You're Standing in my Wave....

As I mentioned in class today, I expect you to be checking this blog on a daily basis to get answers for the HW assignments. This is more of an issue for those of you that are NOT staying for tutoring during 8th or after school.

below you will find a list of the last four digits of the OSIS numbers of people in the class, along with a five letter code word. Please post a comment or send me an email with your codeword to confirm you have checked the blog.

8950 - bbjah
1893 - qpkaj
3853 - zzytn
2615 - jjajs
9409 - quahl
6872 - hhsga
9021 - bnxma
6990 - jqpwo
6352 - nnzna
2531 - mqpao
1679 - mansp
9200 - gsgah
9226 - bbzna
0598 - nnqma
6706 - mpaoq


Q25. The nodes are locations of zero amplitude, so these positions could be touched without disturbing the wave.

Problems:
18. amplitude = 0.3500 meters, frequency = .8754 Hz, period = 1.142 seconds, total energy = .7411 J, KE = .6807 J and PE = .0604 J

51. 440 Hz, 880 Hz, 1320 Hz, 1760 Hz (f_n = nv/2L)
53. 70 Hz, 140 Hz, 210 Hz, 350 Hz
55. nodes are always a half wavelength apart from each other - wavelength is 19.37 cm, so nodes are half of this, 9.685 cm.
56. 87.5 Hz, n = 3 and 4 for the given frequencies.
60. m = 4f²L²*mu/(n²g), so m = 1.421 kg, .355 kg, .0568 kg for (a), (b), and (c).

Web sites from today:

http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference

http://home.austin.rr.com/jmjensen/JeffString.html

Reflected Waves


http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=19 - Waves traveling in opposite directions making standing waves



Standing Wave - nodes and antinodes

http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html - Vibrating string, standing waves fixed at both ends.

http://www.walter-fendt.de/ph11e/stlwaves.htm - open/closed Tube with standing waves

Wave - You're on National Television!

For those of you on the trip today:

The homework is to read P. 322 to 328, and DO Q 10, 18, P 34, 35, 39, 41, 50

We will be having an AP problem wednesday quiz - one EMI problem, another on momentum.

Here are the quiz solutions.

Q10 . The water sloshing back and forth has a specific frequency associated with it because of the size of the pan and the speed of the waves on the surface of the water. The only way to get the water to slosh is if the swinging occurs at this special frequency.

Q18 Hitting across the end creates a transverse wave, hitting perpendicular to the end creates a longitudinal wave.

Problems.
34. 2.83 m/s
35. 1.259 m
39. 0.332 s
41. The speed of sound in water is 1440 m/s. The sound must travel down to the ocean floor and back. The distance is therefore 2160 meters.
50. The string may be flat, but it is not at rest - try one of the sites below to see what happens when two reversed pulses pass each other on a string.

EMI is good for the teeth and bones.

1986B4.
a) 3 V
b) clockwise
c) 0.6 N
d) 1.8 W

1982B5.
a) 0.06 Webers
b) 0.06 V
c) -0.3 A from t = 0 to t = 2 seconds, 0 A from t = 2 to t = 4 seconds, and +0.15 A from t = 4 to t = 6 seconds.

3.
a) final speed is (2gy0)^(1/2)
b) I = Bh/R*(2gy0)^(1/2)
c) flux is increasing into the page, so induced current increases the flux out of the page. Current is CCW.
d) For flux:
Flux increases linearly to whB at w; Stays constant until 3w, and then decreases to zero at 4w.
For current: The magnitude of I is the value given in part (b).
Positive I from 0 to w, 0 from w to 3w, -I from 3w to 4w, and then zero from 4w to 5w.

4.
a) m = ILB/g = B^2L^2v/gR = 0.143 kg
b) 1.51 J
c) 1.51 J

5.
a) 0.3 Volts
b) 0.06 A, counter clockwise
c) 0.018 W
d) F = 0.036 N
e) 20 more turns means potential difference increases by a factor of 20. This also means twice the resistance of the wire. Since current I = deltaV/R, the factors of 20 will divide out, leaving the current the SAME.

Some toys for Tuesday's lesson:

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=18 pulse superposition

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=323.0

http://www.colorado.edu/physics/2000/applets/fourier.html- Constructive/Destructive Interference

http://www.surendranath.org/Applets/Waves/TwaveRefTran/TwaveRefTranApplet.html - simple reflection

EMI solutions and a form!

Hi guys,

Thanks to all who came to robotics this weekend - it was quite a blast. I hope the REACH session was also useful for those that went. Put the next one on your calendar, March 28th at Baruch College.

Please complete the form below to let me know when you will be meeting with me.



Solutions:

I apologize for leaving it out, but please use the image below with Problem 2 from the handout from today's lesson:





Questions:
15. The change in flux causes an induced current within the aluminum. The magnetic field created by the induced current is attracted to the magnetic field of the bar magnet, opposing the force tending to pull the sheet out of the magnet. This induced magnetic field forms because of the induced current, not because the magnetic properties of aluminum.

16. We discussed this in class!

Problems:
3. Counter clock-wise

4. Field lines going out from magnet to the right. Moving it through loop increases the flux going into loop towards the right. By Lenz' law, the induced current will increase the magnetic flux going into the loop towards the left. This will create a current that flows clockwise through the loop, and from right to left through the resistor.

7.
a)The current through the outer loop decreases since the resistance increases. This means that the magnetic field (and therefore the flux) through the inner loop out of the page is decreasing. By Lenz's law, the induced current will attempt to increase the flux out of the page to oppose the change. This will come from a counter-clockwise current.
b) If the loop was moved outside to the left, the flux would be decreasing into the page through the loop. The induced current would try to increase flux into the page through the loop, which would result from a clockwise current.

