Physics 9 Fall 2009 Homework 7  Solutions


 Warren Wesley Marsh
 2 years ago
 Views:
Transcription
1 Physics 9 Fall 009 Homework 7  s 1. Chapter 33  Exercise 10. At what distance on the axis of a current loop is the magnetic field half the strength of the field at the center of the loop? Give your answer as a multiple of R. The magnetic field on the axis of a current loop is given by B (z) = µ 0 IR (z + R ) 3/. We want a distance z, such that B (z) = 1 B (z = 0). The field at the center of the loop is B (z = 0) = µ 0I. So, we want R µ 0 IR (z + R ) 3/ = 1 µ 0 I R. Solving, we find R 3 = (z + R ) 3/, or z = 4 1/3 1R. So, z 0.77R. 1
2 . Chapter 33  Exercise 5. Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of.0 mmdiameter superconducting wire. What size current is needed? The magnetic field of a solenoid is B = µ 0NI BL, and so the current is I =. What s L µ 0 N N? It s just the length of the wire, divided by the diameter of the wire  this is how many times you can wind the coils together for a given length. So, there are N = 1.8/0.00 = 900 turns of the wire. Thus, I = BL µ 0 N = π = 400 A.
3 3. Chapter 33  Exercise 9. Radio astronomers detect electromagnetic radiation at 45 MHz from an interstellar gas cloud. They suspect this radiation is emitted by electrons spiraling in a magnetic field. What is the magnetic field strength inside the gas cloud? The frequency of the emitted radiation is equal to the frequency of the acceleration of the electron (i.e., the frequency of the revolutions). The revolution of the frequency is just the cyclotron frequency, which is f = qb. Solving for the magnetic field gives πm B = πfm. Plugging in the values gives q B = π = T. 3
4 4. Chapter 33  Problem 45. The element niobium, which is a metal, is a superconductor (i.e., no electrical resistance) at temperatures below 9 K. However, the superconductivity is destroyed if the magnetic field at the surface of the metal reaches or exceeds 0.10 T. What is the maximum current in a straight, 3.0 mmdiameter superconducting niobium wire? The magnetic field of a long straight wire is B = µ 0I, where r is the distance away πr from the center. So, at the surface of the wire, when r = d, the current in the wire is I = πbd µ 0. Plugging in the values gives the maximum current as I = πbd µ 0 = π π 10 7 = 750 A. 4
5 5. Chapter 33  Problem 46. (a) Find an expression for the magnetic field at the center (point P) of the circular arc in the figure. (b) Does your result agree with the magnetic field of a current loop when θ = π? (a) Recall the BiotSavart law, B = µ 0 I s ˆr, where s points along the current, and 4π r ˆr points long the direction from the current to the point P. Now, because the current and ˆr point along the same line in both straight segments of the wire, their magnetic field contributions to the point P are zero. So, we only need to find the magnetic field from the curved part of the wire. Along the arc, the current points along the wire, which is always at right angles to ˆr. So, along the arc, s ˆr = s ˆr = s, since ˆr is a unit vector. So, along the arc the little bit of the magnetic field from the little section s is B = µ 0I 4π s r. Now, the segment is always a distance r = R from the point P. So, we just have to add up the distance, s, along the arc, which is s = Rθ. Thus, B = µ 0Iθ 4πR. (b) Recall that the magnetic field of a loop of wire is B = µ 0 IR (z + R ) 3/, which gives the field at the center, when z = 0, as B center = µ 0I, which is precisely R the correct result in part a when θ = π. So, our results agree. 5
6 6. Chapter 33  Problem 56. An electron orbits in a 5.0 mt field with angular momentum kg m /s. What is the diameter of the orbit? Recall that the electron makes an orbit in a magnetic field of radius r = mv eb. The angular momentum of a particle in an orbit is L = mvr. So, multiplying the expression for the radius of the orbit by r gives and so the radius can be expressed as r = mvr eb = L eb, r = L eb. With numbers, we find that r = 6 = 0.01 m So, the electron orbits at a distance of about 1 cm. 6
7 7. Chapter 33  Problem 61. An antiproton (same properties as a proton except that q = e) is moving in the combined electric and magnetic fields of the figure. (a) What are the magnitude and direction of the antiproton s acceleration at this instant? (b) What would be the magnitude and direction of the acceleration if v were reversed? (a) The antiproton is negatively charged, so the electric force is F E = ee, and points up, while the magnetic force is F M = e v B, and points down. So the two forces work ( against) each other. The net force is just the sum of the forces, = e E v B. The magnitude of the force is F net = e (E vb) = F net ( ) N. Because the magnetic force is stronger, the force points down. Finally, since a = F/m, where m is the mass of the proton, the acceleration is a = m/s, and still points down. (b) If v was reversed, then both forces would point up. The forces would add, such that F ( ) net = e E + v B. In this case, the magnitudes would add, instead of subtract, giving F net = N, and points up. Again, the acceleration is just a = F/m, giving a = m/s upwards. 7
