EM Waves, & Their Speed Derived from Maxwell’s Equations

EM Waves, & Their Speed Derived from Maxwell’s Equations The Differential (Local) Form of Maxwell’s Equations

Electromagnetic Waves & Their Speed Derived from Maxwell’s Equations The Differential (Local) Form of Maxwell’s Equations

Electromagnetic Waves & Their Speed Derived from Maxwell’s Equations In the absence of currents & charges, the integral forms of Maxwell’s equations are: B(2r) = EA B(2r) = BA The Integral (Global) Form of Maxwell’s Equations B(2r) = B Eℓ t E B(2r) t

electromagnetic wave of wavelength λ & frequency f. The electric This figure shows an electromagnetic wave of wavelength λ & frequency f. The electric & magnetic fields are given by: Figure 31-10. Applying Faraday’s law to the rectangle (Δy)(dx). where

The magnitude of this speed is 3.0  108 m/s: Precisely equal to the It can be shown that the equations for B and E and the relationship v = f λ give: v is the velocity of the wave. Substituting, The magnitude of this speed is 3.0  108 m/s: Precisely equal to the measured speed of light in vacuum.

Example: Determining E & B in EM waves. Assume a f = 60-Hz EM wave is a sinusoidal wave propagating in the z direction with E pointing in the x direction, and E0 = 2.0 V/m. Write vector expressions for E and B as functions of position and time. Solution: The wavelength is c/f = 5.0 x 106m. The wave number is 2π/λ = 1.26 x 10-6 m-1. The angular frequency is 2πf = 377 rad/s. Finally, B0 = E0/c = 6.7 x 10-9 T. B must be in the y direction, as E, v, and B are mutually perpendicular. Now substitute in equations 31-7.

Light as an Electromagnetic Wave & The Electromagnetic Spectrum The frequency of an electromagnetic wave is related to its wavelength & to the speed of light: ; Electromagnetic waves can have any wavelength; we have given different names to different parts of the wavelength spectrum.

Example: Wavelengths of EM Waves. Calculate the Wavelength: c = fλ so λ = (c/f) (a) of a f = 60-Hz EM wave (b) of a f = 93.3-MHz FM radio wave, (c) of a beam of visible red light from a laser at frequency f = 4.74 x 1014 Hz. Solution: The wavelength is the speed of light multiplied by the frequency. a. 5.0 x 106 m b. 3.22 m c. 6.33 x 10-7 m

Example: Wavelengths of EM Waves. Calculate the Wavelength: c = fλ so λ = (c/f) (a) of a f = 60-Hz EM wave λ = (c/f) = (3  108)/(60) = 5  106 m (b) of a f = 93.3-MHz FM radio wave, (c) of a beam of visible red light from a laser at frequency f = 4.74 x 1014 Hz. Solution: The wavelength is the speed of light multiplied by the frequency. a. 5.0 x 106 m b. 3.22 m c. 6.33 x 10-7 m

Example: Wavelengths of EM Waves. Calculate the Wavelength: c = fλ so λ = (c/f) (a) of a f = 60-Hz EM wave λ = (c/f) = (3  108)/(60) = 5  106 m (b) of a f = 93.3-MHz FM radio wave, λ = (c/f) = (3  108)/(93.3  106) = 3.22 m (c) of a beam of visible red light from a laser at frequency f = 4.74 x 1014 Hz. Solution: The wavelength is the speed of light multiplied by the frequency. a. 5.0 x 106 m b. 3.22 m c. 6.33 x 10-7 m

Example: Wavelengths of EM Waves. Calculate the Wavelength: c = fλ so λ = (c/f) (a) of a f = 60-Hz EM wave λ = (c/f) = (3  108)/(60) = 5  106 m (b) of a f = 93.3-MHz FM radio wave, λ = (c/f) = (3  108)/(93.3  106) = 3.22 m (c) of a beam of visible red light from a laser at frequency f = 4.74 x 1014 Hz. λ = (c/f) = (3  108)/(4.74  1014) = 6.33  10-7 m = 633 nm Solution: The wavelength is the speed of light multiplied by the frequency. a. 5.0 x 106 m b. 3.22 m c. 6.33 x 10-7 m

Example: Cell Phone Antenna. The antenna of a cell phone is often ¼ wavelength (L = ¼ λ) long. A particular cell phone has an L = 8.5-cm long straight rod for its antenna. Estimate the operating frequency of this phone. Solution: The wavelength is the speed of light multiplied by the frequency. a. 5.0 x 106 m b. 3.22 m c. 6.33 x 10-7 m

Example: Cell Phone Antenna. The antenna of a cell phone is often ¼ wavelength (L = ¼ λ) long. A particular cell phone has an L = 8.5-cm long straight rod for its antenna. Estimate the operating frequency of this phone. λ = 4 L = 4(8.5cm) = 34 cm = 0.34 m f = c/λ = (3  108)/(0.34) = 8.8  108 Hz = 880 MHz Solution: The wavelength is the speed of light multiplied by the frequency. a. 5.0 x 106 m b. 3.22 m c. 6.33 x 10-7 m

Measuring the Speed of Light In the early 20th Century, the speed of light was known to be very large, although careful studies of the orbits of Jupiter’s moons showed that it is finite. One important measurement, by Michelson, used a rotating mirror. Over the years, measurements have become more & more precise; now the speed of light is defined to be c = 2.99792458 × 108 m/s. This is then used to define the meter. Figure 31-14. Michelson’s speed-of-light apparatus (not to scale).