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my short report about magnetism.

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- 1. Magnetism Prepared and Presented by: Victor R. Oribe Ph.D. Sci. Ed.-Student
- 2. Magnetism• Is the study of magnetic fields and their effect on materials.• The effect is due to unbalanced spin of electrons in atom.• It is readily observed every day – from the simple magnet that attracts nails and other metals to cassette tapes to magnet-driven trains.
- 3. Magnetism• In terms of applications, magnetism is one of the most important fields in physics.• Large electromagnets are used to pick up heavy loads.• Magnets are used in such devices as meters, motors, and loudspeakers.• Magnetic tapes and disks are used routinely in sound-and video-recording equipment and to store computer data.• Intense magnetic fields are used in magnetic resonance imaging (MRI) devices to explore the human body with better resolution and greater safety than x-rays can provide.
- 4. Magnetism• Giant superconducting magnet are used in the cyclotrons that guide particles into targets at nearly the speed of light.• Magnetism is closely linked with electricity.• Magnetic fields affect moving charges, and moving charges produce magnetic fields.• Changing magnetic field can even create electric fields.• These phenomena signify an underlying unity of electricity and magnetism, which James Clerk Maxwell first described in the 19th century.• The ultimate source of any magnetic field is electric current.
- 5. Nature of Magnetism• In the ancient country of Lydia, in western Asia Minor, now Turkey, was a city called Magnesia.• The Greeks discovered that certain iron ores found in the place could attract other pieces or iron, they called it magnetites.• Magnetites are classified as natural magnet.
- 6. Nature of Magnetism It is now believed that magnetism is due to the spin of electrons within the atoms. Since the electron is a charged particle, the concept implies that magnetism is a property of a charged particle in motion.
- 7. Nature of Magnetism• The power of attraction of a magnet depends on the arrangement of the atoms.• All atoms are in themselves tiny magnet formed into groups called DOMAINS.• The magnetic strength is increased if the domains are induced to fall into line by the action of another magnet.
- 8. General Properties of Magnet• The properties of naturally occurring magnets (magnetites) have been known for over 2,000 years.• Several studies on magnetism were made, but the first thorough investigation was done by William Gilbert in 1600.• Experimental results led to the discovery of the many properties of natural and artificial magnets.
- 9. General Properties of Magnet1. Magnets usually have two poles.• The end of the magnet which points north when magnet is free to turn on a vertical axis is the north-seeking pole, simply the N pole.
- 10. General Properties of Magnet• The opposite end which points south is the south-seeking pole or S pole.• Magnets come in many shapes and sizes, but each has at least two poles.• If you cut a magnet into pieces, every piece will still have at least two poles.
- 11. General Properties of Magnet2. Like Magnetic poles repel andunlike poles attract.
- 12. General Properties of MagnetCharles Augustine de Coulomb,a French physicist, was the firstrecognized scientist to studyquantitatively the force exertedby magnets.The result of his experiments aresummarized in what is known asCoulomb’s Law of Magnetism:“The force of attraction/repulsionbetween two magnetic poles is directlyproportional to the strength of thepoles and inversely proportional to thesquare of the distance between them”.
- 13. The result of his experiments aresummarized in what is known asCoulomb’s Law of Magnetism:“The force of attraction/repulsionbetween two magnetic poles is directlyproportional to the strength of thepoles and inversely proportional to thesquare of the distance between them”.
- 14. Coulomb’s Law• In the MKS system of units, the unit of charge is the coulomb, the force is expressed in newtons and the distance in meter.• A coulomb is a very large unit of charge. A smaller unit is the statcoulomb. 1 coul = 3 x109 statcoul 1 coul = 106 microcoul
- 15. Coulomb’s Law• Experiments have been undertaken to calculate the approximate value of k. This was found to be:• The smallest quantity of charge is the charge on the electron. This is called electronic charge.• Robert Millikan was able to obtain the value of this charge to be e = 1.6 x 10-19 coul.• This is also the charge of proton.
