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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 • 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: • c = fl = 3 108 m/s
Electric Dipole Radiation Example: I(r = 1.0m, angle = 90) is 12 W/m2. I at 2.0m and angle of 30 degrees is:
antennas • can respond to E or B
EM Waves • carry energy and momentum, shared equally between the electric and magnetic fields.
Energy and Momentum in EM Waves • Intensity: energy/area/time [watts/sq.meter]
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:
Polarization • Unpolarized light is the superposition of many waves with random direction of E. • Linearly Polarized light has only one direction of E.
Polarizing Filters • Polarizing material only allows the passage of only one direction of E • Malus’ Law:
Two Filters (incident light unpolarized) • 1st reduces intensity by 1/2. • 2nd reduces according to Malus’ Law
Polarization by Scattering and Reflection • Light scattered at 90 degrees is 100% polarized.
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
EM waves • accelerating charges produce ‘waves’ of E and B • can be ‘pulse’ or ‘harmonic wave’
Momentum • momentum = U/c The total energy received in the time by an area A The momentum received The average force Radiation pressure
Example (cont.) Part (b) 2. Use to find 3. Use to find
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