Physics Physics 2102


 Benjamin Townsend
 4 years ago
 Views:
Transcription
1 Physics 2102 Jonathan Dowling Physics 2102 Exam 2: Review Session Chapters / HW0406 Some links on exam stress:
2 Exam 2 (Ch24) Sec.11 (Electric Potential Energy of a System of Point Charges); Sec.12 (Potential of Charged Isolated Conductor) (Ch 25) Capacitors: capacitance and capacitors; caps in parallel and in series, dielectrics; energy, field and potential in capacitors. (Ch 26) Current and Resistance: current, current density and drift velocity; resistance and resistivity; Ohm s law. (Ch 27) Circuits: emf devices, loop and junction rules; resistances in series and parallel; DC single and multiloop circuits, power; RC circuits. (Ch 28) Magnetic Fields: F=vxB, Right Hand Rule, Circular Motion, Force on Wire, Magnetic Dipole.
3 Potential V of Continuous Charge Distributions r = R! 2 + z 2 dv = kdq r V =! dv Straight Line Charge: dq=λdx λ=q/l Curved Line Charge: dq=λds λ=q/2πr Surface Charge: dq=σda σ=q/πr 2 da=2πr dr
4 Potential V of Continuous Charge Distributions Curved Line Charge: dq=λds λ=q/2πr Straight Line Charge: dq=λdx λ=q/l
5 Potential V of Continuous Charge Distributions Surface Charge: dq=σda σ=q/πr 2 da=r dr dϕ Straight Line Charge: dq=λdx λ=bx is given to you.
6 Capacitors E = σ/ε 0 = q/aε 0 E = V d q = C V C = ε 0 A/d Connected to Battery: V=Constant Disconnected: Q=Constant C = κ ε 0 A/d C=ε 0 ab/(ba)
7 Current and Resistance i = dq/dt Junction rule R = ρl/a V = i R E = J ρ ρ = ρ 0 (1+α(T T 0 ))
8 DC Circuits Loop rule V = ir P = iv Single loop Multiloop
9 Resistors and Capacitors Resistors Capacitors Key formula: V=iR Q=CV In series: same current R eq = R j same charge 1/C eq = 1/C j In parallel: same voltage same voltage 1/R eq = 1/R j C eq = C j
10 Capacitors and Resistors in Series and in Parallel What s the equivalent resistance (capacitance)? What s the current (charge) in each resistor (capacitor)? What s the potential across each resistor (capacitor)? What s the current (charge) delivered by the battery?
11 RC Circuits Time constant: RC Charging: q( t) = CE 1! e Discharging: q( t) =! t / RC ( ) q e 0 i(t)=dq/dt! t / RC
12 Capacitors: Checkpoints, Questions
13 Problem When switch S is thrown to the left, the plates of capacitor 1 acquire a potential V 0. Capacitors 2 and 3 are initially uncharged. The switch is now thrown to the right. What are the final charges q 1, q 2, and q 3 on the capacitors? Series
14 Series SeriQ
15
16 Current and Resistance: Checkpoints, Questions
17 Problem A cylindrical resistor of radius 5.0mm and length 2.0 cm is made of a material that has a resistivity of 3.5x105 Ωm. What are the (a) current density and (b) the potential difference when the energy dissipation rate in the resistor is 1.0W?
18
19 Circuits: Checkpoints, Questions
20 Problem: 27.P.018. [406649] Figure shows five 5.00 resistors. (Hint: For each pair of points, imagine that a battery is connected across the pair.) Fig (a) Find the equivalent resistance between points F and H. (b) Find the equivalent resistance between points F and G.
21
22 Problem: 27.P.046. [406629] In an RC series circuit, E = 17.0 V, R = 1.50 MΩ, and C = 1.80 µf. (a) Calculate the time constant. (b) Find the maximum charge that will appear on the capacitor during charging. (c) How long does it take for the charge to build up to 10.0 µc?
