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