- 16. Problems to Illustrate Coulomb’s Law
- 17. Problem # 1 Two point charges of 2.5 x 10-10 coul and -3.0 x 10-10 coul are 10 cm apart in air. Calculate the magnitude and direction of the force on each charge. 10 cm or .10 m-3.0 x 10-10 coul 2.5 x 10-10 coul N.m2 (-3.0 x 10-10 coul)(2.5 x 10-10 coul) F = 9 X109 Coul2 (.10m)2 F = -6.75 x 10-8 N
- 18. • The result (F= -6.75 x 10-8 N) means that:• F12 has a magnitude of 6.75 x10-8 N and is directed toward q1.• F21 has a magnitude of -6.75 x 10-8 N and is directed toward q2. Therefore: F12 = - F21
- 19. Problem # 2Calculate the force between two pointcharge of +4.5 x 10-6 coul and -5.0 x10-6 coul which are 15 cm apart in air.Answer: - 9.0NThe negative sign in the answerindicates that the force between thepoint charges is an attractive force.
- 20. Problem # 3Given the three point charges q1, q2, and q3 asshown in the figure. If q1 = +1.5 x 10-6 coul, q2 =+2.5 x 10-6 coul and q3 = -2.0 x 10-6 coul and 300 ,d12 =10cm, d13 = 15cm.Find: a) the force between q1 and q3 b)The resultant force on q1 q3 = -2.0x10-6 coul F13 = ? Ø d13 = 0.15m d12 =0.10m F12 =? q1 = +1.5x10-6 coul q2 = +2.5 x10-6 coul
- 21. Solution: a) the force between q1 and q3 k (q1 q3) F13 = r2 9 x 109 N.m2 / coul2 (+1.5x10-6 coul) (-2.0x10-6 coul) F13 = (0.15m)2 F13 = -1.2N The negative value of F13 indicates an attractive force on q1 due to q3. k (q1 q2) F12 = r2 q3 = -2.0x10-6 coul 9 x 109 N.m2 / coul2 (+1.5x10-6 coul) (-2.0x10-6 coul) (0.01m)2 F12 = 3.375N F13 = ? d13 = 0.15mThe positive value of F12indicates a repulsive force q2 = +1.5x10-6 coul F12 = ?on q1 due to q2. d12 =0.10m q1 = +1.5x10-6 coul
- 22. Solution: b)The resultant force on q1To find the resultant force (Fnet ) on q1, add F12 and F13vectorially.The sum of the x-component is:∑X = -3.375N + 1.2 cos 600∑X = -2.775The sum of the y-component is:∑Y = -1.2 sin 600∑X = 1.039 FNet = ? 300Fnet =√(-2.775)2 + (1.039)2 F13 = 1.2 NFnet = 2.96 N F12 = -3.375N
- 23. Solve the following problems:1. Find the force between two eqaul charges of 15 x 10-8 coul if they are 20 cm apart in air. -3 5.06 X10 N2. Find the magnitude and direction of the force in each of two charges of 3.5 x 10-6 coul and -2.8 x 10-6 coul, 10 cm apart. 8.82 X102 N, attractive3. Find the magnitude and kind of force between two charges of 20 x 10-6 coul which are 25 cm apart in air. 432 N, repulsive4. A charge of 2.8 x 10-6 coul is at a distance of 12 cm from another charge of -8.4 x10-6 coul. Find the magnitude and direction of the force on the first charge that is due to the second charge. +14.7 N5. An unknown charge is attracted by a force of 25N when it is at a distance of 10 cm from another charge of 25 x 10-6 coul. What is the magnitude of the un known charge? q1 = 1.11 x 10-6 coul
- 24. General Properties of Magnet3. A piece ofmagnetite, whenmade to hang andswing freely,would align itselfwith the magneticfield of the earthfollowing a north-south direction.
- 25. General Properties of Magnet4. Permanent magnets aremagnets made from alloys ofcobalt and nickel. These magnets retain their magnetism for a long time.
