Exam 3: Tuesday, April 18, 5:00-6:00 PM Test rooms: • Instructor Sections Room • Dr. Hale F, H 104 Physics • Dr. Kurter B, N 125 BCH • Dr. Madison K, M 199 Toomey • Dr. Parris J, L B-10 Bertelsmeyer • M r. Upshaw A, C, E, G G-3 Schrenk • Dr. Waddill D 120 BCH
• Special Accommodations Testing Center (Contact me a.s.a.p. if you need accommodations different than for exam 2)
Exam 3 will cover chapters 28 to 32 (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: q ∫ E⋅= dA enclosed ∫ B⋅= dA 0 εo Φ dΦ d B ⋅ E E⋅=− ds ∫ B ds=μ0 I encl +μ 0 ε 0 ∫ dt dt
These four equations provide a complete description of electromagnetism. Maxwell’s equations can also be written in differential form: ρ ∇⋅E = ∇⋅B0 = ε0 dB 1 dE ∇−×E= ∇×B= +μJ dt c2 dt 0
Missouri S&T Society of Physics Students T-Shirt! Maxwell’s Equations
q ∫ E⋅= dA enclosed ∫ B⋅= dA 0 εo Φ dΦ d B ⋅ E E⋅=− ds ∫ B ds=μ0 I encl +μ 0 ε 0 ∫ dt dt
• oscillating electric and magnetic fields can “sustain each other” away from source charges and fields
d d Faraday’s law BE→ Ampere’s law EB→ dt dt
• result: electromagnetic waves that propagates through space • electromagnetic waves always involve both E and B fields
• propagation direction, E field and B field form right- handed triple of vectors
Example: wave propagating in x-direction E field in y-direction B field in z-direction values of E and B depend only upon x and t
y
x direction of propagation z Wave equation
• combine Faraday’s law and Ampere’s law
• for wave traveling in x-direction with E in y-direction and B in z direction:
Wave equation:
∂∂22E E (x,t) ∂∂22B B (x,t) yy= µε zz= µε ∂∂xt2200 ∂∂xt2200
∂E ∂B • E and B are not independent: y =- z ∂∂xt Solutions of the wave equation
E =E sin kx -ω t y max ( ) Emax and Bmax are the electric and magnetic field
Bz =B max sin( kx -ω t) amplitudes
2π Wave number k, wave length λ k= λ
Angular frequency ω, frequency f ωπ=2 f
ω 1 Wave speed fλ = =c c= k µε00 ∂E ∂B y =- z ∂∂xt
∂E sin( kx - ω∂ t) B sin( kx - ω t) max =- max ∂∂xt
Emax k cos( kx -ωωω t) =Bmax cos( kx - t)
E E1ω max = = =c= . Bmax Bk µε00
Ratio of electric field magnitude to magnetic field magnitude in an electromagnetic wave equals the speed of light. y
x direction of propagation z
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 *J. H. Poynting, 1884. 1 S= E× B This is derived from Maxwell’s equations. µ0
• S represents energy current density, i.e., energy per time per area or power per area (units J/(s·m2) =W/m2)
• direction of S is along the direction of wave propagation for EM wave: E× B =EB y 1 EB S= E× B so S= . µ µ0 0 E S x because B = E/c B c E22 cB z S= = . µµ00c
These equations for S apply at any instant of time and represent the instantaneous rate at which energy is passing through a unit area. EB E22 cB S= = = µµ00c µ 0
EM waves are sinusoidal. Ey =E max sin( kx -ω t) EM wave propagating along x-direction Bz =B max sin( kx -ω t)
The average of S over one or more cycles is called the wave intensity I.
The time average of sin2(kx - ωt) is ½, so
22 Emax B max E max cB max Notice the 2’s in I=Saverage = S = = = this equation. 2µ0 2c µµ 00 2
½ This equation is the same as 32-29 in your text, using c = 1/(µ0ε0) . 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) 2 1 2 1B u=E0ε E u=B 2 2 µ0 ½ Using B = E/c and c = 1/(µ0ε0) :
2 2 E 2 1B 1( c ) 1µε00E 1 2 u=B0 = = =ε E 22µµ00 2 µ 0 2
2 12 1B remember: E and B are uBE =u =ε 0 E = sinusoidal functions of time 22µ0 total energy density: 2 2 B instantaneous energy densities u=uBE0 +u =ε E = (E and B vary with time) µ0
• average over one or more cycles of electromagnetic wave gives factor ½ from average of sin2(kx - ωt).
