1. Copyright © 2009 Pearson Education, Inc. Chapter 27 Magnetism

2. Copyright © 2009 Pearson Education, Inc. 27-4 Force on an Electric Charge Moving in a Magnetic Field Conceptual Example 27-10: Velocity selector, or filter: crossed E and B fields. Some electronic devices and experiments need a beam of charged particles all moving at nearly the same velocity. This can be achieved using both a uniform electric field and a uniform magnetic field, arranged so they are at right angles to each other. Particles of charge q pass through slit S 1 and enter the region where B points into the page and E points down from the positive plate toward the negative plate. If the particles enter with different velocities, show how this device “selects” a particular velocity, and determine what this velocity is.

3. ConcepTest 27.5 Velocity Selector In what direction would a B field have to point for a beam of electrons moving to the right to go undeflected through a region where there is a uniform electric field pointing vertically upward? 1) up (parallel to E ) 2) down (antiparallel to E ) 3) into the page 4) out of the page 5) impossible to accomplish electrons E v B = ?

4. downward upward negatively chargedB field pointing out of the page Without a B field, the electrons feel an electric force downward. In order to compensate, the magnetic force has to point upward. Using the right-hand rule and the fact that the electrons are negatively charged leads to a B field pointing out of the page. ConcepTest 27.5 Velocity Selector In what direction would a B field have to point for a beam of electrons moving to the right to go undeflected through a region where there is a uniform electric field pointing vertically upward? 1) up (parallel to E ) 2) down (antiparallel to E ) 3) into the page 4) out of the page 5) impossible to accomplish electrons E v B = ?

5. Copyright © 2009 Pearson Education, Inc. The forces on opposite sides of a current loop will be equal and opposite (if the field is uniform and the loop is symmetric), but there may be a torque. The magnitude of the torque is given by 27-5 Torque on a Current Loop; Magnetic Dipole Moment

6. Copyright © 2009 Pearson Education, Inc. The quantity NIA is called the magnetic dipole moment, μ : 27-5 Torque on a Current Loop; Magnetic Dipole Moment The potential energy of the loop depends on its orientation in the field:

7. Copyright © 2009 Pearson Education, Inc. 27-5 Torque on a Current Loop; Magnetic Dipole Moment Example 27-11: Torque on a coil. A circular coil of wire has a diameter of 20.0 cm and contains 10 loops. The current in each loop is 3.00 A, and the coil is placed in a 2.00-T external magnetic field. Determine the maximum and minimum torque exerted on the coil by the field.

8. Copyright © 2009 Pearson Education, Inc. 27-5 Torque on a Current Loop; Magnetic Dipole Moment Example 27-12: Magnetic moment of a hydrogen atom. Determine the magnetic dipole moment of the electron orbiting the proton of a hydrogen atom at a given instant, assuming (in the Bohr model) it is in its ground state with a circular orbit of radius r = 0.529 x 10 -10 m. [This is a very rough picture of atomic structure, but nonetheless gives an accurate result.]

9. If there is a current in the loop in the direction shown, the loop will: 1) move up 2) move down 3) rotate clockwise 4) rotate counterclockwise 5) both rotate and move N S NS B field out of North B field into South ConcepTest 27.7b Magnetic Force on a Loop II

10. right out of the page up downward clockwise Look at the north pole: here the magnetic field points to the right and the current points out of the page. The right-hand rule says that the force must point up. At the south pole, the same logic leads to a downward force. Thus the loop rotates clockwise. N S F F 1) move up 2) move down 3) rotate clockwise 4) rotate counterclockwise 5) both rotate and move If there is a current in the loop in the direction shown, the loop will: ConcepTest 27.7b Magnetic Force on a Loop II

11. Copyright © 2009 Pearson Education, Inc. 27-6 Applications: Motors, Loudspeakers, Galvanometers An electric motor uses the torque on a current loop in a magnetic field to turn magnetic energy into kinetic energy.

12. Copyright © 2009 Pearson Education, Inc. A galvanometer takes advantage of the torque on a current loop to measure current; the spring constant is calibrated so the scale reads in amperes. 27-6 Applications: Motors, Loudspeakers, Galvanometers

13. Copyright © 2009 Pearson Education, Inc. 27-8 The Hall Effect When a current-carrying wire is placed in a magnetic field, there is a sideways force on the electrons in the wire. This tends to push them to one side and results in a potential difference from one side of the wire to the other; this is called the Hall effect. The emf differs in sign depending on the sign of the charge carriers; this is how it was first determined that the charge carriers in ordinary conductors are negatively charged.