- 26. General Properties of Magnet5. Other metals like iron can bemagnetized by Induction.When a piece of iron nailstouches a permanent magnet,the nails becomes a magnet.It retains in this condition for aslong as it is within the magneticfield.The nail is a temporary magnetand its magnetism is describedas induced magnetism.
- 27. Magnetic Field of Force• Experiment show that a stationary charged particle doesn’t interact with a static magnetic field.• When a charged particle is moving through a magnetic field, however, a magnetic force acts on it.• The force has its maximum value when the charge moves in a direction perpendicular to the magnetic field line, decreases in value at other angles, and becomes zero when moves along the field of lines.
- 28. Magnetic Field of Force• Magnetic force on a moving charge is directed perpendicular to the magnetic field.• It is found experimentally that the strength of the magnetic force on the particle is proportional to the magnitude of the charge q, the magnitude of the velocity v, the strength of the external magnetic field B, and the sine of the angle between the direction of v and the direction of B. F = qvB sin θ• This expression is used to define the magnitude of the magnetic field as: F B= qv sin θ
- 29. Earth’s Magnetic Field• A small bar magnet is said to have north and south poles, but is it more accurate to say it has a “north-seeking” pole and “south-seeking” pole.• By this expressions, we mean that if such a magnet is used as a compass, one end will “seek” or point to, the geographic North Pole of Earth and the other end will “seek” or point to, the geographic South Pole of Earth.
- 30. Earth’s Magnetic Field• We therefore conclude that:The Geographic North Pole of Earth correspondsto a magnetic south pole, and the geographicSouth Pole of Earth corresponds to a magneticnorth pole.
- 31. Earth’s Magnetic Field• The magnetic field pattern of Earth is similar to the pattern that would be set up by a bar magnet placed at its center.• An interesting fact concerning Earth’s magnetic field is that its direction reverses every few million years.
- 32. Earth’s Magnetic Field and its Present Application – Labeling Airport Runways• The magnetic field of Earth is used to label runways at airports according to their direction.• A large number is painted on the end of the runway so that it can be read by the pilot of an incoming airplane.
- 33. Earth’s Magnetic Field and its Present Application – Labeling Airport Runways• This number describes the direction in which the airplane is traveling, expressed as the magnetic heading, in degrees measured clockwise from magnetic north divided by 10.• A runway marked 9 would be directed toward the east (900 divided by 10), whereas a runway marked 18 would be directed toward magnetic south.
- 34. Magnetic Field of Force• If F is in newton, q in coulombs, and v in meter per second, the SI unit of magnetic field is the tesla (T), also called the weber (Wb) per square meter (1 T = 1 Wb/m2).• The unit of B is: B = T = Wb/m2 = N/C.m/s = N/A.m• The cgs unit of magnetic field is the gauss (G). 1 T = 104 G
- 35. Magnetic Field of Force• From equation F=qvB sin θ, we see that the force on a charged particle moving in a magnetic field has its maximum value when the particle’s motion is perpendicular to the magnetic field, corresponding to θ=900 , so that sin θ = 1.• The magnitude of this maximum force has the value: Fmax = qvB• Experiment also show that the direction of the magnetic force is always perpendicular to both v and B. To determine the direction of force, we employ the right-hand rule # 1.
- 36. Right-Hand Rule #1The implications of this expression include:1. The force is perpendicular to both the velocity v of the charge q and the magnetic field B.2. The magnitude of the force is F = qvB sinθ where θis the angle < 180 degrees between the velocity andthe magnetic field. This implies that the magneticforce on a stationary charge or a charge movingparallel to the magnetic field is zero.3. The direction of the force is given by the right handrule.
- 37. Right-Hand Rule # 1Right-Hand Rule #1 determines the directionsof magnetic force, conventional current and themagnetic field. Given any two of these, thethird can be found.Using your right-hand:1. point your index finger in thedirection of the chargesvelocity, v, (recall conventionalcurrent).2. Point your middle finger in thedirection of the magnetic field,B.3. Your thumb now points inthe direction of the magneticforce, Fmagnetic.
