1. Energy Carried by E-M Waves:
The energy carried by an electromagnetic (E-M) wave with field intensities E, B is given by the Poynting vector, S:
S = 1/mo [E X B]
where mo denotes the magnetic permeability, mo = 4 π x 10-7 H/m
Physically, the Poynting vector is the rate at which energy flows through a unit surface perpendicular to the flow. We already saw it, at least those who worked Problem (2) did, from the previous installment. This was given as: P = E y H z / c = 1300 Wm-2 .
Recall now, for any two vectors A, B
A X B = A B sin Θ
where Θ is the angle between them. If then A is perpendicular to B, then:
sin Θ = sin (90) = 1 so that:
A B sin Θ = AB
By the same token, taking the vectors E, B perpendicular to each other, we find:
S = EB / mo
which has units of Wm-2
Now, recall from earlier, E / B = c, so:
B = E /c
Then: S = EB / mo = (E/c) E/ mo = E 2 / c mo
B = mo H or: S = (c /mo) B
The "time average" is also of interest and entails taking the time average of the function:
cos 2 (kx - wt)
Which yields:
T(av) {cos2 (kx - wt) } = ½
The average value of S (or Intensity) can then be obtained from the maximum vector amplitudes, viz.
I = S(av) = E(max) B(max)/ 2 mo
or: S(av) = E(max) 2/ 2 mo c = cB(max)2 / 2 mo
Note that mo c is a very important quantity known as the "impedance of free space", or:
mo c = Ömo / Ö εo
mo c = 377 ohms
The respective contributions of the two field energies (associated with the electric intensity, E, and magnetic intensity B) can easily be shown to be:
U(E) = ½ εo E2
U(M)= ½ (B2/ mo)
In a given volume the energy is equally shared by the two fields such that:
U(E) = U(M) = ½ εo E2 = (B2 /2 mo)
The total, instantaneous energy density of the fields is then:
U = U(E) + U(M) = 2(½ εo E2) = εo E2 = B2 /mo
Averaged over one or more cycles this leads to the total average energy per unit volume:
U(av) = [εo E2 ]av = ½ εo E(max)2 = B(max)2 /2 mo
The intensity of an EM wave is then:
I = S (av) = c U(av) = PA
where P denotes the radiation pressure, or P = S/c (for complete absorption) or P = 2S/c for complete reflection of the wave.
(In direct sunlight, one finds P R = 5 x 10-6 N/m2)
Suggested Problems:
1) An E-M wave in a vacuum has an electric field amplitude E = 220 V/m. Compute the magnitude of the corresponding magnetic field, B.
S = 1/mo [E X B]
where mo denotes the magnetic permeability, mo = 4 π x 10-7 H/m
Physically, the Poynting vector is the rate at which energy flows through a unit surface perpendicular to the flow. We already saw it, at least those who worked Problem (2) did, from the previous installment. This was given as: P = E y H z / c = 1300 Wm-2 .
Recall now, for any two vectors A, B
A X B = A B sin Θ
where Θ is the angle between them. If then A is perpendicular to B, then:
sin Θ = sin (90) = 1 so that:
A B sin Θ = AB
By the same token, taking the vectors E, B perpendicular to each other, we find:
S = EB / mo
which has units of Wm-2
Now, recall from earlier, E / B = c, so:
B = E /c
Then: S = EB / mo = (E/c) E/ mo = E 2 / c mo
B = mo H or: S = (c /mo) B
The "time average" is also of interest and entails taking the time average of the function:
cos 2 (kx - wt)
Which yields:
T(av) {cos2 (kx - wt) } = ½
The average value of S (or Intensity) can then be obtained from the maximum vector amplitudes, viz.
I = S(av) = E(max) B(max)/ 2 mo
or: S(av) = E(max) 2/ 2 mo c = cB(max)2 / 2 mo
Note that mo c is a very important quantity known as the "impedance of free space", or:
mo c = Ömo / Ö εo
mo c = 377 ohms
The respective contributions of the two field energies (associated with the electric intensity, E, and magnetic intensity B) can easily be shown to be:
U(E) = ½ εo E2
U(M)= ½ (B2/ mo)
In a given volume the energy is equally shared by the two fields such that:
U(E) = U(M) = ½ εo E2 = (B2 /2 mo)
The total, instantaneous energy density of the fields is then:
U = U(E) + U(M) = 2(½ εo E2) = εo E2 = B2 /mo
Averaged over one or more cycles this leads to the total average energy per unit volume:
U(av) = [εo E2 ]av = ½ εo E(max)2 = B(max)2 /2 mo
The intensity of an EM wave is then:
I = S (av) = c U(av) = PA
where P denotes the radiation pressure, or P = S/c (for complete absorption) or P = 2S/c for complete reflection of the wave.
(In direct sunlight, one finds P R = 5 x 10-6 N/m2)
Suggested Problems:
1) An E-M wave in a vacuum has an electric field amplitude E = 220 V/m. Compute the magnitude of the corresponding magnetic field, B.
2) A radio wave transmits 25 W/m 2 of power per unit area. A plane surface of area 2.4 m x 0.7 m is perpendicular to the direction of propagation of the wave. Calculate the radiation pressure P R on the surface if it is assumed to be a perfect absorber.
3) An AM radio station broadcast isotropically with an average power of 4 kW. A dipole receiving antenna 65 cm long is located 4 miles from the transmitter. Find the Emf induced by this signal between the ends of the receiving antenna.
4) A community plans to build a solar power conversion station, i.e. to convert solar radiation into electrical power. They require 1 MW (megawatt) of power, and the final system is assumed to have an efficiency of 30% (30% of the solar energy incident on the surface is converted to electrical energy). What must be the effective area, A, of an assumed perfectly absorbing surface to be used in such an installation? Assume a constant solar energy flux incident of 1000 W/m2.
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