PHYA22 FSGs
Wednesday, January 25th 2017
Session 5
Superposition of Waves: Superposition, Interference
Q1: Briefly state the principle of superposition.
Q2: A long string has a linear density of
28.27. Solve: (a) Since J = E and J = I A , the electric field is
E= I I 0.020 A = = = 1.64 10 3 N / C A r 2 (0.25 10 3 m ) 2 (6.2 10 7 1 m 1 )
(b) Since the current density is related to vd by
28.14. Visualize:
r The direction of the current I in a material is opposite to the direction of motion of the negative charges and is the same as the direction of motion of positive charges. Solve:
28.7. Solve: (a) Each gold atom has one conduction electron. Using Avogadros number and n as the number of moles, the number of atoms is r 2 L m V N = nN A = NA = NA = NA MA MA MA The density of gold
28.4. Solve: Equation 28.2 is N e = nAvd t . Using Table 28.1 for the electron density, we get
A= Ne D 2 = nvd t 4
D=
4(1.0 1016 ) 4 Ne = = 9.26 10 4 m = 0.926 mm nvd t (5.8 10 28 m 3 )(8.0 10
28.23. Solve: The current density is J = E. Using Equation 28.18 and Table 28.2, the current in the wire is
I = EA = (3.5 10 7 1 m 1 )(0.012 N / C)( 4 10 6 m 2 ) = 1.68 A
28.22. Model: We will use the model of conduction to relate the mean time between collisions to conductivity.
Solve: From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions f
28.20. Visualize:
Solve:
The current-carrying cross section of the wire is
A = r12 r22 = (0.0010 m ) (0.0005 m )
2
[
2
] = 2.356 10
6
m2
The current density is
J=
10 A = 4.24 10 6 A /
28.21. Model: Use the model of conduction to relate the mean time between collisions to conductivity.
Solve: From Equation 28.17, Table 28.1, and Table 28.2, the mean time between collisions for alumi
28.12. Solve: Equation 28.10 is Q = It. The amount of charge delivered is
60 s Q = (10.0 A ) 5.0 min = 3000 C 1 min
The number of electrons that flow through the hair dryer is Q 3000 C N= = = 1.88
28.36. Solve: (a) A current of 1.8 pA for the potassium ions means that a charge of 1.8 pC flows through the potassium ion channel per second. The number of potassium ions that pass through the ion ch
PHYA22 FSGs
Friday January 13th 2017
Session 2
Travelling Waves: Sinusoidal Waves
Q1. Find the Amplitude (A), wavelength (), frequency (f), period (T), and identify the
crest and trough of this wave.
PHYA22 FSGs
Friday January 20th 2017
Session 4
Travelling Waves: Sound Intensity Level, Doppler Effect
Q1: Four trumpet players are playing the same note. If three of them suddenly
stop, the sound in
PHYA22 FSGs
Friday January 27th 2017
Session 6
Superposition of Waves: Standing Waves
Q1: The ear canal, which transmits sound vibrations, is about 2.5 cm in length.
What frequency standing waves can
PHYA22 FSGs
Wednesday February 8thd 2017
Session 9
Topic Wrap-up Questions: Travelling Waves, Superposition of Waves, and Wave Optics
Q1: Light from a lamp passes through a diffraction grating having
Wednesday January 18th 2017
PHYA22 FSGs
Session 3
Travelling Waves: Waves in 2D/3D, Sound Waves, EM Waves, Power and Intensity,
Sound Intensity Levels
Q1. Label the following diagram (fill in the dark
28.31. Solve: (a) Let the new current be I such that I = 2I where I is the original current. The current density is J = I A . Since the area of cross section of the wire remains the same, we have
I I
28.40. Solve: Equation 28.13 defines the current density as J = I A . This means
A= I D 2 = D= J 4 4I = J 4(1.0 A ) = 0.050 cm = 0.50 mm (500 A / cm 2 )
Assess: Fuse wires are usually thin.
28.3. Solve: Using Equation 28.3 and Table 28.1, the electron current is
i = nAvd = (5.9 10 28 m 3 ) (0.5 10 3 m ) 5.0 10 5 m / s = 2.3 1018 s 1
2
(
)
The time for 1 mole of electrons to pass t
28.18. Solve: (a) From Equation 28.13 and Table 28.1, the current density in the gold wire is
J = nevd = (5.9 10 28 m 3 )(1.60 10 19 C)(3.0 10 4 m / s) = 2.83 10 6 A / m 2
(b) The current is
I =
28.17. Solve: (a) The current density is
J= I I 0.85 A 7 2 = = 2 = 1.73 10 A / m A R 2 [ 1 (0.00025 m )] 2
(b) The electron current, or number of electrons per second, is
Ne I 0.85 A 0.85 C / s =
28.11. Model: A battery is a charge escalator.
Solve: When a wire is connected to a battery, there is a sustained motion of electrons. A current of 1.5 A means that a charge of 1.5 C flows through a c
28.10. Visualize:
Solve: (a) The charge density is not uniform along the wire. If it was, there would be no electric field inside the wire. The charge density is most positive near the positive termi
28.6. Solve: For L = 1.0 cm = 1.0 10 2 m , the surface area of the wire is
A = (2r)L = DL = (1.0 103 m)(1.0 102 m) = (1.0 105 m2) The surface charge density of the wire is
1 19 Q (1000 cm 1 cm )1
current through this wire is 2.0 1019 s 1 . Using Table 28.1 for the electron density of iron and Equation 28.3, the drift velocity is
28.1. Solve: The wires cross-sectional area is A = r 2 = (1.0
28.2. Solve: We estimate a distance of 5 ft from the wall switch to the ceiling and then a distance of 8 ft to the
center of the room. This yields a total length of approximately L 4.0 m that an elec