- 38. Right-Hand Rule # 1When the magnetic force relationship is appliedto a current-carrying wire, the right-hand rulemay be used to determine the direction of forceon the wire.
- 39. Right-Hand Rule # 2Right-Hand Rule #2 determines the directionof the magnetic field around a current-carrying wire and vice-versaUsing your right-hand: Curl your fingers intoa half-circle aroundthe wire, they point inthe direction of themagnetic field, BPoint your thumb inthe direction of theconventional current.
- 40. Problem• A proton moves with a speed of 1.00 x 105 m/s through Earth’s magnetic field, which has a value of 55.0µT at a particular location. When the proton moves eastward, the magnetic force acting on it is directed straight upward, and when it moves northward, no magnetic force act on it.a) What is the direction of the magnetic field?b) What is the strength of the magnetic force when the proton moves eastward?c) Calculate the gravitational force on the proton and compare it with the magnetic force. If there were an electric field with a magnitude equal to E=1.50x102 N/C at that location, a common value at Earth’s surface. Note the mass of the proton is 1.67 x10-27 kg.
- 41. Solution:a) What is the direction of the magnetic field?No magnetic force act on the proton when it’s goingNorth, so the angle such a proton makes with themagnetic field direction must be either o0 or 1800.Therefore, the magnetic field B must point eithernorth or south.Now apply the right hand rule. When the particletravels east, the magnetic force is directed upward.Point your thumb in the direction of the force andyour finger in the direction of the velocity eastward.When you curl your finger, they point north, whichmust therefore be the direction of the magneticfield.
- 42. Solutiona) What is the strength of the magnetic force when the proton moves eastward?Substitute the given values and the charge of a proton intoequation F=qvB sin θ. From part (a), the angle between thevelocity v of the proton and magnetic field B is 900F= qvB sin θF= (1.60x10-19 C)(1.00X105 m/s) x (55.0x10-6 T) sin (900 )F= 8.80 x 10-19 N
- 43. Solution:Calculate the gravitational force on the proton andcompare it with the magnetic force and also withthe electric force if E=1.50 X102 N/C.Fgrav. = mg = (1.67x10-27 kg.) (9.8 m/s2 ) = 1.64 x10-26 NFelec = qE = (1.60x10-19 C) (1.50 X 102 N/C) = 2.40 X 10-17 N
- 44. Problem # 2A Proton Moving in a Magnetic FieldA proton moves at 8.00 x 106 m/s along the x-axis. Itenters a region in which there is a magnetic field ofmagnitude 2.50 T, directed at an angle of 600 withthe x-axis and lying in the xy-plane.a)Find the initial magnitude and direction of themagnetic force on the proton.b) Calculate the proton’s initial acceleration.Hint: Finding the magnitude and direction of the magneticforce requires substituting values into the equation formagnetic force, (a) (F=qvB sin θ), and using the right-hand rule. (b)Applying Newton’s second law.
- 45. Solution:a) Find the magnitude and direction of the magnetic force on the proton.F=qvB sin θ= (1.60x10-19C) (8.00X106 m/s) (2.50T) (sin 600)= 2.77 x10-12 NPoint the fingers of the right hand in the x-direction(direction of v) and then curl them toward B. Thethumb points (upward, in the positive z-direction.
- 46. Solution:b) Calculate the proton’s initial acceleration.am = F(1.67 X10-27 kg) ( a ) = 2.77 x 10-12 Na = 1.66 x 1015 m/s2
- 47. Magnetic Force on a Current- Carrying Conductor• If a straight conductor of length ℓ carries current, the magnetic force on that conductor when it is placed in a uniform external magnetic field B is: F = BIℓ sin θ• Where Ɵ is the angle between the direction of the current and the direction of the magnetic field. B is the direction of the current. I is the current ℓ is the length of the wire
- 48. Magnetic Force on a Current- Carrying Conductor• Right-hand rule #1 also gives the direction of the magnetic force on the conductor. In this case, however, you must point your fingers in the direction of the current rather than in the direction of v.
