*»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»> *WHAT?: E&M help desk *HOW?: Collaborative problem solving *WHY?: Book promo http://minireference.com *WHO?: Ivan Savov the author of the above *WHEN?: I will check this page on April 21, 22, and 23 2013 * the evening, like say, 9PM to 3:30AM. *FB? Yes, here http://on.fb.me/17L4ICF plz share… *»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»»>
* *Practice exams
http://minireference.com/static/exams/PHYS102_practice_exam_final.pdf http://minireference.com/static/exams/PHYS242_final_Fall_2009.pdf http://minireference.com/static/exams/PHYS242_final_Fall_2010.pdf http://minireference.com/static/exams/PHYS242_final_Fall_2011.pdf (INTENSE STUFF!!!)
A nice summary of Magnetism: http://www.telescopictext.org/text/pFjkqQY9bmfvQ
*Your questions *
OK GOOD LUCK TO YOU ALL YOU DON'T NEED to study till the last minute go out and enjoy the sun for a bit. It will be all good in the end (assuming you consume adequate amounts of coffee). KICK ASS at 6pm! Take no prizonners. … … make the teacher sorry he ever tried to mess with you ;)
OK IVAN OUT Tue 23 Apr 2013 17:52:09 EDT
How'd it go?
Q9
Question: An electron is moving with velocity 4.0x10^6 m/s along the positive x-xis.
At t = 0, the electron is at position (x,y) = (0,0) and it enters a region where the electric field is 1mV/m in the +y direction.
A9
This is a coulmob force (F_e = qE) question followed by (I guess this is what is coming next) a kinematics question.
> How can you determine the direction the electron will be moving in once it enters the electric field? (PHYS102)
\vec{F}_e = q\vec{E}
So if \vec{E} = 0.001 j [V/m] ( j = +y direction)
then \vec{F}_e will be in the -y direction (since electorn is negatively charged)
So the electron will feel a -y direction force.
It will keep its +x velocity (no forces in x direction means no change in vel),
but also accelerate in the -y direction not too dissimilar to the effects of gravity.
Q8
In the circuit shown on the right, R = 30?, L = 1mH, and C = 80pF. A sinusoidal voltage v with a r.m.s. amplitude of 100 mV is applied.
a) At what frequency will the voltage across R be largest? Cal- culate the size of this voltage, and the power dissipated by R at this frequency.
b) If C is increased to 120 pF, and the frequency remains the same as in part a, what will be the new voltage across R? What power will be dissipated by R?
c) Based on your results from a and b, can you explain how one might construct a simple radio tuner?
for 8 is it you use p=I_rms*V_rms and getting I_rms from I_max, I_max from V_max/R.
Also for the the frequency is it just f=w/2pi, w = 1/((LC)^(1/2)) ? thank you
A8 Ok this is clearly a RLC circtuit. The general KVL eqn is: v_R+v_L+v_C=v_batt(t)= 0.1 [V]
a) They want the resonance freq. At this freq. the + reactance of the L cancels exactly the - reactance of the C.
The formula fro resnance freq is w = 1/sqrt(L*C) = 1.0/sqrt(1e-3*80e-12)
f resonance = 562 697.69 = 562[kHz] (sound like a lot...)
at that time, the circuit is effecticle only a resistance so we can use find the power dissip in R as: P = (V_rms)^2/R = 0.1^2/30 = 0.000333 [W]
b)
Ok so given freq. f, (angular freq w) in the circuit, the reactances of the L and the C are as follows:
At resonsnace:
X_C = -1/(wC) = - 3535.53390593274
X_L = wL = + 3535.53390593274
so you see... they cancel.
If we change C-->C' however now we get:
X_C = -1/(w*C') = - 1/(w*120e-12) = - 2357 [Ohms (reactive)]
X_L = +wL = + 3535.53390593274
Ok so they want us to find the voltage v_R now.
TO do so we must use the mega formula for finding the total impedance (resistive and reactive combined) of the circuit.
Z = sqrt( R^2 + (\sum X_i)^2 ) = But this doesn't make much sense since R=50 ohms, which is compl. negligible relative to the other impedences.
