Exam 3: Tuesday, April 18, 5:-6: PM Test rooms: Instructor Sections Room Dr. Hale F, H 14 Physics Dr. Kurter, N 15 CH Dr. Madison K, M 199 Toomey Dr. Parris J, L -1 ertelsmeyer Mr. Upshaw A, C, E, G G-3 Schrenk Dr. Waddill D 1 CH Special Accommodations (Contact me a.s.a.p. if you need accommodations different than for exam ) Testing Center Exam 3 will cover chapters 8 to 3 (From magnetic field of a current to electromagnetic waves)
Today s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.
Maxwell s Equations Recall: E da = q E ds = enclosed ε o dφ dt ds=μ I da = +μ ε encl dφ dt E These four equations provide a complete description of electromagnetism.
Maxwell s equations can also be written in differential form: ρ E = ε = d 1 de E= = +μ J dt c dt Missouri S&T Society of Physics Students T-Shirt!
Maxwell s Equations E da = q enclosed da = εo E ds = dφ dt ds=μ I +μ ε encl dφ dt E oscillating electric and magnetic fields can sustain each other away from source charges and fields Faraday s law d dt E Ampere s law d E dt result: electromagnetic waves that propagates through space
electromagnetic waves always involve both E and fields propagation direction, E field and field form righthanded triple of vectors Example: wave propagating in x-direction E field in y-direction field in z-direction values of E and depend only upon x and t z y x direction of propagation
Wave equation combine Faraday s law and Ampere s law for wave traveling in x-direction with E in y-direction and in z direction: Wave equation: E x E (x,t) y y = µε t x z(x,t) t z = µε E and are not independent: E y x =- t z
Solutions of the wave equation y max ( ω ) E =E sin kx - t z max ( ω ) = sin kx - t E max and max are the electric and magnetic field amplitudes Wave number k, wave length λ Angular frequency ω, frequency f k= ω π λ =πf Wave speed ω f λ = =c k c= 1 µε
E y x =- t z max ( ) ( ) E maxsin kx - ωt maxsin kx - ωt =- x t ( ω ) ω ( ω ) E k cos kx - t = cos kx - t max Emax E ω 1 = = =c= k µε max. Ratio of electric field magnitude to magnetic field magnitude in an electromagnetic wave equals the speed of light.
z y x direction of propagation This static image doesn t show how the wave propagates. Here are animations, available on-line: http://phet.colorado.edu/en/simulation/radio-waves (shows electric field only) http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=35 Here is a movie.
Types of electromagnetic waves enormous range of wave lengths and frequencies spans more than 15 orders of magnitude
Applications of electromagnetic waves
Today s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.
Energy Carried by Electromagnetic Waves rate of energy flow: Poynting vector* S 1 S= E µ This is derived from Maxwell s equations. *J. H. Poynting, 1884. S represents energy current density, i.e., energy per time per area or power per area (units J/(s m ) =W/m ) direction of S is along the direction of wave propagation
y 1 S= E µ for EM wave: E so S=. µ E =E z E S c x because = E/c E c S= =. µ c µ These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area.
E E c S= = = µ µ c µ EM waves are sinusoidal. ( ω ) ( ω ) E =E sin kx - t y max = sin kx - t z max EM wave propagating along x-direction The average of S over one or more cycles is called the wave intensity I. The time average of sin (kx - ωt) is ½, so E E c I=S average = S = = = µ µ c µ max max max max Notice the s in this equation. This equation is the same as 3-9 in your text, using c = 1/(µ ε ) ½.
Energy Density so far: energy transported by EM wave now: energy stored in the field in some volume of space energy densities (energy per volume) 1 1 u E = εe u = µ Using = E/c and c = 1/(µ ε ) ½ : ( ) E µε 1 1 1 E 1 u = = c = = εe µ µ µ 1 1 u =u E = εe = µ remember: E and are sinusoidal functions of time
total energy density: u=u +u = ε E = E µ instantaneous energy densities (E and vary with time) average over one or more cycles of electromagnetic wave gives factor ½ from average of sin (kx - ωt). 1 u = ε E 4 E max, 1 max u =, and 4 µ 1 1 u= εe max = µ max Recall: intensity of an EM wave 1Emax 1 cmax S average =S= = = cu µ c µ
Help! E or individually: At time t: Average: 1 1 (t) u (t) = u E(t) = εe (t) = µ 1 u = ε E 4 E max 1 u = 4 µ max, Total: At time t: Average: (t) u(t) = εe (t) = µ 1 1 u= εe max = µ max
Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 5 kw.* Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 1 km from the antenna. Strategy: we want E max, max. We are given average power. From average power we can calculate intensity, From the and average from power intensity we we can can calculate intensity, E max and and max. from intensity we can calculate E max and max. Station Satellite *In problems like this you need to ask whether the power is radiated into all space or into just part of space.
Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 5 kw.* Assuming the wave is radiated equally in all directions above the ground, find the amplitude of the electric and magnetic fields detected by a satellite 1 km from the antenna. All the radiated power passes through the hemispherical surface* so the average power per unit area (the intensity) is ( ) Area=4πR / R Station Satellite 4 power P 5. 1 W I= = = = 7.96 1 W m area π 5 average R π 1. 1 m ( ) -7 Today s lecture is brought to you by the letter P. *In problems like this you need to ask whether the power is radiated into all space or into just part of space.
1 Emax I= S = µ c R Satellite E max = µ ci ( π )( )( ) = 4 1 3 1 7.96 1 =.45 1 - V m -7 8-7 V ( -.45 1 V ) Emax -11 max = = m = 8.17 1 T 8 c ( 3 1 m s) You could get max from I = c max /µ, but that s a lot more work m Station
Example: for the radio station in the previous example, calculate the average energy densities associated with the electric and magnetic field at the location of the satelite. 1 u = ε E 4 E max 1 u = 4 max µ 1 ( -1 )( - ) J u = 8.85 1.45 1 4 m E 3 u =1.33 1 J m -15 E 3 ( -11 8.17 1 ) ( -7 4π 1 ) 1 J u = 4 m 3 u =1.33 1 J m -15 3 If you are smart, you will write <u > = <u E > = 1.33x1-15 J/m 3 and be done with it.
Today s agenda: Electromagnetic Waves. Energy Carried by Electromagnetic Waves. Momentum and Radiation Pressure of an Electromagnetic Wave.
Momentum of electromagnetic wave EM waves carry linear momentum as well as energy momentum stored in wave in some volume of space momentum density (momentum per volume): dp S I = = dv c c dp is momentum carried in volume dv momentum transported by EM wave: momentum current density (momentum per area and time) dp S I c = = dv c c
Radiation Pressure if EM radiation is incident on an object for a time dt and if radiation is entirely absorbed: object gains momentum S d p = A dt c Newton s nd Law (F = dp/dt): force incident S F = c A Radiation exerts pressure F S I P rad = = = A c c (for total absorption)
if radiation is totally reflected by object, then magnitude of momentum change of the object is twice that for total absorption. d p = S c A dt incident reflected Newton s nd Law (F = dp/dt): force F = S c A Radiation exerts pressure F S I P rad = = = A c c (for total reflection)
I P rad = (total absorption) c incident absorbed Using the arguments above it can also be shown that: I P rad = (total reflection) c incident reflected If an electromagnetic wave does not strike a surface, it still carries momentum away from its emitter, and exerts P rad =I/c on the emitter.
Example: a satellite orbiting the earth has solar energy collection panels with a total area of 4. m. If the sun s radiation is incident perpendicular to the panels and is completely absorbed, find the average solar power absorbed and the average force associated with the radiation pressure. The intensity (I or S average ) of sunlight prior to passing through the earth s atmosphere is 1.4 kw/m. ( W )( ) 3 3 Power = IA = 1.4 1 4. m = 5.6 1 W = 5.6 kw m Assuming total absorption of the radiation: ( ) 3 1.4 1 W Saverage I -6 P rad = = = m = 4.7 1 Pa c c ( 8 3 1 m ) s ( N )( ) -6-5 F =PradA = 4.7 1 4. m =1.9 1 N m Caution! The letter P (or p) has been used in this lecture for power, pressure, and momentum!
Light Mill (Crookes radiometer) airtight glass bulb, containing a partial vacuum vanes mounted on a spindle (one side black, one silver) vanes rotate when exposed to light This is NOT caused by radiation pressure!! (if vacuum is too good, mill does not turn) Mill is heat engine: black surface heats up, detailed mechanism leading to motion is complicated, research papers are written about this!
New starting equations from this lecture: 1 S= E µ Emax E 1 = =c= εµ max 1Emax 1 c S average = = µ c µ max 1 1 u =u E = εe = µ π ω k=, ω=πf, f λ= =c λ k 1 1 u= εe max = µ max U p = or c U c I P rad = or c I c There are even more on your starting equation sheet; they are derived from the above!