Physics comprehensive

Up and Down Quarks

The fact that the free neutron decays and nuclei decay by beta decay in processes like is thought to be the result of a more fundamental quark process

Intrinsic spin of most objects (electron, proton, neutron, quarks, leptons) is described by the label (quantum number) s = ½ . The length is the intrinsic spin vector is calculated as with possible z-components of , where can have the values + ½ or – ½ . These z-components are often referred to as spin-up and spin-down, respectively.

Mediator	Spin	Force
gluon	1	quark-quark
photon	1	charge-charge
Z, Wbosons	1	weak decay
graviton	2	mass-mass

Ex. 33) For atoms in a simple cubic structure, the maximum percentage of the total available volume that can be occupied by the atoms is approximately

(A) 

( 

(D)60%

(E)70%

The idea is to stack solid spheres in the prescribed pattern and to compute the fraction of the total volume that is in the spheres. A cube has eight corners and a side length s. A sphere of radius ½ s can be centered on each corner. One eighth of each sphere will be in the cube times eight corner or one sphere of radius ½s per cube of side s. One complete sphere of radius ½ s per cube of volume s³.

(Prepare a sketch!)

C

The closest packing of hard spheres (74%) is achieved in the hexagonal close-packed and face-centered cubic configurations. Body-centered cubic is about 68 %.

simple cublic body-centered cubic face-centered cubic

hexangonal c.p.

(A) 2 x 10⁰

(2 x 10² 

(C)2 x 10⁴

(D)2 x 10⁶

(E)2 x 10⁸

The sun is 4 light years from the nearest star Proxima Centauri.

Galaxies have lots of stars so Galaxies should be thousands of light years in dimension (40000).

During a 1965 Star Trek episode, science officer Spock remarked that it was 40,000 light years to the galactic center (Milky Way, our galaxy). Google says that the main disk of the Milky Way is about 10⁵ light years in diameter. Galaxies should be separated by 10 to 100 galaxy diameters. Look for 10-100 x 10⁵  D

The universe is 14 x 10⁹ years of age and so is about 14 x 10⁹ light years in extent.

The separation of galaxies should be pretty small compared to the extent of the universe

(< 10^-3*Universe  10⁷ lt. yr.) so response E is not attractive. Possible range: 40,000 x 10 < answer < 10^-3 (14 x 10⁹)_. The geometric mean:= 2.37 x 10⁶ lt. yr.
(35) Math Methods (Hardly a challenge for a 351/2 survivor.)
Ex. 35) Which of the following is the derivative with respect to x of the function x² cos[3 x⁴ + 1]?

The function is the product of x² and cos[3 x⁴ + 1] so by the product rule,

Using the chain rule, ^d/_dx(cos[3 x⁴ + 1]) = - sin[3 x⁴ + 1] (12 x³)

Combining terms: ^df/_dx = 2x cos[3 x⁴ + 1] – 12 x⁵ sin[3 x⁴ + 1] C

All students should be intimately familiar with every concept and detail presented in SP351/2.
Astrophysics Annex (Jeff2)
While the MFT contains "astrophysics" questions, don't think that you cannot answer them without any classes in the subject! Yes, last year some questions involved stars, but stars were simply a convenient real-world blackbody for Wein's Law and the Stefan Boltzmann Relation -- actually just thermodynamics! This section is a quick introduction to the types of "astrophysics" questions which may be expected and some explanation of some of the exotic units which might be used in the problem.

Angles: Astronomical measurements typically use angles to report the apparent size or separation between two objects (as well as to establish a coordinate system on the sky). The formula for angular size or separation is:
tan = ^s/_d
where d is the distance to the object and

s is either a physical size of an object or the distance separating two objects.

