1. 1 30 Outline Maxwell’s Equations and the Displacement Current Electromagnetic Waves Polarization

2. 2 Maxwell’s Equations Gauss’ Law: E & B Faraday’s Law Ampere’s Law

3. 3 displacement current ‘explains’ existence of B around E

4. 4 EM waves accelerating charges produce ‘waves’ of E and B can be ‘pulse’ or ‘harmonic wave’

5. 5 dipole radiation

6. 6 EM waves transverse: E perpendicular to B E and B are in phase speed: c = f = 3  10 8 m/s

7. 7 Electric Dipole Radiation I(r = 1.0m, angle = 90) is 12 W/m 2. I at 2.0m and angle of 30 degrees is: Example:

8. 8 antennas can respond to E or B

9. 9 Light Light is an electromagnetic wave c = f ≈ 3  10 8 m/s As light waves travel through space they: »transport energy and/or information »transport momentum

10. 10

11. 11 EM Waves carry energy and momentum, shared equally between the electric and magnetic fields.

12. 12 Energy and Momentum in EM Waves Intensity: energy/area/time [watts/sq.meter]

13. 13 Example 50W Bulb Assume that 5.00% of the electrical power consumed by the bulb is converted to light. The average intensity at a distance of r = 1.00m: The rms value of E:

14. 14 Polarization Unpolarized light is the superposition of many waves with random direction of E. Linearly Polarized light has only one direction of E.

15. 15 Polarizing Filters Polarizing material only allows the passage of only one direction of E Malus ’ Law:

16. 16 Two Filters (incident light unpolarized) 1 st reduces intensity by 1/2. 2 nd reduces according to Malus’ Law

17. 17 Polarization by Scattering and Reflection Light scattered at 90 degrees is 100% polarized.

18. 18 Summary displacement current added to Ampere’s Law: completes Maxwell Eqs., which explain ‘light’ properties (transverse EM wave with speed c) visible light small segment of spectrum energy density and pressure polarization by reflection/scattering Malus’ Law

19. 19 30-4 The Wave Equation for Electromagnetic Waves Omit End

20. 20 Momentum momentum = U/c The total energy received in the time by an area A The momentum received The average force Radiation pressure

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23. 23 Example (cont.) Part (b) 2. Use to find 3. Use to find

24. 24 Example: Consider a laser that puts out an average power of P=1.0 milliwatt in a beam having a diameter of 1.0 mm. What is the peak amplitude of the electric field? The area of the laser beam is The electromagnetic flux S is Recall so that Substitution of the numerical values yields and thus