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Lecture 19 Electromagnetic energy, the Poynting vector and radiation pressure

Lecture 19 Electromagnetic energy, the Poynting vector and radiation pressure. Introduction

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Lecture 19 Electromagnetic energy, the Poynting vector and radiation pressure

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  1. Lecture 19 Electromagnetic energy, the Poynting vector and radiation pressure

  2. Introduction In previous lectures we have seen that energy can be associated with both electric and magnetic fields. Equations were derived for the energy densities of static E- and B-fields. An electromagnetic wave has both E- and B-field components and, in general, transports energy.

  3. Electromagnetic energy and the Poynting vector We require an expression which gives us the energy flow per unit area per unit time for an electromagnetic wave. We will consider the special case of an unbounded plane wave travelling along the x-axis, as considered in the previous lecture.

  4. If we assume that the E-field of the wave is polarised along the y-axis (only non-zero component is Ey) then we showed that the B-field must be polarised along the z-axis (only non-zero component Bz). Because we have an unbounded plane wave Ey and Bz can only be functions of x (and t).

  5. The components of E and B must satisfy what is left of Maxwell's equations (A) (B) where c is the speed of light in a vacuum (c=(00)-1/2).

  6. The assumption is now made that the energy density, uEM, of the electromagnetic wave is that derived for static E- and B-fields. (C)

  7. We now consider the total energy, UEM, contained within the box shown in the figure below. This is given by the integral of the energy density, uEM, over the volume of the box where A is the cross-sectional area of the box and the final term follows because, as the wave is plane and unbounded, there is no dependence upon the co-ordinates y and z.

  8. Because we are interested in the transport of energy by the wave we need to calculate the rate at which energy is leaving the box. This is given by -UEM/t. Using (C) above for uEM

  9. The derivative with respect to time can be taken inside the integral

  10. Now using the relationships between the components of E and B given by (A) and (B) to eliminate the time dependence As 00=1/(c2) 0c2=1/0

  11. where the relationship Hz=Bz/0 has been used. Finally evaluating the integral where denotes the value of EyHz evaluated at the position x=x2 etc. Indicates that the rate at which energy leaves the box is given by a power EyHzAexiting through the face at x=x2 minus a power EyHzA which enters the box through the face at x=x1 (in general these two fluxes will be equal so there will be no overall change in the energy contained within the box).

  12. Hence we can associate a power flux (power/area) with the present wave given by the quantity EyHz. Using vector form E=Eyj, H=Hzk then the vector N=ExH = (Eyj)x(Hzk)=EyHzi This result demonstrates that for the present case the vector N (given by ExH) points in the direction in which energy is being transported (the propagation direction) and has a magnitude equal to the power density of the wave.

  13. This result can be shown to be a general one and the vector N is known as the Poynting vector. The Poynting vector gives the power density (power per unit area) of an electromagnetic wave and points in the direction for which this power is transported.

  14. Relationships between Power Density and E- and B-field strengths The magnitude of the Poynting vector is EH. Because in vacuum H=B/0 and for an electromagnetic wave E=cB, it follows that the power density of an electromagnetic wave equals E2/(c0).

  15. Worked example Calculate the average and maximum E- and B-fields due to the radiation from a 150W electric light bulb at a distance of 1m.

  16. Worked example What power density is required in a laser beam if the E-field is to breakdown air (dielectric strength 1x106Vm-1)?

  17. Radiation Pressure An electromagnetic wave transports energy. However because energy has mass (from E=mc2) mass is also transported and hence so is momentum. When an electromagnetic wave is incident on a surface this momentum results in a pressure being exerted, this is known as radiation pressure.

  18. N is the magnitude of the Poynting vector and this gives the energy of the wave per unit area per unit time. Using E=mc2 this corresponds to a mass transport per unit area per unit time of N/c2 Momentum is given by p=mc so that the momentum per unit area per unit time is N/c

  19. If the wave is fully absorbed by a surface then this momentum is transferred to the surface and hence exerts a force per unit area (or alternatively a pressure) equal to the momentum per unit area per unit time. Hence Radiation Pressure = N/c If the light is totally reflected by the surface instead of being absorbed then the change in momentum and, hence the pressure, is twice this value.

  20. Worked example What radiation force acts on an absorbing 1 cm diameter disk placed 1m from a 150W light bulb?

  21. Conclusions • The Poynting vector N=ExH • Magnitude of the Poynting vector in terms of the E-field of the waveE2/(c0) • Radiation pressure = N/c for an absorbing surface, = 2N/c for a reflecting surface.

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