Homework 2 solutions

are here. I'll add Quiz 2 & 3 solutions tonight or this weekend.

Solutions and such

I have solutions for quiz 1 and homework 1. You'll get your graded HW1 back tomorrow (Wed), I suspect the TAs will have the quizzes back to you shortly as well. 

Old exam answers

Exam 1, summer 2009, problem 4: this was also on the summer 2008 exam 2.

Exam 2, summer 2010: only the first three problems are material related to your upcoming exam.

1. Just walk around the middle loop - that gives 12-6I=0, and determines the current in the 6Ω resistor to be 2A.

2. Use the right-hand rule to determine directions. Two of the currents give fields pointing to the lower right corner, and two of them give fields going to the lower left corner. All fields have the same magnitude (same current, same distance; watch the geometries). All but the vertical components will cancel; find those and add together. The net field is in the -y direction, and has magnitude (field of 1 wire at center)*(sin 45)*(4 wires)~5e-6T.

3. (a) into the plane. (b) set r=mv/qB equal for both particles using the known electron and proton masses. q is the same for both, and will cancel, as will B. This gives (v_e/v_p)=(m_p/m_e)~1.8e3

4. (a,b) This filter passes speech - the inductors let low frequencies through, and the capacitor shorts out any remaining high frequencies. (c) Add all inductors together, they sum to 10.26mH. The cutoff frequency is when the inductive and capacitive reactances are equal, so f=1/(2*pi*sqrt(LC))~10.6kHz

5. This is motionally-induced voltage, like the conducting bar. (a) V=Blv~2.5mV (b) Right hand rule gives the force on a positive charge toward the drivers door, so it is the positive side.

Solutions to Sum10 exam 1

Some very quick solutions to the Summer 2010 Exam 1.

Quiz 7 solution

Quiz 7 now has a solution posted. I'll be updating some of the homework solutions tonight.

HW5 solutions

A solution set to HW5 is out, though the solutions to the last couple of problems are a bit terse.

Quiz 5 & 6 solutions

Quizzes 5 & 6, along with their solutions, are now up.

HW 3 solutions are out

Here you go.

Quiz 3 and 4 solutions are up

Find them here.

HW 2 solutions are up

Find them here. I did parts of them quickly, so let me know if you find any mistakes.

HW1 solutions

HW1 solutions are out, you'll get them back (along with quiz 2) during Wednesday's class.

Quiz solutions

Quizzes 1 and 2 and their solutions are up, find them here.

Yet more

RE: the last quiz, second problem.

You have only 3 possible energy levels. Photons can only be emitted when you have an electron changing from a higher level to a lower one. Here we have

E1 = 1.2eV
E2 = 2.4eV
E3 = 4.8eV

The only photons that can be emitted must correspond to differences between energy levels:

E3-E2 = 2.4eV
E2-E1 = 1.2eV
E3-E1 = 3.6eV

So three types of photons can be emitted: when an electron jumps from level 3 to 2, from 2 to 1, or from 3 to 1. Given these energies, convert to joules:

E=(2.4eV)(1.6e-19 J/eV) = 3.84e-19 J

Once you have that, use E = hc/(wavelength) and solve for wavelength. You should get about 1e-6, 5.17e-7, and 3.44e-7 meters.

Another question

Could you direct me (like HW so and so) to an example of the kind of refraction problem I should be looking at? They range from easy to REALLY hard and I don't know. HELP.

Study problems more like the easier ones - the quiz 9 problems, for example. Not the pathological ones like atmospheric refraction or prisms.

A couple of other examples that are useful to look at:

PH102 Fall 2007, Exam 2, problems #9 & 10
PH102 Fall 2007 Final, problem #9
PH102 Spring 2008 Exam 2 #6 & #7

If you understand these problems, you basically have things under control.

Collected questions so far this evening.

So far, two questions this evening. Here they are, along with my responses. If I get any more interesting questions by email this evening, I'll try to post the answers here for all to see.

1. Quick question on #7 from fall 2007 final: In part b do you use the E=mc^2 equation?

On #7 you do want to use E=mc^2 - I reused this question on a homework from Spring 2008 (HW12, here). The result is as silly as you would expect - half of the mass of the bullet would have to be turned directly into energy for this to work, which can't be done with any known technology.

2. An FM radio transmitter has a power output of 130kW and operates at a frequency of 98.3MHz. How many photons per second does the transmitter emit?

This is one like the HW question, where you need to convert power and energy. In fact, it is from spring 2008 HW11, here.

First, 130kW means 130e3=1.3e5 Joules per second, since a watt is a joule per second.

The frequency 98.3Mhz means 98.3e6 Hz. If the transmitter has this frequency, than means each photon emitted has an energy of

E = hf = (6.6e-34 J*sec)*(98.3e6 sec^-1) = 6.5e-26 Joules (per photon)

So, if the transmitter puts out 1.3e5 joules per second, and each photon is worth 6.25e-26 joules, that means the transmitter must put out

# photons = (1.3e5 J/sec) / (6.5e-26 J/photon) = 2e30 photons per second.

Basically, find the total energy being emitted per second (just the power given) and divide by how much energy a single photon has to figure out how many photons per second must be coming out. In this case, the intermediate step of converting to electron volts isn't really necessary - the power and energy per photon both need to have the same units, so you can convert to eV or just use Joules.

Exam II and a partial solution

Here. Answers for all problems, solutions for some.

HW 8 Solutions

The key for the crossword puzzle is now up.

Answers for exam 2 will be up once I finish with some grading later tonight.

HW 7 solutions

Here.

Quiz solutions

All quiz solutions are now available (along with the original quizzes). Right over here.

All homework solutions are also available, except for HW 7 that was due about 4 minutes ago. I should have that one posted around class time tomorrow, or at least a partial solution if nothing else.

HW6 Solutions

Here.