Lecture 34: MON 13 APRLecture 34: MON 13 APR Ch.33.1 Ch.33.1––3,53,5––7: E&M Waves7: E&M Waves

James Clerk Maxwell (1831-1879)

Physics 2102

Jonathan Dowling

MT03 Avg: 65/100

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A: 90-100 B: 80-89 C: 60-79 D: 50-59

Problem 1 [18 points] In the figure below, two semicircular arcs (I & II) have radii R1 = 4.10 cm and R2 = 6.60 cm, carry current i = 0.281 A, and share the same center of curvature C. The same current i also flows through the straight sections of wire labeled III & IV. (a) (5 pts) What is the contribution to the magnitude of the magnetic field at point C from the two straight sections of wire III and IV? Explain your answer! (b) (8 pts) Calculate the magnitude of the magnetic field at the point C due to all four sections of wire. (c) (5 pts) What is the direction of the total magnetic field at the point C due to all four sections of wire? Circle one: Out of the page )(• . I nto the page ⊗ . Up towards the top of the page ↑ . D own towards the bottom of the page ↓ . To the right of the page → . T o the left of the page ← . The total magnetic field at C is zero and has no direction.

BIII= B

IV= 0

d!B ! d

!s "!r = 0

Biot-Savart

BTOT

=!BI!!BII=µ

0i"

4#1

RI

!1

RII

$

%&'

()= 8.14 *10

!7T

" = # = half a circle

Right Hand Rule & BI>BII Since I is Closer!

Question 2 [8 points]

The figure below shows four arrangements in which long parallel wires carry equal

currents i directly into or out of the page at the corners of identical squares.

(i) (4 pts) For square B, what is the direction of the magnetic field, with respect to the

page, at the center of the square? Circle one:

Out )(• In ! U p " D o wn#

Left $ R i g h t % Field is Zero )0( =B!

(ii) (4 pts) Rank the sections according to the magnitude of the magnitude of the

magnetic field at the center of each square, greatest first. Circle one:

B A > BB > BC BC > BB > BA BA > BC > BB

B B > BC > BA B A = BB = BC

Right Hand Rule & Vector Addition!

B=µ0i

2!r!;!!!!!!B

A= 0;!!!!!B

B= 2 2B;!!!!!B

C= 2B;

A solution to the Maxwell equations in empty space isa “traveling wave”…

c =1

µ0!0

= 3"108m/s

The “electric” waves travelat the speed of light!Light itself is a wave of electricity and magnetism!

Maxwell, Waves and LightMaxwell, Waves and Light

d2E

dx2= !µ

0"0

d2E

dt2# E = E

0sin k(x ! ct)

electric and magnetic “forces” can travel!

!! •=•SC

dAEdt

ddsB

00"µ !! •"=•

SC

dABdt

ddsE

A solution to Maxwell’s equations in free space:

)sin( txkEEm

!"=

)sin( txkBBm

!"=

n.propagatio of speed ,ck=

!

c =Em

Bm

=1

µ0!

0

=299,462,954 m

s!=!187,163 miles/sec

Visible light, infrared, ultraviolet,radio waves, X rays, Gammarays are all electromagnetic waves.

Electromagnetic WavesElectromagnetic Waves

Radio waves are reflected by the layer of the Earth’satmosphere called the ionosphere.

This allows for transmission between two points which arefar from each other on the globe, despite the curvatureof the earth.

Marconi’s experiment discovered the ionosphere! Expertsthought he was crazy and this would never work.

Fig. 33-1

The wavelength/frequency range in which electromagnetic (EM) waves(light) are visible is only a tiny fraction of the entire electromagneticspectrum.

Maxwell’s Rainbow

Fig. 33-2

(33-2)

An LC oscillator causes currents to flow sinusoidally, which in turn producesoscillating electric and magnetic fields, which then propagate through spaceas EM waves.

Fig. 33-3Oscillation Frequency:

1

LC! =

Next slide

The Traveling Electromagnetic (EM) Wave, Qualitatively

(33-3)

Fig. 33-5

Mathematical Description of Traveling EM Waves

Electric Field: ( )sinm

E E kx t!= "

Magnetic Field: ( )sinm

B B kx t!= "

Wave Speed:0 0

1c

µ !=

Wavenumber:

k =2!

"=#

c

Frequency: ! =

2"

T= 2" f

Vacuum Permittivity:0

!

Vacuum Permeability: 0µ

All EM waves travel a c in vacuum

Amplitude Ratio: m

m

Ec

B= Magnitude Ratio:

( )

( )

E tc

B t=

EM Wave Simulation

(33-5)

Electromagnetic waves are able to transport energy from transmitterto receiver (example: from the Sun to our skin).

