"I'm Sorry, I'll Need to Search Your Bag"...and other funny things said after Mass Spectroscopy....

Q12. Magnetic force requires a charged particle to have velocity. Thus the magnetic force will always be zero unless the particle starts with some velocity. An electric field exerts a force on any charged particle of magnitude F = qE, and therefore, can make an electron at rest accelerate.

P5. Using the left hand rule - the force is directed South. The magnitude will be 7.45 x 10^-13 N

P9. To produce a circular path, the magnetic field is perpendicular to the velocity. The magnetic force
provides the centripetal acceleration:
qvB = mv2/r, or
r = mv/qB;
0.25 m = (6.6 ´ 10–27 kg)(1.6 ´ 107 m/s)/2(1.60 ´ 10–19 C) B, which gives B = 1.3 T.

P14. The equation we derived yesterday comes up in this problem: qvB = mv^2/R. The difficult part is that we can write speed in terms of the radius: 2*pi*R/T = v. Making this substitution results in the expression .5mv^2 = 2*pi*qBR^2/T where T is the period of revolution.

17. skip this - we will do this in class Friday.

55. This is very similar to our result from class, but we don't know the speed of the particle when it enters the magnetic field. We DO know the particle is accelerated through a potential difference V...How can we relate a change in potential of a charge Q to the final speed it has if it starts at rest?

How Could She Not Like you? You have a Magnetic Personality!

Hi everyone,

1. As a recommendation, you might want to check out questions 7 and 8 for more practice with the right/left hand rule. I'll post the solutions below.


As promised, here is a video of a magnetically levitated frog.


You can search on Google for more interesting videos if you enter 'magnetic levitation'.

The magnetic force simulator is located at
http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_feb_24_2003.shtml

You must stop and reset the simulation each time you change something to see the effect of the change.

Here are the HW solutions from the handout:
1. 3.84 x 10^-14N
2. into page, towards left, towards top of page, towards right, force is zero!
3. For this problem, you have to say which direction you are assuming each vector is pointing towards. If velocity is towards the right side of the page, and B is 30 degrees above the velocity towards the top of the page, then you must use the perpendicular component of one vector along another and the right hand rule. This results in a 3.2 x 10^-16N force directed out of the page for (a).
b) 1.92 x 10¹¹m/s^2
c) The force has the same magnitude as (a) but the force and acceleration are directed into the page. The acceleration of the electron would be 3.51 x 10¹¹m/s^2
4. 9.375 T towards top of page

Solutions to 7,8:


7. To find the direction of the force on the electron, we point our fingers in the direction of v and curl them
into the magnetic field B. Our thumb points in the direction of the force on a positive charge. Thus the
force on the electron is opposite to our thumb.
(a) Fingers out, curl down, thumb right, force left.
(b) Fingers down, curl back, thumb right, force left.
(c) Fingers in, curl right, thumb down, force up.
(d) Fingers right, curl up, thumb out, force in.
(e) Fingers left, but cannot curl into B, so force is zero.
(f) Fingers left, curl out, thumb up, force down.

8. We assume that we want the direction of B that produces the maximum force, i. e., perpendicular to v.
Because the charge is positive, we point our thumb in the direction of F and our fingers in the direction
of v. To find the direction of B, we note which way we should curl our fingers, which will be the direction
of the magnetic field B.
(a) Thumb out, fingers left, curl down.
(b) Thumb up, fingers right, curl in.
(c) Thumb down, fingers in, curl right.

From the handout used on Thursday:

1.
a) 8kQ^2/(5^3/2)a^2
b)E = kQ/4a^2 directed to the right
c) -9kQ/2a
d)very similar to the graph of y = x*e^(-x^2)
e)v = (9kQq/m)^(1/2)

2.
a) left (-), right (+) from E field lines
b) 100 V
c) 1.3 x 10^-10F
d) 8.0 x 10^-16N directed to the right
e) by conservation of energy, v = 4.2 x 10⁶m/s

3.
#1.
i. 480 ohms, .25 A
ii. 360 ohms, .33 A
#2.
The resistances are the same as above. Since they are in series, the current rhough both is 0.143 A.
#3. in order, moving down the list: 2, 1, 3, 4
d) parallel: 70 W, series, 17.2 W

4.
a) 20 V
b) Q = 3.0 x 10^-8C
c)
i. 30 V since current is zero!
ii. E = 0 inside any conductor.
iii. With 30 V over the two gaps, using V = Ed, E = 60,000 N/C

5.
a) t = L/v₀
b) a = Dv₀²/L²
c) E = mDv₀²/qL²
d) V = mD²v₀²/qL²

Electric Current exam tomorrow. You might also plan to meet with your partner to get an idea of what you will be trying during the break for the battery project.

