Chapter 4

Maxwells Equations

The main aim of this chapter is to provide sufficient background and concepts on Maxwell’s equations. They include:

• Maxwell’s equations for static and time varying fields in free space and conductive media in differential and integral form
• meaning and proof of Maxwell’s equations
• comprehensive boundary conditions in scalar and vector form
• time varying and retarded potentials
• Helmholtz theorem and Lorentz gauge condition
• points/formulae to remember, objective and multiple choice questions and exercise problems.

4.1 INTRODUCTION

Static electric field has applications in cathode ray oscilloscopes for deflecting charged particles, in ink-jet printers to increase the speed of printing and print quality, in sorting of materials in mining industry and in the development of electrostatic voltmeters.

Static magnetic field has applications in magnetic separators to separate magnetic materials from non-magnetic materials, in cyclotrons for imparting high energy to charged particles, in velocity selector and mass separators. It is also used in magnetohydrodynamic generators.

On the other hand, electromagnetic fields in their time varying form constitute electromagnetic waves. These fields/waves are useful in all communication and radar systems. In fact, the EM waves are carriers of information, mainly in free space, between the transmitter and the receiver. The electric and magnetic fields of the EM waves are related through Maxwell’s equations.

Most of the problems related to antennas can be solved with the help of Maxwell’s equations and boundary conditions. In view of this, in this chapter, Maxwell’s equations are presented in detail. Maxwell’s equations are very popular and important and hence they are also referred to as electromagnetic field equations. The fields E, H, D and B in static form are represented in Cartesian coordinates as

 

and

 

In cylindrical coordinates, they are:

 

and

 

and in spherical coordinates, they are:

 

and

 

The time varying fields E (t, x, y, z), H (t, x, y, z) constitute electromagnetic waves which have wide applications in all communications, radars and also in bio-medical engineering.

In these applications, time varying fields are of more practical value than static electric and magnetic fields.

It may be noted that electrostatic fields are usually produced by static charges. Magnetostatic fields are produced from the motion of electric charges with uniform velocity (direct current) or static magnetic charges namely magnetic dipoles.

On the other hand, time varying fields constituting EM waves are produced by time varying currents, that is, any pulsating current produces radiation fields which are nothing but time varying fields. In brief, note that:

 

4.2 EQUATION OF CONTINUITY FOR TIME VARYING FIELDS

Equation of continuity in point form is

          ∇.J = −_v

where

 

	J	=	conduction current density (A/m2)
	ρυ	=	volume charge density (c/m3),
	∇	=	vector differential operator
		=

Proof     Consider a closed surface enclosing a charge Q. There exists an outward flow of current given by

This is equation of continuity in integral form.

Here, I is the current flowing away through a closed surface, dS is the differential area on the surface whose direction is always outward normal to the surface. As there is outward flow of current, there will be a rate of decrease of charge given by , where Q is the enclosed charge. From the principle of conservation of charge, we have

From divergence theorem, we have

Thus,          

By definition,          

where           ρυ = volume charge density (C/m³)

So,

where          

Two volume integrals are equal only if their integrands are equal.

Thus,            Hence proved.

4.3 MAXWELL’S EQUATIONS FOR TIME VARYING FIELDS

These are basically four in number.

Maxwell’s equations in differential form are given by

             ∇ × H = Ḋ + J

             ∇ × E = −Ḃ

                ∇.D = ρυ

                ∇.B = 0

Here

 

	H	=	magnetic field strength (A/m)
	D	=	electric flux density, (C/m2)
		=	displacement electric current density (A/m2)
	J	=	conduction current density (A/m2)
	E	=	electric field (V/m)
	B	=	magnetic flux density wb/m2 or Tesla
		=	time-derivative of magnetic flux density (wb/m2-sec)

Ḃ is called magnetic current density (V/m²) or Tesla/sec

 

Maxwell’s equations for time varying fields in integral form are given by

Here, dL is the differential length and dS is the differential area whose direction is always outward normal to the surface.

4.4 MEANING OF MAXWELL’S EQUATIONS

It is easy to understand the meaning of Maxwell’s equations from their integral form.

