AGENDA:
* Corrected Moving Charges Worksheet #1-10
* Read Magnetic Formulae handout
* Chapter 27 Review #1-7 for homework
The Moving Charges Worksheet was difficult to do at first, which is why Ms.K suggested on doing the questions in reverse order. Then it’s just a matter of knowing how to use your formulas correctly. Also, visualizing the problem (i.e. drawing a picture) helps a lot too.
MOVING CHARGES WORKSHEET
10. An electron is moving east to west at 5.0 x 10^5 m/s through a magnetic field. A force of 4.0 x 10-6 N north is acting on the electron. What is the magnitude and direction of the magnetic field?
F = Bqv
(4.0 x 10^6) = B (1.6 x 10^-19) (5.0 x 10^8)
B = 5 x 10^7 (into page)
9. What is the magnitude and direction of a magnetic force on a proton moving horizontally to the north at 8.6 x 10^4 m/s, as it enters a magnetic field of 1.2 T pointing vertically upward?
F = Bqv = (1.2 N/(c*m/s)) (1.6 x 10^19 C) (8.6 x 10^4 m/s) = 1.7 x 10^014 [right]
8. An electron experiences the greatest force by a magnetic field as it travels at 2.1 x 10^5 m/s when it is moving south. The force is 5.6 x 10^-13 N and up. Find the magnitude and direction of the magnetic field.
F = Bqv
(5.6 x 10^-13)/ [(1.6 x 10^-29) (2.1 x 10^5)] = B
B = 16.7 T [left]
7. A 10.0 m long high tension power line carries a current of 20.0 A perpendicular to the earth’s magnetic field of 5.5 x 10^-5 T. What is the magnetic force experienced by the power line?
F = BIL = (5.5 x 10^-5 T) (20.0 A) (10.0 m) = 0.011 N
6. A wire connecting a taillight to a motorcycle battery is 0.50 m long and is perpendicular to the Earth’s magnetic field. It experiences a force of 6.0 x 10^-5 N when carrying a current of 1.5 A. What is the magnitude of the Earth’s magnetic field at that location?
F = BIL
6.0 x 10^-5 N = B (1.5 A) (0.50 m)
B = 8 x 10^-5 T
5. A 15 cm length of wire carries a current of 20.0 A. It is perpendicular to a uniform magnetic field. If it experiences a magnetic force of 0.40 N, what is the magnetic field intensity?
B = F/(IL) = 0.4 N / (20.0 A x 0.15 m) = 0.13 T
4. Using the Millikan apparatus, a sphere of mass 3.2 x 10^-15 kg is held motionless. The plates produce an electric field strength of 19 600 N/C.
a) What forces are acting on the sphere?
electric force, gravitational force
b) What type of charge must be on the sphere?
positive
c) Determine the amount of charge on the sphere.
F_g = F_e
mg = Eq
q = mg/E = (3.2 x 10^-15 kg) (9.8 N/kg) / (19 600 N/C) = 1.6 x 10^-18 C
d) How many electrons does this correspond to?
(1.6 x 10^-18 C) x (1 electron)/(1.6 x 10^-19 C) = 10 electrons
3. The electric field between these charged parallel plates is 23 800 N/C. The drop’s mass is 2.4 x 10^-14 kg. What force would be experienced by the drop if it had the charge of three excess electrons?
F_e = Eq = (23 800 N/C) (3 x 1.6 x 10^-19 C) = 1.14 x 10^-14 N [up]
F_g = mg = (2.4 x 10^-14 kg) (9.8 N/kg) = 2.35 x 10^-13 N [down]
F_net = 2.35 x 10^-13 N - 1.14 x 10^-14 N = 2.24 x 10^-13 [down]
2. The plates are 5.0 cm apart and produce an electric field strength of 1200 N/C in the region between them. An electron enters the electric field at 1.00 x 10^2 m/s. With what velocity does the electron strike the positive plate? (Ignore gravity.)
F_e = qE = (-1.6 x 10^-19 C) (1200 N/C) = 1.92 x 10^-15 N
ma = qE
a = qE/m = (1.92 x 10^-15 N) / (9.11 x 10^-3 kg) = 2.11 x 10^14 m/s^2
v_f^2 = v_i^2 + 2ad
v_f = [(1.00 x 10^2 m/s)^2 + 2 (2.11 x 10^14 m/s^2) (0.05 m)]^0.5 = 4.6 x 10^6 m/s
1. This oil drop of mass 4.5 x 10^-15 kg has had two electrons removed. The electric field strength is 3.0 x 10^4 N/C.
a) What will happen to the drop?
F_g = mg = (4.5 x 10^-15 kg) (9.8 N/kg) = 4.4 x 10^-14 N
F_e = Eq = (3.0 x 10^4 N/C) (2 x 1.6 x 10^-19 C) = 9.5 x 10^-15 N
F_g > F_e, so drop will fall
b) What is the total force experienced by the drop?
F_net = 4.4 x 10^-14 N + 9.5 x 10^-15 N = 3.45 x 10^-14 N [down]
Next scribe is DANA.
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