1 Parsec: The distance from the Sun to a star with an annual parallax angle of 1": 3.09 X 10¹³ km

1 Light X (where X= a unit of time like year, day, second) distance is from light travel time: Use d=ct, where t is the appropriate X in seconds. 1 Parsec  3.26 LtYr
Escape Velocity: An object trapped in a gravitational potential well of a larger primary object has a negative gravitational potential energy. In order to escape (to infinite distance) where the potential energy is zero and the kinetic energy must be  0. If the mass of the object is m, the mass of the primary object is M and the separation between the two is r, the escape velocity (v_e) is determined by:


Kepler's Laws: Gravity drives the universe. Planetary motions are typically described in terms of Kepler's laws:
1) All planets move in elliptical orbits with the Sun at one focus.

2) The line that joins a planet to the sun sweeps out equal areas of their orbit in equal times. (Closer to the Sun you move more quickly -- angular momentum conservation; m r v_ )

3) For objects of planet size or smaller orbiting the Sun, if we express the period (P) in years and the average distance (a, semi-major axis) in astronomical units then P²  a³.

How can years squared equal astronomical units cubed? There is a hidden scaling! Kepler didn't know Newton's laws, just what he could observe which was periods and astronomical units (both defined in terms of the Earth's 1 year and 1 AU). Try approaching Kepler's Third Law by deriving it from Newton. Consider a planet (mass m) in circular motion at distance R around the Sun (mass M, M >> m). The only force acting on the planet is gravity and it creates a centripetal acceleration:

Okay, but that also means that , so for two planets with distances R₁and R₂ and periods P₁ and P₂, . Cancelling, or . Choosing planet 1 to be the earth, R₁ = 1 AU and P₁ = 1 year, and

Wein's law: _peak T_Kelvin = 2897 m or f_peak  T_Kelvin

Stefan Boltzmann: irradiance (radiative energy flux density) =  T⁴;  = 5.67 x 10^{-8 J}/_{(s m}2_K4₎
Stellar Spectra: The composition of stars can be determined because the inner core of a star radiates like a blackbody. The outer cooler parts of the star absorb certain colors of light due to radiative excitation and the rest of the spectrum makes it to us minus those colors. The exact wavelengths of light removed are determined using the techniques in section (21). Each element has a unique set of energy levels, therefore we can tell what elements are present based on their spectral lines.
Stellar Structure and Evolution: Stars are held together by gravity but are prevented from collapsing by fusion in their cores. This balance of pressures keeps them stable. For the majority of their lives, stars turn H into He by fusion and these stars are called "main sequence". It is possible to fuse heavier elements (most stars are massive enough to generate sufficient pressures and temperatures to fuse He into C late in their lives when they have depleted the H supply in their cores).
Stars are in hydrostatic equilibrium which you will remember from your general physics class.

In this equation, P is pressure (force/area), r is distance from the center, M(r) is the net mass within r or the center (net mass located inside a sphere of radius r concentric with the , ρ(r) is the density at distance r and G is the universal gravitational constant.

1 light year apart so we live in an uncrowded neighborhood. That's good, because interactions with stars are relatively disruptive to planetary orbits... Also stars are small compared to their separations -- a star is a light second across but a light year away from its neighbor.

billion light years across. I'd personally bet on the lower side of

that...

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Time for a Sun-sized pulsar to rotate: 1 millisecond Time for the Earth to go around the Sun - 365.25 days Time for Pluto to go around the Sun - 248 years. Time for the Sun to go around the Galaxy - 200 million years Time when Andromeda collides with the Milky Way - 3 billion years

- The sun is still a hydrogen burning main sequence star!

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Alternately try the "Galaxy Song" by Monty Python from the meaning of life (which can be downloaded in mp3 from http://www.mwscomp.com/sound.html):
Just remember that you're standing on a planet that's evolving And revolving at nine hundred miles an hour, That's orbiting at nineteen miles a second, so it's reckoned, A sun that is the source of all our power. The sun and you and me and all the stars that we can see Are moving at a million miles a day. In an outer spiral arm, at forty thousand miles an hour, Of the galaxy we call the 'Milky Way'.
Our galaxy itself contains a hundred billion stars. It's a hundred thousand light years side to side.

It bulges in the middle, sixteen thousand light years thick, But out by us, it's just three thousand light years wide. We're thirty thousand light years from galactic central point. We go 'round every two hundred million years, And our galaxy is only one of millions of billions In this amazing and expanding universe. The universe itself keeps on expanding and expanding In all of the directions it can whizz As fast as it can go, at the speed of light, you know, Twelve million miles a minute, and that's the fastest speed there is.

