W13D2: Displacement Current, Maxwell s Equations, Wave Equations. Today s Reading Course Notes: Sections

Transcription

1 W13D2: Displacement Current, Maxwell s Equations, Wave Equations Today s Reading Course Notes: ections

2 Announcements Math Review Tuesday May 6 from 9 pm-11 pm in Pset 10 due May 6 at 9 pm 2

3 Outline Displacement Current Poynting Vector and Energy Flow Maxwell s Equations 3

4 Maxwell s Equations E ˆn da = 1 ρ dv V (Gauss's Law) B ˆn da = 0 (Magnetic Gauss's Law) E d s B ˆn da = d dt C B d s = µ 0 J ˆn da C (Faraday's Law) (Ampere's Law quasi - static) Is there something missing? 4

5 Maxwell s Equations One Last Modification: Displacement Current 5

6 Ampere s Law: Capacitor Consider a charging capacitor: Use Ampere s Law to calculate the magnetic field just above the top plate B d s I enc = = µ 0 I enc J ˆn da 1) urface 1 : I enc = I 2) urface 2 : I enc = 0 What s Going On? 6

7 Displacement Current We don t have current between the capacitor plates but we do have a changing E field. Can we make a current out of that? E = Q A Q = EA = Φ E dq dt = dφ E dt I dis This is called the displacement current. It is not a flow of charge but proportional to changing electric flux 7

8 Displacement Current: I dis = d dt E ˆn da = dφ E dt If surface 2 encloses all of the electric flux due to the charged plate then I dis = I 8

9 Maxwell-Ampere s Law B d s = µ 0 J ˆnda C + µ 0 d = µ 0 (I enc + I dis ) dt E ˆnda flow of electric charge I enc = J ˆnda changing electric flux I dis = d dt E ˆnda 9

10 Concept Question: Capacitor If instead of integrating the magnetic field around the pictured Amperian circular loop of radius r we were to integrate around an Amperian loop of the same radius R as the plates (b) then the integral of the magnetic field around the closed path would be 1. the same. 2. larger. 3. smaller. 10

11 ign Conventions: Right Hand Rule B d s = µ 0 J ˆnda C + µ 0 d dt E ˆnda Integration direction clockwise for line integral requires that unit normal points into page for surface integral. Current positive into the page. Negative out of page. Electric flux positive into page, negative out of page. 11

12 ign Conventions: Right Hand Rule B d s = µ 0 J ˆnda C + µ 0 d dt E ˆnda Integration direction counter clockwise for line integral requires that unit normal points out page for surface integral. Current positive out of page. Negative into page. Electric flux positive out of page, negative into page. 12

13 Concept Question: Capacitor Consider a circular capacitor, with an Amperian circular loop (radius r) in the plane midway between the plates. When the capacitor is charging, the line integral of the magnetic field around the circle (in direction shown) is 1. Zero (No current through loop) 2. Positive 3. Negative 4. Can t tell (need to know direction of E) 13

14 Concept Question: Capacitor The figures above shows a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the charge Q is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time. 14

15 Group Problem: Capacitor A circular capacitor of spacing d and radius R is in a circuit carrying the steady current i shown. At time t = 0, the plates are uncharged 1. Find the electric field E(t) at P vs. time t (mag. & dir.) 2. Find the magnetic field B(t) at P 15

16 Energy Flow 16

17 Poynting Vector Power per unit area: Poynting vector = E B µ 0 Power through a surface P = open surface ˆn da 17

18 Energy Flow: Capacitor = E B µ 0 What is the magnetic field? 18

19 Concept Question: Capacitor The figures above show a side and top view of a capacitor with charge Q and electric and magnetic fields E and B at time t. At this time the energy stored in the electric field is: 1. Increasing in time. 2. Constant in time. 3. Decreasing in time. 19

20 Energy Flow: Capacitor = ( E B) / µ 0 = (E ˆk B ˆθ) / µ 0 = (EB / µ 0 ) ˆr P = ˆn out da = (EB / µ 0 )2π Rh P = = (EB / µ 0 ) ˆr ˆr da cylindrical shell EB µ 0 = E µ 0 µ 0 2 2π Rh de dt R = E de dt π R2 h = d dt 1 2 E 2 π R2 h = du E dt (Volume) cylindrical body 2π Rh de B2π R = µ 0 dt π R2 B = µ 0 2 de dt R! 20