 

4.5 CONVERSION OF DIFFERENTIAL FORM OF MAXWELL’S EQUATION TO INTEGRAL FORM

1. Consider the first Maxwell’s equation

    

   ∇ × H = Ḋ + J

   Take surface integral on both sides.

   

   Applying Stoke’s theorem to LHS, we can write

   

   Hence

2. Consider the second Maxwell’s equation

    

   ∇ × E = − Ḃ

   Take surface integral on both sides.

   

   Applying Stoke’s theorem to LHS, we get

   

   Therefore

3. Consider the third Maxwell’s equation

    

   ∇.D = ρυ

   Take volume integral on both sides.

   

   Applying divergence theorem to LHS, we get

   

   Therefore

4. Consider the fourth Maxwell’s equation

    

   ∇. B = 0

   Take volume integral on both sides.

   

   Applying divergence theorem to LHS, we get

   

   Therefore,

4.6 MAXWELL’S EQUATIONS FOR STATIC FIELDS

Maxwell’s equations for static fields are:

As the fields are static, all the field terms which have time derivatives are zero, that is, Ḃ = 0, Ḋ = 0.

4.7 CHARACTERISTICS OF FREE SPACE

 

4.8 MAXWELL’S EQUATIONS FOR FREE SPACE

4.9 MAXWELL’S EQUATIONS FOR STATIC FIELDS IN FREE SPACE

4.10 PROOF OF MAXWELL’S EQUATIONS

1. From Ampere’s circuital law, we have

    

   ∇ × H =J

   Take dot product on both sides

    

   ∇ . ∇ × H = ∇ . J

   As the divergence of curl of a vector is zero,

    

   RHS = ∇. J = 0

   But the equation of continuity in point form is

   

   This means that if ∇ × H = J is true, it is resulting in ∇. J = 0.

   As the equation of continuity is more fundamental, Ampere’s circuital law should be modified. Hence we can write

    

   ∇ × H = J + F

   Take dot product on both sides

    

   ∇ . ∇ × H = ∇ . J + ∇.F

   that is,

    

   ∇ . ∇ × H = 0 = ∇ . J + ∇.F

   Substituting the value of ∇ .J from the equation of continuity in the above expression, we get

    

   ∇.F + (−_υ) = 0

   or,

   ∇.F = _υ

   The point form of Gauss’s law is

    

   ∇ . D = ρ_υ

   or,

    

   ∇ . Ḋ = _υ

   From the above expressions, we get

    

   ∇ . F = ∇ . Ḋ

   The divergence of two vectors are equal only if the vectors are identical,

   that is,                 F = Ḋ

   So,              ∇ × H = Ḋ + J        Hence proved.

2. According to Faraday’s law,
   

   and by definition,

   

   But

   

   Applying Stoke’s theorem to LHS, we get

   

   Two surface integrals are equal only if their integrands are equal,

   that is,           ∇ × E = − Ḃ     Hence proved.

3. From Gauss’s law in electric field, we have
   

   Applying divergence theorem to LHS, we get

   

   Two volume integrals are equal if their integrands are equal,

   that is        ∇ . D = ρ_υ

   Hence

4. We have Gauss’s law for magnetic fields as
   

   RHS is zero as there are no isolated magnetic charges and the magnetic flux lines are closed loops.

   Applying divergence theorem to LHS, we get

   

   or,           ∇.B = 0     Hence proved.

4.11 SINUSOIDAL TIME VARYING FIELD

In practice, electric and magnetic fields vary sinusoidally. It is well-known that any periodic variation can be described in terms of sinusoidal variations with fundamental and harmonic frequencies.

The fields can be represented by

 

or,

 

where ω = 2π f, f = frequency variation of the field, E_m is the maximum field strength.

It means that a sinusoidal time factor is attached to the field. It is also possible to represent the fields using phasor notation. The time varying field Ẽ (r, t) is related to phasor field E (r) as

 

or,

 

 

It may be noted that cosinusoidal variation is also considered to be sinusoidal in usage.