So remember, when you're feeling very small and insecure, How amazingly unlikely is your birth, And pray that there's intelligent life somewhere up in space, 'Cause there's bugger all down here on Earth.

Additions:

1. Give definitions of the following terms in prose form. Use complete sentences.

Message: Gather more general knowledge about the topic. *** Be absolutely sure that you know everything that earn the designation LAW. In particular know Maxwell’s equations.

What fraction of these question can be answered using only chapter summary material?

a.) Write down the equation that represents the Biot-Savart Law:

b.) Are there magnetic charges? Write down the Maxwell equation that provides the answer to this question.

No. shows that electric field lines are started and stopped by electric charges. The equation shows that magnetic field lines do not start or stop; that is there are no isolated magnetic charges (monopoles). The simplest magnetic field pattern is a dipole pattern.
c.) List a device that works by using an induced (not a motional) emf.

Transformer, inductor, wire loop antenna for a radio.

d.) Define the current density . The current density is the current per cross sectional area in a conductor; it has the direction of the current. ;

e.) Identify a device that operates in a manner described by the equation .
This is the generator equation. N turn coil of area A spinning at  in a field of strength B.

2. In the circuit shown, a current of 3 A is directed up through the 15 V battery. The reference potential is set to zero at the ground symbol.

1. 
   Find the potential at points A, B and C.
   

The electrical power delivered to an element is the current through it times the potential drop across the element in the direction of the current. The potential drops across a resistor so power is flowing into the resistor in which it is dissipate (converted to thermal forms). The potential rises across the battery (in the direction of the current through it) to it is providing power to the circuit rather than receiving power from it. (If current is forced through a battery in the reverse direction, the battery is being charged.)

2. 
   Give the power delivered by the 15 V battery.
   

3. 
   Compute the power dissipated in the 4  resistor.
   

4. 
   Use the voltages found in part a.) to find the currents in the 3 and 6 resistors. Point B is 3 V above ground. 3V across 3  I_3 = 1A down.
   

***** The value of a current is not complete without its direction. ******

Pick your own branch currents and give three Kirchhoff’s rules equations for this circuit that could be solved to find the currents. All three have been chosen to be “up”. If the current in a branch is “down”, the current found for that branch will be negative with the convention chosen.

Voltage Rule Left Loop from lower left: + 15 V – 4 I₁ + 3  I₂ = 0

Voltage Rule Right Loop from lower left: - 3  I₂ + 9 V + 6 I₃ = 0

Substitute “– I_n” for each current that you chose to be in the opposite direction.

3 A + I₂ + I₃ = 0

+ 15 V – 4 (3A) + 3  I₂ = 0

The equation above immediately yields I₂ = - 1 A. The node equations then gives I₃ = - 2A. Solving the equations in this manner is equivalent to walking the circuit.

Walk the circuit values:

3. Use Lenz’s law to answer the following questions. a.) What is the direction of the induced current in the resistor as the magnet is moved to the left? b.) What is the direction of the induced current in the resistor R immediately after the switch is closed? c.) What is the direction of the induced current in the resistor R as the current I decreases rapidly to zero? d.) A rod is perpendicular to the magnetic field and it is moved to the right as shown. The result is the charge separation illustrated. What is the direction of the magnetic field?

a.) The flux from the magnet is to the right and it decreases as the magnet is moved to the left. The change in flux is directed to the left. To fight the change, the induced current must cause a field to the right. By the RHR, the current must come out of the page at the top of the coil.  the current is to the right through the resistor. initial + change = final

b.) After the switch is closed, the current builds up to cause a B field to the left. The field goes from zero to something to the left. The change is to the left. The induced current must cause a field to the right. By the RHR the induced current must be into the page at the bottom side of the rod around which the coil is wrapped. The current comes out of the page through the resistor (or a little left to right).

c.) By the RHR, the B field circulates around the current so that it is into the loop containing the resistor. As the current decreases, the flux into the loop decreases so the change in out of the page through the loop. To fight the out of the page change, the induced current causes a field into the page. The induced current must be clockwise  to the right through the resistor.

d.) As a first try, we look for B perpendicular to the rod and to the velocity. Check the action for B into and out of the page. If B is into the page, is up the page so that is the direction that positive charges are forced. Negative charges would be forced down the page. Look no further!