The power transported by the wave and itsdirection is quantified by the Poynting vector. John Henry Poynting (1852-1914)

211|| E

cEBS

00

==µµ

The The Poynting Poynting Vector:Vector:Points in Direction of Power FlowPoints in Direction of Power Flow

E

BS

Units: Watt/m2

For a wave, sinceE is perpendicular to B: BES

!!!!=

0µ

1

In a wave, the fieldschange with time.Therefore the Poyntingvector changes too!!

The direction is constant,but the magnitude changesfrom 0 to a maximumvalue.

I = S =1

cµ0

___

E2 =

1

cµ0

Em

2sin

2(kx !"t)

____________

=1

2cµ0

Em

The average of sin2 overone cycle is ½:

2

2

1

mEc

I

0

=µ

21

rmsEc

I

0

=µ

Both fields have the same energy density.

2 22 2

0

1 1 1 1( )

2 2 2 2E B

B Bu E cB u! ! !

! µ µ0 0

0 0 0

= = = = =

or,

EM Wave Intensity, Energy DensityEM Wave Intensity, Energy DensityA better measure of the amount of energy in an EM wave is obtainedby averaging the Poynting vector over one wave cycle.The resulting quantity is called intensity. Units are also Watts/m2.

The total EM energy density is then0

22

0/ µ! BEu ==

Em

2

Erms

2

Solar EnergySolar EnergyThe light from the sun has an intensity of about 1kW/m2.What would be the total power incident on a roof ofdimensions 8m x 20m ?I = 1kW/m2 is power per unit area.P = IA = (103 W/m2) x 8m x 20m = 0.16 MegaWatt! !

The solar panel shown (BP-275) has dimensions 47in x29in. The incident power isthen 880 W. The actualsolar panel delivers 75W(4.45A at 17V): less than10% efficiency….

The electric meter on a solar homeruns backwards — Entergy Pays YOU!

The intensity of a wave is power per unit area. If one has asource that emits isotropically (equally in all directions) thepower emitted by the source pierces a larger and largersphere as the wave travels outwards: 1/r2 Law!

24 r

PI

s

!

=

So the power perunit area decreasesas the inverse ofdistance squared.

EM Spherical WavesEM Spherical Waves

ExampleExampleA radio station transmits a 10 kW signal at a frequencyof 100 MHz. At a distance of 1km from the antenna, findthe amplitude of the electric and magnetic fieldstrengths, and the energy incident normally on a squareplate of side 10cm in 5 minutes.

I =Ps

4!r2=

10kW

4! (1km)2= 0.8mW/m

2

I =1

2cµ0

Em

2! E

m= 2cµ0I = 0.775V/m

Bm= E

m/ c = 2.58nT

S =P

A=!U / t

A" !U = SAt = 2.4mJ

Receivedenergy:

Radiation PressureRadiation PressureWaves not only carry energy but also momentum. The effect is very small (we don’t ordinarily feel pressure from light). If lightis completely absorbed during an interval Dt, the momentum transferred is given by

c

up

!=!

t

pF

!

!=Newton’s law:

Now, supposing one has a wave that hits a surfaceof area A (perpendicularly), the amount of energy transferred to that surface in time Dt will be

tIAU !=! thereforec

tIAp

!=!

I

A

c

IAF =

pr=I

c (total absorption), p

r=

2I

c (total reflection)Radiation

pressure:

and twice as much if reflected.

[N/m2]

F

Radiation Pressure & Comet Tails

Solar Sails: Photons Propel Spacecraft!

StarTrek DS9 NASA Concept

NASA Demo

Radio transmitter:

If the dipole antennais vertical, so will bethe electric fields. Themagnetic field will behorizontal.

The radio wave generated is said to be “polarized”.

In general light sources produce “unpolarizedwaves”emitted by atomic motions in randomdirections.

EM waves: polarizationEM waves: polarization

When polarized light hits a polarizing sheet,only the component of the field aligned with thesheet will get through.

)= !cos(EEy

And therefore: !2

0 cosII =

Completely unpolarized light will haveequal components in horizontal and verticaldirections. Therefore running the lightthrougha polarizer will cut the intensity in half:I=I0/2

EM Waves: PolarizationEM Waves: Polarization

ExampleExampleInitially unpolarized light ofintensity I0 is sent into a systemof three polarizers as shown.What fraction of the initialintensity emerges from thesystem? What is the polarizationof the exiting light?

• Through the first polarizer: unpolarized to polarized, so I1=½I0.• Into the second polarizer, the light is now vertically polarized. Then,I2 = I1cos260o = 1/4 I1 = 1/8 I0.• Now the light is again polarized, but at 60o. The last polarizer ishorizontal, so I3 = I2cos230o = 3/4 I2 =3 /32 I0 = 0.094 I0.• The exiting light is horizontally polarized, and has 9% of the originalamplitude.