Lecture 42: FRI 04 DECLecture 42: FRI 04 DECFinal Exam Review IIFinal Exam Review II

Physics 2113

Jonathan Dowling

Final Exam

• 3:00PM – 5:00PM MON 07DEC

PHYS 2113-1 (Moreno) is scheduled to take their exam in Lockett 9

PHYS 2113-2 (Gaarde) is scheduled to take their exam in Lockett 10

PHYS 2113-3 (Dowling) is scheduled to take their exam in Lockett 6

PHYS 2113-4 (O'Connell) is scheduled to take their exam in Lockett 15

PHYS 2113-5 and 6 (Abdelwahab) is scheduled to take their exam in Lockett 2

PHYS 2113-7 (Hansen) Last name starts with A - K: take their exam in Lockett 10, last name starts with L - Z: take their exam in Lockett 2

Final Exam

• 100 PTS: CH 13, 21–30 / HW01-11

This part will be 11 multiple choice

questions one from each chapter.

• 100 PTS: CH 31–33 / HW12-14

This part will be three multiple choice

questions and three word problems, one

each from each chapter.

B !

E

i

B

i

B

Displacement “Current”Displacement “Current”Maxwell proposed it based on symmetry and math — no experiment!

Changing E-field Gives Rise to B-Field!

32.3: Induced Magnetic Fields:

Here B is the magnetic field induced along a closed loop by the changing electric flux E in the region encircled by that loop.

Fig. 32-5 (a) A circular parallel-plate capacitor, shown in side view, is being charged by a constant current i. (b) A view from within the capacitor, looking toward the plate at the right in (a).The electric field is uniform, is directed into the page (toward the plate), and grows in magnitude as the charge on the capacitor increases. The magnetic field induced by this changing electric field is shown at four points on a circle with a radius r less than the plate radius R.

Example, Magnetic Field Induced by Changing Electric Field:

Example, Magnetic Field Induced by Changing Electric Field, cont.:

32.4: Displacement Current:

Comparing the last two terms on the right side of the above equation shows that the term must have the dimension of a current. This product is usually treated as being a fictitious current called the displacement current id:

in which id,enc is the displacement current that is encircled by the integration loop.

The charge q on the plates of a parallel plate capacitor at any time is related to the magnitude E of the field between the plates at that time by in which A is the plate area.

The associated magnetic field are:

AND

Example, Treating a Changing Electric Field as a Displacement Current:

32.5: Maxwell’s Equations:

Fig. 33-5

Mathematical Description of Traveling EM Waves

Electric Field:

Magnetic Field:

Wave Speed:

Wavenumber:

Angular frequency:

Vacuum Permittivity:

Vacuum Permeability:

All EM waves travel a c in vacuum

Amplitude Ratio: Magnitude Ratio:

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)

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

Units: Watt/m2

For a wave, sinceE is perpendicular to B:

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.

The average of sin2 overone cycle is ½:

Both fields have the same energy density.

EM Wave Intensity, Energy DensityEM Wave Intensity, Energy DensityA 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

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!

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

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. Assume a spherical wave. 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.

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 Δt, the momentum Transferred Δp is given by

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 Δt will be

therefore

I

A

Radiation pressure:

and twice as much if reflected.

[Pa=N/m2]

F

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.

And therefore:

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

Completely unpolarized light will have equal components in horizontal and vertical directions. Therefore running the light through a 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 =

I1cos2(60o)= 1/4 I1 = 1/8 I0.

• Now the light is again polarized, but at 60o. The last polarizer is

horizontal, so I3 = I2cos2(30o) = 3/4 I2 =3 /32 I0 = 0.094 I0.

• The exiting light is horizontally polarized, and has 9% of the original

amplitude.

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

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

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

Reflection and RefractionReflection and Refraction

Law of reflection (Light Bounces): the angle of incidence θ1 equals the angle of reflection θ’1. θ1 = θ’1

Law of reflection (Light Bounces): the angle of incidence θ1 equals the angle of reflection θ’1. θ1 = θ’1

When light finds a surface separating two media (air and water, for example), a beam gets reflected (bounces) and another gets refracted (bends).

n is the index of refraction of the medium. In vacuum, n = 1. In air, n ~ 1. In all other media, n > 1.

33.8: Reflection and Refraction:

Toward✔Can’t go past normal✖

Away✖

In each case going from less to greater.

Chromatic DispersionChromatic DispersionThe index of refraction depends on the wavelength (color) of the light.