4.) The switch S in the circuit to the right has been closed for a long time before it is opened at time t = 0 s. a. What is the current through R just before the switch is opened? b. What was the charge on the capacitor just before the switch was opened?  = 50V ; R = 250 

L = 10 mH ; C = 100 F

Pure LC circuit (no damping) after switch opened.

c. What was the current through the inductor just after the switch is opened?

d. What is the oscillation frequency of the circuit after the switch is opened?

e. The charge on the capacitor oscillates for t > 0. What is the maximum charge on C at any time after the switch is opened?

f. Give an expression for Q(t), the charge on the capacitor for t > 0.

a.) After a long time, ^dI/_dt is zero so the voltage across the inductor is zero. All the voltage is across the resistor so I = ^/_R = 0.20 A to the right through R and down through L. As the capacitor is in parallel with the inductor, its voltage is zero so it is (b.) uncharged . c.) The current through the inductor is a continuous function of time. It was I = ^/_R just before so it is I = 0.20 A down through L just after. d.) The oscillation angular frequency is  = [LC]^-½ = [(10 x 10^-3) (100 x 10^-6)]^-½ = 1000 ^rad/_s­. The frequency is /2 or 159 Hz .

e.) All the energy stored in the inductor initially will be stored in the capacitor whenever the capacitor has it maximum charge.

f.) Q(t) = Q_max cos[t + ] I = + ^dQ/_dt = -  Q_max sin[t + ] NOTE: I flows onto +Q.

Q(0) = 200C cos[] = 0. Try  = ½  Q(t) = 200C cos[t + ½ ]  Q(t) = 200C sin[t]

Q(t) = 200C sin[^rad_st]

General method Q(t) = Q_max cos[t + ] I = ^dQ/_dt = -  Q_max sin[t + ]

Q(0) = Q_max cos[] = 0 I(0) = ^dQ/_dt = -  Q_max sin[] = 0.2 A

Solve: Q_max cos[] = 0 and -  Q_max sin[] = 0.2 A with ^rad_s

5.) A conductor consists of a circular loop of radius R and two long straight sections as shown in the figure.

a.) Find the magnitude and direction of the magnetic field due to the current in the conductor at the center of the loop.
The field is the sum of that due to a long straight wire at a distance R and the field due to a single turn circular loop at its center. SUPERPOSITION: B_total = B_lsw + B_loop

b.) Evaluate the magnitude of the field for I = 10 A and R = 2 cm.

By the RHR, the field at the center is into the page. 414 T into the page

6.) Note the direction of each integration path. Give the value of for each path illustrated. CCW: out is positive; CW: in is positive C1: CCW; C2: CW; C3: CW = o Ienc = o Ienc = o [5 A - 2 A] = m = o [- 1 A - 5 A] = - 7.54 T m = o [- 1 A + 2 A] = 1.26 T m

The notation C = A means that C is the curve (path) than bounds the area A. The RHR subscript means that the normal to the area is in the direction dictated by the RHR. In the cases above, an arrow indicating a CCW path means that the normal and hence the positive sense for currents is out of the page. For CW paths, the positive direction is into the page.

7.) Motion of charges in fields: a,b,c, d.) Give the direction in which the charge is deflected by the magnetic field in each situation illustrated to the right. e.) An electron traveling at 2.00 x 106 m/s is in a region in which the electric field is 100 V/m directed northwest. Give the acceleration of the electron.

Parts with a magnetic field and no electric field:

a.) is up the page so a positive charge is deflected up the page.

b.) is into the page so a negative charge is deflected out of the page.

c.) vanishes so the charge is not deflected.

d.) is into the page so the positive charge is deflected into the page.