Solved Problems of Electromagnetic Waves

ANNA UNIVERSITY SOLVED PROBLEMS

Problem 2.1: 

A circular cross section conductor of radius 2 mm carries a current Ic = 2.5 sin (5 × 108t) μ A. What is the amplitude of the displacement current density if σ = 35 M Ω-1'm-1 and εr = 1.

Given data

Ic = 2.5 sin (5 × 108t) μ A

= 2.5 sin (5 × 108t) × 10-6A

σ = 35 MS m-1= 3.5 × 106 S m-1 

εr = 1

ε0 = 8.85 ×10-12 Fm-1

ω = 5 × 108 rads-1 

radius a = 2 mm = 2 × 10-3 m

Solution

We know that

JC = σE

Taking only the amplitude IC

Problem 2.2: 

The conduction current flowing through a wire with conductivity σ = 3 × 107 S/m and relative permittivity εr = 1 is given by Ic = 3 sin ωt (mA). If ω = 108 rad/sec. Find the displacement current.

Given data

Conductivity σ = 3 × 107Sm-1 

Conducting current Ic = 3 sin ωt mA

= 3 × 10-3 sin ωt A

Angular frequency ω = 108 rad s-1

Solution:

Displacement current density

Since Ic is given, E is found from conduction current density as

Jc = σE 

Multiplying both sides by ε

substituting for JD

Problem 2.3: 

A poor conductor is characterised by a conductivity σ = 100 (S/m) and permittivity ε = 4 ε0. At what angular frequency ω is the amplitude of the conduction current density Jc equal to the amplitude of the displacement current density JD?

Given data

Conductivity of the conductor σ = 100 Sm-1

Permittivity of the conductor ε = 4 ε0

Solution:

Jc = σ E

Equating conduction current density and displacement current density, we have

σ E = ε ωe

σ = ω ε

100 = ω x 4 ε 0

Problem 2.4: 

In a material for which σ = 5 S/m and εr = 1 the electric field intensity is E = 250 sin 1010t V/m. Find the conduction and displacement current densities and the frequency at which they have equal magnitudes. 

Given data

Conductivity of medium σ = 5 Sm-1

Relative permittivity of medium εr = 1

Solution 

Conduction current density

Jc = σ E

Jc = 5 × 250 sin 1010t

= 1250 sin 1010t Am-2 

Displacement current density

Problem 2.5: 

Find the conduction and displacement current densities in a material having conductivity of 10-3S/m and εr = 2.5 if the electric field in the material is E = 5.0 × 10-6 sin 9.0 × 109t + V/m.

Given data

Conductivity of the medium σ = 10-3 Sm-1 

Relative permittivity εr = 2.5

Solution

Conduction current density

Jc = σE = 10-3 × 5 × 10-6 sin 9 × 109t Am-2 

= 5 ×10-9 sin 9.0 × 109t

Displacement current density

Problem 2.6 :

A light bulb of 20 W radiates energy isotropically, mostly in infrared region. Assuming it to be a point source find the irradiance (average energy per unit area per unit time or intensity) at a distance 1 m away. Calculate the strength of -field associated with the radiation at this distance.

Given data:

Power radiated by bulb P = 20 W

Distance r = 1 m

Solution:

Irradiance is average energy per unit area per unit time at a distance 'r’

Intensity at a point is nothing but poynting vector at that point.

ie., I = Savg

substituting the given values, we have

= 24.5 Vm-1

Problem 2.7 : 

The intensity of sunlight reaching the earth's surface is about 1300 W m-2 . Calculate the strength of electric and magnetic fields of the incoming sunlight.

Solution:

The time average Poynting's vector is

But in electromagnetic wave

from above relation

Problem 2.8 :

(i) Show that for a good conductor skin depth δ = λc/2 π, where λc, is the wavelength of electromagnetic waves in the conductor. 

(ii) Show that for an electromagnetic wave incident on a good conductor the electric vector reduces to about 1% at a depth of 0.73 % λc 

(iii) Find the wavelength and the propagation speed in copper for radio waves at 1 MHz. For copper assume μ = μ0, ε = ε0 and conductivity σ = 5.8 × 107Ω - m-1

Solution:

(i) Suppose the wave is incident normally on the surface of the conductor along z-axis, then the electric field inside the conductor is given by

k - wave vector

For a good conductor

and skin depth

The wavelength of the wave in the conductor is

(ii) Let us write

taking in on both sides

(iii) The required wavelength at 1 MHz = 1 × 10⁶ Hz

and the propagation speed

Problem 2.9 : 

Find the depth of penetration of a plane wave in copper at a power frequency of 60 Hz and at microwave frequency 1010 Hz. Given σ = 5.8 × 107 mho m-1 

Given data

σ = 5.8 x 107 mho m-1

Depth of penetration

(i) f = 60 Hz, σ = 5.8 × 107 mho m-1, μr = 1

μ0 = 4π × 10-7Hm-1

ω = 2πf

Problem 2.10: 

A sinusoidal plane wave is transmitted through a medium whose electric field is 10 kV/m and relative permittivity of the medium is 4. Determine the mean rms power flow/unit area.

Solution:

E = 10 KVm-1 = 10 × 103 V m-1

εr = 4, μr = 1

Power flow/unit area is poynting vector

Problem 2.11: 

Find the conducting behaviour of ground at 1 KHz, 10 MHz and 10 GHz. Given εr = 10 and σ = 5 × 10-3 mho/m.

Solution:

The ratio of conduction current to displacement current is σ/ωε which determines conducting behaviour

For f = 1 kHz = 1 × 103 Hz

ω = 2 πf

σ = 5 × 10-3 mho m-1

εr = 10

For f = 1 MHz = 1 ×106 Hz

For f = 10 GHz = 10 × 109 Hz

= 9 × 10-4

At 1 KHz the ground acts like a good conductor and at 10 GHz it acts like an insulator. For 10 MHz ground acts like a quasi-conductor.

Problem 2.12 : 

Find the skin depth 8 at a frequency of 1 MHz for copper where σ = 5.8 × 107S/m and μ = μ0. Also find the value of skin depth at 50 Hz.

Given data

Conductivity of the medium σ =5.8 × 107 Sm-1

Frequency f = 1 MHz = 1 × 106 Hz

Solution:

μ = μ0 μr and f = 1 MHz

Substituting the given values, we have

= 0.0661 × 10-3 m

= 6.61 × 10-5m 

Also at f = 50 Hz,

= 9.348 × 10-3 m

Problem 2.13: 

Find the velocity of a plane wave in a lossless medium having a relative permittivity of 4 and a relative permeability of 1.2.

Given data

εr = 4; μr = 1.2

Solution:

Velocity of EM wave in a medium is,

Problem 2.14: 

Find the characteristic impedance of the medium whose relative permittivity is 3 and relative permeability is 1.

Given data

εr = 3; μr = 1

Solution:

Problem 2.15: 

Find the velocity of a plane wave in a lossless medium having a relative permittivity of 5 and relative permeability of unity.

Given data

εr = 5, μr = 1

Solution:

Velocity of propagation in the medium