4.12 MAXWELL’S EQUATIONS IN PHASOR FORM

1. Consider the first Maxwell’s equation
   

   If

   

   it becomes,

   or, Re {∇ × H − jωD − J) e^jωt} = 0

   or,                 ∇ × H = jωD + J

   This is the first Maxwell’s equation in phasor form.

2. Consider the second Maxwell’s equation
   

   If              Ẽ = Re {E e^jωt}

   and             

   the second Maxwell’s equation becomes,

   

   or,     Re [(∇ × E + jωB) e^jωt] = 0

   or,           ∇ × E = − jωB

   This is the second Maxwell’s equation in phasor form.

3. Similarly, consider the third Maxwell’s equation,
   

   that is,        ∇.Re (D e^jωt) = ρ_υ

   or,              ∇.D = ρ_υ

4. Consider the fourth Maxwell’s equation
   

   that is,        ∇.Re (B e^jωt) = 0

   or,              ∇.B = 0

   In summary, Maxwell’s equations in phasor form are:

          ∇ × H = jωD + J

          ∇ × E = −jωB

             ∇.D = ρ_υ

             ∇.B = 0

4.13 INFLUENCE OF MEDIUM ON THE FIELDS

When the sources of electric and magnetic fields exist in a medium, the medium has influence on the characteristics of the fields. The constitutive relations, namely,

 

describe the characteristics.

Here ∈ = permittivity (F/m), μ = permeability (H/m) and σ = conductivity (mho/m) of the medium.

4.14 TYPES OF MEDIA

Medium can be divided into five types:

1. Homogeneous medium
2. Non-homogeneous medium
3. Isotropic medium
4. Anistropic medium
5. Source-free regions

Problem 4.1     Given E = 10sin (ωt –βz) a_y V/m in free space, determine D, B, H.

Solution

Second Maxwell’s equation is

 

that is,

or,

As

Now ∇ × E becomes

or,

and

Problem 4.2     If the electric field strength, E of an electromagnetic wave in free space is given by V/m find the magnetic field, H.

Solution     We have

or,

or,

Thus,

Problem 4.3     The parallel plates in a capacitor have an area of 4 x 10^–4 m² and are separated by 0.4 cm. A voltage of 10 sin 10³ t volts is applied to the capacitor. Find the displacement current when the dielectric material between the plates has a relative permittivity of 4.

Solution     We have

 

or,

The displacement current density, J_d is

But

where d = plate separation = 0.4 cm = 0.004 m.

So,

The displacement current, I_d is

But           V = 10 sin 10³ t volt

So,

Problem 4.4     In free space, the magnetic field of an EM wave is given by H = 0.4ω∈₀ cos (ωt – 50x) a_z A/m. Find the electric field, E and displacement current density, Ḋ.

Solution             H = 0.4ω∈₀ cos (ωt – 50x) a_z A/m

We have       ∇ × H = Ḋ + J

But J = 0 for free space.

So,

But

[H is not a function of y]

So,

that is,

or,

The displacement current density, J_d

Problem 4.5     If there is a magnetic field represented by

 

in a medium where ρ_υ = 0, σ = 0 and J = 0, find the electric field. Assume ∈_r = 1, μ_r = 1.

Solution     We have ∇ × H = Ḋ + J

But              J = 0

as                    D = ∈₀E, B = μ₀ H

We can write    

or,          

or,

As B_x and B_y are independent of y and z,

But

or,

Problem 4.6     An electric field in a medium which is source-free is given by E = 1.5 cos (10⁸t – βz)a_x V/m, where E_m is the amplitude of E, ω is the angular frequency and β is the phase constant. Obtain B, H, D. Assume ∈_r = 1, μ_r = 1, σ = 0.

Solution     We have

As E_x is not a function of y,

or,

As           B = μ H, μ = μ₀ = 4 π × 10⁻⁷ H/m

But           D = ∈ E = ∈₀E, ∈₀ = 8.854 × 10⁻¹² F/m

So,

Problem 4.7     Verify whether the following fields

satisfy Maxwell’s equations in free space.

