝑁𝐼 ⃗ . 𝑑𝑆 = 𝐵 ∫𝑑𝑆 = 𝐵𝑆 = 𝜇 𝜓 = ∫B 𝑆 𝑙 𝑆 𝑆 2 𝑁𝜓 𝑁 𝐼 𝜇𝑁 2 𝑆 𝐿= =𝜇 𝑆= 𝐼 𝑙𝐼 𝑙  Hence the inductance per unit length is 𝐿 = 𝜇𝑁 2 𝑆 90

Magnetic energy ⃗ as shown  Consider a differential volume Δv in a magnetic field of flux density 𝐵 below.  Let the top and bottom faces be covered with thin metal foils that carry a current ΔI

 The inductance of this differential volume is ∆𝜓 𝐵∆𝑥∆𝑧 𝜇𝐻∆𝑥∆𝑧 ∆𝐿 = = = ⋯ (1) ∆𝐼 ∆𝐼 ∆𝐼  The energy stored in this elemental inductance is 1 ∆𝑊𝑚 = ∆𝐿∆𝐼 2 ⋯ (2) 2  Substituting (1) in (2) 1 ∆𝑥∆𝑧 2 1 ∆𝑊𝑚 = 𝜇𝐻 ∆𝐼 = 𝜇𝐻∆𝑥∆𝑧∆𝐼 2 ∆𝐼 2  But Δ𝐼 = 𝐻Δ𝑙 = 𝐻Δ𝑦 1 1 ∆𝑊𝑚 = 𝜇𝐻2 ∆𝑥Δ𝑦∆𝑧 = 𝜇𝐻2 ∆v 2 2  Magnetic energy density 𝑤𝑚 is defined as ∆𝑊𝑚 1 𝜇𝐻2 1 𝑤𝑚 = lim = lim ∆v = 𝜇𝐻2 ∆v→∞ ∆v ∆v→∞ 2 ∆v 2 1 1 1 𝐵 𝐵2 ⃗ ⃗ 𝑤𝑚 = 𝜇𝐻𝐻 = 𝐵 ∙ 𝐻 or 𝑤𝑚 = 𝐵 ∙ = 2 2 2 𝜇 2𝜇 1 𝑤𝑚 = 𝜇𝐻2 2

1 ⃗ ∙𝐻 ⃗ 𝑤𝑚 = 𝐵 2

𝑤𝑚 =

𝐵2 2𝜇

 The total energy in a magnostatic field is 𝑊𝑚 = ∫V 𝑤𝑚 𝑑v 1 𝑊𝑚 = ∫ 𝜇𝐻2 𝑑v 𝑉2

1 ⃗ ∙𝐻 ⃗ 𝑑v 𝑊𝑚 = ∫ 𝐵 2 𝑉 91

𝐵2 𝑊𝑚 = ∫ 𝑑v 𝑉 2𝜇

Chapter 6: Maxwell's Equations Faraday’s law  Whenever the magnetic flux linking with a closed circuit varies with time an emf is induced in that circuit and the induced emf is equal to the time rate of change of the magnetic flux linkage by the circuit. 𝑉𝑒𝑚𝑓 = −𝑁

𝑑𝜓 𝑑𝑡

N ⇒ Number of turns in the circuit 𝜓 ⇒ Flux through each turn  The negative sign shows that the induced emf acts in such a way as to oppose the flux producing it. This is called Lenz’s law.  It shows that the direction of current in the circuit is such that the induced magnetic field produced by the induced current opposes the original magnetic field.  For a single turn 𝑉𝑒𝑚𝑓 = −

𝑑𝜓 𝑑 ⃗ ∙ 𝑑𝑆 = − ∫𝐵 𝑑𝑡 𝑑𝑡 𝑆

⃗ . 𝑑𝑙 = − 𝑉𝑒𝑚𝑓 = ∮E 𝐿

since

⃗ ∙ 𝑑𝑆 𝜓 = ∫𝐵 𝑆

𝑑 ⃗ ∙ 𝑑𝑆 ∫𝐵 𝑑𝑡 𝑆

 The variation of flux with time may be produced in the following ways:  A stationary loop area in time-varying magnetic field.  A time-varying loop area in a static magnetic field.  A time-varying loop area in a time varying magnetic field

92

Stationary loop in time-varying magnetic field (Transformer emf) 𝑑 ⃗ ∙ 𝑑𝑆 ∫𝐵 𝑑𝑡 𝑆

⃗ . ⃗⃗⃗ 𝑉𝑒𝑚𝑓 = ∮E 𝑑𝑙 = − 𝐿

 In the case of stationary loop and time-varying magnetic field, we can modify the above equation. ⃗ 𝜕𝐵 ⃗ . 𝑑𝑙 = − ∫ 𝑉𝑒𝑚𝑓 = ∮E ∙ 𝑑𝑆 𝜕𝑡 𝐿 𝑆  This equation is called transformer equation since the induced emf is due to transformer action.  Applying Stokes’s theorem ⃗ 𝜕𝐵 ⃗ ⃗ ∫(∇ × 𝐸 ) ∙ 𝑑𝑆 = − ∫ ∙ 𝑑𝑆 𝑆 𝑆 𝜕𝑡 ⃗ × 𝐸⃗ = − ∇

⃗ 𝜕𝐵 𝜕𝑡

 This is one of Maxwell’s equations for time-varying fields.

Moving loop in a static magnetic field (Motional emf)  When a conducting loop is moving in a static magnetic field, an emf is induced in the loop. ⃗ is  The force on a charge moving with uniform velocity 𝑢 ⃗ in a magnetic field B ⃗ 𝐹𝑚 = 𝑄𝑢 ⃗ ×B Motional electric field ⃗⃗⃗⃗𝐸𝑚 is ⃗⃗⃗⃗𝐸𝑚 = 𝐹𝑚 = ⃗⃗⃗⃗𝑢 × ⃗⃗⃗⃗B 𝑄  The emf induced in the moving loop is ∎

⃗ . 𝑑𝑙 = ∮(𝑢 ⃗ ) . 𝑑𝑙 𝑉𝑒𝑚𝑓 = ∮E ⃗ ×B 𝐿

𝐿

This type of emf is called motional emf.  Applying Stokes's theorem ⃗ × 𝐸⃗𝑚 ) ∙ 𝑑𝑆 = ∫ (∇ ⃗ × (𝑢 ⃗ )) ∙ 𝑑𝑆 ∫(∇ ⃗ ×B 𝑆

𝑆

⃗ × 𝐸⃗𝑚 = ∇ ⃗ × (𝑢 ⃗) ∇ ⃗ ×B 93

Moving loop in a time-varying magnetic field  In this case both the transformer and motional emf are present. ⃗ 𝜕𝐵 ⃗ ). 𝑑𝑙 ∙ 𝑑𝑆 + ∮(𝑢 ⃗ ×B 𝑆 𝜕𝑡 𝐿

⃗ . 𝑑𝑙 = − ∫ 𝑉𝑒𝑚𝑓 = ∮E 𝐿

 Applying Stokes’s theorem ⃗ 𝜕𝐵 ⃗ × 𝐸⃗𝑚 ) ∙ 𝑑𝑆 = − ∫ ⃗ × (𝑢 ⃗ )) ∙ 𝑑𝑆 ∫(∇ ∙ 𝑑𝑆 + ∫ (∇ ⃗ ×B 𝜕𝑡 𝑆 𝑆 𝑆

⃗ × 𝐸⃗ = − ∇

⃗ 𝜕𝐵 ⃗ × (𝑢 ⃗) +∇ ⃗ ×B 𝜕𝑡

Displacement current  Maxwell’s equation for magnetic fields derived from Ampere’s circuital law is given by ⃗ ×𝐻 ⃗ =𝐽 ∇  For any vector, divergence of the curl is always zero. ⃗ ∙ (∇ ⃗ ×𝐻 ⃗)=∇ ⃗ ∙ 𝐽 = 0 ⋯ (1) ∇  But the continuity equation states that 𝜕𝜌 ⃗⃗⃗ . 𝐽 = − v ⋯ (2) ∇ 𝜕𝑡  Equations 1 and 2 are incompatible for time-varying conditions. Therefore, we have to modify equation 1 to agree with equation 2.  For this we will add an additional term to equation 1 so that it becomes ⃗ ×𝐻 ⃗ = 𝐽 + 𝐽𝑑 ⋯ (3) ∇  Applying the condition that divergence of the curl is always zero to equation 3, ⃗ ∙ (∇ ⃗ ×𝐻 ⃗)=∇ ⃗ ∙ (𝐽 + 𝐽𝑑 ) = 0 ∇ ⃗ ∙ (𝐽 + 𝐽𝑑 ) = ∇ ⃗ ∙𝐽+∇ ⃗ ∙ 𝐽𝑑 = 0 ∇ ⃗ ∙ 𝐽𝑑 = −∇ ⃗⃗⃗ . J = ∇

⃗ 𝜕𝜌v 𝜕 𝜕𝐷 ⃗ .𝐷 ⃗ )=∇ ⃗. = (∇ 𝜕𝑡 𝜕𝑡 𝜕𝑡

⃗ 𝜕𝐷 𝜕𝑡 Equation 3 is now modified as 𝐽𝑑 =

94

⃗ ×𝐻 ⃗ =𝐽+ ∇

⃗ 𝜕𝐷 𝜕𝑡

. . . . . . (4)

 Equation 4 is Maxwell’s equation derived from ampere’s law for a time varying field. ⃗ 𝜕𝐷

 The term is known as displacement current density. 𝜕𝑡  Based on displacement current density, we can define the displacement current as ⃗ 𝜕𝐷 𝐼𝑑 = ∫ 𝐽𝑑 ∙ 𝑑𝑆 = ∫ . 𝑑𝑆 𝜕𝑡  At low frequencies Jd is very small when compared with J  At high frequencies the two terms are comparable.  Without the term Jd electromagnetic wave propagation would be impossible.  Displacement current is a result of time varying electric field.

