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 PH0101 UNIT 2 LECTURE 3

2. Representation of EM Waves in free space: In free space, the volume charge density (ρ) = 0 and conduction current density (J1) = 0 (since  = 0 ) PH0101 UNIT 2 LECTURE 3

3. Maxwell’s equations in free space • In free space the Maxwell’s Equations becomes,  = 0 (1)  0 = (2) = (μ0 H)  = (3)  = (4) = PH0101 UNIT 2 LECTURE 3

4. Differentiating above eqn with respect to time, From eqn (4)  = = [Since D = ε0 E] (5) PH0101 UNIT 2 LECTURE 3

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

6. Using equation (7) in = 2E [since  E= 0] (7) 2 E =  (8) PH0101 UNIT 2 LECTURE 3

7. The above equation is free space electromagnetic equation. In one dimension, (9) Comparing (9) with standard mechanical wave equation, (10) PH0101 UNIT 2 LECTURE 3

8. m/s (11) The velocity of electromagnetic wave in free space. Similarly, the wave equation in terms of H can be written as, 2H = (12) PH0101 UNIT 2 LECTURE 3

9. In a medium of magnetic permeability  and electric permittivity , the wave equation becomes, 2 H= (13) 2 E = (14) The velocity of electromagnetic wave in any medium is, C = (15) PH0101 UNIT 2 LECTURE 3

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

11. 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 3

12. Therefore the change in wavelength = = = i.e. the wavelength decreased by 50 m. PH0101 UNIT 2 LECTURE 3

13. 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 3

14. Characteristic Impedance For magnetic component, Hz = HOsin (ct – x) (2) The solution of the equation for the electric component in the electromagnetic wave is, Ey = Eosin (ct – x) (1) Differentiating equation (1) with respect to time, (3) PH0101 UNIT 2 LECTURE 3

15. For three dimensional variation of H H = (4) PH0101 UNIT 2 LECTURE 3

16. 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 3

17. Substituting in Eqn (6) (7) Integrating with respect to x, Hz = (8) PH0101 UNIT 2 LECTURE 3

18. Hz = c0E0 sin (ct – x) Hz = (ct – x) 0E0sin Hz = E0sin (ct – x) Ey Hz = (9) PH0101 UNIT 2 LECTURE 3

19. (10) Characteristic Impedance of the medium = 376.8  For free space, Z = For any medium, Z = ohm PH0101 UNIT 2 LECTURE 3

20. 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. Solution: Electric field intensity = E =86.6 V m – 1 Magnetic field intensity = H = 0.23 A m – 1 Characteristic impedance = Z = Z = 376. 52 ohm PH0101 UNIT 2 LECTURE 3

21. 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 3

22. 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/m2 PH0101 UNIT 2 LECTURE 3

23. Representation of Poynting vector PH0101 UNIT 2 LECTURE 3

24. Expression for energy density We know (1) Taking divergence on (1) = (2) PH0101 UNIT 2 LECTURE 3

25. . == (3) == PH0101 UNIT 2 LECTURE 3

26. == (4) Integrating the equation (4) over the volume V, we get (5) PH0101 UNIT 2 LECTURE 3

27. Applying divergence theorem to the LHS of Eqn (5), we get (6) • 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. PH0101 UNIT 2 LECTURE 3

28. 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 3

29. Significance ofP 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 HE X H . i.e., in the direction of the vector PH0101 UNIT 2 LECTURE 3

30. Have a nice day. PH0101 UNIT 2 LECTURE 3