Chapter 29: Maxwell s Equation and EM Waves Slide 29-1
Equations of electromagnetism: a review We ve now seen the four fundamental equations of electromagnetism, here listed together for the first time. But one is incomplete: Ampère s law needs refining Slide 29-2
Clicker Question Slide 29-3
Maxwell s Adjustment to Ampere s Law B dr = µ 0 I enc The result of Ampere s law depends on which surface is used to determine the encircled current! Can t have contradictory results either there is a B field or there isn t! Notice that electric field is changing inside conductor. Ampere postulated that a changing electric flux induces a magnetic field (similar to how a changing magnetic flux induces an electric field) Slide 29-4
Maxwell s Adjustment to Ampere s Law Need to add a term to the right side of Ampere s law to account for the changing electric flux. This is the displacement current. Notice, for a parallel plate capacitor, the electric flux is Φ E = EA = σ 0 = Q 0 dφ The the rate of change of the flux is E = I dt 0 So Maxwell added a displacement current term to Ampere s law: dφ E B d = µ 0 (I + I d ) enc = µ 0 I + 0 dt (Ampere-Maxwell Law) enc Corresponding displacement current density: J d = 0 d E dt Slide 29-5
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Determine the electric field between the plates: B d = µ 0 0 dφ E dt 2πrB = µ 0 0 πr 2 I 0 πr 2 B = µ 0 I r (r <R) 2πR2 Slide 29-8
Consider a large parallel plate capacitor as shown, charging so that Q = Q 0 +βt on the positively charged plate. Assuming the edges of the capacitor and the wire connections to the plates can be ignored, what is the magnitude of the magnetic field B halfway between the plates, at a radius r? µ 0! s A. 2"r a z B. µ 0!r 2d 2 I r I C. µ 0!d 2a 2 Q -Q d D. µ 0!a 2"r 2 E. None of the above Slide 29-9
Maxwell s equations The four complete laws of electromagnetism are collectively called Maxwell s equations. They describe all electromagnetic fields in the universe, outside the realm of quantum physics. Slide 29-10
Maxwell s equations in vacuum In a vacuum there s no electric charge and therefore also no electric current. Maxwell s equations in vacuum A changing electric field is a source for a magnetic field, and a changing magnetic field is a source for an electric field. These equations infer the possibility of electromagnetic waves! Slide 29-11
Wave Equation Maxwell s equations can be manipulated to derive the following wave equations: 2 E = µ0 0 2 E t 2 2 B = µ0 0 2 B t 2 2 = 2 2 ˆx + x2 y 2 ŷ + 2 z 2 ẑ Very similar to wave equation for sound waves! Except here it is the electric and magnetic field that is wiggling. For plane waves traveling in one direction (say the x direction): 2 E x 2 = µ 0 0 2 E t 2 2 B x 2 = µ 0 0 2 B t 2 Slide 29-12
Plane electromagnetic waves A plane electromagnetic wave are waves propagating in one direction with one wavelength. (E and B do not vary with respect to the other two dimensions) The fields are perpendicular to each other and to the direction of propagation. ( E B gives direction of propagation) Mathematically r E( x, t) = Ep sin( kx ωt) ˆj r B x, t = B sin kx ωt kˆ ( ) ( ) p k = 2π λ ω = 2π T Slide 29-13
Clicker Question Two traveling waves 1 and 2 are described by the equations. y ( x, t) = 2 sin( 2x! t) 1 y ( x, t) = 4 sin( x! 2t) 2 All the numbers are in the appropriate SI (mks) units. Which wave has the higher speed? A) Wave 1 B) Wave 2 C) Both have the same speed. Slide 29-14
Plane Waves as Solutions E d = d dt = E x = B t x B(x, t) =B 0 sin(kx ωt) ke p = ωb p kb p = 0 µ 0 ωe p plane wave expression is a solution if ω k = 1 ε 0 µ 0 This also implies that S B d A fλ = c B d = µ 0 0 d dt = c =3.0 10 8 m/s = B x = E 0µ 0 t E x = t ( 0µ 0 E t ) 2 E x 2 0µ 0 2 E t 2 =0 E(x, t) =E 0 sin(kx ωt) B p = E p c S E d A Slide 29-15
General Results for EM Radiation Transverse waves (E and B perpendicular to direction of propagation): direction of propagation: E and B perpendicular to each other. E = cb; E and B oscillate in phase E B ˆk Propagation speed is the speed of light in a vacuum independent of wavelength: c = λ f = ω/k radio waves, light, infrared radiation, X-rays are all the same phenomena! In matter, the speed of light is v = 1 = c µ n No medium required for propagation (no ether) Slide 29-16