8 8. Chapter 33  Problem 73. In the semiclassical Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius m with speed m/s. According to this model, what is the magnetic field at the center of a hydrogen atom? Hint: Determine the average current of the orbiting electron. The moving charge orbiting the nucleus constitutes a current, I. The magnitude of the current is just the charge on the electron, divided by the time it takes to orbit the nucleus, i.e., it s orbital period, T. The period is the total distance around πr, divided by the speed of the electron, v. So, T = πr/v, which gives I = e T = ev πr. The magnetic field at the center of the loop is, as we ve seen, B = µ 0I, which gives R upon plugging in for the current, B = µ 0ev 4πR. With numbers, we find B = µ 0ev 4πR = ( ) = 1.5 T, which is a huge field! 8
9 9. Chapter 33  Problem 79. A flat, circular disk of radius R is uniformly charged with total charge Q. The disk spins at angular velocity ω about an axis through its center. What is the magnetic field strength at the center of the disk? The plate is uniformly charged, so it s surface charge density, η = Q = Q, is constant. A πr Consider a small bit of charge, dq. From the BiotSavart law, the tiny bit of magnetic field from the this charge is db = µ 0 ˆr dq v = µ 0 4π r 4π dq v r ˆk, since v points along the direction of spin, which is tangent to the disk, while ˆr points towards the z axis. This means that v and ˆr are perpendicular. Now, for an object spinning in a circle, v = rω, where ω is the angular velocity. Therefore, db = µ 0 4π dq ω r. Next, we integrate, writing dq = ηda = πrηdr, since we are on a disk, as we ve seen before. Then, db = µ 0ωη dr. Then, integrating both sides gives B = µ 0ωη R 0 dr = µ 0ωηR. We can express this in terms of the total charge, Q, on the disk by recalling that η = Q, which gives πr B = µ 0ωQ πr, and points up. 9
10 10. Chapter 33  Problem 81. A long, straight conducting wire of radius R has a nonuniform current density J = J 0 r/r, where J 0 is a constant. The wire carries total current I. (a) Find an expression for J 0 in terms of I and R. (b) Find an expression for the magnetic field strength inside the wire at radius r. (c) At the boundary, r = R, does your solution match the known field outside a long, straight currentcarrying wire? (a) For a varying current density, the total current in the wire is I = JdA. For a disk, da = πrdr, as we ve seen before. Integrating over the total surface area of the disk gives the total current, I, on the wire. So, I = and so J 0 = 3 πr I. JdA = πj 0 R R 0 r dr = πj 0 R R 3 3 = πr 3 J 0, (b) Here we will use Ampere s law, which says that B d s = µ0 I encl. Because of the symmetry of the wire, the magnetic field circles around inside the wire. So, we choose a circle of radius r R for our Amperian loop. Since the magnetic field is constant along that loop, and points along d s, the lefthand side of Ampere s law says B d s = B (πr). Now we just need the enclosed current. We can use the same method as in part (a) to determine the enclosed current  only now we integrate out to radius r, since we only want the enclosed current. Thus, I encl = JdA = πj 0 R r 0 ( ) r dr = πr r 3 J 0 = I, R where in the last step we inserted our value for J 0. Thus, Ampere s law gives B (πr) = µ 0 Ir /R, or B = µ 0I πr r. (c) The field outside a long straight wire with current I is given by B = µ 0I. At the πr boundary, we have r = R, and so B = µ 0I, which is precisely the correct answer πr that we get from part (b) at the boundary. 10
Nowadays we know that magnetic fields are set up by charges in motion, as in
6 Magnetostatics 6.1 The magnetic field Although the phenomenon of magnetism was known about as early as the 13 th century BC, and used in compasses it was only in 1819 than Hans Oersted recognised that
More informationEssential Physics II. Lecture 8:
Essential Physics II E II Lecture 8: 161215 News Schedule change: Monday 7th December ( ) NO CLASS! Thursday 26th November ( ) 18:1519:45 This week s homework: 11/26 (next lecture) Next week s homework:
More informationMagnetism. d. gives the direction of the force on a charge moving in a magnetic field. b. results in negative charges moving. clockwise.
Magnetism 1. An electron which moves with a speed of 3.0 10 4 m/s parallel to a uniform magnetic field of 0.40 T experiences a force of what magnitude? (e = 1.6 10 19 C) a. 4.8 10 14 N c. 2.2 10 24 N b.