23
24 Magnetic Forces and Torques!!!! v F F = q v! B + q E mv r = qb!!! df = i dl! B L! "! = µ! B!
25 Top view! = iab sin" net CF = net 0 Side view C Magnetic Torque on a Current Loop Consider the rectangular loop in fig. a with sides of lengths a and b and that carries a current i. The loop is placed in a magnetic field so that the normal nˆ to the loop! forms an angle! with B. The magnitude of the magnetic force on sides 1 and 3 is F = F = iab sin 90 = iab. The magnetic force on sides 2 and 4 is 1 3 F = F = ibb sin(90 "! ) = ibb cos!. These forces cancel in pairs and thus F = The torque about the loop center C of F and F is zero because both forces pass 2 4 through point C. The moment arm for F and F is equal to ( b / 2)sin!. The two 1 3 torques tend to rotate the loop in the same (clockwise) direction and thus add up. The net torque # = # 1+ # 3=( iabb / 2) sin! + ( iabb / 2) sin! = iabb sin! = iab sin!. (2813) net
26 U = µ B U =!µ B Magnetic Dipole Moment The torque of a coil that has N loops exerted U by a uniform magnetic field B and carries a current i is given by the equation! = NiAB.! We define a new vector µ associated with the coil, which is known as the magnetic dipole moment of the coil.!! = µ " B!! =! µ " B! The magnitude of the magnetic dipole moment is µ = NiA. Its direction is perpendicular to the plane of the coil.! The sense of µ is defined by the righthand rule. We curl the fingers of the right hand in the direction of the current. The thumb gives us the sense. The torque can be!! expressed in the form! = µ B sin " where " is the angle between µ and B.!!! In vector form: = µ # B.!! The potential energy of the coil is: U = $ µ B cos" = $ µ % B. U has a minimum value of $ µ B for " = 0 (position of stable equilibrium). U has a maximum value of µ B for " = 180 (position of unstable equilibrium). Note : For both positions the net torque is! = 0. (2814)
27 Ch 28: Checkpoints and Questions
28 Circular Motion: Since magnetic force is perpendicular to motion, the movement of charges is circular. F v out F centrifugal = ma = mr! 2 = m v 2 r r B into blackboard. in F magnetic = qvb F B = F C! qv B = mv 2 r Solve : r = mv qb In general, path is a helix (component of v parallel to field is unchanged).
29 C. v! F! r.! B electron r = mv qb! = v r = qb m Radius of Circlcular Orbit Angular Frequency: Independent of v T! 2"r v = 2"mv qbv = 2"m qb Period of Orbit: Independent of v f! 1 T = qb 2"m Orbital Frequency: Independent of v
30 Problem: 28.P.024. [566302] In the figure below, a charged particle moves into a region of uniform magnetic field, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 160 ns in the region. (a) What is the magnitude of B? (b) If the particle is sent back through the magnetic field (along the same initial path) but with 3.00 times its previous kinetic energy, how much time does it spend in the field during this trip?
31
32
33
Lecture 31: MON 30 MAR Review Session : Midterm 3
Physics 2113 Jonathan Dowling Lecture 31: MON 30 MAR Review Session : Midterm 3 EXAM 03: 8PM MON 30 MAR in Cox Auditorium The exam will cover: Ch.26 through Ch.29 The exam will be based on: HW07 HW10.
More informationPhysics Jonathan Dowling. Final Exam Review
Physics 2102 Jonathan Dowling Physics 2102 Final Exam Review A few concepts: electric force, field and potential Electric force: What is the force on a charge produced by other charges? What is the force
More informationLecture 32: MON 09 NOV Review Session A : Midterm 3
Physics 2113 Jonathan Dowling Lecture 32: MON 09 NOV Review Session A : Midterm 3 EXAM 03: 6PM WED 11 NOV in Cox Auditorium The exam will cover: Ch.27.4 through Ch.30 The exam will be based on: HW08 11
More informationLouisiana State University Physics 2102, Exam 2, March 5th, 2009.
PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 2, March 5th, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),
More informationPHY 131 Review Session Fall 2015 PART 1:
PHY 131 Review Session Fall 2015 PART 1: 1. Consider the electric field from a point charge. As you move farther away from the point charge, the electric field decreases at a rate of 1/r 2 with r being
More informationLouisiana State University Physics 2102, Exam 3 April 2nd, 2009.
PRINT Your Name: Instructor: Louisiana State University Physics 2102, Exam 3 April 2nd, 2009. Please be sure to PRINT your name and class instructor above. The test consists of 4 questions (multiple choice),
More informationPhysics 212 Midterm 2 Form A
1. A wire contains a steady current of 2 A. The charge that passes a cross section in 2 s is: A. 3.2 1019 C B. 6.4 1019 C C. 1 C D. 2 C E. 4 C 2. In a Physics 212 lab, Jane measures the current versus
More informationThe next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d:
PHYS 102 Exams Exam 2 PRINT (A) The next two questions pertain to the situation described below. Consider a parallel plate capacitor with separation d: It is connected to a battery with constant emf V.