Maxwell’s Equations in final (Generalized) forms Differential form

Integral form

Remarks

⃗ ∙𝐷 ⃗ = 𝜌𝑉 ∇

⃗ ∙ 𝑑𝑆 = ∫𝜌𝑉 𝑑𝑉 ∮𝐷

⃗ ∙𝐵 ⃗ =0 ∇

⃗ ∙ 𝑑𝑆 = 0 ∮𝐵

𝑆

Nonexistence of magnetic monopole

𝑆

⃗ × 𝐸⃗ = − ∇

⃗ 𝜕𝐵 𝜕𝑡

⃗ ×𝐻 ⃗ =𝐽+ ∇

⃗ 𝜕𝐷 𝜕𝑡

Gauss’ s Law

𝑉

⃗ 𝜕𝐵 . 𝑑𝑆 L 𝑆 𝜕𝑡 ⃗ 𝜕𝐷 ⃗ ∙ 𝑑𝑙 = ∫ (𝐽 + ∮𝐻 ) ∙ 𝑑𝑆 𝜕𝑡 𝐿 𝑆 ∮𝐸⃗ ∙ 𝑑𝑙 = − ∫

Faraday ' s Law Ampere’s Law modified by continuity equation

Time-harmonic fields  A time-harmonic field is one that varies periodically or sinusoidally with time.  In most practical applications EM fields are time-harmonic.  Also, any wave form can be expressed in terms of sinusoids using Fourier transform techniques.  Sinusoids are easily expressed in phasors, which are more convenient to work with.  A phasor z is a complex number that can be written as 𝑧 = 𝑥 + 𝑗𝑦 ⇒ Rectangular form 95

𝑧 = 𝑟∠𝜙

⇒ Polar form

𝑧 = 𝑟𝑒 𝑗𝜙 = 𝑟(cos 𝜙 + sin 𝜙) where 𝑗 = √−1 , x is the real part of z, y is the imaginary part of z, r is the magnitude of z, given by 𝑟 = |𝑧| = √𝑥 2 + 𝑦 2 𝑦 𝑥  Addition and subtraction of phasors are better performed in rectangular form while multiplication and division are better done in polar form. and 𝜙 is the phase of z, given by 𝜙 = tan−1

Let 𝑧1 = 𝑥1 + 𝑗𝑦1

𝑧2 = 𝑥2 + 𝑗𝑦2

(𝑧1 + 𝑧2 ) = (𝑥1 + 𝑥2 ) + 𝑗(𝑦1 + 𝑦2 ) (𝑧1 − 𝑧2 ) = (𝑥1 − 𝑥2 ) + 𝑗(𝑦1 − 𝑦2 ) 𝑧1 𝑧2 = 𝑟1 𝑟2 ∠(𝜙1 + 𝜙1 ) 𝑧1 𝑟1 = ∠(𝜙1 − 𝜙1 ) 𝑧2 𝑟2 √𝑧 = √𝑟∠(𝜙/2)  To introduce the time element let 𝜙 = ωt + 𝜃, where 𝜃 may be a function of time or space coordinates or a constant.  The real (Re) and imaginary (Im) parts of 𝑟𝑒 𝑗𝜙 = 𝑟𝑒 𝑗𝜃 𝑒 𝑗𝜔𝑡 are given by Re(𝑟𝑒 𝑗𝜙 ) = 𝑟cos(𝜔𝑡 + 𝜃) Im(𝑟𝑒 𝑗𝜙 ) = 𝑟sin(𝜔𝑡 + 𝜃)  A sinusoidal current 𝐼(𝑡) = 𝐼0 cos(𝜔𝑡 + 𝜃) equals the real part of 𝐼0 𝑒 𝑗𝜃 𝑒 𝑗𝜔𝑡  The sinusoidal current 𝐼(𝑡) = 𝐼0 sin(𝜔𝑡 + 𝜃) equals the imaginary part of 𝐼0 𝑒 𝑗𝜃 𝑒 𝑗𝜔𝑡  The complex term 𝐼0 𝑒 𝑗𝜃 , which results from dropping the time factor 𝑒 𝑗𝜔𝑡 in I (t) is called the phasor current and is denoted by 𝐼𝑠 . 𝐼𝑠 = 𝐼0 𝑒 𝑗𝜃 = 𝐼0 ∠𝜃 𝐼(𝑡) = 𝐼0 cos(𝜔𝑡 + 𝜃) can be expressed as 𝐼(𝑡) = Re(𝐼𝑆 𝑒 𝑗𝜔𝑡 )  A phasor can be a scalar or a vector.  If a vector ⃗A = (𝑥, 𝑦, 𝑧, 𝑡) is a time-harmonic field, the phasor form of ⃗A, is ⃗A𝑠 = (𝑥, 𝑦, 𝑧, 𝑡) 96

⃗A = 𝑅𝑒(A ⃗ 𝑠 𝑒 𝑗𝜔𝑡 ) ⋯ (1) ⃗ = A0 cos(𝜔𝑡 − 𝛽𝑥)𝑎̂𝑦  As an example let A ⃗A = Re(A0 𝑒 −𝑗𝛽𝑥 𝑎̂𝑦 𝑒 𝑗𝜔𝑡 ) ⋯ (2)  Comparing (1) and (2), the phasor form of 𝐴 is ⃗A𝑠 = A0 𝑒 −𝑗𝛽𝑥 𝑎̂𝑦 [

⃗A = 𝑅𝑒(A ⃗ s 𝑒 𝑗𝜔𝑡 ) ⃗ 𝜕A 𝜕 ⃗ s 𝑒 𝑗𝜔𝑡 ) = 𝑅𝑒(𝑗𝜔A ⃗ s 𝑒 𝑗𝜔𝑡 ) = 𝑅𝑒(A 𝜕𝑡 𝜕𝑡  We can apply the concepts of phasors to time-varying EM fields.  The following field quantities and their derivatives can be expressed in phasor form. 𝐸⃗ (𝑥, 𝑦, 𝑧, 𝑡)

⃗ (𝑥, 𝑦, 𝑧, 𝑡) 𝐻

𝐽(𝑥, 𝑦, 𝑧, 𝑡)

⃗ (𝑥, 𝑦, 𝑧, 𝑡) 𝐷

⃗ (𝑥, 𝑦, 𝑧, 𝑡) 𝐵

𝜌𝑉 (𝑥, 𝑦, 𝑧, 𝑡)

 Based on these quantities we can rewrite the Maxwell’s equations for time-harmonic EM fields.

Maxwell’s Equations for time-harmonic fields Differential (Point) form

Integral form

⃗ ∙𝐷 ⃗ 𝑠 = 𝜌v𝑠 ∇

⃗ 𝑠 ∙ 𝑑𝑆 = ∫ 𝜌v𝑠 𝑑𝑉 ∮𝐷

⃗ ∙𝐵 ⃗𝑠 = 0 ∇

⃗ 𝑠 ∙ 𝑑𝑆 = 0 ∮𝐵

⃗ × 𝐸⃗𝑠 = −𝑗𝜔𝐵 ⃗𝑠 ∇

⃗ 𝑠 ∙ 𝑑𝑆 ∮ 𝐸⃗𝑠 ∙ 𝑑𝑙 = −𝑗𝜔 ∫ 𝐵

⃗ ×𝐻 ⃗ 𝑠 = 𝐽𝑠 + 𝑗𝜔𝐷 ⃗𝑠 ∇

⃗ 𝑠 ∙ 𝑑𝑙 = ∫(𝐽𝑠 + 𝑗𝜔𝐷 ⃗ 𝑠 ) ∙ 𝑑𝑆 ∮𝐻

97

Chapter 7: Electromagnetic Wave Propagation Maxwell’s Equations for lossless or nonconducting medium In lossless medium, current density J and charge density ρ are zero (𝜎 = 0), Maxwell’s equations are simplified as below. Differential form

Integral form

⃗ ∙ 𝑑𝑆 = 0 ∮𝐷

⃗ ∙𝐷 ⃗ =0 ∇

Equation

A

𝑆

⃗ ∙ 𝑑𝑆 = 0 ∮𝐵

⃗ ∙𝐵 ⃗ =0 ∇

B

𝑆

⃗ × 𝐸⃗ = − ∇ ⃗ ×𝐻 ⃗ = ∇

⃗ 𝜕𝐵 𝜕𝑡

⃗ 𝑑𝐵 ∙ 𝑑𝑆 𝑑𝑡 𝑆

∮𝐸⃗ ∙ 𝑑𝑙 = − ∫ L

⃗ 𝜕𝐷 𝜕𝑡

⃗ 𝜕𝐷 ∙ 𝑑𝑆 𝑆 𝜕𝑡

⃗ ∙ 𝑑𝑙 = ∫ ∮𝐻 𝐿

C D

 Differentiating equation (D) with respect to t ⃗ 𝜕 𝜕 𝜕𝐷 ⃗ ×𝐻 ⃗ ) = ( ) ⋯ (1) (∇ 𝜕𝑡 𝜕𝑡 𝜕𝑡  The order of differentiation on the LHS may be changed as the curl operation itself is a differentiation ⃗⃗ ⃗ ∂H ∂2 E = ε 2 ⋯ (2) ∂t ∂t  Taking the curl of equation (C) ⃗ ∂B ⃗∇ × ∇ ⃗ × ⃗E = −∇ ⃗ × ∂t ⃗⃗ ∂H ⃗ ×∇ ⃗ × ⃗E = −𝜇 (∇ ⃗ × ) ⋯ (3) ∇ ∂t  Substituting equation (2) in equation (3) ∂2 ⃗E ⃗∇ × ∇ ⃗ × ⃗E = −𝜇 (𝜀 ) ∂t 2 ⃗ × ∇

⃗ ×∇ ⃗ ×E ⃗ = −𝜇𝜀 ∇

∂2 ⃗E ⋯ (4) ∂t 2 98


⃗ ∙ 𝐶 )B ⃗ ∙B ⃗ × 𝐶 = (A ⃗ − (A ⃗ )𝐶 But ⃗A × B ⃗ × 𝐸⃗ = (∇ ⃗ ∙ 𝐸⃗ )∇ ⃗ − (∇ ⃗ ∙∇ ⃗ )𝐸⃗ So, ⃗∇ × ∇

⇒

⃗ ×∇ ⃗ × 𝐸⃗ = (∇ ⃗ ∙ 𝐸⃗ )∇ ⃗ −∇ ⃗ 2 𝐸⃗ ∇

⃗ =0 But ⃗∇ ⋅ 𝐷

⇒

⃗ ⋅ 𝐸⃗ = 0 ∇

Then,

⇒

⃗ ∙ 𝜀𝐸⃗ = 0 ∇

⃗ ×∇ ⃗ × 𝐸⃗ = −∇2 𝐸⃗ ∇

 Substituting in equation (4) 𝜕 2 𝐸⃗ 2⃗ ∇ 𝐸 = 𝜇𝜀 2 ⋯ (5) 𝜕𝑡

∂2 𝐸⃗ ∇ 𝐸 = 𝜇𝜀 2 ∂t 2⃗

 Equation (5) represents the wave equation in terms of E.  Differentiating equation (C) with respect to t ⃗ 𝜕 𝜕 𝜕𝐵 ⃗ ⃗ (∇ × 𝐸 ) = (− ) ⋯ (6) 𝜕𝑡 𝜕𝑡 𝜕𝑡  The order of differentiation on the LHS may be changed as the curl operation itself is a differentiation ⃗ 𝜕𝐸⃗ 𝜕2𝐻 ⃗∇ × = −𝜇 2 ⋯ (7) 𝜕𝑡 𝜕𝑡  Taking the curl of equation (D) ⃗ 𝜕𝐷 ⃗∇ × ∇ ⃗ ×𝐻 ⃗ =∇ ⃗⃗⃗ × 𝜕𝑡 𝜕𝐸⃗ ⃗ ×∇ ⃗ ×𝐻 ⃗ = 𝜀 (∇ ⃗⃗⃗ × ) ⋯ (8) ∇ 𝜕𝑡  Substituting equation (7) in equation (8) ⃗ 𝜕2𝐻 ⃗∇ × ∇ ⃗ ×𝐻 ⃗ = 𝜀 (−𝜇 ) 𝜕𝑡 2 ⃗ 𝜕2𝐻 ⃗∇ × ∇ ⃗ ×𝐻 ⃗ = −𝜇𝜀 ⋯ (9) 𝜕𝑡 2 ⃗ ×𝐻 ⃗ = (∇ ⃗ ∙𝐻 ⃗ )∇ ⃗ −∇ ⃗ 2𝐻 ⃗ But ⃗∇ × ∇ ⃗ =0 ⃗ ∙ 𝜇𝐻 ⃗ =0 But, ⃗∇ ⋅ B ⇒ ∇ ⇒ ⃗ ×∇ ⃗ ×𝐻 ⃗ = −∇ ⃗ 2𝐻 ⃗ Then, ∇