RC Cola Isn't Just A Soft Drink....

3.
a) 12-kOhm: .333 mA, 15 kOhm: .333 mA, 3 kOhm: 0 A
b) 50 microCoulombs

4. 2A

5.
a) 0.4 seconds
b) zero (capacitor and resistor are in series!)
c) 60 microCoulombs

6. 72 microCoulombs

From the book:
51.
a) 4.129 x 10-5C
b) through .5, 6, and 5 ohm resistors: I = .635 A, through 8 ohm = .212 A, through 4 ohm = .424 A.

52.
a) C = 2.33 x 10-9F

54.
a) 8V
b) 16 V
c) 8 V
d) 5.76 microcoulombs

Capacitors - the NEW resistors? More on Page 7!

40. minimum C = 1.36 x 10-9F connected in series, maximum C = 1.95 x 10-8F connected in parallel.

41. 300 pF connected in parallel.

42. C1 + (C2C3/(C2+C3)), Q on C1 = 562.5 microcoulombs, Q on C2 = Q on C3 = 375 microcoulombs

44
(a) For 0.4 microfarad capacitor, Q = 2 microcoulombs, V = 5 V
For 0.5 microfarad capacitor, Q = 2 microcoulombs, V = 4 V
(b) Both capacitors have 9V across them, For 0.4 microfarad capacitor, Q = 3.6 microcoulombs, 0.5 microfarad capacitor, Q = 4.5 microcoulombs

72. (a) When the capacitors are connected in parallel, we find the equivalent capacitance from
Cparallel = C1 + C2 = 0.40 µF + 0.60 µF = 1.00 µF.
The stored energy is
Uparallel = 0.5*Cparallel*V^2 = (1.00 x 10–6 F)(45 V)^2 = 1.0 x 10^–3 J.
(b) When the capacitors are connected in series, we find the equivalent capacitance from
1/Cseries = (1/C1) + (1/C2) = [1/(0.40 µF)] + [1/(0.60 µF)], which gives Cseries = 0.24 µF.
The stored energy is
Useries = 0.5*CseriesV^2 = 0.5*(0.24 x 10–6 F)(45 V)^2 = 2.4 ´ 10^–4 J.
(c) We find the charges from
Q = CeqV;
Qparallel = CparallelV = (1.00 µF)(45 V) = 45 µC.
Qseries = CseriesV = (0.24 µF)(45 V) = 11 µC.

There was some internal resistance to the revolution - Batteries & Internal Resistance

2.
a) 16 ohm current = .106 mA
b) battery current = 6.37 mA
c) New Req = 1413.8 ohms, terminal voltage = 8.987 V

3.
a) 2 A
b) 8V

From Textbook:
18 a) 8.406 V, b) 8.491 V
20. r = .4068 ohms
21. .06 ohms
29.
For the conservation of current at point c, we have
Iin = Iout ;
I1 = I2 + I3 .
When we add the internal resistance terms for the two
loops indicated on the diagram, we have
loop 1: V1 – I2R2 – I1R1 – I1r1 = 0;
+ 9.0 V – I2(15 W) – I1(22 W) – I1(1.2 W) = 0;
loop 2: V3 + I2R2 + I3r3 = 0;
+ 6.0 V + I2(15 W) + I2(1.2 W) = 0.
When we solve these equations, we get
I1 = 0.60 A, I2 = – 0.33 A, I3 = 0.93 A.