14. Copyright © 2009 Pearson Education, Inc. 27-8 The Hall Effect Example 27-13: Drift velocity using the Hall effect. A long copper strip 1.8 cm wide and 1.0 mm thick is placed in a 1.2-T magnetic field. When a steady current of 15 A passes through it, the Hall emf is measured to be 1.02 μV. Determine the drift velocity of the electrons and the density of free (conducting) electrons (number per unit volume) in the copper.

15. Copyright © 2009 Pearson Education, Inc. A mass spectrometer measures the masses of atoms. If a charged particle is moving through perpendicular electric and magnetic fields, there is a particular speed at which it will not be deflected, which then allows the measurement of its mass: 27-9 Mass Spectrometer

16. Copyright © 2009 Pearson Education, Inc. All the atoms reaching the second magnetic field will have the same speed; their radius of curvature will depend on their mass. 27-9 Mass Spectrometer

17. Copyright © 2009 Pearson Education, Inc. 27-9 Mass Spectrometer Example 27-14: Mass spectrometry. Carbon atoms of atomic mass 12.0 u are found to be mixed with another, unknown, element. In a mass spectrometer with fixed B ′, the carbon traverses a path of radius 22.4 cm and the unknown’s path has a 26.2-cm radius. What is the unknown element? Assume the ions of both elements have the same charge.

18. Copyright © 2009 Pearson Education, Inc. Magnets have north and south poles. Like poles repel, unlike attract. Unit of magnetic field: tesla. Electric currents produce magnetic fields. A magnetic field exerts a force on an electric current: Summary of Chapter 27

19. Copyright © 2009 Pearson Education, Inc. A magnetic field exerts a force on a moving charge: Summary of Chapter 27 Torque on a current loop: Magnetic dipole moment:

20. Copyright © 2009 Pearson Education, Inc. Chapter 28 Sources of Magnetic Field

21. Copyright © 2009 Pearson Education, Inc. Magnetic Field Due to a Straight Wire Force between Two Parallel Wires Definitions of the Ampere and the Coulomb Ampère’s Law Magnetic Field of a Solenoid and a Toroid Biot-Savart Law Magnetic Materials – Ferromagnetism Electromagnets and Solenoids – Applications Units of Chapter 28

22. Copyright © 2009 Pearson Education, Inc. Magnetic Fields in Magnetic Materials; Hysteresis Paramagnetism and Diamagnetism Units of Chapter 28

23. Copyright © 2009 Pearson Education, Inc. The magnetic field due to a straight wire is inversely proportional to the distance from the wire: The constant μ 0 is called the permeability of free space, and has the value μ 0 = 4π x 10 -7 T·m/A. 28-1 Magnetic Field Due to a Straight Wire

24. Copyright © 2009 Pearson Education, Inc. 28-1 Magnetic Field Due to a Straight Wire Example 28-1: Calculation of B near a wire. An electric wire in the wall of a building carries a dc current of 25 A vertically upward. What is the magnetic field due to this current at a point P 10 cm due north of the wire?

25. Copyright © 2009 Pearson Education, Inc. 28-1 Magnetic Field Due to a Straight Wire Example 28-2: Magnetic field midway between two currents. Two parallel straight wires 10.0 cm apart carry currents in opposite directions. Current I 1 = 5.0 A is out of the page, and I 2 = 7.0 A is into the page. Determine the magnitude and direction of the magnetic field halfway between the two wires.

26. Copyright © 2009 Pearson Education, Inc. 28-1 Magnetic Field Due to a Straight Wire Conceptual Example 28-3: Magnetic field due to four wires. This figure shows four long parallel wires which carry equal currents into or out of the page. In which configuration, (a) or (b), is the magnetic field greater at the center of the square?

27. ConcepTest 28.1 Magnetic Field of a Wire P 1 2 3 4 If the currents in these wires have the same magnitude but opposite directions, what is the direction of the magnetic field at point P? 1) direction 1 2) direction 2 3) direction 3 4) direction 4 5) the B field is zero

28. Adding them up as vectors downward Using the right-hand rule, we can sketch the B fields due to the two currents. Adding them up as vectors gives a total magnetic field pointing downward. P 1 2 3 4 ConcepTest 28.1 Magnetic Field of a Wire If the currents in these wires have the same magnitude but opposite directions, what is the direction of the magnetic field at point P? 1) direction 1 2) direction 2 3) direction 3 4) direction 4 5) the B field is zero

29. Two straight wires run parallel to each other, each carrying a current in the direction shown below. The two wires experience a force in which direction? 1) toward each other 2) away from each other 3) there is no force ConcepTest 28.2 Field and Force

30. attractive force currents are parallel The current in each wire produces a magnetic field that is felt by the current of the other wire. Using the right-hand rule, we find that each wire experiences a force toward the other wire (i.e., an attractive force) when the currents are parallel (as shown). ConcepTest 28.2 Field and Force Two straight wires run parallel to each other, each carrying a current in the direction shown below. The two wires experience a force in which direction? 1) toward each other 2) away from each other 3) there is no force Follow-up: What happens when one of the currents is turned off?

