Applications of the Magnetic Force

This site has a great description of how electric motors work.
http://electronics.howstuffworks.com/motor1.htm

Solutions:
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.

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.

15. qvB = mv^2/R, so the momentum mv = qBR.

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

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?

Magnetic Force on particles HW

Hi everyone,

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
5. 78.3 cm moving counter clockwise
6. The electron will have the smaller radius because its mass is less than that of the proton.
7. f = qB/(2*pi*m)

Review Packet Answers

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

A reminder that we will have our Electric Current exam on Monday. The answers to the lab questions will be due as you enter class on Tuesday.

RC Circuits

From handout:

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 in Series/Parallel

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

48. 5.128 pF

Kirchoff's Rules

1981B4.

a) 3A
b) 108 W
c) 6V

1989B3.

a) i. 40 W ii. 20 W iii. 60 W
b) i. 20 V ii. 10 V iii. 30 V
c) 15 V
d) 7.5 ohms

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

Batteries & Internal Resistance

From Handout:

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
24. current = .227 A

Circuit Problem Solutions

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

Parallel and Combination Circuit HW

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.

Ohm's Law and Series Circuits HW

From Handout:

2. 1200 Ohms
3. 10 V, 20 Ohms
4.
b)60 Ohms,
c)0.2 A
d) 15 Ohm - 3 V, 21 Ohm - 4.2 V, 24 Ohm - 4.8 V
e) .6 W, .84 W, .96 W
f) .6W + .84 W _ ,96 W = 2.4 W OR (.2 A * 12 V) = 2.4 W
g) thickest filament - smallest resistance, since it has a larger cross sectional area. This must be the 15 Ohm resistor.
h) The brightest bulb uses the most power - this is the 24 Ohm bulb.

From book:
p. 551:
5. 35.5 Ohms
7. 2.1 x 10²¹ electrons
9. 24 Ohms, 360 W
26. Since P = V²/R, and V is the same for both, the resistance must be less for the higher power. Assuming a wall voltage of 120 V, the resistances are 24 Ohms and 12 Ohms for the 600 W and 1200 W settings.
31.
a) 8.57 Ohms, 1.05 W
b) 4 times the power. The bulb would probably not last for long!

p. 581
10. 27 Ohms

Electric Current and Resistance HW

Solutions to Friday Worksheet:

1. 600 C, 600 W.
2. 4 hours, 10 hours
3. 133.3 A
4. A flow of current through one hand and out the other passes through the heart. The current running through the hand and out the elbow would be painful, but not necessarily fatal.
5. Power of the hairdryer is 12,0000J/45 seconds = 266.7 W, voltage = P/I = 88.9 Volts
6. .034 .34 Ohms
7. B, resistance of a cylindrical wire = p*L/A with p = resistivity, L = length of the wire, and A = cross sectional area.
8. 24 Ohms
9. 60/7 Ohms

34. Three lightbulbs.
35. efficiency = 373 W/528 W (1 hp = 746 W. What is the actual power of the motor? What is the electrical power provided to the motor? How can you use efficiency from thermodynamics to relate these two?)