8.
a) counter-clockwise
b) clockwise
c) zero (no change in flux)
d) counter-clockwise

EMI Mr. Weinberg, I DID need to know that....

Question 2:
Magnetic field describes how the influence of magnetic force is transmitted through space. Magnetic force describes how magnetic flux passes through a specific region of space, specifically a loop or area through which magnetic field lines travel. It is possible to have magnetic field and not have magnetic flux (if area = 0 or if the angle between B and the normal vector is 90 degrees), but it is not possible to have magnetic flux without magnetic field.

Problems:
2. 0.147 Volts

10.
a) 0.169 Volts

12.
b) 4.29 x 10^-2 Volts
c) 17.2 mA

13.
a) .841 m/s
b) .757 N/C

89. 7.27 x 10^-3 J

Here are a couple of sites with fun EMI tools to play with:

http://www.ngsir.netfirms.com/englishhtm/Induction.htm
http://higheredbcs.wiley.com/legacy/college/halliday/0471320005/simulations6e/index.htm?newwindow=true

Field, Force, Wire, Charge - Performing This Thursday at 12:25 PM

Hmm...tomorrow is Wednesday, isn't it?

I think it's time for an AP question...think Thermo.


1.
a) 5 x 10^-5T
b) West
c) Net B-field is of magnitude 7.07 x 10^-5T directed at 45 degrees North of West. (It is not enough to say 7.07 x 10^-5T at 45 degrees!)

2. a = 0.0076 m/s^2

3. Since the force is attractive between the two wires, the current must be going in the same direction. k'I1I2/d = 6 x 10^-4 N/m...solving for the unknown current gives I = 48 A



4.
a) Remember that force/unit length means that from F = IBL, force per length is F/L = IB. Thus F/L = 2.5 A * 1.10 T * sin(90) = 2.75 N/m.

b) F/L = 2.5 A * 1.10 T * sin(45) = 1.944 N/m

5. F = IBL so I = F/(BL) = 11.67 A

6. Let's define a magnetic field OUT of the page to be positive.

a) B_total = kI1/d = (2 x 10^-7 T*m/A*10 A/.05 m = 4 x 10^-5T

b) B_total = k'I1/d + k'I2/d = (2 x 10^-7 T*m/A*10 A/.05 m + (2 x 10^-7 T*m/A*5 A/.05 m= 6 x 10^-5T

c) B_total = k'I1/d - k'I2/d = (2 x 10^-7 T*m/A*10 A/.05 m - (2 x 10^-7 T*m/A*5 A/.05 m= 2 x 10^-5T

7. F = IBL sin(60) = 1.212 N

8. This is an AP problem - don't be scared off by it! Remember to set a direction (in or out of the page) as positive and then sum all fields together, being sure to set fields positive or negative depending on the direction of each. We will use out of the page as positive.

Sum of fields at P:
-k'(1A)/(0.5 m) + k'(3A)/(1.0 m) = 2 x 10^-7 T

This shows that the field is directed out of the page (a) at the given magnitude of 2 x 10^-7 T.

c) Force is to the left, B is out of the page, and v is towards the top of the page. This only matches the left hand rule, which means the charge must be negative.

d) Fmag = qvB sin(90) so q = Fmag/(vB) = 5 x 10^-7 C

e) An electric force to the right of magnitude 10^-7 N would make the net force zero. Since Felec = qE, the field would have magnitude 0.2 N/C. Since the charge is negative, the electric field would need to be directed to the left for the force to be to the right.

Wire You Forcing Me To Stay Current?

Q11. Since the North pole is on the left, and the South pole is on the right, the B field lines go towards the right. Since the current is going into the page, and B is going to the right, by the RHR the force will be towards the top of the page.

Q17. The kinetic energy will stay constant since the magnetic force (which is also the net force acting on the particle) is always perpendicular to the velocity. This means acceleration is always perpendicular to velocity, which only occurs when an object is in uniform circular motion. Uniform circular motion is characterized by a constant speed, and by KE = 1/2mv^2, a constant kinetic energy.

17.You can find the acceleration first using kinematics, since you know displacement (1 m), final velocity (30 m/s), and initial velocity of zero since it starts at rest.

a = 450 m/s^2

Using Newton's 2nd and a FBD, you can also find that the magnetic force is the only force acting to accelerate the bar.

Fnetx = IBL = ma so I = ma/BL

I = 1.985 Amperes, magnetic field is directed into the page through the loop formed by the bars and the rails.

66.
a)
Remember that old thing called the Free Body Diagram?

We need to use it here - the magnetic force on the current accelerates the sliding wire down the rails.

Fnetx = Fmag = IBl = ma so a = IBl/m.

We can use v = vo + at to write the velocity as a function of time. The rod starts at rest, so vo = 0.

Thus v(t) = IBLt/m

b) With friction acting, the y-direction (which is directed out of the page in the photo above) becomes more important.

Fnety = Fn - mg = 0 so Fn = mg.

Since friction force Fk = mu*Fn, Fk = mu*mg

The new Fnetx = Fmag - Fk = ma
IBl - mu*mg = ma
a = (IBL/m) - mu*g

Using v = vo + at again, v(t) = 0 + ((IBL/m) - mu*g))t

c) Using the right hand rule, we can point our fingers North, and palm out of the page to go along with B. This leaves our thumb pointing to the right. This is then the direction of force.

Try it - don't guess!