More information1 of 7 4/13/2010 8:05 PM
Chapter 33 Homework Due: 8:00am on Wednesday, April 7, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy [Return to Standard Assignment View] Canceling a Magnetic Field
More informationMagnetism Conceptual Questions. Name: Class: Date:
Name: Class: Date: Magnetism 22.1 Conceptual Questions 1) A proton, moving north, enters a magnetic field. Because of this field, the proton curves downward. We may conclude that the magnetic field must
More informationFall 12 PHY 122 Homework Solutions #8
Fall 12 PHY 122 Homework Solutions #8 Chapter 27 Problem 22 An electron moves with velocity v= (7.0i  6.0j)10 4 m/s in a magnetic field B= (0.80i + 0.60j)T. Determine the magnitude and direction of the
More informationMagnetic Forces and Magnetic Fields
1 Magnets Magnets are metallic objects, mostly made out of iron, which attract other iron containing objects (nails) etc. Magnets orient themselves in roughly a north  south direction if they are allowed
More informationPHYS 155: Final Tutorial
Final Tutorial Saskatoon Engineering Students Society eric.peach@usask.ca April 13, 2015 Overview 1 2 3 4 5 6 7 Tutorial Slides These slides have been posted: sess.usask.ca homepage.usask.ca/esp991/ Section
More informationPhysics 6C, Summer 2006 Homework 1 Solutions F 4
Physics 6C, Summer 006 Homework 1 Solutions All problems are from the nd edition of Walker. Numerical values are different for each student. Chapter Conceptual Questions 18. Consider the four wires shown
More informationMagnetic Forces cont.
Magnetic Fields Magnetism Magnets can exert forces on each other. The magnetic forces between north and south poles have the property that like poles repel each other, and unlike poles attract. This behavior
More informationChapter 19 Magnetism Magnets Poles of a magnet are the ends where objects are most strongly attracted Two poles, called north and south Like poles
Chapter 19 Magnetism Magnets Poles of a magnet are the ends where objects are most strongly attracted Two poles, called north and south Like poles repel each other and unlike poles attract each other Similar
More informationSolution: (a) For a positively charged particle, the direction of the force is that predicted by the right hand rule. These are:
Problem 1. (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields as shown in the figure. (b) Repeat part (a), assuming the moving particle is
More informationMagnetic Fields; Sources of Magnetic Field
This test covers magnetic fields, magnetic forces on charged particles and currentcarrying wires, the Hall effect, the BiotSavart Law, Ampère s Law, and the magnetic fields of currentcarrying loops
More information1. Units of a magnetic field might be: A. C m/s B. C s/m C. C/kg D. kg/c s E. N/C m ans: D
Chapter 28: MAGNETIC FIELDS 1 Units of a magnetic field might be: A C m/s B C s/m C C/kg D kg/c s E N/C m 2 In the formula F = q v B: A F must be perpendicular to v but not necessarily to B B F must be
More information** Can skip to Problems: 6, 7, 9, 15, 19, 28, 29, 35, 36, 39, 42 **
Walker, Physics, 3 rd Edition Chapter 22 ** Can skip to Problems: 6, 7, 9, 15, 19, 28, 29, 35, 36, 39, 42 ** Conceptual Questions (Answers to oddnumbered Conceptual Questions can be found in the back
More informationDr. Todd Satogata (ODU/Jefferson Lab) Monday, March
Vector pointing OUT of page Vector pointing N to page University Physics 227N/232N Review: Gauss s Law and Ampere s Law Solenoids and Magnetic nductors Lab rescheduled for Wednesday, March 26 in ScaleUp
More informationHandout 7: Magnetic force. Magnetic force on moving charge
1 Handout 7: Magnetic force Magnetic force on moving charge A particle of charge q moving at velocity v in magnetic field B experiences a magnetic force F = qv B. The direction of the magnetic force is
More informationMagnetic Field and Magnetic Forces
Chapter 27 Magnetic Field and Magnetic Forces PowerPoint Lectures for University Physics, Thirteenth Edition Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Goals for Chapter 27 Magnets
More informationPhysics Notes for Class 12 Chapter 4 Moving Charges and Magnetrism
1 P a g e Physics Notes for Class 12 Chapter 4 Moving Charges and Magnetrism Oersted s Experiment A magnetic field is produced in the surrounding of any current carrying conductor. The direction of this
More informationChapter 21. Magnetic Forces and Magnetic Fields
Chapter 21 Magnetic Forces and Magnetic Fields 21.1 Magnetic Fields The needle of a compass is permanent magnet that has a north magnetic pole (N) at one end and a south magnetic pole (S) at the other.