More informationPhysics 24 Exam 2 March 18, 2014
Exam Total / 200 Physics 24 Exam 2 March 18, 2014 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. You need to store electrical
More informationAP Physics C. Electric Circuits III.C
AP Physics C Electric Circuits III.C III.C.1 Current, Resistance and Power The direction of conventional current Suppose the crosssectional area of the conductor changes. If a conductor has no current,
More informationExam 2 Solutions. ε 3. ε 1. Problem 1
Exam 2 Solutions Problem 1 In the circuit shown, R1=100 Ω, R2=25 Ω, and the ideal batteries have EMFs of ε1 = 6.0 V, ε2 = 3.0 V, and ε3 = 1.5 V. What is the magnitude of the current flowing through resistor
More informationPhysics 112. Study Notes for Exam II
Chapter 20 Electric Forces and Fields Physics 112 Study Notes for Exam II 4. Electric Field Fields of + and point charges 5. Both fields and forces obey (vector) superposition Example 20.5; Figure 20.29
More informationPhysics 2135 Exam 2 March 22, 2016
Exam Total Physics 2135 Exam 2 March 22, 2016 Key Printed Name: 200 / 200 N/A Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. B 1. An airfilled
More informationCircuits Capacitance of a parallelplate capacitor : C = κ ε o A / d. (ρ = resistivity, L = length, A = crosssectional area) Resistance : R = ρ L / A
k = 9.0 x 109 N m2 / C2 e = 1.60 x 1019 C ε o = 8.85 x 1012 C2 / N m2 Coulomb s law: F = k q Q / r2 (unlike charges attract, like charges repel) Electric field from a point charge : E = k q / r2 ( towards
More informationDiscussion Question 7A P212, Week 7 RC Circuits
Discussion Question 7A P1, Week 7 RC Circuits The circuit shown initially has the acitor uncharged, and the switch connected to neither terminal. At time t = 0, the switch is thrown to position a. C a
More informationPhysics 2135 Exam 2 October 20, 2015
Exam Total / 200 Physics 2135 Exam 2 October 20, 2015 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. A straight wire segment
More informationExam 2 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses.
Exam 2 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 Part of a long, straight insulated wire carrying current i is bent into a circular
More informationName (Print): 4 Digit ID: Section:
Physics 11 Sample Common Exam 3: Sample 5 Name (Print): 4 Digit ID: Section: Honors Code Pledge: As an NJIT student I, pledge to comply with the provisions of the NJIT Academic Honor Code. I assert that
More informationa. Clockwise. b. Counterclockwise. c. Out of the board. d. Into the board. e. There will be no current induced in the wire
Physics 1B Winter 2012: Final Exam For Practice Version A 1 Closed book. No work needs to be shown for multiplechoice questions. The first 10 questions are the makeup Quiz. The remaining questions are
More informationP202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova
P202 Practice Exam 2 Spring 2004 Instructor: Prof. Sinova Name: Date: (5)1. How many electrons flow through a battery that delivers a current of 3.0 A for 12 s? A) 4 B) 36 C) 4.8 10 15 D) 6.4 10 18 E)
More information= e = e 3 = = 4.98%
PHYS 212 Exam 2  Practice Test  Solutions 1E In order to use the equation for discharging, we should consider the amount of charge remaining after three time constants, which would have to be q(t)/q0.
More informationPHYS 1444 Section 02 Review #2
PHYS 1444 Section 02 Review #2 November 9, 2011 Ian Howley 1 1444 Test 2 Eq. Sheet Terminal voltage Resistors in series Resistors in parallel Magnetic field from long straight wire Ampére s Law Force on
More informationPhysics 2135 Exam 2 October 18, 2016
Exam Total / 200 Physics 2135 Exam 2 October 18, 2016 Printed Name: Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. 1. A light bulb having
More informationA) 4 B) 3 C) 2 D) 5 E) 6
Coordinator: Saleem Rao Monday, January 01, 2018 Page: 1 Q1. A standing wave having three nodes is set up in a string fixed at both ends. If the frequency of the wave is doubled, how many antinodes will
More informationSUMMARY Phys 2523 (University Physics II) Compiled by Prof. Erickson. F e (r )=q E(r ) dq r 2 ˆr = k e E = V. V (r )=k e r = k q i. r i r.
