Presentation is loading. Please wait. # 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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PH0101 UNIT 2 LECTURE 31 PH0101 Unit 2 Lecture 3  Maxwell’s equations in free space  Plane electromagnetic wave equation  Characteristic impedance  Poynting vector  Physical significance

PH0101 UNIT 2 LECTURE 32 Representation of EM Waves in free space: In free space, the volume charge density (ρ) = 0 and conduction current density (J 1 ) = 0 (since  = 0 )

PH0101 UNIT 2 LECTURE 33 Maxwell’s equations in free space  In free space the Maxwell’s Equations becomes,  0 =  = 0 (1) (2) = (μ 0 H) =    = = (3) (4)   

PH0101 UNIT 2 LECTURE 34 Differentiating above eqn with respect to time, [Since D = ε 0 E] (5)    == From eqn (4)

PH0101 UNIT 2 LECTURE 35    = (6) Taking curl on both sides of above eqn From eqn (3)

PH0101 UNIT 2 LECTURE 36 Using equation (7) in (7)  2 E =  (8) =   2 E [since   E= 0]

PH0101 UNIT 2 LECTURE 37 The above equation is free space electromagnetic equation. In one dimension, Comparing (9) with standard mechanical wave equation, (9) (10)

PH0101 UNIT 2 LECTURE 38 m/s The velocity of electromagnetic wave in free space. (11) Similarly, the wave equation in terms of H can be written as,  2 H = (12)

PH0101 UNIT 2 LECTURE 39 In a medium of magnetic permeability  and electric permittivity , the wave equation becomes,  2 H =  2 E = (13) (14) The velocity of electromagnetic wave in any medium is, C = (15)

PH0101 UNIT 2 LECTURE 310 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 μ r =1. Solution  Frequency of the em wave = f = 3.0 MHz = 3  10 6 Hz  Relative permittivity of the non – magnetic medium = ε r = 4  Relative permeability of the non – magnetic medium = μ r = 1  Velocity of em wave in vacuum = C =

PH0101 UNIT 2 LECTURE 311  Wavelength of the EM wave in vacuum = λ = Velocity of em wave in non- magnetic medium =  Wavelength of the em wave in non-magnetic medium =

PH0101 UNIT 2 LECTURE 312  Therefore the change in wavelength = = = i.e. the wavelength decreased by 50 m.

PH0101 UNIT 2 LECTURE 313 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.

PH0101 UNIT 2 LECTURE 314 Characteristic Impedance The solution of the equation for the electric component in the electromagnetic wave is, E y = E o sin (ct – x) (1) For magnetic component, (ct – x)H z = H O sin(2) Differentiating equation (1) with respect to time, (3)

PH0101 UNIT 2 LECTURE 315 For three dimensional variation of H  H = (4)

PH0101 UNIT 2 LECTURE 316 Since H varies only in the Z – direction and wave traveling along X – axis, the component of H other than becomes zero in equation (4) From the fourth law of free space Maxwell’s equation, (5) From equations (4) and (5), (6)

PH0101 UNIT 2 LECTURE 317 Substituting in Eqn (6) (7) Integrating with respect to x, H z = (8)

PH0101 UNIT 2 LECTURE 318 H z =c  0 E 0 sin (ct – x)  0 E 0 sin (ct – x) H z = E 0 sin(ct – x) EyEy Hz =(9)

PH0101 UNIT 2 LECTURE 319 Characteristic Impedance of the medium (10) For free space, Z = = 376.8  For any medium, Z = ohm

PH0101 UNIT 2 LECTURE 320 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. 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:

PH0101 UNIT 2 LECTURE 321 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: and

PH0101 UNIT 2 LECTURE 322 Poynting vector  Poynting vector represent the rate of energy flow per unit area in a plane electromagnetic wave. The direction of gives the direction in which the energy is transferred. Unit: W/m 2

PH0101 UNIT 2 LECTURE 323 Representation of Poynting vector

PH0101 UNIT 2 LECTURE 324 Expression for energy density We know Taking divergence on (1) = (1) (2)

PH0101 UNIT 2 LECTURE 325. == (3)

PH0101 UNIT 2 LECTURE 326 == Integrating the equation (4) over the volume V, we get (4) (5)

PH0101 UNIT 2 LECTURE 327  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)

PH0101 UNIT 2 LECTURE 328 Energy associated with the electric field and that of the magnetic field [as and  which shows that instantaneous energy density associated with electric field i.e. energy is equally shared by the two fields. ]

PH0101 UNIT 2 LECTURE 329 Significance of P. 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

PH0101 UNIT 2 LECTURE 330

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