Oh, and here's something interesting I found:

February break clue #2:

February Break Assignment - Clue #1

Clue #1

Solutions will be posted tomorrow. Here are the solutions to Friday's work:

Problem 3.
a) This is a standard problem where you have to find Req. The trick here is to realize that the 3 ohm and 6 ohm resistors are in parallel since their ends are connected directly together. You should obtain that Req = 6 ohms for the entire circuit. The battery current is therefore 1 ampere, and since the 4 ohm resistor is in series with the battery, it also has a current of 1 ampere.

b) To show that the junction rule holds, you must demonstrate that the current into the junction (which is the current through the 3 ohm and 6 ohm resistors) is equal to the current out (the current going through the 4 ohm resistor). This emans the only thing missing is the current through the 3 ohm and 6 ohm resistors.

You can use Ohm's law to find that the voltage drop across the 4 ohm resistor is (4 ohm)*(1 A) = 4 V, which means the voltage left to drop across the 3 ohm and 6 in parallel is 2 V. Since both resistors are in parallel, they BOTH must have 2 volts across them. This allows you to calculate the current through them by Ohm's law, 2/3 A and 1/3 A respectively. Since 2/3 A + 1/3 A (current in) = 1 A (current out) we show that the node rule holds.

c) The only thing you must show is that for the large loop, the sum of potential differences adds to zero. Going clockwise from the negative terminal of the battery:

6 V - (2/3 A)(3 ohms) - (1 A)(4 ohms) = 0.

The rule holds!


1989B3.

a)
i. 40 W (direct application of P = IV)
ii. 20 W (calculation of power for a constant force, P = F*v)
iii. 60 W (the battery is supplying ALL of the power to the circuit, so it must be equal to the sum of the power used by the other circuit elements.)

b)
i. 20 V (Ohm's law)
ii. 10 V (You know the power is 20 W, and the current through the motor is 2 A. Using P = IV, the voltage must be 10 V)
iii. 30 V (Similar to part (a), the battery is the source of all potential difference in the circuit, so it must be the sum of the potential differences for the motor and resistor.)


c) 15 V (You can get this by calculating the new power required to lift the mass and using P = IV, or by using the fact that it says the voltage is directly proportional to the speed of the motor. )

d) 7.5 ohms (15 V potential difference across the motor, and the battery potential difference is still 30 V. Thus the resistor has 15 V across it, with the same 2 A current. This makes the resistance 7.5 ohms by Ohm's law.)

1983B3

a) 5 ohms
b) i. 4/3 A ii. 2/3 A
c) At point B: 10 V, at point C: -10 V, at point D: -2V
d) 40 W

1982B4

a) clock in parallel with the battery, radio in series with a resistor, and together in parallel with the battery.
b) 600 ohms
c) P = .45 W, energy = 27 J for a minute

Also, here are a couple more circuit problems to make sure you're where you need to be. The answers are included. Click to see them full size (unless your eyes are really THAT good....)

Dr. StrangeCircuit, or How I Learned To Stop Worrying And Love Finding Req.

59. 99.8 Ohms
60. 6.762 Ohms
61. 4.601 Ohms
62. I = .083 A, P = .833 W
63. P = 2.222 W
64. R = 24.94 Ohms
65. current through 20 Ohm resistor = .75 A, current through 9 Ohm resistor = 2.11 A

Series? Parallel? Neither? This question offends me.

Equivalent Resistance HW:
1.
a) 2/3 R
b) 2 R
c) 15/2 R

2.
a) Req = 6 Ohms
b) Req = 5 Ohms

Book HW:

Questions:
2. Parallel lights - more difficult to connect, but all lights will stay on even if one goes out.
series lights - easy to connect lights together, but if one goes out, they all do!
4. For the bulbs in series, the current is the same. By P = I²R, the bulb with the largest resistance will have the highest power usage, and will therefore be the brightest.

If the bulbs are connected in parallel, the voltage is the same for each. The power equation becomes P = V²/R, which means the larger power will be used by the bulb with less resistance.

Problems.
2.
a) 360 Ohms
b) 8.89 Ohms
4. Maximum 2800 Ohms, minimum 261.4 Ohms
7. 4550 Ohms
13. 105.2 Ohms
24. I = 0.409 A.

MC Solutions and Results

Hi everyone,

Here is a PDF of the multiple choice solutions and the class results sorted from the worst to the best questions for our class.

Please go through and start looking at the questions so you come in Tuesday with some ideas for what did/didn't go well.

See you all Tuesday,

EMW