SUMMARY Phys 53 (University Physics II) Compiled by Prof. Erickson q 1 q Coulomb s Law: F 1 = k e r ˆr where k e = 1 4π =8.9875 10 9 N m /C, and =8.85 10 1 C /(N m )isthepermittivity of free space. Generally,
More informationSolutions to PHY2049 Exam 2 (Nov. 3, 2017)
Solutions to PHY2049 Exam 2 (Nov. 3, 207) Problem : In figure a, both batteries have emf E =.2 V and the external resistance R is a variable resistor. Figure b gives the electric potentials V between the
More informationName: Class: Date: Multiple Choice Identify the letter of the choice that best completes the statement or answers the question.
Name: Class: _ Date: _ w9final Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. If C = 36 µf, determine the equivalent capacitance for the
More informationGeneral Physics (PHYC 252) Exam 4
General Physics (PHYC 5) Exam 4 Multiple Choice (6 points). Circle the one best answer for each question. For Questions 13, consider a car battery with 1. V emf and internal resistance r of. Ω that is
More informationP114 University of Rochester NAME S. Manly Spring 2010
Exam 2 (March 23, 2010) Please read the problems carefully and answer them in the space provided. Write on the back of the page, if necessary. Show your work where indicated. Problem 1 ( 8 pts): In each
More informationPhysics 6B Summer 2007 Final
Physics 6B Summer 2007 Final Question 1 An electron passes through two rectangular regions that contain uniform magnetic fields, B 1 and B 2. The field B 1 is stronger than the field B 2. Each field fills
More informationPHYS General Physics for Engineering II FIRST MIDTERM
Çankaya University Department of Mathematics and Computer Sciences 20102011 Spring Semester PHYS 112  General Physics for Engineering II FIRST MIDTERM 1) Two fixed particles of charges q 1 = 1.0µC and
More informationPH 2222C Fall Magnetic Field. Lecture 13. Chapter 28 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition)
PH 2222C Fall 2012 Magnetic Field Lecture 13 Chapter 28 (Halliday/Resnick/Walker, Fundamentals of Physics 8 th edition) 1 Chapter 28 Magnetic Fields In this chapter we will cover the following topics:
More information21 MAGNETIC FORCES AND MAGNETIC FIELDS
CHAPTER 1 MAGNETIC FORCES AND MAGNETIC FIELDS ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1 (d) RightHand Rule No 1 gives the direction of the magnetic force as x for both drawings A and B In drawing C, the
More informationPHY102 Electricity Course Summary
TOPIC 1 ELECTOSTTICS PHY1 Electricity Course Summary Coulomb s Law The magnitude of the force between two point charges is directly proportional to the product of the charges and inversely proportional
More informationInduction and Inductance
Welcome Back to Physics 1308 Induction and Inductance Michael Faraday 22 September 1791 25 August 1867 Announcements Assignments for Tuesday, November 6th:  Reading: Chapter 30.630.8  Watch Videos:
More informationFigure 1 A) 2.3 V B) +2.3 V C) +3.6 V D) 1.1 V E) +1.1 V Q2. The current in the 12 Ω resistor shown in the circuit of Figure 2 is:
Term: 13 Wednesday, May 1, 014 Page: 1 Q1. What is the potential difference V B V A in the circuit shown in Figure 1 if R 1 =70.0 Ω, R=105 Ω, R 3 =140 Ω, ε 1 =.0 V and ε =7.0 V? Figure 1 A).3 V B) +.3
More informationQuestions A hair dryer is rated as 1200 W, 120 V. Its effective internal resistance is (A) 0.1 Ω (B) 10 Ω (C) 12Ω (D) 120 Ω (E) 1440 Ω
Questions 441 36. Three 1/ µf capacitors are connected in series as shown in the diagram above. The capacitance of the combination is (A).1 µf (B) 1 µf (C) /3 µf (D) ½ µf (E) 1/6 µf 37. A hair dryer is
More informationPhysics 42 Exam 2 PRACTICE Name: Lab
Physics 42 Exam 2 PRACTICE Name: Lab 1 2 3 4 Conceptual Multiple Choice (2 points each) Circle the best answer. 1.Rank in order, from brightest to dimmest, the identical bulbs A to D. A. C = D > B > A
More informationb) (4) How large is the current through the 2.00 Ω resistor, and in which direction?