Anyone see a mistake? Or is this question just bad?
Q7
how do you do question 7 and question 8 on the phys102 final ?
Q7 A horizontal wire is free to slide on the vertical rails of a conducting frame, as seen below.
The wire has a mass m and a length l, and the resistance of the circuit is R.
If a uniform magnetic field is directed perpendicular to the frame, what is the terminal speed of the wire as it falls under the force of gravity?
A7
This is a magnetic induction question AND a magnetic force on a wire question AND a force quesiton.
The relevant formula ar
- dPhi/dt = EMF (the change in the magnewtic flux through the loop = voltage around the loop)
- F_b = L*I*B The force felt by a wire of length L, carrying a current I in a magn field of str. B
(wow TAB did NOT do what I though it woudl do....) but now I am back.
Ok so imagine this thing falling down. If it reaches "terminal speed" it is because at some poitn
the force of gravity will be balanced by the F_b felt by the wire.
We want to find at what speed F_b + F_g = 0
So we need to brake down the problem a little before we can solve this.
1. Assume some v, find what the induced EMG will be
2. From the EMF and R, find I in loop => I = EMF/R
3. From I in loop find F_b and set it equal to mg
ok lets get crackin'
1. Phi = L * height, therefore dPhi/dt = L* v (where v is v falling), EMF = L* v [V]
2. I = L * v / R
3. F_b = F_g => L*(L * v / R)* B = mg
v terminal is the v in the above eqn:
so v = m*g*R / (L^2*B)
tough question.... lots of stuff going on. Braking down into pieces = good apprach.
Q5
Question 6 on PHYSICS 102 PRACTICE FINAL EXAM
The circuit to the right has an inductor of L = 200 mH and a resistor of R = 50 ?.
The switch is closed, and the current measured to be 0.1 A and rising at a rate of 35 A/s.
a) What is the size of the battery?
b) What is the time constant of the circuit?
c) What is the steady-state current through the circuit?
d) After the current has reached the steady state, the battery is accidentally short-circuited by a clumsy teaching as- sistant.
How long does it take for the current to drop 1 ?A?
http://i.imgur.com/t2qHoV7.png
A5
This is a charging RL circuit. Recall that inductor must have the current flowing
through them vary smoothly (cannot have jumps in current).
Recall also v_L = L*di/dt (where i is the current through the L)
a) So they tell you at t=0, di/dt=35 [A/s]
Using v_L = L*di/dt, we find that v_L=0.2*35 = 7V, at t=0.
Note that v_L is acting AGAINST the battery.
The KVL around the loop (at t=0) gives us
v_Batt - I*R - 7 = 0
Since they told us I=0.1, we find that v_Batt = 0.1*50 + 7 = 57 [V]. // is this suppose to be 5 + 7?
b) L = 200m [H]. R = 50 [Ohm] , so tau = L/R = time constant = 200m/50 = 4m = 0.004 [sec]
c) At t=infty, the inductor acts like a piece of wire
so steady state I = 57/50 [A]. Methinks.
d) Inductor discharge eqn. (ok can't find a good picture on google images, but imagine
and falling exponential curve...
i_L(t) = I_max e^{ -t /tau}
I_max = I steady state --- we established that was 57/50 [A].
So we want to solve for t in:
1e-6 = (57/50)*exp(-t/0.004)
Done---Tue 23 Apr 2013 01:47:58 EDT
Q4
> How do I approach this kind of problem?
http://i.imgur.com/ECr2t12.jpg
OK let's think in terms of current.
THe power in a light bulb is P = I^2*R, where I is the current flowing though the bulb.
And it seems to me (not sure yet) that all the current has to pass through B,
so it should be the brighter.
Does this make sense?
> -Everytime a question gets explained it seems to be so easy :P
:)
Bonus question. How bright is D?
> Should I use the loop law?
THink as if you were the current. Would you choose to go throuhg D or not?
>Would it be the same brightness as C since they are in series?
nope.
check out the junction in the circuit right before it slpits to go to D.
Would you choose to go hourizontal into D or...
>I would go down since there is no resistance there?
Exactly.
