PH0101 UNIT 2 LECTURE 31 PH0101 Unit 2 Lecture 3 Maxwell’s equations in free space Plane...

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PH0101 UNIT 2 LECTURE 3 1 PH0101 Unit 2 Lecture 3 Maxwell’s equations in free space Plane electromagnetic wave equation Characteristic impedance Poynting vector Physical significance

Transcript of PH0101 UNIT 2 LECTURE 31 PH0101 Unit 2 Lecture 3 Maxwell’s equations in free space Plane...

Page 1: PH0101 UNIT 2 LECTURE 31 PH0101 Unit 2 Lecture 3  Maxwell’s equations in free space  Plane electromagnetic wave equation  Characteristic impedance

PH0101 UNIT 2 LECTURE 3 1

PH0101 Unit 2 Lecture 3

Maxwell’s equations in free spacePlane electromagnetic wave equationCharacteristic impedance Poynting vectorPhysical significance

Page 2: PH0101 UNIT 2 LECTURE 31 PH0101 Unit 2 Lecture 3  Maxwell’s equations in free space  Plane electromagnetic wave equation  Characteristic impedance

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Representation of EM Waves in free space:In free space, the volume charge density (ρ) = 0 and conduction current density (J1) = 0 (since = 0 )

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Maxwell’s equations in free space

In free space the Maxwell’s Equations becomes,

D

0 =

B

= 0

(1) (2)

E t

B

= t

(μ0 H) =

H t

D

=

= t

E

0

(3)

(4)

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Differentiating above eqn with respect to time,

t

22

2

0

2

t

E

t

D)H(

[Since D = ε0 E]

(5)

Ht

D

t

E

0==

From eqn (4)

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E

H0

2

2

0t

E

t

H)H(

=

)H()E(

0 (6)

Taking curl on both sides of above eqn

From eqn (3)

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Using equation (7) in

(7)

)H()E(

0

2 E = )H(

0

2

2

0002

t

E)H(E

(8)

E)E()E( 2

= 2E [since E= 0]

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The above equation is free space electromagnetic equation.In one dimension,

2

2

002

2

2

2

002

2

.1

)(x

E

t

Eor

t

E

x

E

22

22

2

x

yC

t

y

Comparing (9) with standard mechanical wave equation,

(9)

(10)

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m/s

8

0000

2 10311

C)or(C

The velocity of electromagnetic wave in free space.

(11)

Similarly, the wave equation in terms of H can be written as,

2

2

00t

H

2 H = (12)

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In a medium of magnetic permeability and electric permittivity , the wave equation becomes,

2

2

t

H

2

2

t

E

2 H =

2 E =

(13)

(14)

1

The velocity of electromagnetic wave in any medium is,

C = (15)

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Worked Example 2.1: An electromagnetic wave of frequency f = 3.0 MHz passes from vacuum into a non – magnetic medium with relative permittivity 4. Calculate the increment in its wavelength. Assume that for a non – magnetic medium μr=1.

SolutionFrequency of the em wave = f = 3.0 MHz = 3 10 6 Hz

Relative permittivity of the non – magnetic medium = εr = 4Relative permeability of the non – magnetic medium = μr = 1

00

1

Velocity of em wave in vacuum = C =

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Wavelength of the EM wave in vacuum = λ =

00

1.

1

ff

C

Velocity of em wave in non- magnetic medium =

rr

C 00

11

Wavelength of the em wave in non-magnetic medium =

rrff

C

00

1.

1

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Therefore the change in wavelength =

1

11.

1

00 rrf

=

m5014

1

103

1036

8

=

i.e. the wavelength decreased by 50 m.

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Worked Example 2.2 Prove that the current density is irrotational.

We know that , J = σ E

curl J = curl (σ E) = σ (curl E) [ since σ is a constant] = σ curl ( - grad V ) [E = - grad V] = - σ curl ( grad V) = 0 [ as curl ( grad V) = 0]

i.e. the current density is irrotational.