General Physics II Exam 2  Chs. 19 21  Circuits, Magnetism, EM Induction  Sep. 29, 2016 Name Rec. Instr. Rec. Time For full credit, make your work clear. Show formulas used, essential steps, and results
More informationExam 2 Solutions. = /10 = / = /m 3, where the factor of
PHY049 Fall 007 Prof. Yasu Takano Prof. Paul Avery Oct. 17, 007 Exam Solutions 1. (WebAssign 6.6) A current of 1.5 A flows in a copper wire with radius 1.5 mm. If the current is uniform, what is the electron
More informationChapter 28. Direct Current Circuits
Chapter 28 Direct Current Circuits Circuit Analysis Simple electric circuits may contain batteries, resistors, and capacitors in various combinations. For some circuits, analysis may consist of combining
More information, where the sum is over all pairs of charges (q 1, q 2 ) that are
February 12, 2010 PHY2054 Solutions Exam I 13. Particle 1 has a mass of 1.0 g and charge 6.0 µc, and particle 2 has a mass of 2.0 g and charge (3.0 5.0 8) µc. While particle 2 is held in place, particle
More informationExam 2 Solutions. Answer: 3.0 W Solution: The total current is in the series circuit is 1 A, so the power dissipated in R 2 is i 2 R 2
Exam 2 Solutions Prof. Pradeep Kumar Prof. Paul Avery Mar. 21, 2012 1. A portable CD player does not have a power rating listed, but it has a label stating that it draws a maximum current of 159.0 ma.
More informationWhere k = 1. The electric field produced by a point charge is given by
Ch 21 review: 1. Electric charge: Electric charge is a property of a matter. There are two kinds of charges, positive and negative. Charges of the same sign repel each other. Charges of opposite sign attract.
More informationPHYS 2135 Exam II March 20, 2018
Exam Total /200 PHYS 2135 Exam II March 20, 2018 Name: Recitation Section: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. For questions 69, solutions must
More informationPhysics 2212 G Quiz #4 Solutions Spring 2018 = E
Physics 2212 G Quiz #4 Solutions Spring 2018 I. (16 points) The circuit shown has an emf E, three resistors with resistance, and one resistor with resistance 3. What is the current through the resistor
More informationPhys102 Final132 Zero Version Coordinator: A.A.Naqvi Wednesday, May 21, 2014 Page: 1
Coordinator: A.A.Naqvi Wednesday, May 1, 014 Page: 1 Q1. What is the potential difference V B V A in the circuit shown in Figure 1 if R 1 =70.0 Ω, R =105 Ω, R 3 =140 Ω, ε 1 =.0 V and ε =7.0 V? A).3 V
More informationPhysics 2135 Exam 2 October 21, 2014
Exam Total Physics 2135 Exam 2 October 21, 2014 Key Printed Name: 200 / 200 N/A Rec. Sec. Letter: Five multiple choice questions, 8 points each. Choose the best or most nearly correct answer. D 1. An airfilled
More informationPHYS 241 EXAM #2 November 9, 2006
1. ( 5 points) A resistance R and a 3.9 H inductance are in series across a 60 Hz AC voltage. The voltage across the resistor is 23 V and the voltage across the inductor is 35 V. Assume that all voltages
More informationPHYS 212 Final Exam (Old Material) Solutions  Practice Test
PHYS 212 Final Exam (Old Material) Solutions  Practice Test 1E If the ball is attracted to the rod, it must be made of a conductive material, otherwise it would not have been influenced by the nearby
More informationPhysics 106 Sections 1 & 2 Midterm Exam #1 Fall 2011
Physics 106 Sections 1 & 2 Midterm Exam #1 Fall 2011 Instructor: Lawrence Rees 3digit CID: This test has 25 problems. The CID is only used to return your test. If you forgot it, make one up. Do NOT write
More informationName (Last, First): You may use only scientific or graphing calculators. In particular you may not use the calculator app on your phone or tablet!
Final Exam : Physics 2113 Fall 2014 5:30PM MON 8 DEC 2014 Name (Last, First): Section # Instructor s name: Answer all 6 problems & all 8 questions. Be sure to write your name. Please read the questions
More informationPhysics 4. Magnetic Forces and Fields. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Physics 4 Magnetic Forces and Fields What creates a magnetic field? Answer: MOVING CHARGES What is affected by a magnetic field? Answer: MOVING CHARGES We have a formula for magnetic force on a moving
More informationPhysics 420 Fall 2004 Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers.