⃗ ⋅ ⃗H ⃗ =0 ∇

 Substituting in equation (9) ⃗ = 𝜇𝜀 ∇2 𝐻

⃗ ∂2 𝐻 ∇ 𝐻 = 𝜇𝜀 2 ∂t 2⃗

2⃗

𝜕 𝐻 ⋯ (10) 𝜕𝑡 2 99


 Equation (10) represents the wave equation in terms of H.  Equations (5) and (10) are known as the time -dependent Helmholtz's equations. ⃗ and 𝐵 ⃗ may be obtained by multiplying the wave  The wave equations for 𝐷 ⃗ by 𝜇 equation for 𝐸⃗ by 𝜀 and equation for 𝐻 𝜕 2 𝐸⃗ 2⃗ ∇ 𝐸 = 𝜇𝜀 2 𝜕𝑡 𝜕 2 (𝜀𝐸⃗ ) 2 ⃗ ∇ (𝜀𝐸 ) = 𝜇𝜀 𝜕𝑡 2

⃗ ∂2 𝐷 ∇ 𝐷 = 𝜇𝜀 2 ∂t 2⃗

 Similarly, ⃗ 𝜕2𝐻 ∇ 𝐻 = 𝜇𝜀 2 𝜕𝑡 ⃗) 𝜕 2 (𝜇𝐻 2 ⃗ ∇ (𝜇𝐻) = 𝜇𝜀 𝜕𝑡 2 2⃗

⃗ ∂2 𝐵 ∇ 𝐵 = 𝜇𝜀 2 ∂t 2⃗

 In rectangular coordinate system, the wave equations assumes the form of scalar wave equations in terms of its components, For 𝐸⃗ 𝜕 2 𝐸⃗𝑥 ∇ 𝐸𝑥 = 𝜇𝜀 𝜕𝑡 2 2⃗

,

𝜕 2 𝐸⃗𝑦 ∇ 𝐸𝑦 = 𝜇𝜀 𝜕𝑡 2 2⃗

,

𝜕 2 𝐸⃗𝑧 ∇ 𝐸𝑧 = 𝜇𝜀 𝜕𝑡 2 2⃗

⃗ ,𝐻 ⃗ ,𝐵 ⃗ we can obtain similar equations, the electromagnetic field do not For 𝐷 decay as they propagate in a lossless medium. For this reason, a nonconducting medium (perfect dielectric) is called a lossless medium.

Wave equations for lossless medium: sinusoidal time variations  If the electric field intensity is varying harmonically with time 𝜕𝐸⃗ 𝐸⃗ = 𝐸⃗s 𝑒 𝑗𝜔𝑡 ⇒ = 𝑗𝜔𝐸⃗s 𝑒 𝑗𝜔𝑡 = 𝑗𝜔𝐸⃗ 𝜕𝑡 2⃗ 𝜕 𝐸 = 𝑗 2 𝜔2 𝐸⃗s 𝑒 𝑗𝜔𝑡 = −𝜔2 𝐸⃗ 𝜕𝑡 2  Using these in wave equation for 100


∂2 𝐸⃗ ∇ 𝐸 = 𝜇𝜀 2 = −𝜔2 𝜇𝜀𝐸⃗ ∂t ∇2 𝐸⃗ = −𝜔2 𝜇𝜀𝐸⃗ 2⃗

∇2 𝐸⃗ = −𝜔2 𝜇𝜀𝐸⃗

 Similarly we may obtain the following wave equations for the other harmonically varying fields ⃗ = −𝜔2 𝜇𝜀𝐷 ⃗ ∇2 𝐷 ⃗ = −𝜔2 𝜇𝜀𝐻 ⃗ ∇2 𝐻 ⃗ = −𝜔2 𝜇𝜀𝐵 ⃗ ∇2 𝐵 Homogeneous vector wave equations in complex time harmonic form for free pace

Wave equations for free space Free space (or vacuum) is a special case of lossless medium in which 𝜇 = 𝜇0 𝑎𝑛𝑑 𝜀 = 𝜀0 . General

Sinusoidal Time Variations

∂2 𝐸⃗ ∇ 𝐸 = 𝜇0 𝜀0 2 ∂t

∇2 𝐸⃗ = −𝜔2 𝜇0 𝜀0 𝐸⃗

⃗ ∂2 𝐷 ∇ 𝐷 = 𝜇0 𝜀0 2 ∂t

⃗ = −𝜔2 𝜇0 𝜀0 𝐷 ⃗ ∇2 𝐷

⃗ ∂2 𝐻 ∇ 𝐻 = 𝜇0 𝜀0 2 ∂t

⃗ = −𝜔2 𝜇0 𝜀0 𝐻 ⃗ ∇2 𝐻

⃗ ∂2 𝐵 ∇ 𝐵 = 𝜇0 𝜀0 2 ∂t

⃗ = −𝜔2 𝜇0 𝜀0 𝐵 ⃗ ∇2 𝐵

2⃗

2⃗

2⃗

2⃗

Solution of wave equations  Standard partial difference equation for wave motion frequently encountered in engineering has the form 1 ∂2 𝑋 ∇ 𝑋= 2 2 𝑣 𝜕𝑡 2

𝑣 ⇒ velocity of the wave

 Comparing with the EM wave equation 101


𝜕 2 𝐸⃗ 1 𝜕 2 𝐸⃗ ∇ 𝐸 = 𝜇0 𝜀0 2 = 2 2 𝜕𝑡 (1/√𝜇0 𝜀0 ) 𝜕𝑡 1 𝑣 ⇒ Velocity of the wave = √𝜇0 𝜀0 2⃗

Putting 𝜇0 = 4𝜋 × 10−7 𝐹/𝑚 𝑣=

1 √𝜇0 𝜀0

= 3 × 108

and 𝜀0 = 8.854 × 10−12 𝐻/𝑚

𝑚 =𝑐 𝑠

where c denotes the velocity of light in free space

Wave equations for conducting medium  In a conducting medium with conductivity σ and charge density ρ, Maxwell’s equations are as given below. Differential form

Integral form

⃗ ∙ 𝑑𝑆 = ∫𝜌𝑉 𝑑𝑉 ∮𝐷

⃗ ∙𝐷 ⃗ =𝜌 ∇

𝑆

A

𝑉

⃗ ∙ 𝑑𝑆 = 0 ∮𝐵

⃗ ∙𝐵 ⃗ =0 ∇

Equation

B

𝑆

⃗ × 𝐸⃗ = − ∇

⃗ 𝜕𝐵 𝜕𝑡

⃗ ×𝐻 ⃗ =𝐽+ ∇

⃗ 𝜕𝐷 𝜕𝑡

⃗ 𝑑𝐵 ∙ 𝑑𝑆 𝑑𝑡 𝑆

∮𝐸⃗ ∙ 𝑑𝑙 = − ∫ L

⃗ ∙ 𝑑𝑙 = ∫ (𝐽 + ∮𝐻 𝐿

𝑆

⃗ 𝜕𝐷 ) ∙ 𝑑𝑆 𝜕𝑡

C D

 From equation (D) ⃗ 𝜕𝐷 𝜕 ⃗ ×𝐻 ⃗ =𝐽+ ⃗ ×𝐻 ⃗ = 𝜎𝐸⃗ + (𝜀𝐸⃗ ) ∇ ⇒ ∇ 𝜕𝑡 𝜕𝑡 𝜕𝐸⃗ ⃗∇ × 𝐻 ⃗ = 𝜎𝐸⃗ + 𝜀 ⋯ (1) 𝜕𝑡 𝜕 𝜕𝐸⃗ ∂2 𝐸⃗ ⃗ ⃗ +𝜀 2 (∇ × 𝐻) = 𝜎 𝜕𝑡 𝜕𝑡 ∂t ⃗ 𝜕𝐻 𝜕𝐸⃗ ∂2 𝐸⃗ ⃗∇ × =𝜎 + 𝜀 2 ⋯ (2) 𝜕𝑡 𝜕𝑡 ∂t 102


 From equation (C) ⃗ 𝜕𝐵 ⃗ × 𝐸⃗ = − ∇ 𝜕𝑡

⇒

⃗ × 𝐸⃗ = −𝜇 ∇

⃗ 𝜕𝐻 𝜕𝑡

 Taking curl of this equation ⃗ 𝜕𝐻 ⃗ ×∇ ⃗ × 𝐸⃗ = −𝜇 (∇ ⃗ × ∇ ) ⋯ (3) 𝜕𝑡  Putting equation (2) in (3) 𝜕𝐸⃗ ∂2 𝐸⃗ ⃗ ×∇ ⃗ × 𝐸⃗ = −𝜇 (𝜎 ∇ + 𝜀 2 ) ⋯ (4) 𝜕𝑡 ∂t ⃗ ×∇ ⃗ × 𝐸⃗ = ∇ ⃗ (∇ ⃗ ∙ 𝐸⃗ ) − ∇ ⃗ 2 𝐸⃗ ⋯ (5) But ∇  Using equation (5) in (4) 𝜕𝐸⃗ ∂2 ⃗E + 𝜇𝜀 2 ⋯ (6) 𝜕𝑡 ∂t  But from equation (A) we have 𝜌 ⃗ .𝐷 ⃗ =𝜌 ⃗∇. 𝐸⃗ = ∇ 𝜀  Putting in equation (6) 𝜌 𝜕𝐸⃗ 𝜕 2 𝐸⃗ 2⃗ ⃗ ∇ 𝐸 = ∇ ( ) + 𝜇𝜎 + 𝜇𝜀 2 ⋯ (7) 𝜀 𝜕𝑡 𝜕𝑡  There is no charge within a conductor, although it may be there on the surface, the charge density 𝜌=0. Therefore, we can rewrite equation (7) as below. 2⃗

⃗ (∇ ⃗ ∙ 𝐸⃗ ) + 𝜇𝜎 ∇ 𝐸=∇

𝜕𝐸⃗ 𝜕 2 𝐸⃗ ∇ 𝐸 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑡 𝜕𝑡  This is the wave equation for conducting medium in terms of E  From equation (D) ⃗ 𝜕𝐷 𝜕 ⃗ ×𝐻 ⃗ =𝐽+ ⃗ ×𝐻 ⃗ = 𝜎𝐸⃗ + (𝜀𝐸⃗ ) ∇ ⇒ ∇ 𝜕𝑡 𝜕𝑡 𝜕𝐸⃗ ⃗∇ × 𝐻 ⃗ = 𝜎𝐸⃗ + 𝜀 ⋯ (1) 𝜕𝑡 2⃗

 Taking curl of this equation 𝜕𝐸⃗ ) ⋯ (2) 𝜕𝑡 ⃗ ×𝐻 ⃗ =∇ ⃗ (∇ ⃗ ∙𝐻 ⃗ )−∇ ⃗ 2𝐻 ⃗ ⋯ (3) But ⃗∇ × ∇ ⃗ ×∇ ⃗ ×𝐻 ⃗ = 𝜎(∇ ⃗ × 𝐸⃗ ) + 𝜀 (∇ ⃗ × ∇

103


 Using equation (3) in (2) ⃗ (∇ ⃗ ∙𝐻 ⃗ )−∇ ⃗ 2𝐻 ⃗ = 𝜎(∇ ⃗ × 𝐸⃗ ) + 𝜀 (∇ ⃗ × ∇  From equation (C) ⃗ 𝜕𝐵 ⃗ × 𝐸⃗ = − ∇ ⋯ (5) 𝜕𝑡 ⃗ 𝜕 ∂2 B ⃗ ⃗ (∇ × 𝐸 ) = − 2 𝜕𝑡 𝜕𝑡