More informationHomework #8 20311721 Physics 2 for Students of Mechanical Engineering. Part A
Homework #8 20311721 Physics 2 for Students of Mechanical Engineering Part A 1. Four particles follow the paths shown in Fig. 3233 below as they pass through the magnetic field there. What can one conclude
More informationThis Set o Slides  Day 20, Friday, Feb 19
This Set o Slides  Day 20, Friday, Feb 19 Magnetic Field of Moving Charge or Current BiotSavart Law Cross Product. BiotSavart Law as cross product. More right hand rules. Three total! Similar but different!
More informationPhys222 Winter 2012 Quiz 4 Chapters 2931. Name
Name If you think that no correct answer is provided, give your answer, state your reasoning briefly; append additional sheet of paper if necessary. 1. A particle (q = 5.0 nc, m = 3.0 µg) moves in a region
More informationChapter 27 Magnetic Field and Magnetic Forces
Chapter 27 Magnetic Field and Magnetic Forces  Magnetism  Magnetic Field  Magnetic Field Lines and Magnetic Flux  Motion of Charged Particles in a Magnetic Field  Applications of Motion of Charged
More informationSolution Derivations for Capa #10
Solution Derivations for Capa #10 1) A 1000turn loop (radius = 0.038 m) of wire is connected to a (25 Ω) resistor as shown in the figure. A magnetic field is directed perpendicular to the plane of the
More information5.Magnetic Fields due to Currents( with Answers)
5.Magnetic Fields due to Currents( with Answers) 1. Suitable units for µ. Ans: TmA 1 ( Recall magnetic field inside a solenoid is B= µ ni. B is in tesla, n in number of turn per metre, I is current in
More informationLecture PowerPoints. Chapter 20 Physics: Principles with Applications, 7 th edition Giancoli
Lecture PowerPoints Chapter 20 Physics: Principles with Applications, 7 th edition Giancoli This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching
More informationPhysics 2220 Module 09 Homework
Physics 2220 Module 09 Homework 01. A potential difference of 0.050 V is developed across the 10cmlong wire of the figure as it moves though a magnetic field perpendicular to the page. What are the strength
More informationMagnetic Fields. David J. Starling Penn State Hazleton PHYS 212
Magnetism, as you recall from physics class, is a powerful force that causes certain items to be attracted to refrigerators.  Dave Barry David J. Starling Penn State Hazleton PHYS 212 Objectives (a) Determine
More informationPHY2049 Exam #2 Solutions Fall 2012
PHY2049 Exam #2 Solutions Fall 2012 1. The diagrams show three circuits consisting of concentric circular arcs (either half or quarter circles of radii r, 2r, and 3r) and radial segments. The circuits
More informationEðlisfræði 2, vor 2007
[ Assignment View ] [ Pri Eðlisfræði 2, vor 2007 28. Sources of Magnetic Field Assignment is due at 2:00am on Wednesday, March 7, 2007 Credit for problems submitted late will decrease to 0% after the deadline
More informationmv = ev ebr Application: circular motion of moving ions In a uniform magnetic field: The mass spectrometer KE=PE magnitude of electron charge
1.4 The Mass Spectrometer Application: circular motion of moving ions In a uniform magnetic field: The mass spectrometer mv r qb mv eb magnitude of electron charge 1 mv ev KEPE v 1 mv ebr m v e r m B m
More informationLecture 13. Magnetic Field, Magnetic Forces on Moving Charges. Outline:
Lecture 13. Magnetic Field, Magnetic Forces on Moving Charges. Outline: Intro to Magnetostatics. Magnetic Field Flux, Absence of Magnetic Monopoles. Force on charges moving in magnetic field. 1 Intro to
More information(b) Draw the direction of for the (b) Draw the the direction of for the
2. An electric dipole consists of 2A. A magnetic dipole consists of a positive charge +Q at one end of a bar magnet with a north pole at one an insulating rod of length d and a end and a south pole at
More informationExam 2 Solutions. PHY2054 Spring Prof. P. Kumar Prof. P. Avery March 5, 2008
Prof. P. Kumar Prof. P. Avery March 5, 008 Exam Solutions 1. Two cylindrical resistors are made of the same material and have the same resistance. The resistors, R 1 and R, have different radii, r 1 and
More informationReview Questions PHYS 2426 Exam 2
Review Questions PHYS 2426 Exam 2 1. If 4.7 x 10 16 electrons pass a particular point in a wire every second, what is the current in the wire? A) 4.7 ma B) 7.5 A C) 2.9 A D) 7.5 ma E) 0.29 A Ans: D 2.
More informationLesson 12: Magnetic Forces and Circular Motion!
