Title: 30 Outline
130 Outline
- Maxwells Equations and the Displacement Current
- Electromagnetic Waves
- Polarization
2Maxwells Equations
- Gauss Law E B
- Faradays Law
- Amperes Law
3displacement current
- explains existence of B around E
4EM waves
- accelerating charges produce waves of E and B
- can be pulse or harmonic wave
5dipole radiation
6EM waves
- transverse E perpendicular to B
- E and B are in phase
- speed c fl 3 ? 108 m/s
7Electric Dipole Radiation
Example
I(r 1.0m, angle 90) is 12 W/m2.
I at 2.0m and angle of 30 degrees is
8antennas
9Light
- Light is an electromagnetic wave
- c lf 3 ? 108 m/s
- As light waves travel through space they
- transport energy and/or information
- transport momentum
-
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11EM Waves
- carry energy and momentum, shared equally between
the electric and magnetic fields.
12Energy and Momentum in EM Waves
- Intensity energy/area/time watts/sq.meter
13Example 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
14Polarization
- Unpolarized light is the superposition of many
waves with random direction of E. - Linearly Polarized light has only one direction
of E.
15Polarizing Filters
- Polarizing material only allows the passage of
only one direction of E - Malus Law
-
16Two Filters (incident light unpolarized)
- 1st reduces intensity by 1/2.
- 2nd reduces according to Malus Law
17Polarization by Scattering and Reflection
- Light scattered at 90 degrees is 100 polarized.
18Summary
- displacement current added to Amperes 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
1930-4The Wave Equation for Electromagnetic Waves
Omit
End
20Momentum
The total energy received in the time by an
area A
The momentum received
The average force
Radiation pressure
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23Example (cont.)
Part (b) 2. Use to find
3. Use to find
24Example Consider a laser that puts out an
average power of P1.0 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