Solution           ∇ × H = Ḋ

 

[as J = 0]

that is,

or,           sin x cost = 2∈₀ sin x cost

or,              μ₀∈₀ = 1

which cannot be satisfied. Therefore, the given fields do not satisfy Maxwell’s equations.

Problem 4.8     In a medium of conduction current density given by

 

find the volume charge density.

Solution     J = 3.0sin (ωt – 10z)a_y + cos (ωt- 10z)a_z

By the equation of continuity, we have

that is,        _υ = −10 sin (ωt − 10z)

or,

Problem 4.9     If the electric field strength of a radio broadcast signal at a TV receiver is given by

 

determine the displacement current density. If the same field exists in a medium whose conductivity is given by 2.0 × 10³ (mho)/cm, find the conduction current density.

Solution    E at a TV receiver in free space

 

Electric flux density,

 

The displacement current density

The conduction current density,

Problem 4.10     Find the electric flux density and volume charge density if the electric field,

 

in a medium whose ∈_r = 2.

Solution

From Maxwell’s equation, we have

4.15 SUMMARY OF MAXWELL’S EQUATIONS FOR DIFFERENT CASES

4.16 CONDITIONS AT A BOUNDARY SURFACE

Relations between the main fields E, D, H, B are expressed in terms of Maxwell’s equations. These are valid at any point in a continuous medium.

As Maxwell’s equations contain space derivatives, they cannot give information at points of discontinuity in the medium. However, Maxwell’s equations in integral form can be used to get the information at the boundary surface between different media.

The electromagnetic fields that are solved using Maxwell’s equations must satisfy boundary conditions at the interface between different media.

The boundary conditions on electric and magnetic fields at any surface of discontinuity are given by:

1. The tangential component of electric field, E is continuous across any discontinuity, that is,

    

   E_tan1 = E_tan2

   Subscript tan 1 is the tangential component of the field at the boundary in medium 1. Subscript tan 2 represents it in medium 2.

2. The tangential component of magnetic field, H is continuous across any surface except at the surface of a perfect conductor. At the surface of a perfect conductor, the tangential component of H is discontinuous by quantity equal to the surface current density (A/m).

    

   H_tan1 − H_tan2 = J_s

    

3. The normal component of B is continuous across any discontinuity.

    

   B_n1 = B_n2

    

4. The normal component of D is continuous except at a surface which has surface charge density. At the surface where surface charge density exists, the normal component of D is discontinuous by a quantity equal to the surface charge density.

    

   D_n1 − D_n2 = ρ_s

   Subscript n1 represents normal component of the field in medium 1 and n2 represents it in medium 2.

4.17 PROOF OF BOUNDARY CONDITIONS ON E, D, H AND B

To Prove     E_tan1 = E_tan2

Consider the rectangular loop on the boundary of two media (Fig. 4.1).

It is well-known that electric field is conservative and hence the line integral of E . dL is zero around a closed path.

So,

From the figure shown above, LHS is written as

As Δy → 0, we get

or,           E_x1 = E_x2

It is obvious that E_x1 and E_x2 are the tangential components of E in medium 1 and 2 respectively.

So,           E_tan1 = E_tan2

Now consider a cylinder across the media 1 and 2 (Fig. 4.2).

According to Gauss’s law,

Applying this to the cylindrical surface on the boundary spreading over medium 1 and medium 2, we get, as Δy → 0

 

or,

To Prove     H_tan1 – H_tan2 = J_s

Consider the rectangular loop on the boundary of two media (Fig. 4.3).

From Ampere’s circuit law, we have

As Δy → 0, we get

or,

Here, H_x1 and H_x2 are nothing but tangential components in medium 1 and 2 respectively.

 

To prove     B_n1 = B_n2

Now consider a cylinder shown in Fig. 4.4.

Gauss’s law for magnetic fields is

Applying LHS to the cylindrical surfaces shown, we get, for Δy → 0

Given the fields in one medium, it is usually required to determine fields in a second medium. This requires the knowledge of both tangential and normal components for each field. The above boundary conditions yield either tangential or normal component. The second unknown component is determined from the constitutive relations between E and D, H and B.