Lesson 12: Magnetic Forces and Circular Motion If a magnet is placed in a magnetic field, it will experience a force. Types of magnets: Direction of the force on a permanent magnet: Direction of the force
More informationChapter 5: Circular Motion
Page 1 Chapter 5: Circular Motion Rotating Objects: Wheels, moon, earth, CDs, DVDs etc. Rigid bodies. Description of circular motion. Angular Position, Angular Displacement θ r s Angle (in radians) θ =
More informationPhysics 121 Sample Common Exam 3 NOTE: ANSWERS ARE ON PAGE 6. Instructions: 1. In the formula F = qvxb:
Physics 121 Sample Common Exam 3 NOTE: ANSWERS ARE ON PAGE 6 Signature Name (Print): 4 Digit ID: Section: Instructions: Answer all questions 24 multiple choice questions. You may need to do some calculation.
More information2. B The magnetic properties of a material depend on its. A) shape B) atomic structure C) position D) magnetic poles
ame: Magnetic Properties 1. B What happens if you break a magnet in half? A) One half will have a north pole only and one half will have a south pole only. B) Each half will be a new magnet, with both
More informationThe Solar Wind. Earth s Magnetic Field p.1/15
The Solar Wind 1. The solar wind is a stream of charged particles  a plasma  from the upper atmosphere of the sun consisting of electrons and protons with energies of 1 kev. 2. The particles escape the
More informationLinear Centripetal Tangential speed acceleration acceleration A) Rω Rω 2 Rα B) Rω Rα Rω 2 C) Rω 2 Rα Rω D) Rω Rω 2 Rω E) Rω 2 Rα Rω 2 Ans: A
1. Two points, A and B, are on a disk that rotates about an axis. Point A is closer to the axis than point B. Which of the following is not true? A) Point B has the greater speed. B) Point A has the lesser
More informationTHE MAGNETIC FIELD. 9. Magnetism 1
THE MAGNETIC FIELD Magnets always have two poles: north and south Opposite poles attract, like poles repel If a bar magnet is suspended from a string so that it is free to rotate in the horizontal plane,
More informationHMWK 3. Ch 23: P 17, 23, 26, 34, 52, 58, 59, 62, 64, 73 Ch 24: Q 17, 34; P 5, 17, 34, 42, 51, 52, 53, 57. Chapter 23
HMWK 3 Ch 23: P 7, 23, 26, 34, 52, 58, 59, 62, 64, 73 Ch 24: Q 7, 34; P 5, 7, 34, 42, 5, 52, 53, 57 Chapter 23 P23.7. Prepare: The connecting wires are ideal with zero resistance. We have to reduce the
More informationIt is the force experienced by a charged particle moving in a space where both electric and magnetic fields exist. F =qe + q(v B )
Moving Charges And Magnetism Moving Charges Moving charges produce magnetic field around them. SI unit of magnetic field is Tesla (T). Lorentz Force It is the force experienced by a charged particle moving
More informationChapter 20. Magnetic Forces and Magnetic Fields
Chapter 20 Magnetic Forces and Magnetic Fields The Motion of a Charged Particle in a Magnetic Field The circular trajectory. Since the magnetic force always remains perpendicular to the velocity, if a
More informationPhysics 2B. Lecture 29B
Physics 2B Lecture 29B "There is a magnet in your heart that will attract true friends. That magnet is unselfishness, thinking of others first. When you learn to live for others, they will live for you."
More informationPhysics 126 Practice Exam #3 Professor Siegel
Physics 126 Practice Exam #3 Professor Siegel Name: Lab Day: 1. Which one of the following statements concerning the magnetic force on a charged particle in a magnetic field is true? A) The magnetic force
More informationChapter 26 Magnetism
What is the fundamental hypothesis of science, the fundamental philosophy? [It is the following:] the sole test of the validity of any idea is experiment. Richard P. Feynman 26.1 The Force on a Charge
More informationRotational Motion. Description of the motion. is the relation between ω and the speed at which the body travels along the circular path.
Rotational Motion We are now going to study one of the most common types of motion, that of a body constrained to move on a circular path. Description of the motion Let s first consider only the description
More informationPhysics 12 Study Guide: Electromagnetism Magnetic Forces & Induction. Text References. 5 th Ed. Giancolli Pg
Objectives: Text References 5 th Ed. Giancolli Pg. 58896 ELECTROMAGNETISM MAGNETIC FORCE AND FIELDS state the rules of magnetic interaction determine the direction of magnetic field lines use the right
More informationChapter 38C  Atomic Physics. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University
Chapter 38C  Atomic Physics A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 007 Objectives: After completing this module, you should be able to:
More information1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 80 µ T at the loop center. What is the loop radius?