Quiz 1 Wednesday This quiz is worth 6 points. Be sure to show your work and label your final answers. 1. A charge q 1 = +5.0 nc is located on the yaxis, 15 µm above the origin, while another charge q
More informationA) I B) II C) III D) IV E) V
1. A square loop of wire moves with a constant speed v from a fieldfree region into a region of uniform B field, as shown. Which of the five graphs correctly shows the induced current i in the loop as
More informationExam 2 Solutions. PHY2054 Spring Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014
Exam 2 Solutions Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014 1. A series circuit consists of an open switch, a 6.0 Ω resistor, an uncharged 4.0 µf capacitor and a battery with emf 15.0 V and internal
More information[1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. Fig. 1.1
1 (a) Define capacitance..... [1] (b) Fig. 1.1 shows a circuit consisting of a resistor and a capacitor of capacitance 4.5 μf. S 1 S 2 6.3 V 4.5 μf Fig. 1.1 Switch S 1 is closed and switch S 2 is left
More informationPhysics 2401 Summer 2, 2008 Exam II
Physics 2401 Summer 2, 2008 Exam II e = 1.60x1019 C, m(electron) = 9.11x1031 kg, ε 0 = 8.845x1012 C 2 /Nm 2, k e = 9.0x10 9 Nm 2 /C 2, m(proton) = 1.67x1027 kg. n = nano = 109, µ = micro = 106, m
More informationNAME: PHYSICS 6B SPRING 2011 FINAL EXAM ( VERSION A )
NAME: PHYSCS 6B SPRNG 2011 FNAL EXAM ( VERSON A ) Choose the best answer for each of the following multiplechoice questions. There is only one answer for each. Questions 12 are based on the following
More information= 8.89x10 9 N m 2 /C 2
PHY303L Useful Formulae for Test 2 Magnetic Force on a moving charged particle F B = q v B Magnetic Force on a current carrying wire F B = i L B Magnetic dipole moment µ = NiA Torque on a magnetic dipole:
More informationLast time. Ampere's Law Faraday s law
Last time Ampere's Law Faraday s law 1 Faraday s Law of Induction (More Quantitative) The magnitude of the induced EMF in conducting loop is equal to the rate at which the magnetic flux through the surface
More informationPhysics 196 Final Test Point
Physics 196 Final Test  120 Point Name You need to complete six 5point problems and six 10point problems. Cross off one 5point problem and one 10point problem. 1. Two small silver spheres, each with
More informationVersion 001 CIRCUITS holland (1290) 1
Version CIRCUITS holland (9) This printout should have questions Multiplechoice questions may continue on the next column or page find all choices before answering AP M 99 MC points The power dissipated
More information1 2 U CV. K dq I dt J nqv d J V IR P VI
o 5 o T C T F 3 9 T K T o C 73.5 L L T V VT Q mct nct Q F V ml F V dq A H k TH TC L pv nrt 3 Ktr nrt 3 CV R ideal monatomic gas 5 CV R ideal diatomic gas w/o vibration V W pdv V U Q W W Q e Q Q e Carnot
More informationPhysics 2220 Fall 2010 George Williams THIRD MIDTERM  REVIEW PROBLEMS
Physics 2220 Fall 2010 George Williams THIRD MIDTERM  REVIEW PROBLEMS Solution sets are available on the course web site. A data sheet is provided. Problems marked by "*" do not have solutions. 1. An
More informationPhysics 111. Exam #1. February 14, 2014
Physics 111 Exam #1 February 14, 2014 Name Please read and follow these instructions carefully: Read all problems carefully before attempting to solve them. Your work must be legible, and the organization
More informationPhysics / Higher Physics 1A. Electricity and Magnetism Revision
Physics / Higher Physics 1A Electricity and Magnetism Revision Electric Charges Two kinds of electric charges Called positive and negative Like charges repel Unlike charges attract Coulomb s Law In vector
More informationn Higher Physics 1B (Special) (PHYS1241) (6UOC) n Advanced Science n Double Degree (Science/Engineering) n Credit or higher in Physics 1A
Physics in Session 2: I n Physics / Higher Physics 1B (PHYS1221/1231) n Science, dvanced Science n Engineering: Electrical, Photovoltaic,Telecom n Double Degree: Science/Engineering n 6 UOC n Waves n Physical
More informationPhysics 2020 Exam 2 Constants and Formulae
Physics 2020 Exam 2 Constants and Formulae Useful Constants k e = 8.99 10 9 N m 2 /C 2 c = 3.00 10 8 m/s ɛ = 8.85 10 12 C 2 /(N m 2 ) µ = 4π 10 7 T m/a e = 1.602 10 19 C h = 6.626 10 34 J s m p = 1.67
More information8. (6) Consider the circuit here with resistors R A, R B and R C. Rank the
General Physics II Exam 2  Chs. 18B 21  Circuits, Magnetism, EM Induction  Oct. 3, 2013 Name Rec. Instr. Rec. Time For full credit, make your work clear. Show formulas used, essential steps, and results
More informationPhysics 126 Fall 2004 Practice Exam 1. Answer will be posted about Oct. 5.
