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Download Chapter 29: Maxwell`s Equation and EM Waves
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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 � � = µ0 Ienc � · dr B • 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 ΦE = EA = σ Q = �0 �0 dΦE I = dt �0 • So Maxwell added a displacement current (Ampere-Maxwell term to Ampere’s law: • The the rate of change of the flux is � � dΦE � � B · d� = µ0 (I + Id )enc = µ0 I + �0 dt • Corresponding displacement current density: � Law) enc � d E J�d = �0 dt Slide 29-5 Slide 29-6 Slide 29-7 • Determine the electric field between the plates: � dΦE � � B · d� = µ0 �0 dt I 2πrB = µ0 �0 πr �0 πR2 2 r B = µ0 I (r < R) 2 2πR Slide 29-8 Consider a large parallel plate capacitor as shown, charging so that Q = Q0+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! A. s z a r I Q I -Q d B. 0 2"r µ 0!r 2d 2 C. µ 0!d 2a 2 D. µ 0!a 2"r 2 E. None of the aboveSlide 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 ∂ 2� ∇ E = µ0 �0 2 ∂t 2� B ∂ 2� ∇ B = µ0 �0 2 ∂t 2 2 2 ∂ ∂ ∂ ∇2 = x̂ + 2 ŷ + 2 ẑ 2 ∂x ∂y ∂z • 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� ∂2B ∂2B ∂2E ∂x2 = µ0 �0 ∂t2 ∂x2 = µ0 �0 ∂t2 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. � ×B � gives direction of propagation) (E • Mathematically r E ( x, t ) = Ep sin ( kx − ωt ) ˆj r B ( x, t ) = Bp sin ( kx − ωt ) kˆ 2π k= λ 2π ω= T Slide 29-13 Clicker Question Two traveling waves 1 and 2 are described by the equations. y1 ( x, t ) = 2 sin(2 x ! t ) y 2 ( x, t ) = 4 sin( x ! 2 t ) 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 � � � � � · d�� = − d E dt S � · dA � B ∂E ∂B =⇒ =− ∂x ∂t � · d�� = µ0 �0 d B dt =⇒ S � · dA � E ∂B ∂E = −�0 µ0 ∂x ∂t ∂ ∂E ∂ ∂E ⇒ = − (−�0 µ0 ) ∂x ∂x ∂t ∂t ∂2E ∂2E − � 0 µ0 2 = 0 2 ∂x ∂t E(x, t) = E0 sin(kx − ωt) B(x, t) = B0 sin(kx − ωt) kEp = ωBp kBp = �0 µ0 ωEp • plane wave expression is a solution if ω k = 1 ε 0 µ0 fλ = c = c = 3.0 × 108 m/s • This also implies that Ep Bp = c Slide 29-15 General Results for EM Radiation • Transverse waves (E and B perpendicular to direction of propagation): direction of propagation: E � ×B � ∝ k̂ • E and B perpendicular to each other. • E = cB; E and B oscillate in phase • 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! 1 c • In matter, the speed of light is v = √�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. + x B. − x C. either + x or − x but you can’t tell which D. − y Slide 29-17 Clicker question • A planar electromagnetic wave is propagating through space. Its electric field vector is given by � = Ep cos(kz − ωt)î E Its magnetic field vector is � = Bp cos(kz − ωt)ĵ 1) B � = Bp cos(ky − ωt)k̂ 2) B � = Bp cos(ky − ωt)ĵ 3) B � = Bp cos(kz − ωt)k̂ 4) B � = Bp sin(kz − ωt)î 5) B 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 Slide 29-21 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/m2 in SI): 2 S = uc = (uE + uB )c = �E �= 1E � ×B � S µ0 • For plane waves (traveling in x direction with E oriented in z direction): c � = 1 Ep Bp cos2 (kx − ωt)î S µ0 • Averaging over the time variations of the oscillating fields gives the average value, also called the average intensity: 1 1 1 Ep2 1 I =< S >= Ep Bp = = �0 Ep2 c 2 µ0 2 2µ0 c 2 • Far from a localized source of radiation, electric field decreases as 1/r. 1 (as required by conservation of energy) Thus, I∝ r2 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/m2. (a) What is the total average power output of the Sun, assuming it to be an isotropic source? ( Pav = IA = I 4πR 2 ( ) 2 )( 11 = 4π 1400 W/m 1.50 ×10 m 2 ) = 4.0 ×10 26 W Slide 29-26 Example continued: (b) What is the intensity of sunlight incident on Mercury, which is 5.8x1010 m from the Sun? Pav Pav I= = A 4πr 2 4.0 ×10 26 W = 4π 5.8 ×1010 m ( 2 ) = 9460 W/m2 (c) What is the maximum electric field (if monochromatic light) Slide 29-27 Flux and Solar Heating Slide 29-28 Polarization Slide Slide25-24 29-29 Light passed through a polarizing filter has an intensity of 2.0 W/m2. How should a second polarizing filter be arranged to decrease the intensity to 1.0 W/m2? Slide Slide25-25 29-30 Clicker Question An unpolarized beam of light passes through 2 Polaroid filters oriented at 45o with respect to each other. The intensity of the original beam is Io. What is the intensity of the light coming I through both filters? o A: (1/1.4)Io B: (1/2)Io C: (1/4)Io D: (1/8)Io 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 Slide 29-33