- 49. Application:• A magnetic force acting on a current-carrying wire in a magnetic field is the operating principle of most speakers in sound systems.
- 50. Application:• Speaker is consists of a coil of wire called the voice coil, a flexible paper cone that acts as the speaker, and a permanent magnet.• The coil of wire is surrounding the north pole of the magnet is shaped so that the magnetic field lines are directed radially outward from the coil’s axis.• When an electrical signal is sent to the coil, producing a current in the coil, a magnetic force to the left acts on the coil. (can be seen by applying right-hand rule #1 to each turn of wire).
- 51. Application:• When the current reverses direction, as it would for a current that varied sinusoidally, the magnetic force on the coil also reverses direction, and the cone, which is attached to the coil, accelerates to the right.• An alternating current through the coil causes an alternating force on the coil, which results in vibrations of the cone.• The vibrating cone creates sound waves as it pushes and pulls on the air in front of it.• In this way, a 1-kHz electrical signal is converted to a 1-kHz sound waves.
- 52. Additional Research Work• Research on the other application of current- carrying conductors, and explain how it works.• What is electromagnetic pumps (artificial heart and kidney). Explain how current-carrying conductors is applied in the manufacture of electromagnetic pump.
- 53. Problem:• A wire carries a current of 22.0 A from west to east. Assume the magnetic field of Earth at this location is horizontal and directed from south to north and it has a magnitude of 0.500 x 10-4 T.• a) Find the magnitude and direction of the magnetic force on a 36.0-m long of wire.• b) Calculate the gravitational force on the same length of wire if it’s made of copper and has a cross-sectional area of 2.50 x 10-6 m2 .
- 54. Solution:• a) Find the magnitude and direction of the magnetic force on a 36.0-m long of wire.F = BIℓ sin θ= (0.500 x 10-4 T) (22.0 A) (36.0 m) sin 900= 3.96 x 10-2 NApplying the right-hand rule #1 to find the directionof the magnetic force: With fingers of your right hand pointing westto east in the direction of the current, curl themnorth in the direction of the magnetic field. Yourthumb points UPWARD.
- 55. Solution:• b) Calculate the gravitational force on the same length of wire if it’s made of copper and has a cross-sectional area of 2.50 x 10-6 m2 . Density of the metal• m= ρv = ρ(Aℓ) (2.50x10-6 m2 . 36.0m) =8.92x103 kg.m3) = 0.803 kg• To get the gravitational force, multiple the mass by the acceleration of gravity:• Fgrav. = mg = (0.803 kg) (9.8 m/s2) = 7.87 N
- 56. Torque on a Current Loop and Electric Motor• The torque Ƭ on a current-carrying loop of wire in a magnetic field B has magnitude: Ƭ = BIA sin θ• Where: I is the current in the loop A is the cross-sectional area• The magnitude of the magnetic moment of the current-carrying coil is defined by µ = IAN N is the number of loops• The magnetic moment is considered vector, µ, that is perpendicular to the plane of the loop.• The angle between B and µ is θ.
- 57. Problem: • A circular wire loop of radius 1.00 m is placed in a magnetic field of magnitude 0.500 T. The normal to the plane of the loop makes an angle of 30.00 with the magnetic field.(see Illustration) The current in the loop is 2.00 A in the direction shown.• a) Find the magnetic moment of the loop and the magnitude of the torque at this instant.• b) The same current is carried by the rectangular 2.00-m by 3.00-m coil with three loops. Find the magnetic moment of the coil and the magnitude of the torque acting on the end of the coil at that instant.