Clicker question At a particular point, the electric field of an electromagnetic wave points in the + y direction, while the magnetic field points in the z direction. Which of the following describes the propagation direction? A. B. C. either or but you can t tell which D. +x x y + x x Slide 29-17
Clicker question A planar electromagnetic wave is propagating through space. Its electric field vector is given by E = E p cos(kz ωt)î Its magnetic field vector is 1) B = B p cos(kz ωt)ĵ 2) B = B p cos(ky ωt)ˆk 3) B = B p cos(ky ωt)ĵ 4) B = B p cos(kz ωt)ˆk 5) B = B p sin(kz ωt)î Slide 29-18
The Electromagnetic Spectrum Slide 29-19
Clicker question Which type of radiation travels with the highest speed? 1. visible light 2. X-rays 3. Gamma-rays 4. radio waves 5. they all have the same speed Slide 29-20
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Producing electromagnetic waves Electromagnetic waves are generated ultimately by accelerated electric charge. Details of emitting systems depend on wavelength, with most efficient emitters being roughly a wavelength in size. Radio waves are generated by alternating currents in metal antennas. Molecular vibration and rotation produce infrared waves. Visible light arises largely from atomic-scale processes. X rays are produced in the rapid deceleration of electric charge. Gamma rays result from nuclear processes. A radio transmitter and antenna Electric fields of an oscillating electric dipole Slide 29-22
Antennae An electric field parallel to an antenna (electric dipole) will shake electrons and produce an AC current. A magnetic dipole antenna (for AM radios) should be oriented so that the B-field passes into and out of the plane of a loop, inducing a current in the loop. Slide 29-23
Energy in EM waves Electromagnetic waves transport energy The Poynting vector describes the rate of energy flow per unit area (W/m 2 in SI): S = uc =(u E + u B )c = E 2 c S = 1 µ 0 E B For plane waves (traveling in x direction with E oriented in z direction): S = 1 E p B p cos 2 (kx ωt)î µ 0 Averaging over the time variations of the oscillating fields gives the average value, also called the average intensity: I =< S>= 1 2 1 µ 0 E p B p = 1 2 E 2 p 2µ 0 c = 1 2 0E 2 pc Far from a localized source of radiation, electric field decreases as 1/r. Thus, (as required by conservation of energy) I 1 r 2 Slide 29-24
Clicker Question Two radio dishes are receiving signals from a radio station which is sending out radio waves in all directions with power P. Dish 2 is twice as far away as Dish 1, but has twice the diameter. Which dish receives more power? A: Dish 1 B: Dish 2 C: Both receive the same power Dish 1 Dish 2 Slide 29-25
Example: The intensity of the sunlight that reaches Earth s upper atmosphere is 1400 W/m 2. (a) What is the total average power output of the Sun, assuming it to be an isotropic source? P av = = = IA = 4π I ( 2 4πR ) ( 2 )( 11 1400 W/m 1.50 10 m) 4.0 10 26 W 2 Slide 29-26
Example continued: (b) What is the intensity of sunlight incident on Mercury, which is 5.8x10 10 m from the Sun? I = = = Pav P = A 4πr 4.0 10 4π av 2 26 ( 10 5.8 10 m) 9460 W/m 2 W 2 (c) What is the maximum electric field (if monochromatic light) Slide 29-27
Flux and Solar Heating Slide 29-28
Polarization Slide 25-24 29-29
Light passed through a polarizing filter has an intensity of 2.0 W/m 2. How should a second polarizing filter be arranged to decrease the intensity to 1.0 W/m 2? Slide 25-25 29-30
Clicker Question An unpolarized beam of light passes through 2 Polaroid filters oriented at 45 o with respect to each other. The intensity of the original beam is I o. What is the intensity of the light coming through both filters? I o A: (1/1.4)I o B: (1/2)I o C: (1/4)I o D: (1/8)I o E: None c Slide 29-31
Clicker question Two polarizers are oriented at right angles, so no light gets through the combination. A third polarizer is inserted between the two, with its preferred direction at 45 to the others. How will this sandwich of polarizers affect a beam of initially unpolarized light? A. All of the initial light will be blocked. B. Half of the initial light is blocked. C. One-quarter of the initial light is blocked. D. None of the initial light will be blocked. Slide 29-32
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