21 Maxwell s Equations E ˆn da = 1 ε ρ dv 0 V (Gauss's Law) B ˆn da = 0 (Magnetic Gauss's Law) C E d s = d dt B ˆn da B d s = µ 0 J ˆn da C + µ 0 d dt E ˆn da (Faraday's Law) (Maxwell - Ampere's Law) 21

22 Electromagnetism Review E fields are associated with: (1) electric charges (Gauss s Law ) (2) time changing B fields (Faraday s Law) B fields are associated with (3a) moving electric charges (Ampere-Maxwell Law) (3b) time changing E fields (Maxwell s Addition (Ampere- Maxwell Law) Conservation of magnetic flux (4) No magnetic charge (Gauss s Law for Magnetism) 22

23 Electromagnetism Review Conservation of charge: closed surface J d A = d dt volume enclosed ρ dv E and B fields exert forces on (moving) electric charges: F q = q( E + v B) Energy stored in electric and magnetic fields U E = u E dv = all space all space 2 E 2 dv U B = u B dv = all space 1 B 2 dv 2µ all space 0 23

24 Maxwell s Equations in Vacua 24

25 Maxwell s Equations E d A = Q 0 in (Gauss's Law) B da = 0 (Magnetic Gauss's Law) C C E d s B d s = dφ B dt 0 dφ = µ 0 I enc + µ 0 ε E 0 dt (Faraday's Law) (Ampere - Maxwell Law) What about free space (no charge or current)? 25

26 26

27 Maxwell: First Colour Photograph James Clerk Maxwell is sometimes credited as being the father of additive color. He had the photographer Thomas utton photograph a tartan ribbon on black-andwhite film three times, first with a red, then green, then blue color filter over the lens. The three black-and-white images were developed and then projected onto a screen with three different projectors, each equipped with the corresponding red, green, or blue color filter used to take its image. When brought into alignment, the three images (a black-and-red image, a black-and-green image and a black-and-blue image) formed a full color image, thus demonstrating the principles of additive color. 27

28 How Do Maxwell s Equations Lead to EM Waves? 28

29 Wave Equation tart with Ampere-Maxwell Eq and closed oriented loop B d s E ˆn da C = µ 0 d dt 29

30 Wave Equation tart with Ampere-Maxwell Eq: Apply it to red rectangle: B d s = B z (x,t)l B z (x + Δx,t)l C d µ 0 dt E ˆn da = µ 0 l Δx E (x + Δx / 2,t) y t C B d s = µ 0 d dt E ˆn da B z (x + Δx,t) B z (x,t) Δx = µ 0 E y (x + Δx / 2,t) t o in the limit that dx is very small: B z x = µ 0 E y t 30

31 Group Problem: Wave Equation Use Faraday s Law E d s = d dt B ˆn da C and apply it to red rectangle to find the partial differential equation in order to find a relationship between E y / x and B z / t 31

32 Group Problem: Wave Equation ol. Use Faraday s Law: and apply it to red rectangle: C E d s d dt = E y (x + Δx,t)l E y (x,t)l B ˆn da = ldx B z t C E d s = d dt B ˆn da E y (x + dx,t) E y (x,t) dx = B z t o in the limit that dx is very small: E y x = B z t 32

33 1D Wave Equation for Electric Field E y x = B z t (1) B z x = µ 0 E y t (2) Take x-derivative of Eq.(1) and use the Eq. (2) 2 E y x 2 = x E y x = x B z t = t B z x = µ 0 2 E y t 2 2 E y x 2 = µ 0 2 E y t 2 33

34 1D Wave Equation for E 2 E y x 2 2 E = µ ε y 0 0 t 2 This is an equation for a wave. Let E y = f (x vt) 2 E y x 2 2 E y t 2 = f ''( x vt) ( ) = v 2 f '' x vt v = 1 µ 0 34

35 Definition of Constants and Wave peed Recall exact definitions of c m s 1 µ 0 4π 10 7 N s 2 C 2 The permittivity of free space is exactly defined by 1 c 2 µ C 2 m -2 N 1 v = c = 1 µ 0 in vacua 35

36 Group Problem: 1D Wave Eq. for B B z t = E y x B z x = µ 0 E y t Take appropriate derivatives of the above equations and show that 2 B z x 2 = 1 c 2 2 B z t 2 36

37 Wave Equations: ummary Both electric & magnetic fields travel like waves: 2 E y x 2 = 1 2 E y c 2 t 2 2 B z x 2 = 1 c 2 2 B z t 2 with speed c = 1 µ 0 But there are strict relations between them: B z t = E y x B z x = µ 0 E y t 37

38 Electromagnetic Waves 38

39 Electromagnetic Radiation: Plane Waves 39