⟹

⃗⃗⃗ × ∇

𝜕𝐸⃗ ) ⋯ (4) 𝜕𝑡

𝜕𝐸⃗ ∂2 ⃗B = − 2 ⋯ (6) 𝜕𝑡 𝜕𝑡

 Putting equation (5) and (6) in (4) ⃗ (∇ ⃗ ∙𝐻 ⃗ ) − ∇2 𝐻 ⃗ = 𝜎 (− ∇

⃗ ⃗ 𝜕𝐵 ∂2 B ) + 𝜀 ( 2 ) ⋯ (7) 𝜕𝑡 𝜕𝑡

⃗ ⃗ 𝜕𝐻 ∂2 ⃗H − 𝜇𝜀 2 ⋯ (8) 𝜕𝑡 𝜕𝑡 ⃗ ⃗ 𝜕𝐻 ∂2 ⃗H 2⃗ ⃗ ⃗ ⃗ ∇ 𝐻 = ∇(∇ ∙ 𝐻) + 𝜇𝜎 + 𝜇𝜀 2 ⋯ (9) 𝜕𝑡 𝜕𝑡 ⃗ .𝐵 ⃗ =𝜌 ⃗ .𝐻 ⃗ =0 But ∇ ∇ ⃗ (∇ ⃗ ∙𝐻 ⃗ ) − ∇2 𝐻 ⃗ = −𝜇𝜎 ∇

⃗ ⃗ 𝜕𝐻 𝜕2𝐻 ∇ 𝐻 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑡 𝜕𝑡 2⃗

 This is the wave equation for conducting medium in terms of H

Wave equations for lossy or conducting medium 𝜕𝐸⃗ 𝜕2𝐸 ∇ 𝐸 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑡 𝜕𝑡 2⃗

⃗ ⃗ 𝜕𝐷 𝜕2𝐷 ∇ 𝐷 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑡 𝜕𝑡 2⃗

⃗ ⃗ 𝜕𝐻 𝜕2𝐻 ∇ 𝐻 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑡 𝜕𝑡 2⃗

⃗ ⃗ 𝜕𝐵 𝜕2𝐵 ∇ 𝐵 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑡 𝜕𝑡 2⃗

104


The presence of the first-order term in a second- order differential equation indicates that the electromagnetic fields decay (lose energy) as they propagate through the medium. For this reason, a conducting medium is called a lossy medium.

Wave equations for lossy or conducting medium: time- harmonic forms ∇2 𝐸⃗ = 𝑗𝜔𝜇𝜎𝐸⃗ − 𝜔2 𝜇𝜀𝐸⃗ ⃗ = 𝑗𝜔𝜇𝜎𝐷 ⃗ − 𝜔2 𝜇𝜀𝐷 ⃗ ∇2 𝐷 ⃗ = 𝑗𝜔𝜇𝜎𝐻 ⃗ − 𝜔2 𝜇𝜀𝐻 ⃗ ∇2 𝐻 ⃗ = 𝑗𝜔𝜇𝜎𝐵 ⃗ − 𝜔2 𝜇𝜀𝐵 ⃗ ∇2 𝐵

Uniform plane wave  An electromagnetic wave originates from a point in free space, spreads out uniformly in all directions, and it forms a spherical wave front.  An observer at a large distance from the source is able to observe only a small part of the wave and the wave appears to him as a plane wave. ⃗ are perpendicular to  For such a wave, the electric field 𝐸⃗ and the magnetic field 𝐻 each other and to the direction of propagation. ⃗ lie in a plane and have the same  A uniform plane wave is one in which 𝐸⃗ and 𝐻 value everywhere in that plane at any fixed instant.  For a uniform plane wave travelling in the z direction, the space variations of 𝐸⃗ ⃗ are zero over a z=constant plane. and 𝐻  This implies the fields have neither x nor y dependence. 𝜕 𝜕 = =0 𝜕𝑥 𝜕𝑦 ⃗ are both perpendicular to the  Plane wave is transverse in nature, that is 𝐸⃗ and 𝐻 direction of propagation.  So they are called transverse electromagnetic waves (TEM waves)

105


 Consider a uniform plane wave propagating in z direction.  It will have Ex and Ey components but no Ez component. Ez = 0.  There is no variation of the field components along x and y direction 𝜕𝐸⃗ 𝜕𝐸⃗ = =0 𝜕𝑥 𝜕𝑦  Wave equations for free space is given by ∂2 𝐸⃗ 2⃗ ∇ 𝐸 = 𝜇0 𝜀0 2 ∂t 2 2 𝜕 𝜕 𝜕2 𝜕2 𝜕2 𝜕2 ( 2 + 2 + 2 ) 𝐸𝑥 𝑎̂𝑥 + ( 2 + 2 + 2 ) 𝐸𝑦 𝑎̂𝑦 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕2 𝜕2 𝜕2 ∂2 + ( 2 + 2 + 2 ) 𝐸𝑧 𝑎̂𝑧 = 𝜇0 𝜀0 2 (𝐸𝑥 𝑎̂𝑥 + 𝐸𝑦 𝑎̂𝑦 + 𝐸𝑧 𝑎̂𝑧 ) ⋯ (1) 𝜕𝑥 𝜕𝑦 𝜕𝑧 𝜕𝑡 𝜕𝐸𝑥 𝜕𝐸𝑦 𝜕 2 𝐸𝑥 𝜕 2 𝐸𝑦 But 𝐸𝑧 = 0 , = =0 = =0 𝜕𝑥 𝜕𝑦 𝜕𝑥 2 𝜕𝑦 2 Putting in equation (1) 𝜕 2 𝐸𝑦 𝜕 2 𝐸𝑥 ∂2 𝑎̂ + 𝑎̂ = 𝜇0 𝜀0 2 (𝐸𝑥 𝑎̂𝑥 + 𝐸𝑦 𝑎̂𝑦 ) 𝜕𝑧 2 𝑥 𝜕𝑧 2 𝑦 𝜕𝑡 Hence 𝜕 2 𝐸𝑦 ∂2 𝐸𝑦 𝜕 2 𝐸𝑥 ∂2 𝐸𝑥 = 𝜇0 𝜀0 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2 𝜕𝑧 2 𝜕𝑡 2 ⃗ we can obtain Similarly for 𝐻 𝜕 2 𝐻𝑥 ∂2 𝐻𝑥 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2

𝜕 2 𝐻𝑦 ∂2 𝐻𝑦 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2

Wave equations for uniform plane wave traveling in z direction in free space – sinusoidal time variations 106


𝜕 2 𝐸𝑥 ∂2 𝐸𝑥 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2 𝜕 2 𝐸𝑦 ∂2 𝐸𝑦 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2 𝜕 2 𝐻𝑥 ∂2 𝐻𝑥 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2 𝜕 2 𝐻𝑦 ∂2 𝐻𝑦 = 𝜇0 𝜀0 𝜕𝑧 2 𝜕𝑡 2

⇒

𝜕 2 𝐸𝑥 = −𝜔2 𝜇0 𝜀0 𝐸𝑥 2 𝜕𝑧

⇒

𝜕 2 𝐸𝑦 = −𝜔2 𝜇0 𝜀0 𝐸𝑦 2 𝜕𝑧

⇒

𝜕 2 𝐻𝑥 = −𝜔2 𝜇0 𝜀0 𝐻𝑥 2 𝜕𝑧

⇒

𝜕 2 𝐻𝑥 = −𝜔2 𝜇0 𝜀0 𝐻𝑦 2 𝜕𝑧

Wave equations for uniform plane wave traveling in z direction in conducting medium – sinusoidal time variations 𝜕 2 𝐸𝑥 𝜕𝐸𝑥 ∂2 𝐸𝑥 = 𝜇𝜎 + 𝜇𝜀 2 𝜕𝑧 2 𝜕𝑥 𝜕𝑡 𝜕 2 𝐸𝑦 𝜕𝐸𝑦 ∂2 𝐸𝑦 = 𝜇𝜎 + 𝜇𝜀 𝜕𝑧 2 𝜕𝑥 𝜕𝑡 2 𝜕 2 𝐻𝑥 𝜕𝐻𝑥 ∂2 𝐻𝑥 = 𝜇𝜎 + 𝜇𝜀 𝜕𝑧 2 𝜕𝑥 𝜕𝑡 2 𝜕 2 𝐻𝑦 𝜕𝐻𝑦 ∂2 𝐻𝑦 = 𝜇𝜎 + 𝜇𝜀 𝜕𝑧 2 𝜕𝑥 𝜕𝑡 2

⇒

𝜕 2 𝐸𝑥 = 𝑗𝜔𝜇𝜎𝐸𝑥 − 𝜔2 𝜇𝜀𝐸𝑥 2 𝜕𝑧

⇒

𝜕 2 𝐸𝑦 = 𝑗𝜔𝜇𝜎𝐸𝑦 − 𝜔2 𝜇𝜀𝐸𝑦 𝜕𝑧 2

⇒

𝜕 2 𝐻𝑥 = 𝑗𝜔𝜇𝜎𝐻𝑥 − 𝜔2 𝜇𝜀𝐻𝑥 2 𝜕𝑧

⇒

𝜕 2 𝐻𝑦 = 𝑗𝜔𝜇𝜎𝐻𝑦 − 𝜔2 𝜇𝜀𝐻𝑦 𝜕𝑧 2

Uniform plane wave traveling in z direction in nonconducting mediumsinusoidal time variations  In this case the wave equations in E are 𝜕 2 𝐸𝑦 𝜕 2 𝐸𝑦 2 = −𝜔 𝜇𝜀𝐸𝑦 ⇒ + 𝜔2 𝜇𝜀𝐸𝑦 = 0 2 2 𝜕𝑧 𝜕𝑧 Hence the solution is 𝐸𝑦 = 𝐸𝑚1 𝑒 −𝑗𝛽𝑧 + 𝐸𝑚2 𝑒 𝑗𝛽𝑧 where 𝐸𝑚1 and 𝐸𝑚2 are arbitrary constants, and 𝛽 = 𝜔√𝜇𝜀  The time domain form, 𝐸𝑦 (𝑧, 𝑡) = 𝐸𝑚1 cos(𝜔𝑡 − 𝛽𝑧) + 𝐸𝑚2 cos(𝜔𝑡 + 𝛽𝑧)

107


Here the first term (𝐸𝑚1 cos(𝜔𝑡 − 𝛽𝑧)) represents the forward traveling wave, and the second term (𝐸𝑚2 cos(𝜔𝑡 + 𝛽𝑧)) is the backward traveling wave 𝜔 1 𝜔 𝛽 = 𝜔√𝜇𝜀 ⇒ = =𝑐 𝑐 = 𝛽 √𝜇𝜀 𝛽 𝐸𝑦 = 𝐸𝑚1 cos(𝜔𝑡 − 𝛽𝑧) + 𝐸𝑚2 cos(𝜔𝑡 + 𝛽𝑧)

or,

𝐸𝑦 = 𝐸𝑚1 sin(𝜔𝑡 − 𝛽𝑧) + 𝐸𝑚2 sin(𝜔𝑡 + 𝛽𝑧)  For the phase of the forward traveling wave to remain constant, ωt − βz = A constant  As t increases z must also increase. This means that the wave travels in the z direction with a constant phase.  Similarly for the backward traveling component ωt + βz = A constant  As t increases z must decrease in order to keep the phase constant. This means that the wave travels in the –z direction with a constant phase.