CHAPTER 3 SOURCES O THE MAGNETC ELD 1. A wire carries 15 A. You form the wire into a singleturn circular loop with magnetic field 8 µ T at the loop center. What is the loop radius? Equation 33, with
More informationChapter 27 Magnetic Fields and Magnetic Forces
Chapter 27 Magnetic Fields and Magnetic Forces In this chapter we investigate forces exerted by magnetic fields. In the next chapter we will study the sources of magnetic fields. The force produced by
More informationChapter 29 Electromagnetic Induction
Chapter 29 Electromagnetic Induction  Induction Experiments  Faraday s Law  Lenz s Law  Motional Electromotive Force  Induced Electric Fields  Eddy Currents  Displacement Current and Maxwell s Equations
More informationPearson Physics Level 30 Unit VI Forces and Fields: Chapter 12 Solutions
Concept Check (top) Pearson Physics Level 30 Unit VI Forces and Fields: Chapter 1 Solutions Student Book page 583 Concept Check (bottom) The northseeking needle of a compass is attracted to what is called
More informationPhysics 112 Homework 5 (solutions) (2004 Fall) Solutions to Homework Questions 5
Solutions to Homework Questions 5 Chapt19, Problem2: (a) Find the direction of the force on a proton (a positively charged particle) moving through the magnetic fields in Figure P19.2, as shown. (b) Repeat
More informationMagnetism, a history. Existence of a Magnetic Field, B. Magnetic Force on a charged particle. The particle in the figure
Existence of a Magnetic Field, B Magnetic field, B, is a vector You may be familiar with bar magnets have a magnetic field similar to electric field of dipoles Amazing experimental finding: there is a
More informationThe Big Idea. Key Concepts
The ig Idea For static electric charges, the electromagnetic force is manifested by the Coulomb electric force alone. If charges are moing, howeer, there is created an additional force, called magnetism.
More informationForce on Moving Charges in a Magnetic Field
[ Assignment View ] [ Eðlisfræði 2, vor 2007 27. Magnetic Field and Magnetic Forces Assignment is due at 2:00am on Wednesday, February 28, 2007 Credit for problems submitted late will decrease to 0% after
More informationTest  A2 Physics. Primary focus Magnetic Fields  Secondary focus electric fields (including circular motion and SHM elements)
Test  A2 Physics Primary focus Magnetic Fields  Secondary focus electric fields (including circular motion and SHM elements) Time allocation 40 minutes These questions were ALL taken from the June 2010
More information6.1: Angle Measure in degrees
6.1: Angle Measure in degrees How to measure angles Numbers on protractor = angle measure in degrees 1 full rotation = 360 degrees = 360 half rotation = quarter rotation = 1/8 rotation = 1 = Right angle
More informationMAGNETISM LAB: The ChargetoMass Ratio of the Electron
Physics 9 Chargetomass: e/m p. 1 NAME: SECTION NUMBER: TA: LAB PARTNERS: MAGNETISM LAB: The ChargetoMass Ratio of the Electron Introduction In this lab you will explore the motion of a charged particle
More informationRotational Mechanics  1
Rotational Mechanics  1 The Radian The radian is a unit of angular measure. The radian can be defined as the arc length s along a circle divided by the radius r. s r Comparing degrees and radians 360
More information2. Consider a dipole AB of dipole moment p placed at an angle θ in an uniform electric field E
1) Field due to an infinite long straight charged wire Consider an uniformly charged wire of infinite length having a constant linear charge density λ (charge per unit length). 2. Consider a dipole AB
More informationMagnetic Field and Magnetic Forces
Chapter 27 Magnetic Field and Magnetic Forces PowerPoint Lectures for University Physics, Thirteenth Edition Hugh D. Young and Roger A. Freedman Lectures by Wayne Anderson Goals for Chapter 27 To study
More informationMagnetic Fields ild and Forces
Magnetic Fields ild and Forces Physics 1 Facts about Magnetism Magnets have 2 poles (north and south) Like poles repel Unlike poles attract Magnets create a MAGNETIC FIELD around them Magnetic Field A
More informationExam 2 Practice Problems Part 2 Solutions
Problem 1: Short Questions MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8. Exam Practice Problems Part Solutions (a) Can a constant magnetic field set into motion an electron, which is initially
More informationphysics 112N magnetic fields and forces
physics 112N magnetic fields and forces bar magnet & iron filings physics 112N 2 bar magnets physics 112N 3 the Earth s magnetic field physics 112N 4 electro magnetism! is there a connection between electricity
More informationANGULAR POSITION. 4. Rotational Kinematics and Dynamics
ANGULAR POSITION To describe rotational motion, we define angular quantities that are analogous to linear quantities Consider a bicycle wheel that is free to rotate about its axle The axle is the axis
More information* Biot Savart s Law Statement, Proof Applications of Biot Savart s Law * Magnetic Field Intensity H * Divergence of B * Curl of B. PPT No.