We have           E_{tan 1} = E_{tan 2}

But                         D = ∈ E

We have        D_n1 − D_n2 = ρ_s

 

We have        B_n1 = B_n2

But

                     B = μH

             μ₁ H_n1 = μ₂ H_n2

We have           H_{tan 1} − H_{tan 2} = J_s

4.18 COMPLETE BOUNDARY CONDITIONS IN SCALAR FORM

          E_{tan 1} − E_{tan 2} = 0

          ∈₁ E_n1 − ∈₂ E_n2 = ρ_s

          H_{tan 1} − H_{tan 2} = J_s

          μr₁ H_n1 − μ_r2 H_n2 = 0

          B_n1 − B_n2 = 0

4.19 BOUNDARY CONDITIONS IN VECTOR FORM

 

 

 

 

 

Problem 4.11     x < 0 defines region 1 and x > 0 defines region 2. Region 1 is characterised by μ_r1 = 3.0 and region 2 is characterised by μ_r2 = 5.0. If the magnetic field in region 1 is given by H₁ = 4.0a_x + 1.5a_y – 3.0a_z, A/m, find H₂ and H₂.

Solution             H₁ = 4.0a_x + 1.5a_y – 3.0a_z, A/m

For the regions given,

and          

As

and

or,          

and,          

The magnitude of H₂ is

Problem 4.12     In a three-dimensional space, divided into region 1 (x < 0) and region 2 (x > 0), σ₁ = σ₂ = 0. E₁ = 1a_x + 2a_y + 3a_z. Find E₂ and D₂.∈_r1 = 1 and ∈_r2

 

Solution	Et1 = 2ay + 3az, V/m
	En1 = ax , ∈r1 = 1
	Dt1 = ∈0 (2ay + 3az ), C/m2
As	Et1 = Et2
	Et2 = 2ay + 3az , V/m
As	Dn1 = Dn2
	∈1 En1 = ∈2En2

or,

or,       

          D₂ = ∈₂E₂

                = 2 ∈₀ (0.5a_x + 2a_y + 3a_z)

or,                

4.20 TIME VARYING POTENTIALS

It is always useful to relate electric and magnetic fields in terms of their sources. However, it is often more convenient to relate potentials in terms of sources and then the fields in terms of potentials.

In the first method potentials already developed for static fields are generalised to obtain time varying fields. Scalar electrostatic potential is given by

and vector magnetic potential is

But the time varying potentials are expressed in the form of

These time varying potentials are due to time varying charge and current distributions.

The above expressions do not take care of propagation delay. To obtain far-field expressions, this delay time must be taken into account.

4.21 RETARDED POTENTIALS

These are defined as the potentials in which a time delay or retarded time is taken into account. They are expressed as:

where retarded time or delay time is

             t_d = (t − r/υ₀)

             υ₀ = velocity of propagation of the EM wave

From the above approach, there is no change in the expressions for E and H, that is,

                E = − ∇V

and          

4.22 MAXWELL’S EQUATIONS APPROACH TO RELATE POTENTIALS, FIELDS AND THEIR SOURCES

By definition, vector magnetic potential, A is related to magnetic flux density as

 

But the second Maxwell’s equation is

 

From the above two expressions, we have

or,              ∇ × E + ∇ × Ȧ = 0

or,              ∇ × (E + Ȧ = 0

This is true only if (E + Ȧ) is the gradient of a scalar. Therefore, setting (E + Ȧ) = – ∇V, we get E as

 

Note that E is not equal to (−∇V) for time varying fields but it is given by the above expression.

So,           E = − ∇V is valid only for static fields

Now consider

 

Take curl on both sides,

 

or,

But        ∇ × H = Ḋ + J = ∈₀ Ė + J

From the above expressions, we get

             ∈₀ Ė + J =− ∈₀ ∇ − ∈₀ Ä + J

             ∇ × ∇ × A = −∈₀ ∇V − ∈₀ Ä + J

But vector identity gives

 

that is,     ∇∇.A − ∇² A = μ₀ ∈₀ ∇ − μ₀ ∈₀ Ä + μ0 J

The third Maxwell’s equation is

 

This becomes        ∇.E = − ∇.∇V − ∇.Ȧ = ρ_υ/∈₀

or,           ∇²V + ∇Ȧ = −ρ_υ/∈₀

and           ∇² A − μ₀ ∈₀ Ä = −μ₀ J + μ₀ ∈₀ ∇ + ∇∇.A

Each of the equations contain both the potentials and one of the sources. It is difficult to solve such type of equations. We should be able to get equations containing one potential and its own source to solve them easily. Helmholtz theorem helps to do this.

