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P36 - 1 W15D2 Poynting Vector and EM Waves Radiation Pressure Final Exam Review.

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P36 - 1 W15D2 Poynting Vector and EM Waves Radiation Pressure Final Exam Review
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P36 - 1

W15D2

Poynting Vector and EM Waves

Radiation Pressure

Final Exam Review

P36 -

Announcements

Final Exam Monday Morning May 20

from 9 am-12 noon

Johnson Athletic Center Track 2nd floor

2

P36 - 3

Before Starting…

All of your grades should now be posted (with possible exception of last problem set). If this is not the case contact your grad TA immediately.

We have given you enough information about how we grade and what the cut lines are that you can estimate what you need on the final to get the letter grade of your heart’s desire.

P36 - 4

Poynting Vector and EM Waves

P36 - 5

Energy in EM Waves

Energy densities:

Consider cylinder:

What is the energy flow per unit area?

0

EB

1 dUS

A dt

P36 - 6

Poynting Vector and Intensity

units: Joules per square meter per sec

Direction of energy flow = direction of wave propagation

Intensity I: 2 2

20 0 0 00 0

0 0 0

1

2 2 2 2

E B E cBI S c E

c

P36 - 7

Momentum & Radiation PressureEM waves transport energy:

This is only for hitting an absorbing surface. For hitting a perfectly reflecting surface the values are doubled:

They also transport momentum:

They exert a pressure:

P36 - 8

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 is the maximum value of the electric field, E , at this same distance from the bulb in V/m? 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.

Group Problem: Radiation

8

1.25664 06

8.84194 12

3 10 /

o

o

E

E

c x m s

P36 - 9

Electromagnetism Review

P36 -10

Circuits

P36 -11

Sign Conventions - BatteryMoving from the negative to positive terminal of a

battery increases your potential

Moving from the positive to negative terminal of a battery decreases your potential

P36 -12

Sign Conventions - Resistor

Moving across a resistor in the direction of current decreases your potential

Moving across a resistor opposite the direction of current increases your potential

P36 -13

Sign Conventions - CapacitorMoving across a capacitor from the negatively to positively charged plate increases the electric potential

Moving across a capacitor from the positively to negatively charged plate decreases the electric potential

P36 -14

(Dis)Charging a Capacitor1. When the direction of current flow is toward

the positive plate of a capacitor, then

2. When the direction of current flow is away from the positive plate of a capacitor, then

P36 -15

Sign Conventions - Inductor

Moving across an inductor in the direction of current contributes

dIL

dt

Moving across an inductor opposite the direction of current contributes

P36 -16

Kirchhoff’s Modified 2nd Rule

If all inductance is ‘localized’ in inductors then our problems go away – we just have:

P36 -17

Steps For Setting Up Circuit Equations for Circuit with N loops and M junctions

1. Simplify resistors in series/parallel

2. Assign current in every branch

3. Choose circulation direction for N-1 loops

4. Assign charges to each side of capacitor.

5. Determine relation between current and charge for each branch containing a capacitor

6. Write M-1 current conservation equations for junctions

7. Write N-1 loop equations.

8. Solve system of equations.

P36 -18

Displacement Current

P36 -19

Displacement Current

So we had to modify Ampere’s Law:

P36 -20

EM Waves

P36 -21

Traveling Sinusoidal Wave: Summary

y(x,t) y0sin(k(x vt))

Spatial period : Wavelength ; Temporal period T .

Two periodicities:

P36 -22

Traveling Sinusoidal Wave

Wave Number : k 2 /

Angular Frequency : 2 / T

Dispersion Relation : vT kv

Frequency : f 1 / T v f

y(x,t) y0sin(k(x vt)) y

0sin(kx t)

Alternative form:

P36 -23

Plane Electromagnetic Waves

http://youtu.be/3IvZF_LXzcc

P36 -24

Electromagnetic Waves: Plane Sinusoidal Waves

http://youtu.be/3IvZF_LXzcc

Watch 2 Ways:

1) Sine wave traveling to right (+x)

2) Collection of out of phase oscillators (watch one position)

Don’t confuse vectors with heights – they are magnitudes of electric field (gold) and magnetic field (blue)

P36 -25

Traveling Plane Sinusoidal Electromagnetic Waves

are special solutions to the 1-dim wave equations

2Ey

x2

1

c2

2Ey

t2

2Bz

x2

1

c2

2Bz

t2

where

P36 -26

1 Dim’l Sinusoidal EM Waves

In order for the fields

to satisfy either condition below

then

B0E

0/ c

P36 -

Properties of 1 Dim’l EM Waves

c 1

0

0

3.0 108 m

s

E0/ B

0c

1. Travel (through vacuum) with speed of light

2. At every point in the wave and any instant of time, electric and magnetic fields are in phase with one another, amplitudes obey

3. Electric and magnetic fields are perpendicular to one another, and to the direction of propagation (they are transverse):

27

P36 -28

Concept Questions:EM Waves

P36 -29

Concept Question: Direction of Propagation

The figure shows the E (yellow) and B (blue) fields of a plane wave. This wave is propagating in the

1. +x direction

2. –x direction

3. +z direction

4. –z direction

P36 -30

Concept Question Answer: Propagation

The propagation direction is given by the (Yellow x Blue)

Answer: 4. The wave is moving in the –z direction

P36 -31

Concept Question: Traveling Wave

The B field of a plane EM wave isThe electric field of this wave is given by

P36 -32

Concept Q. Ans.: Traveling Wave

From the argument of the , we know the wave propagates in the positive y-direction.

Answer: 4.

sin(ky t)

P36 -

Concept Question EM Wave

33

The magnetic field of this wave is given by:

The electric field of a plane wave is:

P36 -34

Concept Q. Ans.: EM Wave

From the argument of the , we know the wave propagates in the negative z-direction.

Answer: 1.

sin(kz t)

P36 -35

Energy Flow

P36 -36

Also in Circuit Elements…

On surface of resistor is INWARD

P36 -37

Concept Questions:Poynting Vector

P36 -38

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 time2. Constant in time.3. Decreasing in time. 4. Not enough information given to determine how Q is changing.

P36 -39

Concept Q. Answer: Capacitor

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.

Answer: 3. The charge Q is decreasing in time

P36 -40

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 in2. Constant in time.3. Decreasing in time.

P36 -41

Concept Q. Answer: Capacitor

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.

Answer: 1. The the energy stored in the electric field is increasing in time

P36 -42

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.

P36 -43

Concept Question Answer: Inductor

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.

Answer: 3. The current I is decreasing in time

P36 -44

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 time2. Constant in time.3. Decreasing in time.

P36 -45

Concept Question Answer: Inductor

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.

Answer: 3. The energy stored in the magnetic field is decreasing in time.


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