Physics 126 Fall 2004 Practice Exam 1. Answer will be posted about Oct. 5. 1. Which one of the following statements best explains why tiny bits of paper are attracted to a charged rubber rod? A) Paper
More informationPhysics 6B. Practice Final Solutions
Physics 6B Practice Final Solutions . Two speakers placed 4m apart produce sound waves with frequency 45Hz. A listener is standing m in front of the left speaker. Describe the sound that he hears. Assume
More informationGravity Electromagnetism Weak Strong
19. Magnetism 19.1. Magnets 19.1.1. Considering the typical bar magnet we can investigate the notion of poles and how they apply to magnets. 19.1.1.1. Every magnet has two distinct poles. 19.1.1.1.1. N
More informationPhysics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS:
Physics 2B Spring 2010: Final Version A 1 COMMENTS AND REMINDERS: Closed book. No work needs to be shown for multiplechoice questions. 1. A charge of +4.0 C is placed at the origin. A charge of 3.0 C
More informationExam 2 Solutions. Prof. Darin Acosta Prof. Greg Stewart March 27, b 5Ω >i 1 < i 2 5Ω. 3V 10Ω 6 V   i 3 d c 10Ω
PHY49 Spring 6 Exam Solutions Prof. Darin Acosta Prof. Greg Stewart March 7, 6 Exam Solutions a b 5Ω >i 1
More informationExam 2, Phy 2049, Spring Solutions:
Exam 2, Phy 2049, Spring 2017. Solutions: 1. A battery, which has an emf of EMF = 10V and an internal resistance of R 0 = 50Ω, is connected to three resistors, as shown in the figure. The resistors have
More informationPH 102 Exam I N N N N. 3. Which of the following is true for the electric force and not true for the gravitational force?
Name Date INSTRUCTIONS PH 102 Exam I 1. nswer all questions below. ll problems have equal weight. 2. Clearly mark the answer you choose by filling in the adjacent circle. 3. There will be no partial credit
More information9. M = 2 π R µ 0 n. 3. M = π R 2 µ 0 n N correct. 5. M = π R 2 µ 0 n. 8. M = π r 2 µ 0 n N
This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 00 0.0 points A coil has an inductance of 4.5 mh, and the current
More informationExam 1 Solutions. The ratio of forces is 1.0, as can be seen from Coulomb s law or Newton s third law.
Prof. Eugene Dunnam Prof. Paul Avery Feb. 6, 007 Exam 1 Solutions 1. A charge Q 1 and a charge Q = 1000Q 1 are located 5 cm apart. The ratio of the electrostatic force on Q 1 to that on Q is: (1) none
More informationNotes and Solved Problems for Common Exam 3 (Does not include Induction)
Notes and Solved Problems for Common Exam 3 (Does not include Induction) 8. MULTI LOOP CIRCUITS Key concepts: Multi loop circuits of batteries and resistors: loops, branches and junctions should be distinguished.
More informationAnswer Key. Chapter 23. c. What is the current through each resistor?
Chapter 23. Three 2.0 resistors are connected in series to a 50.0 power source. a. What is the equivalent resistance of the circuit? R R R 2 R 3 2.0 2.0 2.0 36.0 b. What is the current in the circuit?