Phase velocity  Due to the variation of E with both time and space we may plot E as a function of t by keeping z constant or vice versa.  The plots of E(t, z=constant) and E(z, t=constant) are shown in figure.  The wave takes distance λ to repeats itself, and hence λ is called the wavelength.  The wave takes time T to repeat itself, and hence T is called the period of wave.  Since it takes time T for the wave to travel a distance 𝜆 at the speed v, 𝜆 = v𝑇  Since 𝑇 = 1/𝑓 we may write 𝑣 = 𝑓𝜆 1 2𝜋 𝜔 2𝜋 𝜔 = 2𝜋𝑓 , 𝑇 = = , 𝛽= Substituting , 𝛽 = 𝑓 𝜔 𝑣 𝜆  For every wavelength of distance travelled, the wave undergoes a phase change of 2π radians.  Now consider the forward wave 𝐸𝑦 = 𝐸𝑚1 sin(𝜔𝑡 − 𝛽𝑧)  To prove that this wave travels with a velocity v in the z direction, consider a fixed point P on the wave.  Sketch the above wave equation at times t=T/4 and t=T/2 as in figure.  As the wave advances with time, point P moves along the z direction.  Point P is a point of constant phase, 𝜔𝑡 − 𝛽𝑧 = 𝐴 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 108


𝑑𝑧 𝜔 = =𝑣 𝑑𝑡 𝛽  This proves that the wave is traveling in the z direction with velocity v

Relationship between E and H and Intrinsic impedance 𝐻𝑥 = −𝜀

1 𝜀 𝐸𝑦 = −√ 𝐸𝑦 𝜇 √𝜇𝜀

 In a similar way, we get 𝐻𝑦 = 𝜀

1 𝜀 𝐸𝑥 = √ 𝐸𝑥 𝜇 √𝜇𝜀

𝜇 𝜇 𝐸 = √𝐸𝑥 2 + 𝐸𝑦 2 = √ 𝐻𝑦 2 + 𝐻𝑥 2 𝜀 𝜀 𝜇 𝜇 = √ √𝐻𝑥 2 + 𝐻𝑦 2 = √ 𝐻 𝜀 𝜀 109


𝜂=

𝐸 𝜇 =√ 𝐻 𝜀

E ⇒ Total electric field H ⇒ Total magnetic field  This equation is similar to ohms law R=V/I by the analogy 𝑉⇔𝐸 𝐻⇔𝐼  So 𝜂 is called intrinsic impedance or characteristic impedance of the medium. For free space, intrinsic impedance is 𝜇0 𝜂0 = √ = 377Ω 𝜀0 For lossy dielectric,

𝑗𝜔𝜇 𝜂=√ 𝜎 + 𝑗𝜔𝜀

For fany other medium,

𝜇0 𝜇𝑟 𝜂=√ 𝜀0 𝜀𝑟

𝜂 = 377√

110

𝜇𝑟 Ω 𝜀𝑟


Solution of wave equations: Uniform plane wave traveling in z direction in conducting mediumsinusoidal time variations  In this case the wave equations in E are 𝜕 2 𝐸𝑦 = 𝑗𝜔𝜇𝜎𝐸𝑦 − 𝜔2 𝜇𝜀𝐸𝑦 𝜕𝑧 2  The wave equation for Ey may be written as 𝜕 2 𝐸𝑦 = 𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀)𝐸𝑦 = 𝛾 2 𝐸𝑦 where 2 𝜕𝑧

𝛾 = √𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀)

 Here 𝛾 is defined as the propagation constant.  It is a complex number and can be represented as 𝛾 = 𝛼 + 𝑗𝛽  It has a real part α called attenuation constant and an imaginary part 𝛽 called phase shift constant. Hence the solution of the partial differential equation is 𝐸𝑦 = 𝐸𝑚1 𝑒 −𝛼𝑧 𝑒 −𝛽𝑧 + 𝐸𝑚2 𝑒 𝛼𝑧 𝑒 𝛽𝑧  Similarly by taking the wave equation in terms of Hx and then proceeding we get 𝐻𝑦 = 𝐻𝑚1 𝑒 −𝛼𝑧 𝑒 −𝛽𝑧 + 𝐻𝑚2 𝑒 𝛼𝑧 𝑒𝛽𝑧  The terms 𝐸𝑚1 𝑒 −𝛼𝑧 and 𝐸𝑚2 𝑒 𝛼𝑧 are the amplitudes of the forward-traveling and backward-traveling waves.  The terms 𝑒 ±𝛼𝑧 and 𝐸𝑚2 𝑒 𝛼𝑧 cause the amplitudes of the forward and backward waves to decay as they propagate through the medium. Hence α is termed as the attenuation constant.  The time domain fields are obtained by multiplying 𝐸𝑦 by 𝑒 𝑗𝜔𝑡 and taking the real part of the result. 𝐸𝑦 (𝑥, 𝑡) = 𝐸𝑚1 𝑒 −𝛼𝑧 cos(𝜔𝑡 − 𝛽𝑧) + 𝐸𝑚2 𝑒 𝛼𝑧 cos(𝜔𝑡 + 𝛽𝑧) We know that

𝛾 = √𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀) = 𝛼 + 𝑗𝛽

𝛾 2 = 𝑗𝜔𝜇𝜎 − 𝜔2 𝜇𝜀 ⋯ (1) 𝛾 2 = (𝛼 + 𝑗𝛽)2 = 𝛼 2 − 𝛽 2 + 𝑗2𝛼𝛽 ⋯ (2)  Equating (1) and (2) 𝑗𝜔𝜇𝜎 − 𝜔2 𝜇𝜀 = 𝛼 2 − 𝛽 2 + 𝑗2𝛼𝛽 ⋯ (3) 111


 Equating the real and imaginary parts of (3) −𝜔2 𝜇𝜀 = 𝛼 2 − 𝛽 2 ⋯ (4) 𝜔𝜇𝜎 = 2𝛼𝛽 ⋯ (5)  From equation (5) we have 𝜔𝜇𝜎 𝜔𝜇𝜎 𝛼= ⋯ (6) 𝛽= ⋯ (7) 2𝛽 2𝛼  Putting equation (7) in (4) (𝜔𝜇𝜎) 𝜔𝜇𝜎 2 −𝜔 𝜇𝜀 = 𝛼 − ( ) = 𝛼2 − 2𝛼 4𝛼 2 2

2

2

𝜔𝜇𝜎 2 𝛼 4 + 𝜔2 𝜇𝜀𝛼 2 − ( ) =0 2 𝜔𝜇𝜎 2 𝑝2 + 𝜔2 𝜇𝜀𝑝 − ( ) = 0 ⋯ (8) 2

−𝜔2 𝜇𝜀4𝛼 2 = 4𝛼 4 − (𝜔𝜇𝜎)2 𝑃𝑢𝑡 𝛼 2 = 𝑝

⇒

⇒

 Solving the quadratic equation (8) for p we get 𝜔2 𝜇𝜀 𝜎 2 √ 𝑝= 1+[ ] −1 2 𝜔𝜀

𝛼 = ±ඩ

}

𝜔 2 𝜇𝜀 𝜎 2 ቎√1 + [ ] − 1቏ 2 𝜔𝜀

Taking the positive root only

𝛼 is Attenuation Constant

 Similarly by putting equation (6) in (4) we get another quadratic equation in 𝛽 2 , solving which we get

𝛽 = ±ඩ

𝜔 2 𝜇𝜀 𝜎 2 ቎√1 + [ ] + 1቏ 2 𝜔𝜀

Phase velocity

vp = ±

vp =

𝛽 is Phase shift Constant

𝜔 𝛽

1 𝜇𝜀 𝜎 2 √ ቊට1 + ( ) + 1ቋ 2 𝜔𝜀 112


Phase velocity

λ=

2𝜋 𝛽

2𝜋

λ=±

𝜔 2 𝜇𝜀 ට 𝜎 2 √ ቊ 1 + ( ) + 1ቋ 2 𝜔𝜀

Conductors and dielectrics  Electromagnetic materials are roughly classified as conductors and dielectrics.  Maxwell’s curl equation for sinusoidally varying quantities is given by ⃗ ×𝐻 ⃗ = 𝜎𝐸⃗ + 𝑗𝜔𝜀𝐸⃗ ∇ 𝐽𝐶 ⇒ Conduction current density ⃗ ×𝐻 ⃗ = 𝐽𝐶 + 𝐽𝐷 ∇ |𝐽𝐶 | |𝐽𝐷 |

=

𝐽𝐷 ⇒ Displacement current density

𝜎 |𝜎𝐸⃗ | = |𝑗𝜔𝜀𝐸⃗ | 𝜔𝜀

The ratio

|𝐽𝐶 | |𝐽𝐷 |

=

𝜎 𝜔𝜀

determines the nature of material

𝜎 ≪ 1 ⇒ 𝜔𝜀 ≫ 𝜎 ⇒ |𝐽𝐷 | ≫ |𝐽𝐶 | ⇒ Good dielectrics 𝜔𝜀 𝜎 2. ≅ 1 ⇒ 𝜔𝜀 ≅ 𝜎 ⇒ |𝐽𝐷 | ≅ |𝐽𝐶 | ⇒ Semiconductor 𝜔𝜀 𝜎 3. ≫ 1 ⇒ 𝜔𝜀 ≪ 𝜎 ⇒ |𝐽𝐷 | ≪ |𝐽𝐶 | ⇒ Good conductor 𝜔𝜀 1. When the displacement current is much greater than conduction current the medium behaves like a dielectric. If σ = 0 the medium is a perfect or lossless dielectric. If σ ≠ 0 the medium is a lossy dielectric. 2. When the displacement current is comparable to conduction current the medium behaves like a semiconductor. 3. When the displacement current is much smaller than conduction current the medium behaves like a conductor. If 𝜔𝜀 ≪ 𝜎 the medium is a good conductor. 𝜎 The term is called loss tangent 𝜔𝜀 𝜎 tan𝜃 = ⇒ loss tangent 𝜔𝜀 113 Assist. Prof. Dr. Hamad Rahman Jappor 1.

Wave propagation in good dielectrics  This can be considered as a special case of the wave motion in conducting medium.  All the equations derived for this case is applicable in the case of good dielectrics with appropriate modification of parameters.  For a good dielectric 𝜎/𝜔𝜀 > 𝜔𝜀 so that the value of 𝛼 and 𝛽 as follows 𝛾 = √𝑗𝜔𝜇(𝜎 + 𝑗𝜔𝜀) 𝛾 = √𝑗𝜔𝜇𝜎 (1 + 𝛾 = √𝑗𝜔𝜇𝜎

𝑗𝜔𝜀 ) 𝜎

𝑠𝑖𝑛𝑐𝑒

𝜎 ≫1 𝜔𝜀 114

or

𝜎 ≫ 𝜔𝜀 Assist. Prof. Dr. Hamad Rahman Jappor

𝛼=𝛽=ට

𝜔𝜇𝜎 = √𝜋𝑓𝜇𝜎 2

 The wave equation in this case is of the form 𝐸𝑦 (𝑧, 𝑡) = 𝐸𝑚1 𝑒 −𝛼𝑧 cos(𝜔𝑡 − 𝛽𝑧) + 𝐸𝑚2 𝑒 𝛼𝑧 cos(𝜔𝑡 + 𝛽𝑧)  When we consider the forward traveling wave only 𝐸𝑦 (𝑧, 𝑡) = 𝐸𝑚1 𝑒 −𝛼𝑧 cos(𝜔𝑡 − 𝛽𝑧)  A high frequency uniform plane wave suffers attenuation as it passes through a lossy medium.  Its amplitude gets multiplied by the factor, 𝑒 −𝛼𝑧 where 𝛼 is the attenuation constant.