- 58. Solution:• a) Find the magnetic moment of the circular loop and the magnetic torque exerted on it.• Hint: First, calculate the enclosed area of the circular loop. Calculate the magnetic moment of the loop Substitute values for the magnetic moment, magnetic field, and θ into Ƭ = µB sin θ A = ╥r2 = 1.00m)2 = 3.14m2 µ = IAN = (2.00A) (3.14m2) (1) = 6.28A.m2 Ƭ = µb sin θ = (6.28A.m2) (0.500T) (sin 30.00) = 1.57 N.m
- 59. Solution:• Find the magnetic moment of the rectangular coil and the magnetic torque exerted on it.• Hint: Calculate the area of the coil. Calculate the magnetic moment of the coil Substitute values into equation Ƭ = µB sin θ A= L x H = (2.00m) (3.00m) = 6.00m2 µ= IAN = (2.00 A) (6.00 m2) (3) = 36.0 A.m2 Ƭ = µb sin θ = (.500T) (36.0 A.m2) (sin 300) = 9.00 N. m
- 60. Application:• Its hard to imagine life in the 21st century without motors.• Some of the appliances that contain motors include computer disk drives, CD players, DVD players, food processors and blenders, car starters, furnaces, and air conditioners.• The motors convert electrical energy to kinetic energy of rotation and consist of a rigid current-carrying loop that rotates when placed in the field of a magnet.
- 61. Magnetic Field of a Long, Straight WireDuring the lecturedemonstration in1819, Danish ScientistHans Oersted found thanan electric current in a wiredeflected a nearbycompass needle.This momentous discovery, linking a magneticfield with an electric current for the first time,was the beginning of our understanding of theorigin of magnet.
- 62. Magnetic Field of a Long, Straight Wire• In the simple experiment carried by Oersted in 1820, several compass needles are placed in a horizontal plane near a long vertical wire.
- 63. Magnetic Field of a Long, Straight Wire• When there is no current in the wire, all needles point in the same direction (that of Earth’s field), as one would expect. • When the wire carries a strong, steady current, however, the needles all deflect in directions tangent to the circle.
- 64. Magnetic Field of a Long, Straight Wire• The observation of Oersted show that the direction of B is consistent with the following convenient rule, Right-Hand Rule # 2.
- 65. Magnetic Field of a Long, Straight Wire• The magnetic field at distance ṙ from along, straight wire carrying current I has the magnitude: µ0I B= 2╥ṙ• Where; µ0 = 4╥ X 10-7 T.m/A is the permeability of free space The magnetic field lines around a long, straight line wire are circles concentric with the wire.
- 66. Magnetic Field of a Long, Straight Wire• Ampere’s law can be used to find the magnetic field around certain simple current-carrying conductors. It can be written: ∑Bǁ ∆ℓ = µ0I• Where: Bǁ is the component B tangent to a small current element of length ∆ℓ that is part of a closed path and I is the total current that penetrates the closed path.
- 67. Sample Problem:• A coaxial cable consists of an insulated wire carrying current I1 = 3.00 A surrounded by a cylindrical conductor carrying current I2 = 1.00 A in the opposite direction.( see illustration)a) calculate the magneticfield inside the cylindricalconductor at rint = 0.500cm.Calculate the magnetic fieldoutside the cylindricalconductor at rext = 1.50 cm.
- 68. Solution:• a) Calculate the magnetic field Bint inside the cylindrical conductor at rint = 0.500 cm.• Hint: Write the Ampere’s law The magnetic field is constant on the given path and the total path length is 2╥ṙint : Solve for Bint and substitute values: µ0I1∑Bǁ ∆ℓ = µ0I Bint = 2╥ṙintBint =(2╥ṙint )= µ0I1 (4╥ x 10-7 T. m/A) (3.00A) = 2╥ (0.005m) = 1.20 x 10-4 T
- 69. Solution:• b) Calculate the magnetic field Bext outside the cylindrical conductor at rext = 1.50 cm.• Hints: Write the Ampere’s law The magnetic field is again constant on the given path and the totla path length is 2╥rext Solve for Bext and sustitute values∑Bǁ ∆ℓ = µ0IBext (2╥ṙext )= µ0(I1 – I2) µ0(I1 – I2) (4╥ X 10-7 T.m/A) (3.00 A – 1.00 A)Bext = 2╥ṙext 2╥ (0.015 m) = 2.64 x 10-5 T

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