* Biot Savart s Law Statement, Proof Applications of Biot Savart s Law * Magnetic Field Intensity H * Divergence of B * Curl of B PPT No. 17 Biot Savart s Law A straight infinitely long wire is carrying
More informationThe Charge to Mass Ratio (e/m) Ratio of the Electron. NOTE: You will make several sketches of magnetic fields during the lab.
The Charge to Mass Ratio (e/m) Ratio of the Electron NOTE: You will make several sketches of magnetic fields during the lab. Remember to include these sketches in your lab notebook as they will be part
More informationCHARGE TO MASS RATIO OF THE ELECTRON
CHARGE TO MASS RATIO OF THE ELECTRON In solving many physics problems, it is necessary to use the value of one or more physical constants. Examples are the velocity of light, c, and mass of the electron,
More informationPhysics 1653 Exam 3  Review Questions
Physics 1653 Exam 3  Review Questions 3.0 Two uncharged conducting spheres, A and B, are suspended from insulating threads so that they touch each other. While a negatively charged rod is held near, but
More informationPHYSICS 212 INDUCED VOLTAGES AND INDUCTANCE WORKBOOK ANSWERS
PHYSICS 212 CHAPTER 20 INDUCED VOLTAGES AND INDUCTANCE WORKOOK ANSWERS STUDENT S FULL NAME (y placing your name above and submitting this for credit you are affirming this to be predominantly your own
More informationLecture 13. Magnetic Field, Magnetic Forces on Moving Charges. Outline:
Lecture 13. Magnetic Field, Magnetic Forces on Moving Charges. Outline: Intro to Magnetostatics. Magnetic Field Flux, Absence of Magnetic Monopoles. Force on charges moving in magnetic field. 1 Structure
More informationChap 21. Electromagnetic Induction
Chap 21. Electromagnetic Induction Sec. 1  Magnetic field Magnetic fields are produced by electric currents: They can be macroscopic currents in wires. They can be microscopic currents ex: with electrons
More informationHome Work 9. i 2 a 2. a 2 4 a 2 2
Home Work 9 91 A square loop of wire of edge length a carries current i. Show that, at the center of the loop, the of the magnetic field produced by the current is 0i B a The center of a square is a distance
More information5 Magnets and electromagnetism
Magnetism 5 Magnets and electromagnetism n our modern everyday life, the phenomenon of magnetism is associated with iron that is attracted by permanent magnets that can also be made of iron compounds.
More informationMagnetic Field Lines. Uniform Magnetic Field. Earth s Magnetic Field 6/3/2013
Chapter 33: Magnetism Ferromagnetism Iron, cobalt, gadolinium strongly magnetic Can cut a magnet to produce more magnets (no magnetic monopole) Electric fields can magnetize nonmagnetic metals Heat and
More information4/16/ Bertrand
Physics B AP Review: Electricity and Magnetism Name: Charge (Q or q, unit: Coulomb) Comes in + and The proton has a charge of e. The electron has a charge of e. e = 1.602 1019 Coulombs. Charge distribution
More informationElectromagnetic Induction
. Electromagnetic Induction Concepts and Principles Creating Electrical Energy When electric charges move, their electric fields vary. In the previous two chapters we considered moving electric charges
More informationMASSACHUSETTS INSTINUTE OF TECHNOLOGY ESG Physics. Problem Set 9 Solution
MASSACHUSETTS INSTINUTE OF TECHNOLOGY ESG Physics 8. with Kai Spring 3 Problem 1: 37 and 8 Problem Set 9 Solution A conductor consists of a circular loop of radius R =.1 m and two straight, long sections,
More information2015 Pearson Education, Inc. Section 24.5 Magnetic Fields Exert Forces on Moving Charges
Section 24.5 Magnetic Fields Exert Forces on Moving Charges Magnetic Fields Sources of Magnetic Fields You already know that a moving charge is the creator of a magnetic field. Effects of Magnetic Fields
More informationThe Milky Way Galaxy. Our Home Away From Home
The Milky Way Galaxy Our Home Away From Home Lecture 231 Galaxies Group of stars are called galaxies Our star, the Sun, belongs to a system called The Milky Way Galaxy The Milky Way can be seen as a band
More informationChapter 14 Magnets and Electromagnetism
Chapter 14 Magnets and Electromagnetism Magnets and Electromagnetism In the 19 th century experiments were done that showed that magnetic and electric effects were just different aspect of one fundamental
More informationMOVING CHARGES AND MAGNETISM
MOVING CHARGES AND MAGNETISM 1. A circular Coil of 50 turns and radius 0.2m carries of current of 12A Find (a). magnetic field at the centre of the coil and (b) magnetic moment associated with it. 3 scores
More informationChapter 16 Electric Forces and Fields
Chapter 16 Electric Forces and Fields 2. How many electrons does it take to make one coulomb of negative charge? A. 1.00 10 9 B. 6.25 10 18 C. 6.02 10 23 D. 1.66 10 18 E. 2.24 10 4 10. Two equal point
More informationv 2 = v a(x x 0 ) and v = 0 a = v2 0 2d = K md F net = qe = ma E = ma e = K ed = V/m dq = λ ds de r = de cosθ = 1 λ ds = r dθ E r =
Physics 11 Honors Final Exam Spring 003 Name: Section: Closed book exam. Only one 8.5 11 formula sheet (front and back side) can be used. Calculators are allowed. Use the scantron forms (pencil only!)