31. Copyright © 2009 Pearson Education, Inc. The magnetic field produced at the position of wire 2 due to the current in wire 1 is The force this field exerts on a length l 2 of wire 2 is 28-2 Force between Two Parallel Wires

32. Copyright © 2009 Pearson Education, Inc. Parallel currents attract; antiparallel currents repel. 28-2 Force between Two Parallel Wires

33. Copyright © 2009 Pearson Education, Inc. 28-2 Force between Two Parallel Wires Example 28-4. Force between two current-carrying wires. The two wires of a 2.0-m-long appliance cord are 3.0 mm apart and carry a current of 8.0 A dc. Calculate the force one wire exerts on the other.

34. Copyright © 2009 Pearson Education, Inc. 28-2 Force between Two Parallel Wires Example 28-5: Suspending a wire with a current. A horizontal wire carries a current I 1 = 80 A dc. A second parallel wire 20 cm below it must carry how much current I 2 so that it doesn’t fall due to gravity? The lower wire has a mass of 0.12 g per meter of length.

35. Copyright © 2009 Pearson Education, Inc. 28-3 Definitions of the Ampere and the Coulomb The ampere is officially defined in terms of the force between two current-carrying wires: One ampere is defined as that current flowing in each of two long parallel wires 1 m apart, which results in a force of exactly 2 x 10 -7 N per meter of length of each wire. The coulomb is then defined as exactly one ampere-second.

36. Copyright © 2009 Pearson Education, Inc. Ampère’s law relates the magnetic field around a closed loop to the total current flowing through the loop: 28-4 Ampère’s Law This integral is taken around the edge of the closed loop.

37. Copyright © 2009 Pearson Education, Inc. 28-4 Ampère’s Law Using Ampère’s law to find the field around a long straight wire: Use a circular path with the wire at the center; then B is tangent to dl at every point. The integral then gives so B = μ 0 I/2πr, as before.

38. Copyright © 2009 Pearson Education, Inc. 28-4 Ampère’s Law Example 28-6: Field inside and outside a wire. A long straight cylindrical wire conductor of radius R carries a current I of uniform current density in the conductor. Determine the magnetic field due to this current at (a) points outside the conductor ( r > R ) and (b) points inside the conductor ( r < R ). Assume that r, the radial distance from the axis, is much less than the length of the wire. (c) If R = 2.0 mm and I = 60 A, what is B at r = 1.0 mm, r = 2.0 mm, and r = 3.0 mm?

39. Copyright © 2009 Pearson Education, Inc. 28-4 Ampère’s Law Conceptual Example 28-7: Coaxial cable. A coaxial cable is a single wire surrounded by a cylindrical metallic braid. The two conductors are separated by an insulator. The central wire carries current to the other end of the cable, and the outer braid carries the return current and is usually considered ground. Describe the magnetic field (a) in the space between the conductors, and (b) outside the cable.

40. Copyright © 2009 Pearson Education, Inc. 28-4 Ampère’s Law Example 28-8: A nice use for Ampère’s law. Use Ampère’s law to show that in any region of space where there are no currents the magnetic field cannot be both unidirectional and nonuniform as shown in the figure.

41. Copyright © 2009 Pearson Education, Inc. 28-4 Ampère’s Law Solving problems using Ampère’s law: Ampère’s law is only useful for solving problems when there is a great deal of symmetry. Identify the symmetry. Choose an integration path that reflects the symmetry (typically, the path is along lines where the field is constant and perpendicular to the field where it is changing). Use the symmetry to determine the direction of the field. Determine the enclosed current.

42. Copyright © 2009 Pearson Education, Inc. 28-5 Magnetic Field of a Solenoid and a Toroid A solenoid is a coil of wire containing many loops. To find the field inside, we use Ampère’s law along the path indicated in the figure.

43. Copyright © 2009 Pearson Education, Inc. 28-5 Magnetic Field of a Solenoid and a Toroid The field is zero outside the solenoid, and the path integral is zero along the vertical lines, so the field is ( n is the number of loops per unit length)

44. Copyright © 2009 Pearson Education, Inc. 28-5 Magnetic Field of a Solenoid and a Toroid Example 28-9: Field inside a solenoid. A thin 10-cm-long solenoid used for fast electromechanical switching has a total of 400 turns of wire and carries a current of 2.0 A. Calculate the field inside near the center.

45. Copyright © 2009 Pearson Education, Inc. 28-5 Magnetic Field of a Solenoid and a Toroid Example 28-10: Toroid. Use Ampère’s law to determine the magnetic field (a) inside and (b) outside a toroid, which is like a solenoid bent into the shape of a circle as shown.

