Poynting Vector and Energy Flow W14D1

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1 Poynting Vector and Energy Flow W14D1 1

2 Announcements Week 14 Prepset due online Friday 8:30 am PS 11 due Week 14 Friday at 9 pm in boxes outside Sunday Tutoring 1-5 pm in

3 Outline Poynting Vector and Electromagnetic Waves Poynting Vector and Energy Conservation 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 in EM Waves: The Poynting Vector 6

7 Sinusoidal Plane Waves Summary For the plane wave! E(x,t) = E 0 sin 2π λ! B(x,t) = E 0 c sin 2π λ (x ct) ĵ (x ct) ˆk 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 ε 0 7

8 Group Problem: Energy in EM Waves A plane sinusoidal electromagnetic wave is moving in the +x-direction. At t=0, it passes through one end of a rectangular volume of space of cross sectional area A and length dx=cdt. After the time interval dt, the wave fills the volume of space dv = A cdt. The time averaged energy densities for electric magnetic fields are given by < u E >= 1 2 ε < E 2 >, <u 0 B >= 1 < B 2 > 2µ 0 What is the time average rate of change of electromagnetic energy per time in the volume, du/dt? Express your answer only in terms of A, µ 0, E 0, and B 0.! E(x,t) = E 0 sin 2π λ! B(x,t) = E 0 c sin 2π λ + y c + z (x ct) ĵ (x ct) ˆk cdt A +x 8

9 Energy Flow in EM Waves The time averaged power that flows into the box is given by where of has dimensions and SI units By energy conservation, the time averaged rate that energy increases in the box. < P in > = < S > A = < du > = 1 E 0 B 0 dt 2 µ 0 A < S > = 1 2 < P in > = < S > A < S > = < P in > / A energy sec 1 area 1 E 0 B 0 µ 0 W m 2 + y c + z cdt A +x 9

10 Poynting Vector and Intensity Direction of energy flow = direction of wave propagation S = E B µ 0 : Poynting vector Intensity I: time averaged magnitude of Poynting vector I < S >= E B 0 0 = E 2 0 2µ 2µ 0 0 c = cb 2 0 2µ 0 10

11 Group Problem: Poynting Vector An electric field of a plane wave is given by the expression E( y,t) = E 0 sin(ky + ωt) ˆk a) Find a vector expression for the Poynting vector S( y,t) = 1 µ 0 E( y,t) B( y,t) associated with this plane wave. b) What is the direction of propagation for this palne wave? 11

12 Poynting Vector and Power Power per unit area: Poynting vector S = E B µ 0 Power through a surface P = open surface S ˆn da 12

13 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? For simplicity, you may assume the emitted light is a monochromatic plane wave of wavelength λ. 13

14 Momentum & Radiation Pressure EM waves transport energy:! S =! E! B µ 0 and also transport momentum: p = U c Hence exert a pressure: P rad = F A = 1 A dp dt = 1 ca du dt = S c (perfectly absorbing surface) P rad = F A = 1 A dp dt = 2 1 ca du dt = 2 S c (perfectly reflecting surface) 14

15 Group Problem: Radiation From the Sun As you lie on a beach in the bright midday sun, approximately what force does the light exert on you? Assume light from sun is a plane wave. Facts about the sun: Distance from Earth r ES =1.5 x m Total power output P = 4 x Watts Power per area at surface of earth: 2 P / 4πr ES Speed of light c = 3 x 10 8 m/s 15

16 Energy Flow in Resistors, Capacitors and Inductors 16

17 Energy Flow: Resistor S = E B µ 0 On surface of resistor direction is INWARD 17

18 Group Problem: Resistor a I Power l Consider a cylindrical resistor of length l, radius a, resistance R and current I through the resistor. Recall that there is an electric field in the wire given by E l = ΔV = I R a) What are vector expressions for 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 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? ˆ ˆr r ˆk I 18

19 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 19

20 I +Q(t) P d Group Problem: Capacitor Q(t) R I ˆr ˆ ˆk unit vectors at point P A circular capacitor of spacing d and radius R is in a circuit carrying the steady current I. Neglect edge effects. At time t = 0 it is uncharged. The point P lies a distance R from the central axis. a) Find the electric field E(t) at P (mag. & dir.) b) Find the magnetic field B(t) at P (mag. & dir.) c) Find the Poynting vector S(t) at P (mag. & dir.) d) What is the flux of the Poynting vector into/out of the capacitor? e) How does this compare to the time derivative of the energy stored in the electric field? 20

21 CQ: Capacitor For the capacitor shown in the figure, the direction of the Poynting vector at the point P is in the I +Q(t) P d Q(t) R I ˆr ˆ ˆk unit vectors at point P 1. + ˆr direction 2. + ˆk direction 3. + ˆθ direction 4. ˆr direction 5. ˆk direction 6. ˆθ direction 21

22 Inductors I L LI = Φ Self ε = dφ B dt = L di dt h 22

23 I(t) I(t) z a Group Problem: Inductor A solenoid with N turns, radius a and length h, has a current I(t) that is h N turns; each turn carries a current I 1. Find the magnetic field B(t) at P (dir. and mag.) 2. Find the electric field E(t) at P (dir. and mag.) a P decreasing in time. Consider a point P located at a radial distance a from the center axis of the solenoid at the inner edge of the wires. Neglect edge effects. 3. Find the Poynting vector S(t) at P (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? ˆk ˆ ˆr 23

24 CQ: Inductor I(t) I(t) z a h a N turns; each turn carries a current I P ˆk ˆ ˆr A solenoid has a current I(t) that is increasing in time. The direction of the Poynting vector at the point P is in the 1. + ˆr direction 2. + ˆk direction 3. + ˆθ direction 4. ˆr direction 5. ˆk direction 6. ˆθ direction 24

25 S = E B µ 0 Energy Flow: Inductor Direction on surface of inductor with increasing current is INWARD 25

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

27 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 27

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