More informationInduction_P1. 1. [1 mark]
Induction_P1 1. [1 mark] Two identical circular coils are placed one below the other so that their planes are both horizontal. The top coil is connected to a cell and a switch. The switch is closed and
More informationWelcome back to PHY101: Major Concepts in Physics I. Photo: J. M. Schwarz
Welcome back to PHY101: Major Concepts in Physics I Photo: J. M. Schwarz Announcements In class today we will finish Chapter 18 on circuits and begin Chapter 19 (sections 1 and 8) on magnetic fields. There
More informationLenz s Law (Section 22.5)
Lenz s Law (Section 22.5) : Thursday, 25 of February 7:00 9:00 pm Rooms: Last Name Room (Armes) Seats A  F 201 122 G  R 200 221 S  Z 205 128 20160221 Phys 1030 General Physics II (Gericke) 1 1) Charging
More informationElectric_Field_core_P1
Electric_Field_core_P1 1. [1 mark] An electron enters the region between two charged parallel plates initially moving parallel to the plates. The electromagnetic force acting on the electron A. causes
More informationPhysics Will Farmer. May 5, Physics 1120 Contents 2
Physics 1120 Will Farmer May 5, 2013 Contents Physics 1120 Contents 2 1 Charges 3 1.1 Terms................................................... 3 1.2 Electric Charge..............................................
More information2 Coulomb s Law and Electric Field 23.13, 23.17, 23.23, 23.25, 23.26, 23.27, 23.62, 23.77, 23.78
College of Engineering and Technology Department of Basic and Applied Sciences PHYSICS I Sheet Suggested Problems 1 Vectors 2 Coulomb s Law and Electric Field 23.13, 23.17, 23.23, 23.25, 23.26, 23.27,
More informationExam 3PHYS 202S15
Name: Class: Date: Exam 3PHYS 202S15 Multiple Choice Identify the choice that best completes the statement or answers the question 1 Consider this circuit Which of these equations is correct? 3 Which
More informationwe can said that matter can be regarded as composed of three kinds of elementary particles; proton, neutron (no charge), and electron.
Physics II we can said that matter can be regarded as composed of three kinds of elementary particles; proton, neutron (no charge), and electron. Particle Symbol Charge (e) Mass (kg) Proton P +1 1.67
More informationGen. Phys. II Exam 2  Chs. 21,22,23  Circuits, Magnetism, EM Induction Mar. 5, 2018
Gen. Phys. II Exam 2  Chs. 21,22,23  Circuits, Magnetism, EM Induction Mar. 5, 2018 Rec. Time Name For full credit, make your work clear. Show formulas used, essential steps, and results with correct
More informationCircuits Practice Websheet 18.1
Circuits Practice Websheet 18.1 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. How much power is being dissipated by one of the 10Ω resistors? a. 24
More informationChapter 19. Magnetism
Chapter 19 Magnetism Magnetic Fields When moving through a magnetic field, a charged particle experiences a magnetic force This force has a maximum value when the charge moves perpendicularly to the magnetic
More informationPHYS 2326 University Physics II Class number
PHYS 2326 University Physics II Class number HOMEWORK SET #1 CHAPTERS: 27,28,29 (DUE JULY 22, 2013) Ch. 27.======================================================= 1. A rod of 2.0m length and a square
More informationMagnetic Force Acting on a Current Carrying Conductor IL B
Magnetic Force Acting on a Current Carrying Conductor A segment of a currentcarrying wire in a magnetic field. The magnetic force exerted on each charge making up the current is qvd and the net force
More informationChapter 27. Circuits
Chapter 27 Circuits 1 1. Pumping Chagres We need to establish a potential difference between the ends of a device to make charge carriers follow through the device. To generate a steady flow of charges,
More informationPhys 2025, First Test. September 20, minutes Name:
Phys 05, First Test. September 0, 011 50 minutes Name: Show all work for maximum credit. Each problem is worth 10 points. Work 10 of the 11 problems. k = 9.0 x 10 9 N m / C ε 0 = 8.85 x 101 C / N m e
More informationLecture 29. PHYC 161 Fall 2016
Lecture 29 PHYC 161 Fall 2016 Magnetic Force and Torque on a Current Loop Let s look at the Net force and net torque on a current loop: df Idl B F IaB top and bottom F IbB sides But, the forces on opposite
More informationApplication of Physics II for. Final Exam
Application of Physics II for Final Exam Question 1 Four resistors are connected as shown in Figure. (A)Find the equivalent resistance between points a and c. (B)What is the current in each resistor if
More informationr where the electric constant
1.0 ELECTROSTATICS At the end of this topic, students will be able to: 10 1.1 Coulomb s law a) Explain the concepts of electrons, protons, charged objects, charged up, gaining charge, losing charge, charging
More informationQuestion 1. Question 2. Question 3
Question 1 Switch S in in the figure is closed at time t = 0, to begin charging an initially uncharged capacitor of capacitance C = 18.2 μf through a resistor of resistance R = 22.3 Ω. At what time (in
More information