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30 Outline

30 Outline. Maxwell’s Equations and the Displacement Current Electromagnetic Waves Polarization. displacement current. ‘explains’ existence of B around E. Maxwell’s Equations. Gauss’ Law: E & B Faraday’s Law Ampere’s Law. EM waves. transverse: E perpendicular to B speed:

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30 Outline

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

  2. displacement current • ‘explains’ existence of B around E

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

  4. EM waves • transverse: • E perpendicular to B • speed: • c = fl = 3  108 m/s

  5. dipole radiation

  6. Electric Dipole Radiation Example: I(r = 1.0m, angle = 90) is 12 W/m2. I at 2.0m and angle of 30 degrees is:

  7. antennas • can respond to E or B

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

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

  10. 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:

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

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

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

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

  15. 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, intensity • polarization by reflection/scattering • Malus’ Law

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

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

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

  19. 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

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