More informationChapter 13. Gravitation
Chapter 13 Gravitation 13.2 Newton s Law of Gravitation In vector notation: Here m 1 and m 2 are the masses of the particles, r is the distance between them, and G is the gravitational constant. G = 6.67
More informationConceptual: 1, 3, 5, 6, 8, 16, 18, 19. Problems: 4, 6, 8, 11, 16, 20, 23, 27, 34, 41, 45, 56, 60, 65. Conceptual Questions
Conceptual: 1, 3, 5, 6, 8, 16, 18, 19 Problems: 4, 6, 8, 11, 16, 20, 23, 27, 34, 41, 45, 56, 60, 65 Conceptual Questions 1. The magnetic field cannot be described as the magnetic force per unit charge
More informationForce on a square loop of current in a uniform Bfield.
Force on a square loop of current in a uniform Bfield. F top = 0 θ = 0; sinθ = 0; so F B = 0 F bottom = 0 F left = I a B (out of page) F right = I a B (into page) Assume loop is on a frictionless axis
More informationRelativity II. Selected Problems
Chapter Relativity II. Selected Problems.1 Problem.5 (In the text book) Recall that the magnetic force on a charge q moving with velocity v in a magnetic field B is equal to qv B. If a charged particle
More informationEXPERIMENT IV. FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD (e/m OF ELECTRON ) AND. FORCE ON A CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD (µ o )
1 PRINCETON UNIVERSITY PHYSICS 104 LAB Physics Department Week #4 EXPERIMENT IV FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD (e/m OF ELECTRON ) AND FORCE ON A CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD
More informationChapter 8. Rotational Motion. 8.1 Purpose. 8.2 Introduction. s r 2π (rad) = 360 o. r θ
Chapter 8 Rotational Motion 8.1 Purpose In this experiment, rotational motion will be examined. Angular kinematic variables, angular momentum, Newton s 2 nd law for rotational motion, torque, and moments
More informationObjectives for the standardized exam
III. ELECTRICITY AND MAGNETISM A. Electrostatics 1. Charge and Coulomb s Law a) Students should understand the concept of electric charge, so they can: (1) Describe the types of charge and the attraction
More informationPhysics 2212 GH Quiz #4 Solutions Spring 2015
Physics 1 GH Quiz #4 Solutions Spring 15 Fundamental Charge e = 1.6 1 19 C Mass of an Electron m e = 9.19 1 31 kg Coulomb constant K = 8.988 1 9 N m /C Vacuum Permittivity ϵ = 8.854 1 1 C /N m Earth s
More informationMilky Way Galaxy. Orbital Speed of Solar System: 220 km/s Orbital Period: 225 Million Years
Circular Motion Milky Way Galaxy Orbital Speed of Solar System: 220 km/s Orbital Period: 225 Million Years Mercury: 48 km/s Venus: 35 km/s Earth: 30 km/s Mars: 24 km/s Jupiter: 13 km/s Neptune: 5 km/s
More information1 of 7 10/1/2012 3:17 PM
Assignment Previewer http://www.webassign.net/v4cgijfederici@njit/control.pl 1 of 7 10/1/2012 3:17 PM HW11Faraday (2861550) Question 1 2 3 4 5 6 7 8 9 10 1. Question Details SerPSE8 31.P.011.WI. [1742725]
More informationTIME OF COMPLETION DEPARTMENT OF NATURAL SCIENCES. PHYS 2212, Exam 2 Section 1 Version 1 April 16, 2014 Total Weight: 100 points
TIME OF COMPLETION NAME DEPARTMENT OF NATURAL SCIENCES PHYS 2212, Exam 2 Section 1 Version 1 April 16, 2014 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There
More information