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∫_{closed loop} B ⋅ ds = μ (I + ε dΦ_{E}/dt).
Given that such electromagnetic fields propagate energy in the direction of E ⊗ B, we examine an instant when E is along the y axis. For propagation along the x axis, B will be along the z axis. Applying Faraday's law to a loop of height dy and width dx normal to B:
we have
E(x + dx) * dy - E(x) * dy = - d(dx * dy * B)/dtwhich in the limit dx → 0 becomes
dE/dx = - dB/dt.Applying Ampere's law (with I = 0) to a loop of height dz and width dx normal to E gives us
B(x) * dz - B(x + dx) * dz = μ ε d(dx * dz * E)/dtwhich in the same limit becomes
dB/dx = - μ ε dE/dt.Differentiating the first equation with respect to x and the second with respect to t and equating the results gives us the electromagnetic wave equation for E:
d^{2}E/dx^{2} = d^{2}E/dt^{2} / c^{2}with propagation velocity
c = 1 / √(μ ε) = 2.998 * 10^{8} m/s = the speed of light.Doing the reverse gives us the (same) wave equation for B.
E / B = ω / k = c.
radiation | λ | ν (Hz) | energy (eV) | source |
---|---|---|---|---|
radio | > 1 m | < 3 * 10^{8} | < 1.24 * 10^{-6} | low-energy atomic or molecular motions |
microwave | > .1 mm | < 3 * 10^{12} | < .0124 | rigid molecular motions |
infrared | > 7000 Angstroms | < 4.3 * 10^{14} | < 1.78 | molecular bond motions |
visible light | > 4000 Angstroms | < 7.5 * 10^{14} | < 3.1 | atomic electron transitions |
ultraviolet | > 50 Angstroms | < 6 * 10^{16} | < 248 | atomic electron transitions |
x-rays | > .03 Angstroms | < 10^{20} | < 414 K | electron transitions in heavy atoms |
gamma rays | < .03 Angstroms | > 10^{20} | > 414 K | nuclear decays |
Electromagnetic waves also undergo Doppler shifting, as shown here:
S = E ⊗ B / μ,with units of W / m^{2}. Since the integral of cos^{2} over one wavelength is 1/2, the intensity is
<S> = E^{2} / (2 μ c)with u equal to the sum of the electric and magnetic energy densities:= c <u>
ε E^{2} / 2 + B^{2} / (2 μ) = E^{2} / (2 μ c^{2}) + (E / c)^{2} / (2 μ)("<>" denotes the average over one wavelength, which introduces the factor of 1/2).= E^{2} / (μ c^{2})
This implies that the electromagnetic field exerts a pressure
P = S / c(under conditions of perfect absorption). Pressure is measured in Pascals: 1 Pa ≡ 1 N / m^{2}.
Note that the units of pressure are identical to those of energy density.
P = 2 S / c.
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©2012, Kenneth R. Koehler. All Rights Reserved. This document may be freely reproduced provided that this copyright notice is included.
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