W15D1: Poynting Vector and Energy Flow. Today s Readings: Course Notes: Sections 13.6,

Transcription

1 W15D1: Poynting Vector and Energy Flow Today s Readings: Course Notes: Sections 13.6,

2 Announcements Final Math Review Week 15 Tues from 9-11 pm in Final Exam Monday Morning May 20 from 9 am-12 noon Johnson Athletic Center Track 2 nd floor 2

3 Outline Poynting Vector and Energy Flow Examples 3

4 Maxwell s Equations S E ˆn da = 1 ρ dv ε 0 V (Gauss's Law) B ˆn da = 0 (Magnetic Gauss's Law) S C E d s = d dt S B ˆn da B d s = µ 0 J ˆn da C S + µ 0 ε 0 d dt S E ˆn da (Faraday's Law) (Maxwell - Ampere's Law) 4

5 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 ε 0 2 E 2 dv U B = u B dv = all space 1 B 2 dv 2µ all space 0 5

6 Energy Flow 6

7 Poynting Vector Power per unit area: Poynting vector S = E B µ 0 Power through a surface P = open surface S ˆn da 7

8 Group Problem: Resistor Power I a L Consider the above cylindrical resistor with resistance R and current I flowing as shown. a) What are the electric and magnetic fields on the surface of the resistor? b) Calculate the flux of the Poynting vector through the surface of the resistor (power) in terms of the electric and magnetic fields. c) Express your answer to part b) in terms of the current I and resistance R. Does your answer make sense? 8

9 Energy Flow: Resistor S = E B µ 0 On surface of resistor direction is INWARD 9

10 Energy Flow: Capacitor S = E B µ 0 What is the magnetic field? 10

11 Displacement Current E = Q ε 0 A Q = ε 0 EA = ε 0 Φ E dq dt = ε 0 dφ E dt I dis So we had to modify Ampere s Law: B d d s = µ 0 J ˆnda + µ 0 ε 0 dt C S = µ 0 (I con + I dis ) S E ˆnda 11

12 Sign Conventions: Right Hand Rule B d s = µ 0 J ˆnda C + µ 0 ε 0 d S dt S 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. B field clockwise, line integral negative. B field counterclockwise, line integral positive. 12

13 Concept Questions: Poynting Vector 13

14 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 charge Q is: 1. Increasing in time 2. Constant in time. 3. Decreasing in time. 4. Not enough information given to determine how Q is changing. 14

15 Concept Q. Answer: Capacitor Answer: 3. The charge Q is decreasing in time Use the Ampere-Maxwell Law. Choose positive unit normal out of plane. Because the magnetic field points clockwise line integral is negative hence positive electric flux (out of the plane of the figure on the right) must be decreasing. Hence E is decreasing. Thus Q must be decreasing, since E is proportional to Q. 15

16 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 2. Constant in time. 3. Decreasing in time. 16

17 Concept Q. Answer: Capacitor Answer: 1. The the energy stored in the electric field is increasing in time The direction of the Poynting Flux S (= E x B) inside the capacitor is inward. Therefore electromagnetic energy is flowing inward, and the energy in the electric field inside is increasing. 17

18 Group Problem: Capacitor A circular capacitor of spacing d and radius R is in a circuit carrying the steady current i shown. Ignore edge effects. At time t = 0 it is uncharged 1. Find the electric field E(t) at A (mag. & dir.) 2. Find the magnetic field B(t) at A (mag. & dir.) 3. Find the Poynting vector S(t) at A (mag. & dir.) 4. What is the flux of the Poynting vector into/out of the capacitor? 5. How does this compare to the time derivative of the energy stored in the electric field? 6. Does this make sense? 18

19 Another look at Inductance 19

20 Faraday & Inductors I L LI = Φ Self ε = dφ B dt = L di dt 20

21 Concept Question: Inductor The figures above show a side and top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. The current I is 1. increasing in time. 2. constant in time. 3. decreasing in time. 21

22 Concept Question Answer: Inductor Answer: 3. The current I is decreasing in time Use Faraday s law. Choose positive unit normal out of plane. Because the electric field points counterclockwise line integral is positive, therefore the positive magnetic flux must be decreasing (out of the plane of the figure on the right). Hence B is decreasing. Thus I must be decreasing, since B is proportional to I. 22

23 Concept Question: Inductor The figures above show a side and top view of a solenoid carrying current I with electric and magnetic fields E and B at time t. The energy stored in the magnetic field is 1. Increasing in time 2. Constant in time. 3. Decreasing in time. 23

24 Concept Question Answer: Inductor Answer: 3. The energy stored in the magnetic field is decreasing in time. The Poynting Flux S (= E x B) inside the solenoid is directed outward from the center of the solenoid. Therefore EM energy is flowing outward, and the energy stored in the magnetic field inside is decreasing. 24

25 Group Problem: Inductor A solenoid of radius a and length h has an increasing current I(t) as pictured. Consider a point D inside the solenoid at radius r (r < a). Ignore edge effects. 1. Find the magnetic field B(t) at D (dir. and mag.) 2. Find the electric field E(t) at D (dir. and mag.) 3. Find the Poynting vector S(t) at D (dir. and mag.) 4. What is the flux of the Poynting vector into/out of the inductor? 5. How does this compare to the time derivative of the energy stored in the magnetic field? 6. Does this make sense to you? 25

26 S = E B µ 0 Energy Flow: Inductor Direction on surface of inductor with increasing current is INWARD 26

27 S = E B µ 0 Energy Flow: Inductor Direction on surface of inductor with decreasing current is OUTWARD 27

28 Power & Energy in Circuit Elements P = Surface S da Dissipates Power u E = 1 2 ε 0 E 2 u B = 1 2µ 0 B 2 Store Energy POWER When (dis)charging 28

29 Poynting Vector and EM Waves 29

30 Energy densities: Consider cylinder: Energy in EM Waves u E = 1 2 ε 0 E 2, u B = 1 2µ 0 B 2 du = (u E + u B )Adz = 1 2 ε E 2 + B2 0 Acdt S µ 0 What is the energy flow per unit area? 1 du = c 2 ε EB = = c ceb + 0 A dt 2 ε E 2 + B2 0 µ 0 cµ 0 = EB ( ε 2µ 0 µ 0 c 2 + 1) EB = µ

31 Poynting Vector and Intensity Direction of energy flow = direction of wave propagation S = E B µ 0 : Poynting vector units: Joules per square meter per sec Intensity I: I < S >= E 0 B 0 = E 2 0 2µ 2µ c = cb 2 0 2µ

32 Momentum & Radiation Pressure EM waves transport energy: S = E B µ 0 They also transport momentum: p = U / c They exert a pressure: P = F A = 1 A dp dt = 1 ca du dt = S c This is only for hitting an absorbing surface. For hitting a perfectly reflecting surface the values are doubled: Momentum transfer : p = 2U c ; Radiation pressure : P = 2S c 32

33 Group Problem: Radiation A light bulb puts out 100 W of electromagnetic radiation. What is the time-average intensity of radiation from this light bulb at a distance of one meter from the bulb? What are the maximum values of electric and magnetic fields, E and B, at this same distance from the bulb? What is the pressure this radiation will exert on a very small perfectly conducting plate at 1 meter. For simplicity, you may assume the radiation is a plane wave of wavelength λ. 33