 Skin depth is defined as the distance the wave must travel to have its amplitude reduced by a factor of e-1.  The exponential multiplying factor is unity at z=0 and decreases to 1/e when z=1/α  So skin depth (𝛿) is 𝛿=

1 2 1 =√ =√ 𝛼 𝜔𝜇𝜎 𝜋𝑓𝜇𝜎

 Thus we see that the skin depth decreases with an increase in frequency.  The intrinsic impedance of a medium may be expressed in terms of 𝛿 as following: 𝜂=

1+j 𝛿𝜎

Ω

 The phenomenon by which field intensity in a conductor rapidly decreases with increase in frequency is called skin effect.  At high frequencies the fields and the associated currents are confined to a very thin layer of the conductor surface. 115


Poynting’s theorem In order to find the power flow associated with an electromagnetic wave, it is necessary to develop a power theorem for the electromagnetic field.  Maxwell’s first curl equation states that ⃗ 𝜕𝐷 ⃗ ×𝐻 ⃗ =𝐽+ ∇ ⋯ (1) 𝜕𝑡  Next, we take the scalar product of both sides of equation (1) with 𝐸⃗ ∙ we get, ⃗ 𝜕𝐷 ⋯ (2) 𝜕𝑡  We have the vector identity ⃗ ×𝐻 ⃗ = 𝐸⃗ ∙ 𝐽 + 𝐸⃗ ∙ 𝐸⃗ ∙ ∇

⃗ ∙ (𝐸⃗ × 𝐻 ⃗)=𝐻 ⃗ ∙ (∇ ⃗ × 𝐸⃗ ) − 𝐸⃗ ∙ (∇ ⃗ ×𝐻 ⃗) ∇ ⃗ ×𝐻 ⃗)=𝐻 ⃗ ∙ (∇ ⃗ × 𝐸⃗ ) − ∇ ⃗ ∙ (𝐸⃗ × 𝐻 ⃗ ) ⋯ (3) 𝐸⃗ ∙ (∇  Putting equation (3) in (2) ⃗ 𝜕𝐷 ⋯ (4) 𝜕𝑡 ⃗ ⃗ 𝜕𝐵 𝜕𝐷 ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ −𝐻 ∙ − ∇ ∙ (𝐸 × 𝐻) = 𝐸 ∙ 𝐽 + 𝐸 ∙ or 𝜕𝑡 𝜕𝑡 ⃗ 𝜕𝐸⃗ ∂𝐻 ⃗ ∙ (𝐸⃗ × 𝐻 ⃗ ) = 𝐸⃗ ∙ 𝐽 + 𝜀𝐸⃗ ∙ ⃗⃗ ∙ −∇ + 𝜇H ⋯ (5) 𝜕𝑡 ∂t ⃗H ⃗ ∙ (∇ ⃗ × 𝐸⃗ ) − ∇ ⃗ ∙ (𝐸⃗ × 𝐻 ⃗ ) = 𝐸⃗ ∙ 𝐽 + 𝐸⃗ ∙

But

⃗ ∙ 𝐻

⃗ ∂𝐻 1∂ 2 = 𝐻 ∂t 2 ∂t

and

𝐸⃗ ∙

⃗ ∂E 1∂ 2 = 𝐸 ⋯ (6) ∂t 2 ∂t

 Substituting equation (6) in equation (5), we get ⃗ ∙ (𝐸⃗ × 𝐻 ⃗ ) = ⃗E ∙ J + ε −∇

1∂ 2 1𝜕 2 𝐸 +𝜇 𝐻 2 ∂t 2 𝜕𝑡

⃗ ∙ (𝐸⃗ × 𝐻 ⃗ ) = 𝐸⃗ ∙ 𝜎𝐸⃗ + −∇

∂ 𝜀𝐸 2 𝜕 𝜇𝐻2 ( )+ ( ) ∂t 2 𝜕𝑡 2

,

where 𝐽 = 𝜎𝐸⃗

∂ 𝜀𝐸 2 𝜇𝐻2 + ( ) ⋯ (7) ∂t 2 2  Rearranging terms and taking the volume integral of both sides of equation (7), 𝜕 𝜀𝐸 2 𝜇𝐻2 ⃗ ∙ (𝐸⃗ × 𝐻 ⃗ )𝑑v = − ∫ [ ∫∇ + ] 𝑑v − ∫𝜎𝐸 2 𝑑v 𝜕𝑡 v 2 2 v v  We apply divergence theorem to the left-hand side, convert the volume integral into a surface integral and get 116 Assist. Prof. Dr. Hamad Rahman Jappor ⃗ ∙ (𝐸⃗ × 𝐻 ⃗ ) = 𝜎𝐸 2 + −∇

⃗ ) ∙ 𝑑S = − ∮(𝐸⃗ × 𝐻 𝑆

𝜕 𝜀𝐸 2 𝜇𝐻2 ∫[ + ] 𝑑v − ∫𝜎𝐸 2 𝑑v 𝜕𝑡 v 2 2 v

⋯ (8)

 Equation (8) is known as Poynting’s theorem; it is the "work-energy theorem" of electrodynamics. The various terms in the equation are identified using energy-conservation arguments for EM fields.  The first term on the right-hand side of equation (8) is interpreted as the rate of decrease in energy stored in the electric and magnetic fields. The second term is the power dissipated (Ohmic power dissipated) due to the fact that the medium is conducting (𝜎 ≠ 0)  The left-hand side of equation (8) represent the total power leaving the volume.  Poynting theorem says, then, that the energy that flowed out through the surface is equal to the decrease in energy stored in the field, less the work done on the charges by the electromagnetic force.

Poynting vector  From Poynting theorem it can be seen that the vector product of electric field ⃗ is another vector which is denoted by intensity 𝐸⃗ and magnetic field intensity 𝐻 ⃗ 𝑆 = 𝐸⃗ × 𝐻

watts/m2

 The vector 𝑆 is called Poynting vector. 𝑆 is the energy per unit area passing per unit time through the surface of the volume v in watts/meter square.  It measures the rate of flow of energy, and its direction is the direction of power ⃗ vectors. flow and it is perpendicular to the plane containing 𝐸⃗ and 𝐻  For plane waves the direction of energy flow is the direction of propagation.  The Poynting vector offers a useful coordinate free way of specifying the direction of propagation of plane waves. 117 Assist. Prof. Dr. Hamad Rahman Jappor

Monochromatic plane waves Since different frequencies in the visible range correspond to different color, such waves are called monochromatic. Suppose that the waves are traveling in the z direction and have no x or y dependence; these are called plane waves, because the fields are uniform over every plane perpendicular to the direction of propagation. We are interested, then, in fields of the form 𝐸⃗ (𝑧, 𝑡) = 𝐸⃗0 𝑒 𝑖(𝑘𝑧−𝜔𝑡) ⃗ (𝑧, 𝑡) = 𝐵 ⃗ 0 𝑒 𝑖(𝑘𝑧−𝜔𝑡) 𝐵 ⃗ 0 are the (complex) amplitudes (the physical fields, of course, are the where 𝐸⃗0 and 𝐵 ⃗ ). Here, k is the wave number, and 𝜔 is the angular frequency of real parts of 𝐸⃗ and 𝐵 EM wave.

Monochromatic plane wave, The wave is polarized in the x direction. Now, every solution to Maxwell's equations (in empty space) must obey the wave equation, the converse is not true; Maxwell's equations impose extra constraints on ⃗ 0 . In particular, since ∇ ⃗ ⋅ ⃗E = 0 and ∇ ⃗ ⋅B ⃗ = 0, it follows that: 𝐸⃗0 and 𝐵 ⃗ 0𝑧 = 0 𝐸⃗0𝑧 = 𝐵 That is, electromagnetic waves are transverse: the electric and magnetic fields are ⃗

⃗ × 𝐸⃗ = − 𝜕𝐵 perpendicular to the direction of propagation. Moreover, Faraday's law, ∇ 𝜕𝑡

implies a relation between the electric and magnetic amplitudes, which results in ⃗ 0𝑥 −𝑘𝐸⃗0𝑦 = 𝜔𝐵

,

⃗ 0𝑦 , 𝑘𝐸⃗0𝑥 = 𝜔𝐵

. . . (1)

or, more compactly: 𝑘 1 ⃗ 0 = (𝑎 𝐵 ̂𝑧 × 𝐸⃗0 ) = (𝑎 ̂ × 𝐸⃗0 ) 𝜔 c 𝑧 Evidently, E and B are in phase and mutually perpendicular; their (real) amplitudes are related by 𝐵0 =

𝑘 1 𝐸0 = 𝐸0 𝜔 c 118


⃗ = 𝜇0 𝜀0 The fourth of Maxwell's equations, ∇2 𝐵

∂2 𝐸⃗ ∂t2

, does not yield an independent

condition; it simply reproduces Eq. (1). There is nothing special about the z direction, we can easily generalize to monochromatic plane waves traveling in an arbitrary direction. The notation is facilitated by the introduction of the wave vector, ⃗k pointing in the direction of propagation, whose magnitude is the wave number k. The scalar product ⃗k . 𝑟 is the appropriate generalization of kz, so ⃗ 𝐸⃗ (𝑟, 𝑡) = 𝐸⃗0 𝑒 𝑖(k .𝑟−𝜔𝑡) 𝑎̂𝑛 , . . . (2) 1 1 ⃗ (𝑟, 𝑡) = 𝐸⃗0 𝑒 𝑖(k⃗ .𝑟−𝜔𝑡) (𝑎̂𝑘 × 𝑎̂𝑛 ) = 𝑎̂𝑘 × 𝐸⃗ , 𝐵 . . . (3) c c where 𝑎̂𝑛 is the polarization vector. Because 𝐸⃗ is transverse, (𝑎̂𝑛 . 𝑎̂𝑘 ) = 0

Linear and circular polarizations The plane wave Eq. (2) and Eq. (3) is a wave with its electric field vector always in the direction e. Such a wave is said to be linearly polarized with polarization vector 𝑎̂𝑛1 = 𝑎̂𝑛 . Evidently, there exists another wave which is linearly polarized with polarization vector 𝑎̂𝑛2 = 𝑎̂𝑛 and is linearly independent of the first. Thus the two waves are ⃗ ⃗ 𝐸⃗1 (𝑟, 𝑡) = 𝐸1 𝑎̂𝑛1 𝑒 𝑖(k .𝑟−𝜔𝑡) , 𝐸⃗2 (𝑟, 𝑡) = 𝐸2 𝑎̂𝑛2 𝑒 𝑖(k .𝑟−𝜔𝑡) 1 ⃗ × 𝐸⃗1,2 ) ⃗ 1,2 = (k with 𝐵 c They can be combined to give the most general homogeneous plane wave propagating in the direction ⃗k = 𝑘𝑎̂𝑘 : ⃗ 𝐸⃗ (𝑟, 𝑡) = 𝐸⃗1 (𝑟, 𝑡) + 𝐸⃗2 (𝑟, 𝑡) = (𝐸1 𝑎̂𝑛1 + 𝐸2 𝑎̂𝑛2 )𝑒 𝑖(k .𝑟−𝜔𝑡) . . . (4)