2 2 1 2 1 Bmax 112 Bmax u=Eε 0 E max , u=B , and u=ε0 E max = 4 4 µ0 22µ0
Recall: intensity of an EM wave
22 11Emax cBmax Saverage =S= == cu 2c2µµ00 Help!
E or B individually: 2 12 1 B (t) At time t: uBE (t) = u (t) =ε 0 E (t) = 22µ0 2 1 2 1 Bmax Average: u=Eε 0 E max u=B , 4 4 µ0
Total: 2 2 B (t) At time t: u(t) =ε0 E (t) = µ0 2 112 Bmax Average: u=ε0 E max = 22µ0 Example: a radio station on the surface of the earth radiates a sinusoidal wave with an average total power of 50 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 100 km from the antenna.
Strategy: we want Emax, Bmax. We Satellite are given average power. From average power we can calculate Fromintensity, the andaverage from power intensity we wecan can calculate intensity,Emax and Bandmax .from Station intensity we can calculate Emax and Bmax.
*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 50 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 100 km from the antenna.
Area=4πR2/2 All the radiated power passes Satellite through the hemispherical R surface* so the average power per unit area (the intensity) is Station
4 power P (5.00× 10 W) II= = = = = 7.96× 10-7 W m 2 2 5 2 areaaverage 2π R 2π×( 1.00 10 m) 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 E2 I= S = max Satellite 2c µ0 R
Emax = 2µ0 cI Station
= 2 4π× 10-7 3 × 108 7.96 × 10-7 V ( )( )( ) m
= 2.45× 10-2 V m
-2 V E (2.45× 10 ) B =max = m = 8.17× 10-11 T max c (3× 108 m s)
2 You could get Bmax from I = c Bmax /2µ0, but that’s a lot more work 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.
2 1 2 1 Bmax u=Eε 0 E max u=B 4 4 µ0
-11 2 1J2 1J(8.17× 10 ) -12 -2 u= uE =( 8.85×× 10)( 2.45 10 ) 3 B -7 3 4m4m(4π× 10 )
J J u =1.33× 10-15 u =1.33× 10-15 E m3 B m3
-15 3 If you are smart, you will write
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 = = dp is momentum carried in volume dV dV c22 c 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 incident S d p = A dt c S • Newton’s 2nd Law (F = dp/dt): force F = A c FSI • Radiation exerts pressure P= = = rad A cc (for total absorption) • if radiation is totally reflected by object, then magnitude of momentum change of the object is twice that for total absorption. S d p = 2 A dt incident c reflected
S • Newton’s 2nd Law (F = dp/dt): force F = 2 A c FSI • Radiation exerts pressure P = =2= 2 rad A cc (for total reflection) I incident P = (total absorption) rad c absorbed
Using the arguments above it can also be shown that:
2I incident P = (total reflection) rad c reflected
If an electromagnetic wave does not strike a surface, it still carries momentum away from its emitter, and exerts Prad=I/c on the emitter. Example: a satellite orbiting the earth has solar energy collection panels with a total area of 4.0 m2. 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 Saverage) of sunlight prior to passing through the earth’s atmosphere is 1.4 kW/m2.
Power = IA = 1.4×× 103W 4.0 m 23 = 5.6 10 W = 5.6 kW ( m2 )( )
Assuming total absorption of the radiation: 3 W 1.4× 10 2 Caution! The letter P Saverage I ( m ) -6 (or p) has been used Prad = = = = 4.7× 10 Pa cc3× 108 m in this lecture for ( s ) power, pressure, and momentum! F =P A = 4.7×× 10-6 N 4.0 m2 =1.9 10-5 N rad ( m2 )( ) 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:
22 1 11Emax cBmax S= E× B S=average = µ0 2c2µµ00
2 Emax E1 12 1B = =c= uBE =u =ε 0 E = BBmax εµ00 22µ0
2 2πω 112 Bmax k=, ωπλ =2 f, f = =c u=ε0 E max = λ k 22µ0
∆∆U 2U I 2I ∆p = or P = or cc rad cc
There are even more on your starting equation sheet; they are derived from the above!