HW04 solutions - General Physics: Modern Physics PDF

Preview

Summary

General Physics: Modern Physics
...

Description

Homework(4(solutions( ! Notes:!Please!keep!in!mind!that!most!of!the!problems!in!MasteringPhysics!with!numerical!answers!are! randomized!for!each!student.!The!solutions!below!correspond!to!the!version!of!the!problem!that!is!in!the! paper!copy!of!the!textbook.!There!may!be!some!differences!between!the!textbook!version!and!the!online! version.! 1. Reading!questions! a. Actually,!the!opposite!of!this!is!true.!When!you!add!resistors!in!parallel!with!other!resistors!the! equivalent!resistance!decreases.!Here’s!an!example.!Let’s!say!you!have!a!single!2"Ω!resistor!in!a! circuit!and!you!add!a!4"Ω!resistor!in!parallel!with!it.!The!equivalent!resistance!of!the!two!is! 1 1 -* 1 1 -* 4 𝑅&' = + = Ω! = + 𝑅* 𝑅, 3 2"Ω 4"Ω This!is!less!than!the!resistance!of!the!initial!2"Ω!resistor,!not!greater.! b. This!is!true.!The!magnetic!force!exerted!on!a!moving!charged!particle!always!points!perpendicular! to!the!particle’s!direction!of!motion.!This!means!the!magnetic!force!will!not!speed!up!or!slow! down!the!particle.!The!most!it!can!do!is!change!its!direction.! 2. Tutorial (the problem parts plus the hints in MasteringPhysics walk you through how to do it) 3. Problem 16.41 The total power consumed is Ptotal = Pbulb + PTV + Pcomp + PTO + Pfridge = 60 W + 180 W + 300 W + 1050 W + 240 W = 1830 W

The current in the electrical line is I =

Ptotal ΔV

=

1830 W = 15.25 A. 120 V

4. Problem 16.27 Sketch and translate We label the currents as shown in the figure to the right. Solve and evaluate (a) Applying Kirchhoff’s rule clockwise to the loop gives ε1 − IR3 − ε 2 − IR2 − IR1 = 0

(b) Solving for I, we obtain I =

ε1 − ε2 20 V − 8.0 V = = 0.20 A. R1 + R2 + R3 30 Ω + 20 Ω + 10 Ω

(c) The potential change across each of the elements is as follows:

ΔVR = 2IR1 = 2 (0.20 A )(30 Ω ) = 26.0 V 1

ΔVε = 1ε1 = 120 V 1

ΔVR = 2IR3 = 2 (0.20 A )(10 Ω) = 22.0 V 3

ΔVε = 2ε 2 = 28.0 V 2

ΔVR = 2IR2 = 2 (0.20 A ) ( 20 Ω) = 24.0 V 2

(d) Adding up all the potential changes gives ΔV = ΔVR + ΔVε + ΔVR + ΔVε + ΔVR ∑฀ loop i 1

1

3

2

2

= 26.0 V + 20 V − 2.0 V − 8.0 V − 4.0 V = 0

Our result agrees with Kirchhoff’s loop rule. 5. Problem 16.45 In series connection, the equivalent resistance is Req = 3R, and the power consumed by the resistors is 2

Pseries

⎛ ε ⎞ ε2 ε2 = I Req = ⎜ = ⎟ Req = Req 3R ⎝ Req ⎠ 2

However, if the three resistors are connected in parallel, R′ eq = R/3, and the power consumed is now

Pparallel

=

ε2 3ε 2 = R R′ eq

which is 9 times greater than that in a series connection. Since Pseries = 15 W, we obtain

Pparallel = 9Pseries = 9( 15 W ) = 135 W 6. Tutorial (the problem parts plus the hints in MasteringPhysics walk you through how to do it) 7. Problem 17.10 The increase in scale reading by 6.0 g is due to the magnetic force exerted on the rod. Using FB on R = IlB, we find the magnitude of the magnetic field to be B=

(

)(

)

6.0 × 1023 kg 9.8 m/s2 FB on R (Δm )g = 0.84 T = = Il Il (1.0 A )(0.070 m )

8. Problem 17.42   Represent mathematically A particle with charge q moving with a velocity v through a region where both E fields and B fields are present will experience both electric and magnetic forces. The magnitudes of the forces   are 𝐹0"12"' = 𝑞 E and 𝐹5"12"' = 𝑞 𝑣𝐵 sin 𝜃 where θ is the angle between B and the particle’s velocity v . The two forces can cancel if they have the same magnitude but point in opposite directions. When this happens, the particle can travel in a straight path.

Solve and evaluate (a) The magnetic force points upward and has magnitude

(

)(

)

FB = qvB = 1.6 × 10219 C 8.0 × 106 m/s ( 0.12 T ) = 1.54 × 10213 N (b) To balance  the magnetic force, we want the electric force to have the same magnitude but point downward. Since FE = 2eE (electrons are negatively charged), E points upward and has a magnitude

(

)

E = vB = 8.0 × 106 m/s ( 0.12 T ) = 9.6× 105 N/C 9. Problem 17.26 Sketch and translate When the velocity of a charged particle is perpendicular to the field, it moves in a circular path at constant speed. The magnetic force exerted on the charged particle provides the necessary radial force to keep the particle in circular orbit. Represent mathematically Since the magnetic force 𝐹5"12"' = 𝑞 𝑣𝐵 sin 𝜃 is the same for both the proton and the electron, their paths will be circular with

mp v 2 rp

mev 2 = re

⇒

mp rp

=

me re

Solve and evaluate Given that re = 7.0 mm for the electron, the radius of the proton is

⎛m ⎞ ⎛ 1.67 × 10227 kg ⎞ rp = ⎜ p ⎟ re = ⎜ 231 ⎟ (7.00 mm ) = 12.8 m ⎝ 9.11 × 10 kg ⎠ ⎝ me ⎠ !...