If E1 and E2 have the same phase wave Eq. (4) represents a linearly polarized wave, with its polarization vector making an angle tan𝜃 = 𝐸2 /𝐸1 with respect to 𝑎̂𝑛1 and a magnitude 𝐸 = ට𝐸2 2 + 𝐸1 2 as shown in the following figure:

Electric field of: (a) a linearly polarized wave. 119

(b) a circularly polarized wave. Assist. Prof. Dr. Hamad Rahman Jappor

If E1 and E2 have different phases, the wave Eq.(4) is elliptically polarized. To understand what this means let us consider the simplest case, circular polarization. Then E1 and E2 have the same magnitude, but differ in phase by 90°. The wave Eq.(4) becomes: ⃗ 𝐸⃗ (𝑟, 𝑡) = 𝐸0 (𝑎̂𝑛1 + 𝑖𝑎̂𝑛2 )𝑒 𝑖(k .𝑟−𝜔𝑡) . . . (5) With E0 the common real amplitude. We imagine axes chosen so that the wave is propagating in the positive z direction, while 𝑎̂𝑛1 and 𝑎̂𝑛2 are in the x and y directions, respectively. Then the components of the actual electric field, obtained by taking the real part of (5), are 𝐸𝑥 (𝑟, 𝑡) = 𝐸0 cos(𝑘𝑧 − 𝜔𝑡) 𝐸𝑦 (𝑟, 𝑡) = ∓𝐸0 sin(𝑘𝑧 − 𝜔𝑡) At a fixed point in space, the fields are such that the electric vector is constant in magnitude, but sweeps around in a circle at a frequency ω. For the upper sign (𝑎̂𝑛1 + 𝑖𝑎̂𝑛2 ), the rotation is counterclockwise when the observer is facing into the oncoming wave. This wave is called left circularly polarized in optics. In modern physics, such a wave is said to have positive helicity. The latter description seems more appropriate because such a wave has a positive projection of angular momentum on the z axis. For the lower sign (𝑎̂𝑛1 − 𝑖𝑎̂𝑛2 ), the rotation of 𝐸⃗ is clockwise when looking into the wave; the wave is right circularly polarized (optics): it has negative helicity.

Electric field and magnetic induction for an elliptically polarized wave.

Energy and momentum of electromagnetic waves Energy per unit volume stored in electromagnetic fields is 1 1 𝑢 = (𝜀0 𝐸 2 + 𝐵2 ) 2 𝜇0 Note that here the E and B are real quantities (real pars of complex quantities used in he previous paragraph). In case of monochromatic plane wave 120


1 2 𝐸 = 𝜇0 𝜀0 𝐸0 2 𝑐2 0 Therefore, the magnetic and electric contribution are equal 𝐵0 2 =

As the wave travels, it carriers this energy alone with it. The energy flux density transported by the fields is given by the Poynting vector: 1 ⃗) 𝑆 = (𝐸⃗ × 𝐵 𝜇0 For monochromatic plane waves propagating in the z direction, 1 𝐸2 𝑆= 𝑎 ̂ 𝑧 = 𝑐𝜀0 𝐸 2 𝑎 ̂𝑧 = 𝑐𝜀0 𝐸0 2 cos 2 (𝑘𝑧 − 𝜔𝑡 + 𝛿)𝑎 ̂𝑧 = 𝑐𝑢𝑎 ̂𝑧 𝜇0 𝑐 where u is the energy density. For in a time ∆t, a length c∆t passes through area A, carrying with it an energy uAc∆t. The energy per unit time, per unit area, transported by the wave is therefore uc. Electromagnetic fields not only carry energy, they also carry momentum. The momentum density stored in the fields is 1 𝑃⃗𝑒𝑚 = 2 𝑆 𝑐 For monochromatic plane waves, then, 1 1 𝑃⃗𝑒𝑚 = 𝜀0 𝐸0 2 cos 2 (𝑘𝑧 − 𝜔𝑡 + 𝛿)𝑎 ̂𝑧 = 𝑢𝑎 ̂𝑧 𝑐 𝑐 In the case of light, the wavelength is so short (~ 5 × 10-7 m), and the period so brief (~ 10-15s), that any macroscopic measurement will encompass many cycles. Typically, therefore, we are not interested in the fluctuating cosine-squared term in the energy and momentum densities; all we want is the average value. Now, the average of cosine-squared over a complete cycle is ½ so 1 〈𝑢〉 = 𝜀0 𝐸0 2 2 1 〈𝑆〉 = 𝑐𝜀0 𝐸0 2 𝑎 ̂𝑧 2 1 〈𝑃⃗𝑒𝑚 〉 = 𝜀0 𝐸0 2 𝑎 ̂𝑧 2c We use brackets 〈 〉, to denote the time average over a complete cycle. The average power per unit area transported by an electromagnetic wave is called the intensity: 1 𝐼 = 〈𝑆〉 = 𝑣𝜀0 𝐸0 2 2

. . . (6) 121


Electromagnetic waves in matter In regions of matter where there are no free charges and free currents, Maxwell equations are ⃗ ∙𝐷 ⃗ =0 ∇ ⃗ ∙𝐵 ⃗ =0 ∇ ⃗ 𝜕𝐵 𝜕𝑡 ⃗ 𝜕𝐷

⃗ × 𝐸⃗ = − ∇ ⃗ ×𝐻 ⃗ = ∇

𝜕𝑡 If the matter is linear 1 ⃗ 𝐵 𝜇 and homogeneous (ε and µ are constants), Maxwell equations reduce to ⃗ ∙ 𝐸⃗ = 0 ∇ ⃗ = 𝜀𝐸⃗ 𝐷

,

⃗ = 𝐻

⃗ ∙𝐵 ⃗ =0 ∇ ⃗ 𝜕𝐵 𝜕𝑡 𝜕𝐸⃗ ⃗ ×𝐵 ⃗ = ε𝜇 ∇ 𝜕𝑡 which are different from Maxwell equations in vacuum only by the replacement of 𝜀0 𝜇0 to 𝜀𝜇. Evidently, electromagnetic waves propagate trough a linear homogeneous medium at a speed ⃗ × 𝐸⃗ = − ∇

𝑣=

1 𝑐 = √ε𝜇 𝑛

⟹

ε𝜇 𝑛=√ 𝜀0 𝜇0

Where n is the index of refraction. For most (non-ferromagnetic) materials µ is very close to 𝜇0 so 𝑛 ≅ √𝜀𝑟 In our consideration of EM waves below, we assume for simplicity that 𝜇 = 𝜇0 . In that case all the results we obtained above to EM waves in vacuum are valid with replacement 𝜀0 by 𝜀. Reflection and transmission of EM waves at normal incidence If a wave passes from one transparent medium into another, there is a reflected wave and a transmitted wave. The details depend on the exact nature of the electrodynamics boundary conditions, which are 122 Assist. Prof. Dr. Hamad Rahman Jappor

(𝐢) 𝜀1 𝐸1⊥ = 𝜀2 𝐸2⊥

(𝐢𝐢) 𝐵1⊥ = 𝐵2⊥

(𝐢𝐢𝐢) 𝐸⃗1∥ = 𝐸⃗2∥

(𝐢𝐯)

1 𝜇1

⃗ 1∥ = 𝐵

1 𝜇2

⃗ 2∥ 𝐵

. . . (7)

Suppose the xy plane forms the boundary between two linear media. A plane wave of frequency ω, traveling in the z direction and polarized in the x direction approaches the interface from the left: 𝐸⃗I (𝑧, 𝑡) = 𝐸⃗0I 𝑒 𝑖(𝑘1𝑧−𝜔𝑡) 𝑎 ̂𝑥 , 1 ⃗ I (𝑧, 𝑡) = 𝐸⃗0I 𝑒 𝑖(𝑘1𝑧−𝜔𝑡) 𝑎 𝐵 ̂𝑦 𝑣1 It gives rise to the reflected wave 𝐸⃗𝑅 (𝑧, 𝑡) = 𝐸⃗0𝑅 𝑒 𝑖(−𝑘1 𝑧−𝜔𝑡) 𝑎 ̂𝑥 , 1 ⃗ 𝑅 (𝑧, 𝑡) = − 𝐸⃗0𝑅 𝑒 𝑖(−𝑘1𝑧−𝜔𝑡) 𝑎 𝐵 ̂𝑦 𝑣1

which travels back to the left in medium (1), and a transmitted wave 𝐸⃗𝑇 (𝑧, 𝑡) = 𝐸⃗0𝑇 𝑒 𝑖(𝑘2 𝑧−𝜔𝑡) 𝑎 ̂𝑥 , ⃗ 𝑇 (𝑧, 𝑡) = 𝐵

1 𝐸⃗0𝑇 𝑒 𝑖(𝑘2 𝑧−𝜔𝑡) 𝑎 ̂𝑦 𝑣2

⃗I+𝐵 ⃗ 𝑅 , must join which continues on the right in medium (2). Note that 𝐸⃗I + 𝐸⃗𝑅 and 𝐵 ⃗ 𝑇 , in accordance with the boundary conditions. In this the fields on the right, 𝐸⃗𝑇 and 𝐵 case there are no components perpendicular to the surface, so (i) and (ii) are trivial. However, (iii) and (iv) require that 𝐸⃗0I + 𝐸⃗0𝑅 = 𝐸⃗0T

. . . (8)

1 1 1 1 1 ( 𝐸⃗0I − 𝐸⃗0𝑅 ) = ( 𝐸⃗0𝑇 ) 𝜇1 𝑣1 𝑣1 𝜇2 𝑣2 𝜇1 𝑣1 𝜇1 𝑛2 where 𝛽≡ = 𝜇2 𝑣2 𝜇2 𝑛1 123

or

𝐸⃗0I − 𝐸⃗0𝑅 = 𝛽𝐸⃗0𝑇 . . . (8 ∗)


Equations (8) and (8*) are easily solved for the outgoing amplitudes, in terms of the incident amplitude: 𝐸⃗0𝑅 = (

1−𝛽 ) 𝐸⃗ 1 + 𝛽 0I

𝐸⃗0𝑇 = (

,

2 ) 𝐸⃗ 1 + 𝛽 0I

In terms of the indices of refraction this result is as follows 𝐸0𝑅 = |

𝑛1 − 𝑛2 |𝐸 𝑛1 + 𝑛2 0I

,

𝐸0𝑇 = (

2𝑛1 )𝐸 𝑛1 + 𝑛2 0I

. . . (9)

In order to calculate the fraction of energy which is transmitted and reflected we need to find the intensity (average power per unit area) which is given by Eq. (6) with 𝜀0 replaced by 𝜀: 1 𝐼 = 〈𝑆〉 = 𝜀𝑣𝐸0 2 2 If 𝜇1 = 𝜇2 = 𝜇0 , then the ratio of the reflected intensity to the incident intensity is 𝐼𝑅 𝐸0𝑅 2 𝑛1 − 𝑛2 2 𝑅≡ =( ) =( ) 𝐼I 𝐸0I 𝑛1 + 𝑛2 whereas the ratio of the transmitted intensity to the incident intensity is 𝐼𝑇 𝜀2 𝑣2 𝐸0𝑇 2 4𝑛1 𝑛2 𝑇≡ = ( ) = (𝑛1 + 𝑛2 )2 𝐼I 𝜀1 𝑣1 𝐸0I R is called the reflection coefficient and T the transmission coefficient; they measure the fraction of the incident energy that is reflected and transmitted, respectively. Notice that R+T=1 as conservation of energy, of course, requires. For instance, when light passes from air (n1=1) into glass (n2=1.5), R = 0.04 and T = 0.96. Not surprisingly, most of the light is transmitted. Reflection and Transmission at Oblique Incidence Now we consider the more general case of oblique incidence, in which the incoming wave meets the boundary at an arbitrary angle 𝜃I . The normal incidence considered in the previous section is really just a special case of oblique incidence, with 𝜃I = 0. Suppose, then, that a monochromatic plane wave 𝐸⃗I (𝑟, 𝑡) = 𝐸⃗0I 𝑒 𝑖(𝑘⃗I.𝑟−𝜔𝑡)

