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

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

James Clerk Maxwell (1831-1879)

Physics 2102

Jonathan Dowling

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Q1/P3 K. Schafer Office hours: MW 1:30-2:30 pm 222B Nicholson

P1/Q2 J. Dowling Office hours: MWF 10:30-11:30 am 453 Nicholson

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P3/Q2 C. Buth Office hours: MF 2:30-3:30 pm 222A Nicholson

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 )(•. Into the page ⊗. Up towards the top of the page ↑. Down towards the bottom of the page ↓. To the right of the page →. To the left of the page ←. The total m agnetic field at C is zero and has no direction.

BIII =BIV =0

drB∝ drs× rr=0Biot-Savart

BTOT =rBI −

rBII =

m0if4p

1RI

−1RII

⎛⎝⎜

⎞⎠⎟ =8.14 ×10−7T

f =p =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 ⊗ Up ↑ Down↓ Left ← Right → Field is Zero )0(=Br (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: BA > BB > BC BC > BB > BA BA > BC > BB BB > BC > BA BA = BB = BC

Right Hand Rule & Vector Addition!

B=m0i2pr

; BA =0; BB =2 2B; BC =2B;

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

c =1m0e0

=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

d 2Edx2 =−m0e0

d 2Edt2

⇒ E =E0 sink(x−ct)

electric and magnetic “forces” can travel!

∫∫ •=•SC

dAEdtddsB 00εμ ∫∫ •−=•

SC

dABdtddsE

A solution to Maxwell’s equations in free space:

)sin( txkEE m w−=

)sin( txkBB m w−=n.propagatio of speed ,c

k=w

c =Em

Bm

=1m0e0

=299,462,954 ms = 187,163 m iles/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’s atmosphere called the ionosphere.

This allows for transmission between two points which are far from each other on the globe, despite the curvature of the earth.

Marconi’s experiment discovered the ionosphere! Experts thought 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 electromagnetic spectrum.

Maxwell’s Rainbow

Fig. 33-2

(33-2)

An LC oscillator causes currents to flow sinusoidally, which in turn produces oscillating electric and magnetic fields, which then propagate through space as EM waves.

Fig. 33-3Oscillation Frequency:

1LC

w =

Next slide

The Traveling Electromagnetic (EM) Wave, Qualitatively

(33-3)

Fig. 33-5

Mathematical Description of Traveling EM Waves

Electric Field: ( )sinmE E kx tω= −

Magnetic Field: ( )sinmB B kx tω= −

Wave Speed:0 0

1cm e

=

Wavenumber: k =

2pl

=wc

Frequency: w =

2pT

=2p f

Vacuum Permittivity:0e

Vacuum Permeability:0m

All EM waves travel a c in vacuum

Amplitude Ratio:m

m

E cB

= 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|| Ec

EBS00

==mm

The Poynting Vector: The Poynting 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

rrr×=

0m1

In a wave, the fields change with time. Therefore the Poynting vector changes too!!

The direction is constant, but the magnitude changes from 0 to a maximum value.

I =S =1cm0

___

E2 =1cm0

Em2 sin2(kx−wt)

____________

=1

2cm0

EmThe average of sin2 overone cycle is ½:

2

21

mEc

I0

=m

21rmsE

cI

0=

m

Both fields have the same energy density.

2 22 2

01 1 1 1( )2 2 2 2E B

B Bu E cB ue e ee m m0 00 0 0

= = = = =

or,

EM Wave Intensity, Energy EM Wave Intensity, Energy DensityDensity

A better measure of the amount of energy in an EM wave is obtained by 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 then 022

0 / me BEu ==

Em2

Erms2

Solar EnergySolar EnergyThe light from the sun has an intensity of about 1kW/m2. What would be the total power incident on a roof of dimensions 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 x 29in. The incident power is then 880 W. The actual solar panel delivers 75W (4.45A at 17V): less than 10% efficiency….



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

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

24 rPI s

p=

So the power per unit area decreases as the inverse of distance squared.

EM Spherical WavesEM Spherical Waves

ExampleExampleA radio station transmits a 10 kW signal at a frequency of 100 MHz. At a distance of 1km from the antenna, find the amplitude of the electric and magnetic field strengths, and the energy incident normally on a square plate of side 10cm in 5 minutes.

I =Ps

4pr2=

10kW4p(1km )2

=0.8mW /m 2

I =1

2cm0

Em2 ⇒ Em = 2cm0I =0.775V /m

Bm =Em / c=2.58nT

S =PA=DU /tA

⇒ DU =SAt=2.4m JReceivedenergy:

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

cup D=D

tpF

DD=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 D=D therefore ctIAp D=D

I

A

cIAF =

pr =Ic (total absorption), pr =

2Ic (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 “unpolarized waves”emitted by atomic motions in random directions.

EM waves: polarizationEM waves: polarization

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

)= θcos(EEy

And therefore: θ20 cosII =

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

EM Waves: PolarizationEM Waves: Polarization

ExampleExampleInitially unpolarized light of intensity I0 is sent into a system of three polarizers as shown. What fraction of the initial intensity emerges from the system? What is the polarization of 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 is horizontal, so I3 = I2cos230o = 3/4 I2 =3 /32 I0 = 0.094 I0. • The exiting light is horizontally polarized, and has 9% of the original amplitude.

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