4.23 HELMHOLTZ THEOREM

 

To specify A, we know its curl, that is,

 

But the ∇.A is specified as

 

 

and           ∇² V − μ₀ ∈₀ = −ρ_υ/∈₀

These are the expressions between time varying potentials and their sources.

The corresponding expressions for static fields are:

 

 

4.24 LORENTZ GAUGE CONDITION

It is given by

 

Potential functions for sinusoidal fields are given by

       ∇² A + ω² μ₀ ∈₀ A = −μ₀ J

       ∇² V + ω² μ₀ ∈₀ V = −ρ_υ/∈₀

Problem 4.13     If the retarded scalar electric potential, V = x – υ₀t and vector magnetic potential, where υ₀ is the velocity of propagation,

1. find ∇. A
2. find B, H, E, and D
3. show that

Solution    (a)

But

(b) We have           B = ∇ × A

Here

As and A_x is not function of z,

Therefore,        ∇ × A = 0

and                

As

Here

or,              E = − ∇V − Ȧ

and

(c)             

But          

or,

 

1. E = – ∇ V is valid for all types of fields.	(Yes/No)
2. is magnetic current density.	(Yes/No)
3. Theta polarisation is called linear polarisation.	(Yes/No)
4. B and D can be related.	(Yes/No)
5. The characteristic impedance of a medium is	(Yes/No)
6.	(Yes/No)
7. ∇.E =− Ḃ	(Yes/No)
8. ∇.J =	(Yes/No)
9. In free space, ∇. D = 0.	(Yes/No)
10. In free space, ∇ × H = J.	(Yes/No)
11. For static fields, ∇ × H = D.	(Yes/No)
12. The unit of D is wb/m2.	(Yes/No)
13. The unit of Ḋ is A/m2.	(Yes/No)
14. The unit of J is A/m.	(Yes/No)
15. E and D have the same direction.	(Yes/No)
16. B and H have the same direction.	(Yes/No)
17. Ampere’s circuit law and Maxwell’s first equation are the same.	(Yes/No)
18. Bn1 is always equal to Bn2.	(Yes/No)
19. Dn1 is always equal to Dn2.	(Yes/No)
20. Etan 1 is always equal to Etan2.	(Yes/No)
21. Dt1 is always equal to Dt2.	(Yes/No)
22. Surface current density in dielectrics is zero.	(Yes/No)
23. ρυ = 0 for free space.	(Yes/No)
24. In dielectrics, displacement current density is greater than conduction current density.	(Yes/No)
25. In conductors, Jc > Jd.	(Yes/No)
26. ∇. B = 0 because there exists no isolated magnetic poles.	(Yes/No)
27. The unit of vector magnetic potential is _____.
28. The unit of magnetic current density is _____.
29. If E = cos (6 × 107 t −β z) ax, β is _____.
30. For time varying fields, E = _____.