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Characteristic ImpedanceThe solution of the equation for the electric component in the electromagnetic wave is,

Ey = Eo sin (ct – x) (1)

2

For magnetic component,

2

(ct – x) Hz = HO sin (2)

Differentiating equation (1) with respect to time,

)(2

cos2

0 xctc

Et

Ey

(3)

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zyx HHHzyx

kji

H

For three dimensional variation of H

x

H

y

H

x

Hk

z

H

x

Hj

z

H

y

Hi zxyxzyz

H =

(4)

zyx HHHzyx

kji

H

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Since H varies only in the Z – direction and wave traveling along X – axis, the component of H other than

x

H z

becomes zero in equation (4)

From the fourth law of free space Maxwell’s equation,

t

EH

y

0 (5)

From equations (4) and (5),

t

E

x

H yz

0

(6)

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Substituting )(2

cos2

0 xctc

Et

Ey

in Eqn (6)

)(2

cos2

00 xctc

Ex

Hz

(7)

Integrating with respect to x,

2

)(2

sin2

00 xctc

E

Hz = (8)

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Hz = c0E0 sin

2

(ct – x)

00

1

0E0 sin 2

(ct – x)Hz =

0

0

Hz = E0 sin 2

(ct – x)

Ey Hz =0

0

(9)

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ZH

E

z

y 0

0

Characteristic Impedance of the medium

(10)

For free space, Z =0

0

= 376.8

For any medium, Z =

ohm

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Worked Example 2.3 Electromagnetic radiation propagating in free space

has the values of electric and magnetic fields 86.6 V m – 1 and 0.23 A m – 1 respectively. Calculate the characteristic impedance.

23.0

6.86

H

E

Electric field intensity = E =86.6 V m – 1

Magnetic field intensity = H = 0.23 A m – 1

Z = 376. 52 ohm

Characteristic impedance = Z =

Solution:

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Exercise Problem In a plane electromagnetic wave, the electric field

oscillates sinusoidally at a frequency of 2 10 10Hz and amplitude of 48 V m – 1. What is the wavelength of the wave? What is the amplitude of the oscillation of the magnetic field?

Hint: mf

C 2105.1 T106.1 700

C

EBand

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Poynting vector Poynting vector represent the rate of energy

flow per unit area in a plane electromagnetic wave.

)(P

HEBEP 0

1

The direction of )(P gives the direction in which the energy is

transferred. Unit: W/m2

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Representation of Poynting vector Y

X

Z

Ey

Hz

P

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Expression for energy density

HEP

We know

Taking divergence on (1)

H)E.(E)H.(H).(E

t

DE

t

BH

..=

HEP (1)

(2)

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. H).(E

t

DE

t

BH ..

t

HH

t

EE .. 00 ==

t

HH

t

EE ).2(

2

1).2(

2

100 == (3)

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H).(E

t

H

t

E 2

0

2

0

)(.

2

1)(.

2

1

20

20 2

1

2

1HE

t==

Integrating the equation (4) over the volume V, we get

VV

dVHEt

HE 20

20 2

1

2

1).(

(4)

(5)

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VS

dVHEt

dSHE 20

20 2

1

2

1).(

The term on the RHS within the integral of the equation (6) represents the sum of the energies of electric and magnetic fields.

Hence the RHS of the equation (6) represents the rate of flow of energy over the volume V.

Applying divergence theorem to the LHS of Eqn (5), we get

(6)

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Energy associated with the electric field 2

20 E

U E

and that of the magnetic field

0

220

22 BH

U m

Em UEC

BHU 2

00

22

0

2

1

22

[as CB

E and

00

1

C

which shows that instantaneous energy density associated with electric field i.e. energy is equally shared by the two fields.

]

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Significance of P

HEP

HE

The Vector P = E X H has interpreted as representing the amount of field energy passing through the unit area of surface in unit time normally to the direction of flow of energy.This statement is termed as Poynting’s theorem and the vector P is called Poynting Vector. The direction of flow of energy is perpendicular to vectors E and H E X H

i.e., in the direction of the vector

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