46. Copyright © 2009 Pearson Education, Inc. 28-6 Biot-Savart Law The Biot-Savart law gives the magnetic field due to an infinitesimal length of current; the total field can then be found by integrating over the total length of all currents:

47. Copyright © 2009 Pearson Education, Inc. 28-6 Biot-Savart Law Example 28-11: B due to current I in straight wire. For the field near a long straight wire carrying a current I, show that the Biot-Savart law gives B = μ 0 I/2πR.

48. Copyright © 2009 Pearson Education, Inc. 28-6 Biot-Savart Law Example 28-12: Current loop. Determine B for points on the axis of a circular loop of wire of radius R carrying a current I.

49. Copyright © 2009 Pearson Education, Inc. 28-6 Biot-Savart Law Example 28-13: B due to a wire segment. One quarter of a circular loop of wire carries a current I. The current I enters and leaves on straight segments of wire, as shown; the straight wires are along the radial direction from the center C of the circular portion. Find the magnetic field at point C.

50. Copyright © 2009 Pearson Education, Inc. Ferromagnetic materials are those that can become strongly magnetized, such as iron and nickel. These materials are made up of tiny regions called domains; the magnetic field in each domain is in a single direction. 28-7 Magnetic Materials – Ferromagnetism

51. Copyright © 2009 Pearson Education, Inc. When the material is unmagnetized, the domains are randomly oriented. They can be partially or fully aligned by placing the material in an external magnetic field. 28-7 Magnetic Materials – Ferromagnetism

52. Copyright © 2009 Pearson Education, Inc. A magnet, if undisturbed, will tend to retain its magnetism. It can be demagnetized by shock or heat. The relationship between the external magnetic field and the internal field in a ferromagnet is not simple, as the magnetization can vary. 28-7 Magnetic Materials – Ferromagnetism

53. ConcepTest 28.3 Current Loop P I What is the direction of the magnetic field at the center (point P) of the square loop of current? 1) left 2) right 3) zero 4) into the page 5) out of the page

54. out of the page Use the right-hand rule for each wire segment to find that each segment has its B field pointing out of the page at point P. ConcepTest 28.3 Current Loop P I What is the direction of the magnetic field at the center (point P) of the square loop of current? 1) left 2) right 3) zero 4) into the page 5) out of the page

55. Copyright © 2009 Pearson Education, Inc. Remember that a solenoid is a long coil of wire. If it is tightly wrapped, the magnetic field in its interior is almost uniform. 28-8 Electromagnets and Solenoids – Applications

56. Copyright © 2009 Pearson Education, Inc. 28-9 Magnetic Fields in Magnetic Materials; Hysteresis If a ferromagnetic material is placed in the core of a solenoid or toroid, the magnetic field is enhanced by the field created by the ferromagnet itself. This is usually much greater than the field created by the current alone. If we write B = μI where μ is the magnetic permeability, ferromagnets have μ >> μ 0, while all other materials have μ ≈ μ 0.

57. Copyright © 2009 Pearson Education, Inc. 28-9 Magnetic Fields in Magnetic Materials; Hysteresis Not only is the permeability very large for ferromagnets, its value depends on the external field.

58. Copyright © 2009 Pearson Education, Inc. Furthermore, the induced field depends on the history of the material. Starting with unmagnetized material and no magnetic field, the magnetic field can be increased, decreased, reversed, and the cycle repeated. The resulting plot of the total magnetic field within the ferromagnet is called a hysteresis loop. 28-9 Magnetic Fields in Magnetic Materials; Hysteresis

59. Copyright © 2009 Pearson Education, Inc. 28-10 Paramagnetism and Diamagnetism All materials exhibit some level of magnetic behavior; most are either paramagnetic ( μ slightly greater than μ 0 ) or diamagnetic ( μ slightly less than μ 0 ). The following is a table of magnetic susceptibility χ m, where χ m = μ/μ 0 – 1.

60. Copyright © 2009 Pearson Education, Inc. 28-10 Paramagnetism and Diamagnetism Molecules of paramagnetic materials have a small intrinsic magnetic dipole moment, and they tend to align somewhat with an external magnetic field, increasing it slightly. Molecules of diamagnetic materials have no intrinsic magnetic dipole moment; an external field induces a small dipole moment, but in such a way that the total field is slightly decreased.

61. Copyright © 2009 Pearson Education, Inc. Magnitude of the field of a long, straight current-carrying wire: The force of one current-carrying wire on another defines the ampere. Ampère’s law: Summary of Chapter 28

62. Copyright © 2009 Pearson Education, Inc. Magnetic field inside a solenoid: Biot-Savart law: Summary of Chapter 28 Ferromagnetic materials can be made into strong permanent magnets.