1. The first Maxwell’s equation in free space
   1. ∇ × H = Ḋ + J
   2. ∇ × H = Ḋ
   3. ∇ × H = 0
   4. ∇ × H = J
2. Absolute permeability of free space is
   1. 4π × 10^-7 A/m
   2. 4π × 10^-7 H/m
   3. 4π × 10^-7 F/m
   4. 4π × 10^-7 H/m²
3. For static magnetic field,
   1. ∇ × B = ρ
   2. ∇ × B = μJ
   3. ∇ × B = μ₀J
   4. ∇ × B = 0
4. The electric field in free space
   1. 
   2. 
   3. ∈₀ D
   4. 
5. Displacement current density is
   1. D
   2. J
   3. ∂D/∂t
   4. ∂J/∂t
6. The time varying electric field is
   1. E = − ∇V
   2. E = − ∇V − Ȧ
   3. E = − ∇V − B
   4. E = − ∇V − D
7. A field can exist if it satisfies
   1. Gauss’s law
   2. Faraday’s law
   3. Coulomb’s law
   4. All Maxwell’s equations
8. If σ = 2.0 mho/m, E = 10.0 V/m, the conduction current density is
   1. 5.0 A/m²
   2. 20.0 A/m²
   3. 40.0 A/m²
   4. 20 A
9. Maxwell’s equations give the relations between
   1. different fields
   2. different sources
   3. different boundary conditions
   4. none of these
10. The boundary condition on E is
    1. a_n × (E₁ − E₂) = 0
    2. a_n.(E₁ − E₂) = 0
    3. E₁ = E₂
    4. none of these
11. The boundary conditions on H is
    1. a_n × (H₁ − H₂) = J_s
    2. a_n .(H₁ − H₂) = 0
    3. a_n × (H₁ − H₂) = J_s
    4. a_n .(H₁ − H₂) = 0
12. If E = 2 V/m, of a wave in free space, (H) is
    1. 
    2. 60 πA/m
    3. 120 πA/m
    4. 240 πA/m
13. cosine of the angle between the two vectors is
    1. sum of the products of the directions of the two vectors
    2. difference of the products of the directions of the two vectors
    3. product of the products of the directions of the two vectors
    4. none of these
14. The electric field intensity, E at a point (1, 2, 2) due to (1/9) nc located at (0, 0, 0) is
    1. 33 V/m
    2. 0.333 V/m
    3. 0.33 V/m
    4. zero
15. If E is a vector, then ∇ . ∇ × E is
    1. 0
    2. 1
    3. does not exist
    4. none of these
16. The Maxwell’s equation, ∇. B = 0 is due to
    1. B = μ H
    2. 
    3. non-existence of a mono pole
    4. none of these
17. For free space,
    1. σ = ∞
    2. σ =0
    3. J ≠ 0
    4. none of these
18. The electric field for time varying potentials
    1. E = − ∇V
    2. E = − ∇V − A
    3. E = ∇V
    4. E = − ∇V + A
19. The intrinsic impedence of the medium whose σ = 0, ∈_r = 9, μ_r = 1 is
    1. 40 πΩ
    2. 9 Ω
    3. 120 πΩ
    4. 60 πΩ
20. For time varying EM fields
    1. ∇ × H = J
    2. ∇ × H = Ḋ + J
    3. ∇ × E = 0
    4. none of these
21. The wavelength of a wave with a propagation constant = 0.1π + j 0.2π is
    1. 10 m
    2. 20 m
    3. 30 m
    4. 25 m
22. The electric field just above a conductor is always
    1. normal to the surface
    2. tangential to source
    3. zero
    4. ∞
23. The normal components of D are
    1. continuous across a dielectric boundary
    2. discontinuous across a dielectric boundary
    3. zero
    4. ∞
24. If J_c = 1 mA/m² in a medium whose conductivity is σ = 10 Mho/m, E is
    1. 0.1 V/m
    2. 10μ V/m
    3. 1.0μ V/m
    4. 10 V/m.
25. If J_d = 2 mA/m² in a medium whose ∈_r = 2, σ= 4.95 Mho/m at a frequency of 1 GHz, J_c is
    1. 8.9 mA/m²
    2. 89 mA/m²
    3. 0.89 mA/m²
    4. 89 A/m²

1. If A is the vector magnetic potential, prove
2. If E = 2.0 sinkx cosωta_x in free space, find the volume charge density.
3. If E = 10 cos (ωt – kz)a_x V/m, find D, H, B in free space.
4. If the electric field strength of an EM wave in free space has an amplitude of 5.0 V/m, find the magnetic field strength.
5. When the amplitude of a magnetic field in free space is 10 mA/m, find the magnitude of the electric field.
6. The electric field of an EM wave is Find H.
7. If Ẽ = 2 cos (10⁸t – 20x + 40°)a_z, what is the phasor form of E?