THe choice is zero vs nonzero.
CUrrenty will chooze the zero path (just the piece of wire)
So D is off.
>So D is not even lit even though it is not an open ciruit??
Yep. D is off. (technical term "shunted" -- when you give current
an alternate route with zero resistancw, you essentially turn off the circ. element.
> AHhhhh i didn't know this :D
;) now you know!
> thank you!!!!!
You are welcome. Plz share/like this or something http://on.fb.me/17L4ICF
We need to bring mor ppl herez
Q3
Is there anyway I can upload a picture file or sth? I have a circuit question :)
-Thanks! ...nice!
a) I was wondering why is it that when C is fully charged the I through R3 is zero?
b) I know that there is no potential difference when C is fully charged.
So why isn't R1 current zero too? Thanks for answering!
http://i.imgur.com/ffPcDU0.jpg
Ok.
So C when not charged works like a wire --- current will prefer to go into cap.
a) When C is fully charged, it acts like an open circuit --- no current can go into cap anymore since it is fully charged.
So if S is open, then current can't go into R_4 either so there is nowehre current can flow in ther right-hand side
of the circuit. Hence no current will flow throuhg R_3.
b) Ok so why current in R_1?
The C will charge to the same voltage as the voltage drop accross R_2 which is not equal to the battery V.
define V_C = voltage on cap
so V_{R_2} = V_C
And I_1=I_2 --- just current flowing in the left hand side.
Does this make sense?
>>"So if S is open, then current can't go into R_4 either so there is nowehre current can flow in ther right-hand side
of the circuit. Hence no current will flow throuhg R_3."
COrrect
I get how current can't go into R_4 but I don't get how that relates to no current flowing through R_3?
Ok so if there were an I_3 where would it go?
Ahhhh...I see...:) THanks!!!
np u r welcome
Oh also, finals for 142 and 102 happens tomorrow so I don't think you need to hold this session tomorrow night :D Unless someone wants to check answers i suppose lol..oh
Ya it is a one night only event ;)
Okay cause on the date thing you wrote 23rd as well...
as in AM... but i think tonight will be the "main" show.
I've got to work till late so i w b around.
Thx for thatr imgur thing. I didn't know u can post imgs.
nope nope i will probably ask more questions later :D aaait
Q2 ivan i gotta question what do you recommend me study from now to 7 am my crib sheet is made but dont htink il be able to do all qs frome serway jewett btw guy moore said there will be more from chapters after the mid on final exam and okay thanks A2 hmm... read examples in the book and just see if you would have been able to solve on your own. is the stuff from before the midterm goign to be on the exam? if so you should review that -- since it has been a long time. also, sleep is a good idea ;) you are welcome. Ok. I got to run. ttyl
Q1
Hi, I'm struggling with question #5 on here: http://s3.amazonaws.com/docuum/attachments/2682/Final%202008%20Sample%20Questions%20%231.pdf?1259706487
(This is for PHYS 142)
My attempt at solving the problem:
Relevant equations -
( This one you don't need F = I*L*B*sin(theta) - this is the force on a current-carrying conductor in a magnetic field
B = ((mu_0)*I)/(2*pi*r) = the B field produced at a dist of r from an inf.-long wire carrying I
Not really sure how to proceed from here.
B = (mu)*(10A)/(2*pi*(1 cm)) ?
There's the force acting on the conductor carrying I2,
and then there's the additional effect of the magnetic field.
How do I take these into account?
A1
You nearly got it. Just got to be clear what you are computing.
||\vec{B}_P|| = the strength of the B field at pt P = info they give you
Define the following conventions:
- currents flow to the right to be positive (I_1 is positive current)
- vectors that points out of the plane will be positive
The B field at pt P is the sum of the B field produced by wire 1 and wire 2.
so you have to s to solve for I_2 in
3.5e-5 = - (mu_0)*(10A)/(2*pi*(4 cm)) - (mu_0)*(I_2)/(2*pi*(3 cm))
note the negative signs --- if + current is along the wire to the right (we said so),
the grab-the-wire-w-right-hand rule tells u B field produced will go into page
which for us is negative.
