Physics 24100 Electricity & Optics Lecture 21 – Chapter 30 sec. 1-4 Fall 2012 Semester Matthew Jones Question • An LC circuit has and . = 100 = 100 • If it oscillates with an amplitude of 100 mV, what is the amplitude of the current? (a) 100 (b) 100 (c) 10 (d) 10 Question • The energy bounces back and forth between the capacitor and the inductor. () 1 = 2 1 = 2 10 10 = = = 10 10 = 10 = 0.1 100 Maxwell’s Equations (so far) • ()!*)&+ (Gauss’s law) !" # $ &' = % ,- • (Gauss’s law for magnetism) % !" #. &' = - • & (Faraday’s law) $ # &/ = 0 & 1% !" #.&' • (Ampere’s law) . # &/ = 2- The Problem with Ampere’s Law Current through is 3 Current through is 0! 3 The current through is zero, 3 but the electric flux is not zero. The electric flux changes as charge flows onto the capacitor. Maxwell’s Displacement Current • We can think of the changing electric flux through as if it were a current: 3 67 6 4 = 5 = 5 9 : #;< 6 68 68 => & ? . # &/ = 2- + & = 2-A + 2-,- 9 $ #!" &' & %B Calculating Displacement Current Parallel plate capacitor Gauss’s law: (Lecture 5) C D : = = 5 5 Electric flux: D 7 = : = 5 Change in electric flux: 67 1 6D = = 68 5 68 5 Displacement current: 67 4 = 5 = 68 Coaxial Capacitor In Lecture 8 we calculated the electric field between the inner and outer conductors using Gauss’s law: DEFGE4 ? ;< #: 6 = = 5 We choose to be a cylindrical Gaussian surface of length 3 and radius with . H I JHJI But the left-hand-side is just so the displacement current is 7 67 6DEFGE4 4 = 5 = The real current flows into the68 ends of the68 Gaussian surface, the displacement current flows out the sides. Clicker Question • Suppose charge flows between two spherical conductors. Which surface will have the largest displacement current flowing through it? (d) (a) (b) (c) Clicker Question • Suppose charge flows between two spherical conductors. Which surface will have the largest displacement current flowing through it? (d) (a) (b) (c) 67 6DEFGE4 4 = 5 = 68 68 Maxwell’s Equations (1864) ()!*)&+ ? !" # $ &' = % ,- ? !" #. &' = - % &KL ? $ # &/ = 0 & &K+ ? . # &/ = 2- + 2-M- & Maxwell’s Equations in Free Space In “free space” where there are no electric charges or sources of current, Maxwell’s equations are quite symmetric: ? !" # $ &' = - % ? !" #. &' = - % &KL ? $ # &/ = 0 & &K+ ? . # &/ = 2-M- & Maxwell’s Equations in Free Space &KL ? $ # &/ = 0 & &K+ ? . # &/ = 2-M- & A changing magnetic flux induces an electric field. A changing electric flux induces a magnetic field. Will this process continue indefinitely? Light is an Electromagnetic Wave • Maxwell showed that these equations contain the wave equation: NO 1 NO = where is a functionNP of positionQ N8 and time. O P, 8 • General solution: O P, 8 = S P 0 Q8 + S(P + Q8) • Harmonic solutions: O P, 8 = O sin WP X Y8 • Wavelength, and frequency . Z = 2[/W S = Y/2[ • Velocity, . Q = Z] S = Y] W Light is an Electromagnetic Wave • Faraday’s Law: 4_` bcd ^ : # 6/ = 0 4a = 0 ba ePef ? : # 6/ = :g f hP 0 :g(f)hP ^ x E bjk x Ex hP i bl efeP z1 z2 z By y hf Light is an Electromagnetic Wave • Ampere’s law: 4_n bjk ^ m # 6/ = 5 4a = 5 ba eOef ? m # 6/ = mo f hO 0 mo(f)hO ^ x Ex bcd ef i 0 efeO z z bl 1 2 z B B y y y eO Putting these together… N:g Nmo = 0 Nf N8 Nmo N:g 0 = 5 Nf N8 Differentiate the first with respect to : f N :g N mo = 0 Nf NfN8 Differentiate the second with respect to : 8 N mo N :g 0 = 5 NfN8 N8 B B p $q p $q = 2-,- prB pB Velocity of Electromagnetic Waves NO 1 NO = NP Q N8 B B p $q p $q B = 2-,- B Speed of wave propagationpr is p 1 Q = 5 1 = s[ t 10 uv/ 8.85s t 10 /v # = B. yyz t {-z |/} (speed of light) The electromagnetic spectrum • In 1850, the only known forms electromagnetic waves were ultraviolet, visible and infrared. • The human eye is only sensitive to a very narrow range of wavelengths: Discovery of Radio Waves The Electromagnetic Spectrum Electromagnetic Waves B B p $q p $q B = 2-,- B • A solution is pr p :g f, 8 = : sin Wf 0 Y8 where Y = W~ = 2[~/Z • What is the magnetic field? Nmo N:g = 0 = 0W: cos Wf 0 Y8 N8 Nf W mo P, 8 = :sin WP 0 Y8 Y Electromagnetic Waves x P z f y O • , and are mutually perpendicular. : m Q• • In general, the direction is = : t m Energy in Electromagnetic Waves Energy stored in electric and magnetic fields (lecture 17): 1 1 … = 5: … = m 2 2 For an electromagnetic wave, m = :] ~ = : 5 1 1 … = m = 5: = … 2 2 The total energy density is … = … + … = 5: Intensity of Electromagnetic Waves • Intensity is defined as the average power transmitted per unit area. Intensity = Energy density wave velocity t : = 5~ : = ~ ~ = 377 (Impedance of free space) Poynting Vector • We can construct a vector from the intensity and the direction : = : t m : t m 3• = : 3• = = • This represents the flow of power in the direction • Average electric field: :G = :/ 2 : 3• = 2 • Units: Watts/m2 Question • The magnetic field component of a radio wave is expressed: mo = m sin Wf 0 Y8 • In which directions do the Poynting vector and the -field point? : (a) +z,+x (b) +z,-x (c) -z,+x (d) -z,-x (e) +y,+x.