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Today’s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.

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Presentation on theme: "Today’s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave."— Presentation transcript:

1 Today’s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.

2 The magnitude S represents the rate at which energy flows through a unit surface area perpendicular to the direction of wave propagation (energy per time per area). Energy Carried by Electromagnetic Waves Electromagnetic waves carry energy, and as they propagate through space they can transfer energy to objects in their path. The rate of flow of energy in an electromagnetic wave is described by a vector S, called the Poynting vector.* Thus, S represents power per unit area. The direction of S is along the direction of wave propagation. The units of S are J/(s·m 2 ) =W/m 2. *J. H. Poynting, 1884. This is derived from Maxwell’s equations.

3 z x y c S E B Because B = E/c we can write These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area. For an EM wave so

4 The time average of sin 2 (kx -  t) is ½, so EM waves are sinusoidal. The average of S over one or more cycles is called the wave intensity I. This equation is the same as 32-29 in your text, using c = 1/(  0  0 ) ½. Notice the 2’s in this equation. EM wave propagating along x-direction

5 Thus, The magnitude of S is the rate at which energy is transported by a wave across a unit area at any instant: Note: S average and mean the same thing!

6 The energy densities (energy per unit volume) associated with electric and magnetic fields are: Using B = E/c and c = 1/(  0  0 ) ½ we can write Energy Density remember: E and B are sinusoidal functions of time

7 For an electromagnetic wave, the instantaneous energy density associated with the magnetic field equals the instantaneous energy density associated with the electric field. Hence, in a given volume the energy is equally shared by the two fields. The total energy density is equal to the sum of the energy densities associated with the electric and magnetic fields:

8 When we average this instantaneous energy density over one or more cycles of an electromagnetic wave, we again get a factor of ½ from the time average of sin 2 (kx -  t). so we see that Recall The intensity of an electromagnetic wave equals the average energy density multiplied by the speed of light. and instantaneous energy densities (E and B vary with time)

9 “These factors of ¼, ½, and 1 are making my brain hurt!” Help! It’s really not that bad. It’s really not that bad. These are the energy densities associated with E(t) and B(t) at some time t: Add u B and u E to get the total energy density u(t) at time t:

10 Help! Again, these are the energy densities associated with E(t) and B(t) at some time t: If you average u B and u E over one or more cycles, you get an additional factor of ½ from the time average of sin 2 (kx-  t). The E max and B max come from writing E = E max sin(kx-  t) and B = B max sin(kx-  t), and canceling the sine factors.

11 Help! These are the average energy densities associated with E(t) and B(t) over one or more complete cycles. Add  u E  and  u B  to get the total average energy density over one or more cycles:

12 Help! Summary: At time t: Average: At time t: Average: If you use a starting equation that is not valid for the problem scenario, you will get incorrect results!


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