My guess is that I_2 will be a current in the negative direction.
Yep, http://live.sympy.org/?evaluate=mu_0%20%3D%201.25663706e-6%0A%23--%0Asolve(%20-3.5e-5%20-1*(mu_0)*(10)%2F(2*pi*(0.04))%20%20-%20%20(mu_0)*x%2F(2*pi*(0.03))%2C%20x)%0A%23--%0A
(copy paste the whole link)
I_2 = -12.75[A] --- so I_2 flows to the left.
Will be back later. Thx for ur question. Mon 22 Apr 2013 18:35:39 EDT
* *Final review question * *QF: *Explain the following equation in words: *1. ||F_e®|| = kQq/r^2 – integrate from r=infty to r=R –> U(R) = kQq/R *2. \vec{F}_B = q \vec{v}×\vec{B} * *AF: *1. The electric potential energy of the charges Q and q separated by a distance R * is equal to the Work you would have to do against the electric force \vec{F}_e * to bring the charges, starting from infinitly far apart to dist R apart. *2. The magnetic force on an object is prop. to its charge, times its velocity, times the stength of the B field it is in. * The direction where the magnetic force \vec{F}_b acts is perp. to both the v and the B directions. *
*Solved problems from past year's finals *
PHYSICS 102 PRACTICE FINAL EXAM http://minireference.com/static/exams/PHYS102_practice_exam_final.pdf
Question 1: Determine the location of the field point P for which the net electric field due to the two charges shown in the figure below is zero. Neglect the trivial solution P at ?. Ans: This is an electrostatics equation. Need formula E=kq/r. Let +x point to the right. We want to find d such that: \sum E = 0 = k(2.5)/d - k(6)/(d+1). we can cancel the k, then running solve((2.5)/d - (6)/(d+1), d) on live.sympy gives d=0.714 Question 2: The potential difference between a pair of oppositely charged parallel plates is 400 V. a.) If the spacing between the plates is doubled without changing the charge on the plates, what is the new potential difference between the plates? b.) If the plate spacing is doubled and the potential difference between the plates is kept constant, what is the ratio of the final charge to the original charge on one of the plates?
Ans: This is Gauss' law question. We can also use the formula V=d*E for E field between cap.
a) Q stays the same, therefore strength of E field remans the same too.
However since d --> 2d, the voltage has now doubled V =d*E ---> (2d)*E.
b) On the other hand if V is held fixed, V --> V while d --> 2d, this means E must have
gotten twice weaker inside the cap. Using Gauss' law in reverse we conclude that
the Q on each plate must have halved.
Question 3: Determine the current in each branch of the circuit presented below.
Ans: We have to solve for the three unknowns I_1, I_2, I_3. We will find 2 KVL eqns + 1 KCL eqn.
Label each resistor with + on the side where the current goes in and - where current leaves.
The KVL equation for the loop on the left is:
12 + 1*I_1 + 3*I_1 - 5*I_2 - 1*I_2 - 4 = 0 (starting from below the 12V batt)
The loop on the right is:
+8*I_3 + 4 + 1*I_2 + 5*I_2 = 0 (starting from a)
The KCL eqn is:
I1+I_2 = I_3
The solution can be done via subst, elim or subst. Or using live.sympy.org:
>>> I_1, I_2, I_3 = symbols('I_1 I_2 I_3')
>>> solve((12 + 1*I_1 + 3*I_1 - 5*I_2 - 1*I_2 - 4, +8*I_3 + 4 + 1*I_2 + 5*I_2, I_1+I_2-I_3), I_1, I_2, I_3)
{I_1:17/13, I_2:6/13, I_3:11/13}
Question 4: A deuteron with mass m = 3.34 × 10?27 kg and charge +e is traveling in a circular path with radius 3.48 cm in a magnetic field of magnitude B = 1.5 T , as shown in the figure. a) Indicate the direction that the deuteron is traveling. b) Find its speed. c) How much time does it take for the deuteron to complete one rev- olution?
Ans: This is a radial (centripetal) acceleration question \sum F_r = ma = mv^2/R, and magnetic force question F_b = qv × B. a) ANSWE ME b) c)