; 124

⃗ I (𝑟, 𝑡) = 𝐵

1 ⃗ × 𝐸⃗I ) (k 𝑣1 I


approaches from the left, giving rise to a reflected wave, 𝐸⃗𝑅 (𝑟, 𝑡) = 𝐸⃗0𝑅 𝑒 𝑖(𝑘⃗𝑅.𝑟−𝜔𝑡)

;

⃗ 𝑅 (𝑟, 𝑡) = 𝐵

1 ⃗ × 𝐸⃗𝑅 ) (k 𝑣1 𝑅

and a transmitted wave 1 ⃗ 𝑇 × 𝐸⃗𝑇 ) (k 𝑣1 All three waves have the same frequency ω that is determined once and for all at the source. The three wave numbers are related so that v2 k I v1 = k 𝑅 v1 = k 𝑇 v2 = ω or k I = k 𝑅 v1 = k 𝑇 . . . (10) v1 ⃗I+𝐵 ⃗ 𝑅 must now be joined to the The combined fields in medium (1), 𝐸⃗I + 𝐸⃗𝑅 and 𝐵 ⃗ 𝑇 in medium (2), using the boundary conditions Eq.(7). These all share fields 𝐸⃗𝑇 and 𝐵 𝐸⃗𝑇 (𝑟, 𝑡) = 𝐸⃗0𝑇 𝑒 𝑖(𝑘⃗𝑇.𝑟−𝜔𝑡)

;

⃗ 𝑇 (𝑟, 𝑡) = 𝐵

the generic structure ( )𝑒 𝑖(𝑘⃗I.𝑟−𝜔𝑡) + ( )𝑒 𝑖(𝑘⃗𝑅.𝑟−𝜔𝑡) = ( )𝑒 𝑖(𝑘⃗𝑇.𝑟−𝜔𝑡) ,

𝑎𝑡 𝑧 = 0

The important thing to notice is that the x, y, and t dependence is confined to the exponents. Because the boundary conditions must hold at all points on the plane, and for all times, these exponential factors must be equal at z = 0. The time factors are already equal. As for the spatial terms, evidently when z = 0 ⃗ I. 𝑟 = 𝑘 ⃗ 𝑅. 𝑟 = 𝑘 ⃗ 𝑇 . 𝑟, when 𝑧 = 0 . . . (11) 𝑘 or, more explicitly, 𝑥(𝑘I )𝑥 + 𝑦(𝑘I )𝑦 = 𝑥(𝑘𝑅 )𝑥 + 𝑦(𝑘𝑅 )𝑦 = 𝑥(𝑘 𝑇 )𝑥 + 𝑦(𝑘 𝑇 )𝑦

. . . (12)

for all x and all y. Eq.(12) can only hold if the components are separately equal, for if x = 0, we get (𝑘I )𝑦 = (𝑘𝑅 )𝑦 = (𝑘𝑇 )𝑦 . . . (13) while y=0 gives (𝑘I )𝑥 = (𝑘𝑅 )𝑥 = (𝑘 𝑇 )𝑥

. . . (14) 125


⃗ I lies in the xz plane (i.e. (𝑘I )𝑦 = 0); We may as well orient our axes so that 𝑘 ⃗ 𝑅 and 𝑘 ⃗ 𝑇 . Conclusion: according to Eq. (13), so too will 𝑘 First Law: The incident, reflected, and transmitted wave vectors form a plane (called the plane of incidence), which also includes the normal to the surface (here, the z axis) Meanwhile, Eq. (14) implies that 𝑘I sin 𝜃I = 𝑘𝑅 sin 𝜃𝑅 = 𝑘 𝑇 sin 𝜃𝑇 Where 𝜃I is the angle of incidence, 𝜃𝑅 is the angle of reflection, and 𝜃𝑇 is the angle of transmission, more commonly known as the angle of refraction, all of them measured with respect to the normal. In view of Eq. (10), then, Second Law: The angle of incidence is equal to the angle of reflection, 𝜃I = 𝜃𝑅 This is the law of reflection. As for the transmitted angle, there is Third Law: sin 𝜃𝑇 𝑛1 = sin 𝜃I 𝑛2 This is the law of refraction, or Snell's law. These are the three fundamental laws of geometrical optics. Now that we have taken care of the exponential factors – they cancel, given Eq. (11) the boundary conditions Eq. (7) become: (𝐢) 𝜀1 (𝐸⃗0I + 𝐸⃗0𝑅 ) = 𝜀2 (𝐸⃗0T ) 𝑧 𝑧 ⃗ 0I + 𝐵 ⃗ 0𝑅 ) = (𝐵 ⃗ 0T ) (𝐢𝐢) (𝐵 𝑧 𝑧 (𝐢𝐢𝐢) (𝐸⃗0I + 𝐸⃗0𝑅 )

𝑥,𝑦

(𝐢𝐯)

= (𝐸⃗0T )𝑥,𝑦

1 1 ⃗ 0I + 𝐵 ⃗ 0𝑅 ) = (𝐵 ⃗ ) (𝐵 𝑥,𝑦 𝜇1 𝜇2 0T 𝑥.𝑦

⃗ 0 (r, t) = 1 (𝑎̂𝑘 × ⃗E0 ) in each case. The last two represent pairs of equations, where B 𝑣

one for the x component and one for the y-component. Suppose that the polarization of the incident wave is parallel to the plane of incidence (the xz plane); then the reflected and transmitted waves are also polarized in this plane. Then Eq. (i) reads 𝜀1 (−𝐸⃗0I sin 𝜃I + 𝐸⃗0𝑅 sin 𝜃𝑅 ) = 𝜀2 (−𝐸⃗0T sin 𝜃𝑇 ) . . . (15) 126


Eq. (ii) adds nothing (0 = 0), since the magnetic fields have no z components; Eq. (iii) becomes 𝐸⃗0I cos 𝜃I + 𝐸⃗0𝑅 cos 𝜃𝑅 = 𝐸⃗0T cos 𝜃𝑇

(16)

and Eq. (iv) gives 1 1 𝐸⃗ (𝐸⃗0𝐼 − 𝐸⃗0𝑅 ) = 𝜇1 𝑣1 𝜇2 𝑣2 0𝑇 𝐸⃗0I − 𝐸⃗0𝑅 = 𝛽𝐸⃗0𝑇

. . . (17)

. . . (18)

Eq. (16) says 𝐸⃗0I + 𝐸⃗0𝑅 = 𝛼𝐸⃗0𝑇 𝛼≡

. . . (19)

cos 𝜃𝑇 cos 𝜃I

Solving Eqs. (18) and (19) for the reflected and transmitted amplitudes, we obtain 𝐸⃗0𝑅 = (

𝛼−𝛽 , ) 𝐸⃗ 𝛼 + 𝛽 0I

𝐸⃗0𝑇 = (

2 . . . (20) ) 𝐸⃗ 𝛼 + 𝛽 0I

These are known as Fresnel's equations, for the case of polarization in the plane of incidence. There are two other Fresnel equations, giving the reflected and transmitted amplitudes when the polarization is perpendicular to the plane of incidence. These equations you are asked to derive at home. Notice that the transmitted wave is always in phase with the incident one; the reflected wave is either in phase (“right side up"), if 𝛼 > 𝛽, or 180° out of phase (“upside down”), if 𝛼 < 𝛽. The amplitudes of the transmitted and reflected waves depend on the angle of incidence, because α is a function of 𝜃I : 127


√1 − sin2 𝜃𝑇 √1 − [(𝑛1 /𝑛2 )sin𝜃I ]2 𝛼= = cos 𝜃I cos 𝜃I In the case of normal incidence (𝜃𝐼 = 0 , 𝛼 = 1, and we recover Eq. (9). At grazing incidence 𝜃I = 90𝑜 , 𝛼 diverges, and the wave is totally reflected. Interestingly, there is an intermediate angle, 𝜃𝐵 (called Brewster's angle), at which the reflected wave is completely extinguished. According to Eq. (20), this occurs when 𝛼 = 𝛽 or 𝛽2 sin 𝜃𝐵 ≅ 1 + 𝛽2 2

or equivalently tanθB ≅ β =

n2 n1

The following figure shows a plot of the transmitted and reflected amplitudes as functions of θI , for light incident on glass (n2 = 1.5) from air (n2 = 1). On the graph, a negative number indicates that the wave is 180° out of phase with the incident beam – the amplitude itself is the absolute value.

If the wave is polarized perpendicular to the plane of incidence, there is no Brewster's angle for any n. Therefore, if a plane wave of mixed polarization is incident on a plane interface at the Brewster angle, the reflected radiation is completely planepolarized with polarization vector perpendicular to the plane of incidence. This behavior can be utilized to produce beams of plane-polarized light but is not as efficient as other means employing anisotropic properties of some dielectric media. Even if the unpolarized wave is reflected at angles other than the Brewster angle, there is a tendency for the reflected wave to be predominantly polarized perpendicular to the plane of incidence. 128


The power per unit area striking the interface is ⃗S. 𝑎̂𝑧 . Thus the incident intensity is 𝐼𝐼 =

1 𝜀1 𝑣1 𝐸0𝐼 2 cos 𝜃𝐼 2

while the reflected and transmitted intensities are 𝐼𝑅 =

1 1 𝜀1 𝑣1 𝐸0𝑅 2 cos 𝜃𝑅 , and 𝐼𝑇 = 𝜀2 𝑣2 𝐸0𝑇 2 cos 𝜃𝑇 2 2

The cosines are there because the intensities represent the average power per unit area of interface, and the interface is at an angle to the wave front. The reflection and transmission coefficients for waves polarized parallel to the plane of incidence are 𝐼𝑅 𝐸0𝑅 2 𝛼−𝛽 2 𝑅≡ =( ) =( ) 𝐼I 𝐸0I 𝛼+𝛽 whereas the ratio of the transmitted intensity to the incident intensity is 𝐼𝑇 𝜀2 𝑣2 𝐸0𝑇 2 cos 𝜃𝑇 2 2 𝑇≡ = = 𝛼𝛽 ( ( ) ) 𝐼𝐼 𝜀1 𝑣1 𝐸0I cos 𝜃I 𝛼+𝛽 They are plotted as functions of the angle of incidence in the following figure (for the air/glass interface). R is the fraction of the incident energy that is reflected – naturally, it goes to zero at Brewster's angle; T is the fraction transmitted – it goes to 1 at θB . Note that R+T=1, as required by conservation of energy: the energy per unit time reaching a particular patch of area on the surface is equal to the energy per unit time leaving the patch.

129


Lectures in Electromagnetism (4th Year, Course 1)

Lectures in Electromagnetism (4th Year, Course 1)

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